Solucionario-Boylestad-10ma-Edicion

Instructor’s Resource Manual

to accompany

Electronic Devices and

Circuit Theory

Tenth Edition

Robert L. Boylestad

Louis Nashelsky

Upper Saddle River, New Jersey

Columbus, Ohio

Copyright © 2009 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458.

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Instructors of classes using Boylestad/Nashelsky, Electronic Devices and Circuit Theory, 10 th edition, may

reproduce material from the instructor’s text solutions manual for classroom use.

10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-503865-9

ISBN-10: 0-13-503865-0

Contents

Solutions to Problems in Text 1

Solutions for Laboratory Manual 185

iii

Chapter 1

1. Copper has 20 orbiting electrons with only one electron in the outermost shell. The fact that

the outermost shell with its 29 th electron is incomplete (subshell can contain 2 electrons) and

distant from the nucleus reveals that this electron is loosely bound to its parent atom. The

application of an external electric field of the correct polarity can easily draw this loosely

bound electron from its atomic structure for conduction.

Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent

bonding) of electrons between atoms. Electrons that are part of a complete shell structure

require increased levels of applied attractive forces to be removed from their parent atom.

2. Intrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as

physically possible. That is, one with the fewest possible number of impurities.

3. −

Negative temperature coefficient: materials with negative temperature coefficients have

decreasing resistance levels as the temperature increases.

Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to

form complete outermost shells and a more stable lattice structure.

4. W = QV = (6 C)(3 V) = 18 J

5. 48 eV = 48(1.6 × 10 −19 J) = 76.8 × 10 −19 J

Q = W −19

V = 76.8×

10 J

= 6.40 × 10 −19 C

12 V

6.4 × 10 −19 C is the charge associated with 4 electrons.

6. GaP Gallium Phosphide E g = 2.24 eV

ZnS Zinc Sulfide E g = 3.67 eV

7. An n-type semiconductor material has an excess of electrons for conduction established by

doping an intrinsic material with donor atoms having more valence electrons than needed to

establish the covalent bonding. The majority carrier is the electron while the minority carrier

is the hole.

A p-type semiconductor material is formed by doping an intrinsic material with acceptor

atoms having an insufficient number of electrons in the valence shell to complete the covalent

bonding thereby creating a hole in the covalent structure. The majority carrier is the hole

while the minority carrier is the electron.

8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has

only 3 electrons in the valence shell.

9. Majority carriers are those carriers of a material that far exceed the number of any other

carriers in the material.

Minority carriers are those carriers of a material that are less in number than any other carrier

of the material.

1

10. Same basic appearance as Fig. 1.7 since arsenic also has 5 valence electrons (pentavalent).

11. Same basic appearance as Fig. 1.9 since boron also has 3 valence electrons (trivalent).

12. −

13. −

14. For forward bias, the positive potential is applied to the p-type material and the negative

potential to the n-type material.

15. T K = 20 + 273 = 293

k = 11,600/n = 11,600/2 (low value of V D ) = 5800

(5800)(0.6)

kVD

⎛ ⎞

⎛

⎞

TK

I D = I s e − 1

= 50 × 10 −9 293

⎜e

− 1⎟

⎜ ⎟

⎝

⎠

⎝ ⎠

= 50 × 10 −9 (e 11.877 − 1) = 7.197 mA

16. k = 11,600/n = 11,600/2 = 5800 (n = 2 for V D = 0.6 V)

T K = T C + 273 = 100 + 273 = 373

(5800)(0.6 V)

kV / T K

373 9.33

e = e = e = 11.27 × 10 3

I =

I e − = 5 μA(11.27 × 10 3 − 1) = 56.35 mA

/

( kV T K

s

1)

17. (a) T K = 20 + 273 = 293

k = 11,600/n = 11,600/2 = 5800

kVD

⎛ ⎞

(5800)( −10 V)

T

⎛

⎞

K

293

I D = I s e − 1

= 0.1μA e 1

⎜ ⎟

⎜ − ⎟

⎝ ⎠ ⎝

⎠

= 0.1 × 10 −6 (e −197.95 − 1) = 0.1 × 10 −6 (1.07 × 10 −86 − 1)

≅ 0.1 × 10 −6 0.1μA

I D = I s = 0.1 μA

(b) The result is expected since the diode current under reverse-bias conditions should equal

the saturation value.

18. (a)

x y = e x

0 1

1 2.7182

2 7.389

3 20.086

4 54.6

5 148.4

(b) y = e 0 = 1

(c) For V = 0 V, e 0 = 1 and I = I s (1 − 1) = 0 mA

2

19. T = 20°C: I s = 0.1 μA

T = 30°C: I s = 2(0.1 μA) = 0.2 μA (Doubles every 10°C rise in temperature)

T = 40°C: I s = 2(0.2 μA) = 0.4 μA

T = 50°C: I s = 2(0.4 μA) = 0.8 μA

T = 60°C: I s = 2(0.8 μA) = 1.6 μA

1.6 μA: 0.1 μA ⇒ 16:1 increase due to rise in temperature of 40°C.

20. For most applications the silicon diode is the device of choice due to its higher temperature

capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be

used at temperatures approaching 200 degrees centigrade. Silicon diodes also have a higher

current handling capability. Germanium diodes are the better device for some RF small signal

applications, where the smaller threshold voltage may prove advantageous.

21. From 1.19:

V F

@ 10 mA

I s

−75°C 25°C 125°C

1.1 V 0.85 V 0.6 V

0.01 pA 1 pA 1.05 μA

V F decreased with increase in temperature

1.1 V: 0.6 V ≅ 1.83:1

I s increased with increase in temperature

1.05 μA: 0.01 pA = 105 × 10 3 :1

22. An “ideal” device or system is one that has the characteristics we would prefer to have when

using a device or system in a practical application. Usually, however, technology only

permits a close replica of the desired characteristics. The “ideal” characteristics provide an

excellent basis for comparison with the actual device characteristics permitting an estimate of

how well the device or system will perform. On occasion, the “ideal” device or system can be

assumed to obtain a good estimate of the overall response of the design. When assuming an

“ideal” device or system there is no regard for component or manufacturing tolerances or any

variation from device to device of a particular lot.

23. In the forward-bias region the 0 V drop across the diode at any level of current results in a

resistance level of zero ohms – the “on” state – conduction is established. In the reverse-bias

region the zero current level at any reverse-bias voltage assures a very high resistance level −

the open circuit or “off” state − conduction is interrupted.

24. The most important difference between the characteristics of a diode and a simple switch is

that the switch, being mechanical, is capable of conducting current in either direction while

the diode only allows charge to flow through the element in one direction (specifically the

direction defined by the arrow of the symbol using conventional current flow).

25. V D ≅ 0.66 V, I D = 2 mA

VD

0.65 V

R DC = = = 325 Ω

I 2 mA

D

3

26. At I D = 15 mA, V D = 0.82 V

VD

0.82 V

R DC = = = 54.67 Ω

I

D

15 mA

As the forward diode current increases, the static resistance decreases.

27. V D = −10 V, I D = I s = −0.1 μA

VD

10 V

R DC =

I

= D

0.1 μA

= 100 MΩ

V D = −30 V, I D = I s = −0.1 μA

VD

30 V

R DC = = = 300 MΩ

I 0.1μA

D

As the reverse voltage increases, the reverse resistance increases directly (since the diode

leakage current remains constant).

28. (a) r d =

(b) r d =

ΔV

ΔI

d

d

0.79 V − 0.76 V 0.03 V

= =

15 mA −5 mA 10 mA

26 mV 26 mV

= = 2.6 Ω

10 mA

I D

= 3 Ω

(c) quite close

29. I D = 10 mA, V D = 0.76 V

VD

0.76 V

R DC = = = 76 Ω

I 10 mA

r d =

ΔV

ΔI

D

d

d

R DC >> r d

0.79 V − 0.76 V 0.03 V

≅ =

15 mA −5 mA 10 mA

= 3 Ω

30. I D = 1 mA, r d =

I D = 15 mA, r d =

ΔV

ΔI

d

d

ΔV

ΔI

0.72 V − 0.61 V

=

= 55 Ω

2 mA −0 mA

d

d

0.8 V − 0.78 V

=

= 2 Ω

20 mA −10 mA

⎛26 mV ⎞

31. I D = 1 mA, r d = 2⎜

⎟ = 2(26 Ω) = 52 Ω vs 55 Ω (#30)

⎝ I D ⎠

26 mV 26 mV

I D = 15 mA, r d = = = 1.73 Ω vs 2 Ω (#30)

15 mA

I D

32. r av =

ΔV

ΔI

d

d

0.9 V − 0.6 V

=

= 24.4 Ω

13.5 mA −1.2 mA

4

33. r d =

ΔV

ΔI

d

d

0.8 V − 0.7 V 0.09 V

≅ = = 22.5 Ω

7 mA −3 mA 4 mA

(relatively close to average value of 24.4 Ω (#32))

34. r av =

ΔV

ΔI

d

d

0.9 V − 0.7 V 0.2 V

= =

14 mA −0 mA 14 mA

= 14.29 Ω

35. Using the best approximation to the curve beyond V D = 0.7 V:

ΔVd

0.8 V − 0.7 V 0.1 V

r av = ≅ = = 4 Ω

ΔI

25 mA −0 mA 25 mA

d

36. (a) V R = −25 V: C T ≅ 0.75 pF

V R = −10 V: C T ≅ 1.25 pF

ΔC

ΔV

T

R

1.25 pF − 0.75 pF 0.5 pF

= =

10 V −25 V 15 V

= 0.033 pF/V

(b) V R = −10 V: C T ≅ 1.25 pF

V R = −1 V: C T ≅ 3 pF

ΔC

ΔV

T

R

1.25 pF − 3 pF 1.75 pF

= =

10 V −1 V 9 V

= 0.194 pF/V

(c) 0.194 pF/V: 0.033 pF/V = 5.88:1 ≅ 6:1

Increased sensitivity near V D = 0 V

37. From Fig. 1.33

V D = 0 V, C D = 3.3 pF

V D = 0.25 V, C D = 9 pF

38. The transition capacitance is due to the depletion region acting like a dielectric in the reversebias

region, while the diffusion capacitance is determined by the rate of charge injection into

the region just outside the depletion boundaries of a forward-biased device. Both

capacitances are present in both the reverse- and forward-bias directions, but the transition

capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is

the dominant effect for forward-biased conditions.

5

39. V D = 0.2 V, C D = 7.3 pF

1 1

X C = = = 3.64 kΩ

2π

fC 2 π(6 MHz)(7.3 pF)

V D = −20 V, C T = 0.9 pF

1 1

X C = = = 29.47 kΩ

2π

fC 2 π(6 MHz)(0.9 pF)

40. I f = 10 V

10 kΩ = 1 mA

t s + t t = t rr = 9 ns

t s + 2t s = 9 ns

t s = 3 ns

t t = 2t s = 6 ns

41.

42. As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a

level of about 5 pF with no bias. For reverse-bias potentials in excess of 10 V the capacitance

levels off at about 1.5 pF.

43. At V D = −25 V, I D = −0.2 nA and at V D = −100 V, I D ≅ −0.45 nA. Although the change in I R is

more than 100%, the level of I R and the resulting change is relatively small for most

applications.

44. Log scale: T A = 25°C, I R = 0.5 nA

T A = 100°C, I R = 60 nA

The change is significant.

60 nA: 0.5 nA = 120:1

Yes, at 95°C I R would increase to 64 nA starting with 0.5 nA (at 25°C)

(and double the level every 10°C).

6

45. I F = 0.1 mA: r d ≅ 700 Ω

I F = 1.5 mA: r d ≅ 70 Ω

I F = 20 mA: r d ≅ 6 Ω

The results support the fact that the dynamic or ac resistance decreases rapidly with

increasing current levels.

46. T = 25°C: P max = 500 mW

T = 100°C: P max = 260 mW

P max = V F I F

Pmax

500 mW

I F = = = 714.29 mA

V

F

0.7 V

Pmax

260 mW

I F = = = 371.43 mA

V 0.7 V

F

714.29 mA: 371.43 mA = 1.92:1 ≅ 2:1

47. Using the bottom right graph of Fig. 1.37:

I F = 500 mA @ T = 25°C

At I F = 250 mA, T ≅ 104°C

48.

ΔVZ

49. T C = +0.072% = × 100%

VZ

( T1 − T0)

0.75 V

0.072 =

× 100

10 V( T1

− 25)

7.5

0.072 =

T − 25

1

7.5

T 1 − 25° =

0.072 = 104.17°

T 1 = 104.17° + 25° = 129.17°

50. T C =

=

ΔVZ

× 100%

V ( T − T )

Z

1 0

(5 V − 4.8 V)

× 100% = 0.053%/°C

5 V(100°− 25 ° )

7

51.

(20 V − 6.8 V)

× 100% = 77%

(24 V − 6.8 V)

The 20 V Zener is therefore ≅ 77% of the distance between 6.8 V and 24 V measured from

the 6.8 V characteristic.

At I Z = 0.1 mA, T C ≅ 0.06%/°C

(5 V − 3.6 V)

× 100% = 44%

(6.8 V − 3.6 V)

The 5 V Zener is therefore ≅ 44% of the distance between 3.6 V and 6.8 V measured from the

3.6 V characteristic.

At I Z = 0.1 mA, T C ≅ −0.025%/°C

52.

53. 24 V Zener:

0.2 mA: ≅ 400 Ω

1 mA: ≅ 95 Ω

10 mA: ≅ 13 Ω

The steeper the curve (higher dI/dV) the less the dynamic resistance.

54. V T ≅ 2.0 V, which is considerably higher than germanium (≅ 0.3 V) or silicon (≅ 0.7 V). For

germanium it is a 6.7:1 ratio, and for silicon a 2.86:1 ratio.

55. Fig. 1.53 (f) I F ≅ 13 mA

Fig. 1.53 (e) V F ≅ 2.3 V

56. (a) Relative efficiency @ 5 mA ≅ 0.82

@ 10 mA ≅ 1.02

1.02 − 0.82

× 100% = 24.4% increase

0.82

ratio: 1.02

0.82 = 1.24

(b) Relative efficiency @ 30 mA ≅ 1.38

@ 35 mA ≅ 1.42

1.42 −1.38

× 100% = 2.9% increase

1.38

ratio: 1.42

1.38 = 1.03

(c) For currents greater than about 30 mA the percent increase is significantly less than for

increasing currents of lesser magnitude.

8

57. (a)

0.75

3.0 = 0.25

From Fig. 1.53 (i) ≅ 75°

(b) 0.5 ⇒ = 40 °

58. For the high-efficiency red unit of Fig. 1.53:

0.2 mA 20 mA

=

° C x

20 mA

x =

= 100°C

0.2 mA/ ° C

9

Chapter 2

1. The load line will intersect at I D =

E 8 V

R = 330 Ω = 24.24 mA and V D = 8 V.

(a)

(b)

(c)

V

D Q

≅ 0.92 V

I

D Q

≅ 21.5 mA

V R = E − V

D Q

= 8 V − 0.92 V = 7.08 V

V

D Q

≅ 0.7 V

I

D Q

≅ 22.2 mA

V R = E − V

D Q

= 8 V − 0.7 V = 7.3 V

V

D Q

≅ 0 V

I

D Q

≅ 24.24 mA

V R = E − V

D Q

= 8 V − 0 V = 8 V

For (a) and (b), levels of V

D Q

and I

D Q

are quite close. Levels of part (c) are reasonably close

but as expected due to level of applied voltage E.

E 5 V

2. (a) I D =

R = = 2.27 mA

2.2 kΩ The load line extends from I D = 2.27 mA to V D = 5 V.

V ≅ 0.7 V, I ≅ 2 mA

D Q

D Q

E 5 V

(b) I D =

R = = 10.64 mA


V ≅ 0.8 V, I ≅ 9 mA

D Q

D Q

E 5 V

(c) I D =

R = = 27.78 mA


V ≅ 0.93 V, I ≅ 22.5 mA

D Q

D Q

The resulting values of

3. Load line through

11.25 mA.

I D = 11.25 mA =

with R =

V

D Q

are quite close, while

I

D Q

extends from 2 mA to 22.5 mA.

I

D Q

= 10 mA of characteristics and V D = 7 V will intersect I D axis as

E 7 V

=

R R

7 V

11.25 mA

= 0.62 kΩ

10

E−V 30 V 0.7 V

4. (a) I D = I R = D

−

=

= 13.32 mA

R 2.2 kΩ

V D = 0.7 V, V R = E − V D = 30 V − 0.7 V = 29.3 V

E−V 30 V 0 V

(b) I D = D

−

=

R 2.2 kΩ

V D = 0 V, V R = 30 V

= 13.64 mA

Yes, since E V T the levels of I D and V R are quite close.

5. (a) I = 0 mA; diode reverse-biased.

(b) V 20Ω = 20 V − 0.7 V = 19.3 V (Kirchhoff’s voltage law)

I = 19.3 V

20 Ω = 0.965 A

(c) I = 10 V = 1 A; center branch open

10 Ω

6. (a) Diode forward-biased,

Kirchhoff’s voltage law (CW): −5 V + 0.7 V − V o = 0

V o = −4.3 V

V 4.3 V

I R = I D =

o

R = = 1.955 mA

2.2 kΩ (b) Diode forward-biased,

8 V − 0.7 V

I D =

= 1.24 mA

1.2 kΩ+ 4.7 kΩ V o = V 4.7 kΩ + V D = (1.24 mA)(4.7 kΩ) + 0.7 V

= 6.53 V

7. (a) V o =

2 k Ω(20 V −0.7 V −0.3V)

2 kΩ+ 2 kΩ

= 1 2 (20 V – 1 V) = 1 (19 V) = 9.5 V

2

10 V + 2V − 0.7 V) 11.3 V

(b) I = = = 1.915 mA

1.2 kΩ+ 4.7 kΩ 5.9 kΩ V ′ = IR = (1.915 mA)(4.7 kΩ) = 9 V

V o = V ′ − 2 V = 9 V − 2 V = 7 V

11

8. (a) Determine the Thevenin equivalent circuit for the 10 mA source and 2.2 kΩ resistor.

E Th = IR = (10 mA)(2.2 kΩ) = 22 V

R Th = 2. 2kΩ

(b) Diode forward-biased

I D = 20 V + 5 V − 0.7 V = 2.65 mA

6.8 kΩ

Kirchhoff’s voltage law (CW):

+V o − 0.7 V + 5 V = 0

V o = −4.3 V

Diode forward-biased

22 V − 0.7 V

I D =

= 6.26 mA

2.2 kΩ + 1.2 kΩ V o = I D (1.2 kΩ)

= (6.26 mA)(1.2 kΩ)

= 7.51 V

9. (a)

(b)

I =

V

o 1

= 12 V – 0.7 V = 11.3 V

V

o 2

= 0.3 V

V

o 1

= −10 V + 0.3 V + 0.7 V = −9 V

10 V −0.7 V −0.3 V 9 V

=

1.2 kΩ+ 3.3 kΩ 4.5 kΩ = 2 mA, V

o

= −(2 mA)(3.3 kΩ) = −6.6 V

2

10. (a) Both diodes forward-biased

I R = 20 V − 0.7 V = 4.106 mA

4.7 kΩ

Assuming identical diodes:

I 4.106 mA

I D = R

= = 2.05 mA

2 2

V o = 20 V − 0.7 V = 19.3 V

(b) Right diode forward-biased:

I D = 15 V + 5 V − 0.7 V = 8.77 mA

2.2 kΩ

V o = 15 V − 0.7 V = 14.3 V

11. (a) Ge diode “on” preventing Si diode from turning “on”:

10 V − 0.3 V 9.7 V

I = = = 9.7 mA

1 kΩ

1 kΩ 16 V −0.7 V −0.7 V −12 V 2.6 V

(b) I = = = 0.553 mA

4.7 kΩ

4.7 kΩ V o = 12 V + (0.553 mA)(4.7 kΩ) = 14.6 V

12

12. Both diodes forward-biased:

V = 0.7 V, V = 0.3 V

o 1

o 2

20 V − 0.7 V

I 1 kΩ = = 19.3 V = 19.3 mA

1 kΩ

1 kΩ

I 0.47 kΩ = 0.7 V − 0.3 V = 0.851 mA

0.47 kΩ

I(Si diode) = I 1 kΩ − I 0.47 kΩ

= 19.3 mA − 0.851 mA

= 18.45 mA

13. For the parallel Si − 2 kΩ branches a Thevenin equivalent will result (for “on” diodes) in a

single series branch of 0.7 V and 1 kΩ resistor as shown below:

I 2 kΩ = 6.2 V = 3.1 mA

2 kΩ

I2 k

3.1 mA

I D = Ω

= = 1.55 mA

2 2

14. Both diodes “off”. The threshold voltage of 0.7 V is unavailable for either diode.

V o = 0 V

15. Both diodes “on”, V o = 10 V − 0.7 V = 9.3 V

16. Both diodes “on”.

V o = 0.7 V

17. Both diodes “off”, V o = 10 V

18. The Si diode with −5 V at the cathode is “on” while the other is “off”. The result is

V o = −5 V + 0.7 V = −4.3 V

19. 0 V at one terminal is “more positive” than −5 V at the other input terminal. Therefore

assume lower diode “on” and upper diode “off”.

The result:

V o = 0 V − 0.7 V = −0.7 V

The result supports the above assumptions.

20. Since all the system terminals are at 10 V the required difference of 0.7 V across either diode

cannot be established. Therefore, both diodes are “off” and

V o = +10 V

as established by 10 V supply connected to 1 kΩ resistor.

13

21. The Si diode requires more terminal voltage than the Ge diode to turn “on”. Therefore, with

5 V at both input terminals, assume Si diode “off” and Ge diode “on”.

The result: V o = 5 V − 0.3 V = 4.7 V

The result supports the above assumptions.

Vdc 2 V

22. V dc = 0.318 V m ⇒V m = = = 6.28 V

0.318 0.318

I m =

V m

6.28 V

R = 2.2 kΩ = 2.85 mA

23. Using V dc ≅ 0.318(V m − V T )

2 V = 0.318(V m − 0.7 V)

Solving: V m = 6.98 V ≅ 10:1 for V m :V T

Vdc 2 V

24. V m = = = 6.28 V

0.318 0.318

I

L max

= 6.28 V

6.8 kΩ

= 0.924 mA

14

I max (2.2 kΩ) = 6.28 V = 2.855 mA

2.2 kΩ

I = I + I max (2.2 kΩ) = 0.924 mA + 2.855 mA = 3.78 mA

Dmax

Lmax

25. V m = 2 (110 V) = 155.56 V

V dc = 0.318V m = 0.318(155.56 V) = 49.47 V

26. Diode will conduct when v o = 0.7 V; that is,

10 k Ω( v )

v o = 0.7 V =

i

10 kΩ+ 1 kΩ

Solving: v i = 0.77 V

For v i ≥ 0.77 V Si diode is “on” and v o = 0.7 V.

For v i < 0.77 V Si diode is open and level of v o is determined

by voltage divider rule:

10 k Ω( v )

v o =

i

10 kΩ+ 1 kΩ = 0.909 v i

For v i = −10 V:

v o = 0.909(−10 V)

= −9.09 V

When v o = 0.7 V, vR

= v

max i

− 0.7 V

max

= 10 V − 0.7 V = 9.3 V

9.3 V

I

R

= = 9.3 mA

max

1 kΩ 10 V

I max (reverse) =

= 0.909 mA

1 kΩ+ 10 kΩ 15

27. (a) P max = 14 mW = (0.7 V)I D

14 mW

I D = = 20 mA

0.7 V

(b) 4.7 kΩ || 56 kΩ = 4.34 kΩ

V R = 160 V − 0.7 V = 159.3 V

I max = 159.3 V = 36.71 mA

4.34 kΩ

Imax

36.71 mA

(c) I diode = = = 18.36 mA

2 2

(d) Yes, I D = 20 mA > 18.36 mA

(e) I diode = 36.71 mA I max = 20 mA

28. (a) V m = 2 (120 V) = 169.7 V

V

L m

= V

i m

− 2V D

= 169.7 V − 2(0.7 V) = 169.7 V − 1.4 V

= 168.3 V

V dc = 0.636(168.3 V) = 107.04 V

(b) PIV = V m (load) + V D = 168.3 V + 0.7 V = 169 V

(c) I D (max) =

VL m

168.3 V

R = 1 kΩ L

= 168.3 mA

(d) P max = V D I D = (0.7 V)I max

= (0.7 V)(168.3 mA)

= 117.81 mW

29.

16

30. Positive half-cycle of v i :

Voltage-divider rule:

2.2 k Ω( V i

)

max

V

o max

=

2.2 kΩ+ 2.2 kΩ

1

= ( V

imax

)

2

= 1 (100 V)

2

= 50 V

31. Positive pulse of v i :

Top left diode “off”, bottom left diode “on”

2.2 kΩ || 2.2 kΩ = 1.1 kΩ

1.1 k Ω(170 V)

V

o peak

=

1.1 kΩ + 2.2 kΩ = 56.67 V

Negative pulse of v i :

Top left diode “on”, bottom left diode “off”

1.1 k Ω(170 V)

V

o peak

=

1.1 kΩ + 2.2 kΩ = 56.67 V

V dc = 0.636(56.67 V) = 36.04 V

V dc = 0.636V m = 0.636 (50 V)

= 31.8 V

32. (a) Si diode open for positive pulse of v i and v o = 0 V

For −20 V < v i ≤ −0.7 V diode “on” and v o = v i + 0.7 V.

For v i = −20 V, v o = −20 V + 0.7 V = −19.3 V

For v i = −0.7 V, v o = −0.7 V + 0.7 V = 0 V

Polarity of v o across the 2.2 kΩ

resistor acting as a load is the same.

Voltage-divider rule:

2.2 k Ω( V i

)

max

V

o max

=

2.2 kΩ+ 2.2 kΩ

1

= ( V

imax

)

2

= 1 (100 V)

2

= 50 V

17

(b) For v i ≤ 5 V the 5 V battery will ensure the diode is forward-biased and v o = v i − 5 V.

At v i = 5 V

v o = 5 V − 5 V = 0 V

At v i = −20 V

v o = −20 V − 5 V = −25 V

For v i > 5 V the diode is reverse-biased and v o = 0 V.

33. (a) Positive pulse of v i :

1.2 k Ω(10 V −0.7 V)

V o =

1.2 kΩ+ 2.2 kΩ


diode “open”, v o = 0 V

= 3.28 V

(b) Positive pulse of v i :

V o = 10 V − 0.7 V + 5 V = 14.3 V


diode “open”, v o = 0 V

34. (a) For v i = 20 V the diode is reverse-biased and v o = 0 V.

For v i = −5 V, v i overpowers the 2 V battery and the diode is “on”.

Applying Kirchhoff’s voltage law in the clockwise direction:

−5 V + 2 V − v o = 0

v o = −3 V

(b) For v i = 20 V the 20 V level overpowers the 5 V supply and the diode is “on”. Using the

short-circuit equivalent for the diode we find v o = v i = 20 V.

For v i = −5 V, both v i and the 5 V supply reverse-bias the diode and separate v i from v o .

However, v o is connected directly through the 2.2 kΩ resistor to the 5 V supply and

v o = 5 V.

18

35. (a) Diode “on” for v i ≥ 4.7 V

For v i > 4.7 V, V o = 4 V + 0.7 V = 4.7 V

For v i < 4.7 V, diode “off” and v o = v i

(b) Again, diode “on” for v i ≥ 4.7 V but v o

now defined as the voltage across the diode

For v i ≥ 4.7 V, v o = 0.7 V

For v i < 4.7 V, diode “off”, I D = I R = 0 mA and V 2.2 kΩ = IR = (0 mA)R = 0 V

Therefore, v o = v i − 4 V

At v i = 0 V, v o = −4 V

v i = −8 V, v o = −8 V − 4 V = −12 V

36. For the positive region of v i :

The right Si diode is reverse-biased.

The left Si diode is “on” for levels of v i greater than

5.3 V + 0.7 V = 6 V. In fact, v o = 6 V for v i ≥ 6 V.

For v i < 6 V both diodes are reverse-biased and v o = v i .

For the negative region of v i :

The left Si diode is reverse-biased.

The right Si diode is “on” for levels of v i more negative than 7.3 V + 0.7 V = 8 V. In

fact, v o = −8 V for v i ≤ −8 V.

For v i > −8 V both diodes are reverse-biased and v o = v i .

i R : For −8 V < v i < 6 V there is no conduction through the 10 kΩ resistor due to the lack of a

complete circuit. Therefore, i R = 0 mA.

For v i ≥ 6 V

v R = v i − v o = v i − 6 V

For v i = 10 V, v R = 10 V − 6 V = 4 V

and i R =

4 V

10 kΩ

= 0.4 mA

For v i ≤ −8 V

v R = v i − v o = v i + 8 V

19

For v i = −10 V

v R = −10 V + 8 V = −2 V

−2V

and i R = = −0.2 mA

10 kΩ 37. (a) Starting with v i = −20 V, the diode is in the “on” state and the capacitor quickly charges

to −20 V+. During this interval of time v o is across the “on” diode (short-current

equivalent) and v o = 0 V.

When v i switches to the +20 V level the diode enters the “off” state (open-circuit

equivalent) and v o = v i + v C = 20 V + 20 V = +40 V

(b) Starting with v i = −20 V, the diode is in the “on” state and the capacitor quickly charges

up to −15 V+. Note that v i = +20 V and the 5 V supply are additive across the capacitor.

During this time interval v o is across “on” diode and 5 V supply and v o = −5 V.

When v i switches to the +20 V level the diode enters the “off” state and v o = v i + v C =

20 V + 15 V = 35 V.

20

38. (a) For negative half cycle capacitor charges to peak value of 120 V − 0.7 V = 119.3 V with

polarity

. The output v o is directly across the “on” diode resulting in

v o = −0.7 V as a negative peak value.

For next positive half cycle v o = v i + 119.3 V with peak value of

v o = 120 V + 119.3 V = 239.3 V.

(b) For positive half cycle capacitor charges to peak value of 120 V − 20 V − 0.7 V = 99.3 V

with polarity

. The output v o = 20 V + 0.7 V = 20.7 V

For next negative half cycle v o = v i − 99.3 V with negative peak value of

v o = −120 V − 99.3 V = −219.3 V.

Using the ideal diode approximation the vertical shift of part (a) would be 120 V rather

than 119.3 V and −100 V rather than −99.3 V for part (b). Using the ideal diode

approximation would certainly be appropriate in this case.

39. (a) τ = RC = (56 kΩ)(0.1 μF) = 5.6 ms

5τ = 28 ms

(b) 5τ = 28 ms 2

T =

1 ms

2

= 0.5 ms, 56:1

(c) Positive pulse of v i :

Diode “on” and v o = −2 V + 0.7 V = −1.3 V

Capacitor charges to 10 V + 2 V − 0.7 V = 11.3 V


Diode “off” and v o = −10 V − 11.3 V = −21.3 V

21

40. Solution is network of Fig. 2.176(b) using a 10 V supply in place of the 5 V source.

41. Network of Fig. 2.178 with 2 V battery reversed.

42. (a) In the absence of the Zener diode

180 Ω(20 V)

V L =

180 Ω+ 220 Ω = 9 V

V L = 9 V < V Z = 10 V and diode non-conducting

20 V

Therefore, I L = I R =

220 Ω+ 180 Ω

with I Z = 0 mA

and V L = 9 V

= 50 mA

(b) In the absence of the Zener diode

470 Ω(20 V)

V L =

470 Ω+ 220 Ω = 13.62 V

V L = 13.62 V > V Z = 10 V and Zener diode “on”

Therefore, V L = 10 V and V

R s

= 10 V

I = V / R = 10 V/220 Ω = 45.45 mA

Rs

Rs

s

I L = V L /R L = 10 V/470 Ω = 21.28 mA

and I Z = I − I L = 45.45 mA − 21.28 mA = 24.17 mA

R s

(c)

P

Z max

= 400 mW = V Z I Z = (10 V)(I Z )

400 mW

I Z = = 40 mA

10 V

I

L min

= IR

− I

s Z

= 45.45 mA − 40 mA = 5.45 mA

max

VL

10 V

R L = = = 1,834.86 Ω

I 5.45 mA

Lmin

Large R L reduces I L and forces more of

I

R s

to pass through Zener diode.

(d) In the absence of the Zener diode

RL

(20 V)

V L = 10 V =

R

L

+ 220 Ω

10R L + 2200 = 20R L

10R L = 2200

R L = 220 Ω

22

VL

12 V

43. (a) V Z = 12 V, R L = = = 60 Ω

I

L

200 mA

RV

L i

60 Ω(16 V)

V L = V Z = 12 V = =

RL + Rs 60 Ω+ Rs

720 + 12R s = 960

12R s = 240

R s = 20 Ω

(b)

P

Z max

= V Z I

Zmax

= (12 V)(200 mA)

= 2.4 W

VL VZ

44. Since I L = = is fixed in magnitude the maximum value of

RL

RL

maximum. The maximum level of

of V i .

PZ

400 mW

max

IZ

= = = 50 mA

max

VZ

8 V

VL

VZ

8 V

I L = = = = 36.36 mA

RL

RL

220 Ω

I

R s

= I Z + I L = 50 mA + 36.36 mA = 86.36 mA

Vi

−VZ

I

R

=

s

Rs

or V i = IR

R

s s

+ V Z

= (86.36 mA)(91 Ω) + 8 V = 7.86 V + 8 V = 15.86 V

I

R s

will occur when I Z is a

I

R s

will in turn determine the maximum permissible level

Any value of v i that exceeds 15.86 V will result in a current I Z that will exceed the maximum

value.

45. At 30 V we have to be sure Zener diode is “on”.

RV

L i

1 k Ω(30 V)

∴ V L = 20 V = =

RL + Rs 1 kΩ+

Rs

Solving, R s = 0.5 kΩ

At 50 V,

I ZM =

I RS

50 V − 20 V

=

0.5 kΩ

= 60 mA, I L = 20 V

1 kΩ

I

R S

− I L = 60 mA − 20 mA = 40 mA

= 20 mA

46. For v i = +50 V:

Z 1 forward-biased at 0.7 V

Z 2 reverse-biased at the Zener potential and V

Z 2

= 10 V.

Therefore, V o = V + V = 0.7 V + 10 V = 10.7 V

Z1 Z2

23

For v i = −50 V:

Z 1 reverse-biased at the Zener potential and V

Z 1

= −10 V.

Z 2 forward-biased at −0.7 V.

Therefore, V o = V + V = −10.7 V

Z1 Z2

For a 5 V square wave neither Zener diode will reach its Zener potential. In fact, for either

polarity of v i one Zener diode will be in an open-circuit state resulting in v o = v i .

47. V m = 1.414(120 V) = 169.68 V

2V m = 2(169.68 V) = 339.36 V

48. The PIV for each diode is 2V m

∴PIV = 2(1.414)(V rms )

24

Chapter 3

1. −

2. A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while

unipolar devices utilize either electrons or holes, but not both, in the charge flow process.

3. Forward- and reverse-biased.

4. The leakage current I CO is the minority carrier current in the collector.

5. −

6. −

7. −

8. I E the largest

I B the smallest

I C ≅ I E

1

9. I B =

100 I ⇒ I C C = 100I B

I E = I C + I B = 100I B + I B = 101I B

I 8 mA

I B = E

= = 79.21 μA

101 101

I C = 100I B = 100(79.21 μA) = 7.921 mA

10. −

11. I E = 5 mA, V CB = 1 V: V BE = 800 mV

V CB = 10 V: V BE = 770 mV

V CB = 20 V: V BE = 750 mV

The change in V CB is 20 V:1 V = 20:1

The resulting change in V BE is 800 mV:750 mV = 1.07:1 (very slight)

ΔV

0.9 V −0.7 V

12. (a) r av = =

= 25 Ω

ΔI

8 mA −0

(b) Yes, since 25 Ω is often negligible compared to the other resistance levels of the

network.

13. (a) I C ≅ I E = 4.5 mA

(b) I C ≅ I E = 4.5 mA

(c) negligible: change cannot be detected on this set of characteristics.

(d) I C ≅ I E

25

14. (a) Using Fig. 3.7 first, I E ≅ 7 mA

Then Fig. 3.8 results in I C ≅ 7 mA

(b) Using Fig. 3.8 first, I E ≅ 5 mA

Then Fig. 3.7 results in V BE ≅ 0.78 V

(c) Using Fig. 3.10(b) I E = 5 mA results in V BE ≅ 0.81 V

(d) Using Fig. 3.10(c) I E = 5 mA results in V BE = 0.7 V

(e) Yes, the difference in levels of V BE can be ignored for most applications if voltages of

several volts are present in the network.

15. (a) I C = α I E = (0.998)(4 mA) = 3.992 mA

16. −

(b) I E = I C + I B ⇒ I C = I E − I B = 2.8 mA − 0.02 mA = 2.78 mA

IC

2.78 mA

α dc = = = 0.993

I 2.8 mA

E

⎛ α ⎞ ⎛ 0.98 ⎞

(c) I C = βI B = ⎜ ⎟I B

= ⎜ ⎟(40 μA) = 1.96 mA

⎝1−α

⎠ ⎝1−0.98⎠ I 1.96 mA

I E = C

α = 0.993

= 2 mA

17. I i = V i /R i = 500 mV/20 Ω = 25 mA

I L ≅ I i = 25 mA

V L = I L R L = (25 mA)(1 kΩ) = 25 V

Vo

25 V

A v =

V = 0.5 V

= 50

i

Vi

200 mV 200 mV

18. I i =

Ri

+ R = s

20 Ω+ 100 Ω = 120 Ω

I L = I i = 1.67 mA

V L = I L R = (1.67 mA)(5 kΩ) = 8.35 V

Vo

8.35 V

A v =

V = 0.2 V

= 41.75

i

= 1.67 mA

19. −

20. (a) Fig. 3.14(b): I B ≅ 35μA

Fig. 3.14(a): I C ≅ 3.6 mA

(b) Fig. 3.14(a): V CE ≅ 2.5 V

Fig. 3.14(b): V BE ≅ 0.72 V

26

IC

2 mA

21. (a) β =

I

= B

17 μA

= 117.65

β 117.65

(b) α = =

β + 1 117.65+ 1

= 0.992

(c) I CEO = 0.3 mA

(d) I CBO = (1 − α)I CEO

= (1 − 0.992)(0.3 mA) = 2.4 μA

22. (a) Fig. 3.14(a): I CEO ≅ 0.3 mA

(b) Fig. 3.14(a): I C ≅ 1.35 mA

IC

1.35 mA

β dc =

I

= 10 μA

= 135

B

(c) α =

β 135

=

β + 1 136

= 0.9926

I CBO ≅ (1 − α)I CEO

= (1 − 0.9926)(0.3 mA)

= 2.2 μA

IC

6.7 mA

23. (a) β dc =

I

= B

80 μA

= 83.75

IC

0.85 mA

(b) β dc = = = 170

I

B

5 μA

IC

3.4 mA

(c) β dc =

I

= B

30 μA

= 113.33

(d) β dc does change from pt. to pt. on the characteristics.

Low I B , high V CE → higher betas

High I B , low V CE → lower betas

24. (a) β ac =

(b) β ac =

ΔI

ΔI

ΔI

ΔI

C

B

C

B

V

V

CE

CE

=

= 5 V

7.3 mA − 6 mA 1.3 mA

= = 65

90 μA − 70 μA 20 μA

1.4 mA − 0.3 mA 1.1 mA

= = = 110

= 15 V 10 μA − 0 μA 10 μA

ΔIC

4.25 mA − 2.35 mA 1.9 mA

(c) β ac =

= = = 95

ΔI

B

VCE

= 10 V 40 μA − 20 μA 20 μA

(d) β ac does change from point to point on the characteristics. The highest value was

obtained at a higher level of V CE and lower level of I C . The separation between I B curves

is the greatest in this region.

27

(e) V CE I B β dc β ac I C β dc /β ac

5 V 80 μA 83.75 65 6.7 mA 1.29

10 V 30 μA 113.33 95 3.4 mA 1.19

15 V 5 μA 170 110 0.85 mA 1.55

As I C decreased, the level of β dc and β ac increased. Note that the level of β dc and β ac in

the center of the active region is close to the average value of the levels obtained. In

each case β dc is larger than β ac , with the least difference occurring in the center of the

active region.

IC

2.9 mA

25. β dc = = = 116

I

B

25 μA

β 116

α = =

β + 1 116+ 1

= 0.991

I E = I C /α = 2.9 mA/0.991 = 2.93 mA

α 0.987 0.987

26. (a) β =

1 α = 1 0.987 = = 75.92

− − 0.013

β 120 120

(b) α = = = = 0.992

β + 1 120 + 1 121

I 2 mA

(c) I B = C

= = 11.11 μA

β 180

I E = I C + I B = 2 mA + 11.11 μA

= 2.011 mA

27. −

28. V e = V i − V be = 2 V − 0.1 V = 1.9 V

Vo

1.9 V

A v =

V = i

2 V

= 0.95 ≅ 1

VE

1.9 V

I e =

R = = 1.9 mA (rms)

1 kΩ E

29. Output characteristics:

Curves are essentially the

same with new scales as shown.

Input characteristics:

Common-emitter input characteristics may be used directly for common-collector

calculations.

28

30.

P

C max

= 30 mW = V CE I C

PC

30 mW

max

I C = I

C max

, V CE = =

IC

7 mA = 4.29 V

max

PC

30 mW

max

V CE = V

CE max

, I C = = = 1.5 mA

VCE

20 V

max

PC

30 mW

max

V CE = 10 V, I C = = = 3 mA

VCE

10 V

PC

30 mW

max

I C = 4 mA, V CE = =

IC

4 mA = 7.5 V

PC

30 mW

max

V CE = 15 V, I C = = = 2 mA

V 15 V

CE

31. I C =

PC

max

I

C max

, V CE =

I

Cmax

PC

max

V CB = V

CB max

, I C =

V

CBmax

30 mW

= = 5 V

6 mA

30 mW

= = 2 mA

15 V

PC

30 mW

max

I C = 4 mA, V CB = = = 7.5 V

I 4 mA

C

PC

30 mW

max

V CB = 10 V, I C = = = 3 mA

V 10 V

CB

29

32. The operating temperature range is −55°C ≤ T J ≤ 150°C

°F = 9 C

5 ° + 32°

= 9 ( − 55° C) + 32° = −67°F

5

°F = 9 (150° C) + 32° = 302°F

5

∴ −67°F ≤ T J ≤ 302°F

33.

I

C max

= 200 mA,

V

CE max

= 30 V,

P

P

D max

= 625 mW

D 625 mW

max

I C = I

C max

, V

CE

= =

IC

200 mA = 3.125 V

max

PD

625 mW

max

V CE = V

CE max

, I C = = = 20.83 mA

VCE

30 V

max

PD

625 mW

max

I C = 100 mA, V CE = = = 6.25 V

IC

100 mA

PD

625 mW

max

V CE = 20 V, I C = = = 31.25 mA

V 20 V

CE

34. From Fig. 3.23 (a) I CBO = 50 nA max

βmin + βmax

β avg =

2

50 + 150

= = 200

2 2

= 100

∴I CEO ≅ βI CBO = (100)(50 nA)

= 5 μA

30

35. h FE (β dc ) with V CE = 1 V, T = 25°C

I C = 0.1 mA, h FE ≅ 0.43(100) = 43

↓

I C = 10 mA, h FE ≅ 0.98(100) = 98

h fe (β ac ) with V CE = 10 V, T = 25°C

I C = 0.1 mA, h fe ≅ 72

↓

I C = 10 mA, h fe ≅ 160

For both h FE and h fe the same increase in collector current resulted in a similar increase

(relatively speaking) in the gain parameter. The levels are higher for h fe but note that V CE is

higher also.

36. As the reverse-bias potential increases in magnitude the input capacitance C ibo decreases (Fig.

3.23(b)). Increasing reverse-bias potentials causes the width of the depletion region to

⎛ A ⎞

increase, thereby reducing the capacitance ⎜C

=∈ ⎟

⎝ d ⎠ .

37. (a) At I C = 1 mA, h fe ≅ 120

At I C = 10 mA, h fe ≅ 160

(b) The results confirm the conclusions of problems 23 and 24 that beta tends to increase

with increasing collector current.

39. (a) β ac =

ΔI

ΔI

C

B

V

CE

= 3 V

=

16 mA −12.2 mA 3.8 mA

=

80 μA − 60 μA 20 μA

= 190

(b) β dc =

I

I

C

B

12 mA

= = 201.7

59.5 μA

(c) β ac =

4 mA − 2 mA 2 mA

=

18 μA − 8 μA 10 μA

= 200

(d) β dc =

I

I

C

B

3 mA

= = 230.77

13 μA

(e) In both cases β dc is slightly higher than β ac (≅ 10%)

(f)(g)

In general β dc and β ac increase with increasing I C for fixed V CE and both decrease for

decreasing levels of V CE for a fixed I E . However, if I C increases while V CE decreases

when moving between two points on the characteristics, chances are the level of β dc or

β ac may not change significantly. In other words, the expected increase due to an

increase in collector current may be offset by a decrease in V CE . The above data reveals

that this is a strong possibility since the levels of β are relatively close.

31

Chapter 4

1. (a)

I

BQ

VCC

−VBE

16 V − 0.7 V 15.3 V

= = =

R 470 kΩ

470 kΩ B

= 32.55 μA

(b)

I

CQ

= β I = (90)(32.55 μA) = 2.93 mA

BQ

(c)

V = V − I R = 16 V − (2.93 mA)(2.7 kΩ) = 8.09 V

CEQ

CC CQ

C

(d) V C =

V

CE Q

= 8.09 V

(e) V B = V BE = 0.7 V

(f) V E = 0 V

2. (a) I C = βI B = 80(40 μA) = 3.2 mA

(b) R C =

(c) R B =

VR V 12 V 6 V 6 V

C CC

−VC

−

= = = = 1.875 kΩ

I I 3.2 mA 3.2 mA

C

C

VR B

12 V − 0.7 V 11.3 V

= = = 282.5 kΩ

I 40 μA 40 μA

B

(d) V CE = V C = 6 V

3. (a) I C = I E − I B = 4 mA − 20 μA = 3.98 mA ≅ 4 mA

(b) V CC = V CE + I C R C = 7.2 V + (3.98 mA)(2.2 kΩ)

= 15.96 V ≅ 16 V

(c) β =

I

I

C

B

3.98 mA

= = 199 ≅ 200

20 μA

(d) R B =

VR V 15.96 V 0.7 V

B CC

−VBE

−

= = = 763 kΩ

I I 20 μA

B

B

4.

I

C sat

=

VCC

16 V

R = 2.7 kΩ C

= 5.93 mA

5. (a) Load line intersects vertical axis at I C = 21 V

3 kΩ = 7 mA

and horizontal axis at V CE = 21 V.

VCC

−VBE

21 V − 0.7 V

(b) I B = 25 μA: R B = = = 812 kΩ

I 25 μA

B

(c)

I

C Q

≅ 3.4 mA, V

CE Q

≅ 10.75 V

32

(d) β =

I

I

C

B

3.4 mA

= = 136

25 μA

(e) α =

β 136 136

= =

β + 1 136+

1 137

= 0.992

(f)

I

C sat

=

VCC

21 V

R = 3 kΩ = 7 mA

C

(g) −

(h) P D = V I = (10.75 V)(3.4 mA) = 36.55 mW

CEQ

CQ

(i)

(j)

6. (a)

P s = V CC (I C + I B ) = 21 V(3.4 mA + 25 μA) = 71.92 mW

P R = P s − P D = 71.92 mW − 36.55 mW = 35.37 mW

I

BQ

VCC

−VBE

20 V − 0.7 V

= =

R + ( β + 1) R 510 k Ω+ (101)1.5 kΩ = 19.3 V

661.5 kΩ

B

= 29.18 μA

E

(b)

I

CQ

= β I = (100)(29.18 μA) = 2.92 mA

BQ

(c)

V

CE Q

= V CC − I C (R C + R E ) = 20 V − (2.92 mA)(2.4 kΩ + 1.5 kΩ)

= 20 V − 11.388 V

= 8.61 V

(d) V C = V CC − I C R C = 20 V − (2.92 mA)(2.4 kΩ) = 20 V − 7.008 V

= 13 V

(e) V B = V CC − I B R B = 20 V − (29.18 μA)(510 kΩ)

= 20 V − 14.882 V = 5.12 V

(f)

V E = V C − V CE = 13 V − 8.61 V = 4.39 V

7. (a) R C =

V

CC

−V

I

C

C

12 V − 7.6 V 4.4 V

= = = 2.2 kΩ

2 mA 2 mA

(b) I E ≅ I C : R E =

V

I

E

E

2.4 V

= = 1.2 kΩ

2 mA

(c) R B =

VR V 12 V 0.7 V 2.4 V 8.9 V

B CC

−VBE −VE

− −

= = = = 356 kΩ

I I 2 mA/80 25 μA

B

B

(d) V CE = V C − V E = 7.6 V − 2.4 V = 5.2 V

(e) V B = V BE + V E = 0.7 V + 2.4 V = 3.1 V

33

VE

2.1 V

8. (a) I C ≅ I E =

R = = 3.09 mA

E

0.68 kΩ IC

3.09 mA

β =

I

= B

20 μA

= 154.5

(b) V CC = V

R C

+ V CE + V E

= (3.09 mA)(2.7 kΩ) + 7.3 V + 2.1 V = 8.34 V + 7.3 V + 2.1 V

= 17.74 V

VR V 17.74 V 0.7 V 2.1 V

B CC

−VBE −VE

− −

(c) R B = = =

I I 20 μA

B

= 14.94 V

20 A μ

B

= 747 kΩ

9.

I

Csot

= VCC

20 V 20 V

R + R

= 2.4 kΩ+ 1.5 kΩ = 3.9 kΩ C

E

= 5.13 mA

10. (a)

VCC

24 V

I

C sat

= 6.8 mA = =

RC + RE RC

+ 1.2 kΩ

24 V

R C + 1.2 kΩ = = 3.529 kΩ

6.8 mA

R C = 2.33 kΩ

(b) β =

I

I

C

B

4 mA

= = 133.33

30 μA

VR V 24 V 0.7 V (4 mA)(1.2 k )

B CC

−VBE −VE

− − Ω

(c) R B = = =

I I 30 μA

B

= 18.5 V

30 A μ

B

= 616.67 kΩ

(d) P D = VCE

I

Q CQ

= (10 V)(4 mA) = 40 mW

2

(e) P = ICR C

= (4 mA) 2 (2.33 kΩ)

= 37.28 mW

11. (a) Problem 1:

I

C Q

= 2.93 mA, V

CE Q

= 8.09 V

(b)

I

B Q

= 32.55 μA (the same)

IC

= β I

Q B

= (135)(32.55 μA) = 4.39 mA

Q

V = V − I R = 16 V − (4.39 mA)(2.7 kΩ) = 4.15 V

CEQ

CC CQ

C

34

4.39 mA − 2.93 mA

(c) %ΔI C = × 100% = 49.83%

2.93 mA

%ΔV CE = 4.15 V − 8.09 V × 100% = 48.70%

8.09 V

Less than 50% due to level of accuracy carried through calculations.

(d) Problem 6:

I

C Q

= 2.92 mA,

V = 8.61 V ( I = 29.18 μA)

CE Q

(e)

VCC

−VBE

20 V − 0.7 V

I

B

= =

Q

RB

+ ( β + 1) RE

510 k Ω+ (150 + 1)(1.5 k Ω )

= 26.21 μA

IC

= β I

Q B

= (150)(26.21 μA) = 3.93 mA

Q

V

CE Q

= V CC − I C (R C + R E )

= 20 V − (3.93 mA)(2.4 kΩ + 1.5 kΩ) = 4.67 V

3.93 mA − 2.92 mA

(f) %ΔI C =

2.92 mA

× 100% = 34.59%

%ΔV CE = 4.67 V − 8.61 V

8.61 V

× 100% = 46.76%

(g) For both I C and V CE the % change is less for the emitter-stabilized.

?

12. βR E ≥ 10R 2

(80)(0.68 kΩ) ≥ 10(9.1 kΩ)

54.4 kΩ ≥ 91 kΩ (No!)

(a) Use exact approach:

R Th = R 1 || R 2 = 62 kΩ || 9.1 kΩ = 7.94 kΩ

RV

2

E Th = CC

(9.1 k Ω)(16 V)

=

R2 + R1

9.1 kΩ+ 62 kΩ = 2.05 V

ETh

−VBE

2.05 V − 0.7 V

I

B

= =

Q

RTh

+ ( β + 1) RE

7.94 k Ω+ (81)(0.68 k Ω)

= 21.42 μA

(b) I = β I = (80)(21.42 μA) = 1.71 mA

CQ

BQ

B Q

(c)

V

CE Q

= V CC − I

C Q

(R C + R E )

= 16 V − (1.71 mA)(3.9 kΩ + 0.68 kΩ)

= 8.17 V

(d) V C = V CC − I C R C

= 16 V − (1.71 mA)(3.9 kΩ)

= 9.33 V

(e) V E = I E R E ≅ I C R E = (1.71 mA)(0.68 kΩ)

= 1.16 V

(f)

V B = V E + V BE = 1.16 V + 0.7 V

= 1.86 V

35

13. (a) I C =

V

CC

−V

R

C

C

18 V −12 V

=

4.7 kΩ

= 1.28 mA

(b) V E = I E R E ≅ I C R E = (1.28 mA)(1.2 kΩ) = 1.54 V

(c) V B = V BE + V E = 0.7 V + 1.54 V = 2.24 V

VR

1

(d) R 1 = :

VR

1

R 1 =

I

R1

I

R1

R1

V = V CC − V B = 18 V − 2.24 V = 15.76 V

I

R 1

≅

I

R2

= 15.76 V

0.4 mA

VB

2.24 V

= = = 0.4 mA

R 5.6 kΩ 2

= 39.4 kΩ

14. (a) I C = βI B = (100)(20 μA) = 2 mA

(b) I E = I C + I B = 2 mA + 20 μA

= 2.02 mA

V E = I E R E = (2.02 mA)(1.2 kΩ)

= 2.42 V

(c) V CC = V C + I C R C = 10.6 V + (2 mA)(2.7 kΩ)

= 10.6 V + 5.4 V

= 16 V

(d) V CE = V C − V E = 10.6 V − 2.42 V

= 8.18 V

(e) V B = V E + V BE = 2.42 V + 0.7 V = 3.12 V

(f)

I = I + I

R1 R2

B

= 3.12 V

8.2 kΩ

VCC

−VB

R 1 =

I

R1

+ 20 μA = 380.5 μA + 20 μA = 400.5 μA

16 V − 3.12 V

= = 32.16 kΩ

400.5 μA

15.

I

C sat

=

VCC

16 V

R + R = 3.9 kΩ+ 0.68 kΩ = 16 V

= 3.49 mA

4.58 kΩ

C

E

36

16. (a) βR E ≥ 10R 2

(120)(1 kΩ) ≥ 10(8.2 kΩ)

120 kΩ ≥ 82 kΩ (checks)

RV

2

∴V B = CC

(8.2 k Ω)(18 V)

=

R1 + R2

39 kΩ+ 8.2 kΩ = 3.13 V

V E = V B − V BE = 3.13 V − 0.7 V = 2.43 V

VE

2.43 V

I C ≅ I E =

R = = 2.43 mA

1 kΩ E

(b) V CE = V CC − I C (R C + R E )

= 18 V − (2.43 mA)(3.3 kΩ + 1 kΩ)

= 7.55 V

(c) I B =

I C

2.43 mA

β = 120

= 20.25 μA

(d) V E = I E R E ≅ I C R E = (2.43 mA)(1 kΩ) = 2.43 V

(e) V B = 3.13 V

17. (a) R Th = R 1 || R 2 = 39 kΩ || 8.2 kΩ = 6.78 kΩ

RV

C CC

8.2 k Ω(18 V)

E Th = =

R1 + R2

39 kΩ+ 8.2 kΩ = 3.13 V

ETh

−VBE

3.13 V − 0.7 V

I B =

=

R + ( β + 1) R 6.78 k Ω+ (121)(1 k Ω)

Th

E

2.43 V

=

= 19.02 μA

127.78 kΩ

I C = βI B = (120)(19.02 μA) = 2.28 mA (vs. 2.43 mA #16)

(b) V CE = V CC − I C (R C + R E ) = 18 V − (2.28 mA)(3.3 kΩ + 1 kΩ)

= 18 V − 9.8 V = 8.2 V (vs. 7.55 V #16)

(c) 19.02 μA (vs. 20.25 μA #16)

(d) V E = I E R E ≅ I C R E = (2.28 mA)(1 kΩ) = 2.28 V (vs. 2.43 V #16)

(e) V B = V BE + V E = 0.7 V + 2.28 V = 2.98 V (vs. 3.13 V #16)

The results suggest that the approximate approach is valid if Eq. 4.33 is satisfied.

R2

9.1 k Ω(16 V)

18. (a) V B = VCC

=

R1 + R2

62 kΩ+ 9.1 kΩ = 2.05 V

V E = V B − V BE = 2.05 V − 0.7 V = 1.35 V

VE

1.35 V

I E =

R = = 1.99 mA

E

0.68 kΩ I ≅ I E = 1.99 mA

C Q

37

V

CE Q

= V CC − I C (R C + R E )

= 16 V − (1.99 mA)(3.9 kΩ + 0.68 kΩ)

= 16 V − 9.11 V

= 6.89 V

I

B Q

=

I CQ

1.99 mA

= = 24.88 μA

β 80

(b) From Problem 12:

I = 1.71 mA, V = 8.17 V,

C Q

CE Q

I

B Q

= 21.42 μA

(c) The differences of about 14% suggest that the exact approach should be employed when

appropriate.

19. (a)

I

Csat

VCC

24 V 24 V

= 7.5 mA = = =

RC + RE 3RE + RE 4RE

24 V 24 V

R E = = = 0.8 kΩ

4(7.5 mA) 30 mA

R C = 3R E = 3(0.8 kΩ) = 2.4 kΩ

(b) V E = I E R E ≅ I C R E = (5 mA)(0.8 kΩ) = 4 V

(c) V B = V E + V BE = 4 V + 0.7 V = 4.7 V

RV

2

(d) V B = CC

R2

(24 V)

, 4.7 V =

R + R

R + 24 kΩ

(e) β dc =

I

I

2 1

C

B

R 2 = 5.84 kΩ

= 5 mA

38.5 μA

= 129.8

2

(f) βR E ≥ 10R 2

(129.8)(0.8 kΩ) ≥ 10(5.84 kΩ)

103.84 kΩ ≥ 58.4 kΩ (checks)

20. (a) From problem 12b, I C = 1.71 mA

From problem 12c, V CE = 8.17 V

(b) β changed to 120:

From problem 12a, E Th = 2.05 V, R Th = 7.94 kΩ

ETh

−VBE

2.05 V − 0.7 V

I B =

=

RTh

+ ( β + 1) RE

7.94 k Ω + (121)(0.68 k Ω)

= 14.96 μA

I C = βI B = (120)(14.96 μA) = 1.8 mA

V CE = V CC − I C (R C + R E )

= 16 V − (1.8 mA)(3.9 kΩ + 0.68 kΩ)

= 7.76 V

38

(c)

1.8 mA −1.71 mA

% Δ I C

= × 100% = 5.26%

1.71 mA

7.76 V − 8.17 V

% Δ V CE

= × 100% = 5.02%

8.17 V

(d) 11c 11f 20c

%ΔI C 49.83% 34.59% 5.26%

%ΔV CE 48.70% 46.76% 5.02%

Fixed-bias

Emitter

feedback

Voltagedivider

(e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in β.

21. I.(a) Problem 16: Approximation approach:

Problem 17: Exact analysis:

I

C Q

= 2.28 mA,

I

C Q

= 2.43 mA, V

CE Q

= 7.55 V

V

CE Q

= 8.2 V

The exact solution will be employed to demonstrate the effect of the change of β. Using

the approximate approach would result in %ΔI C = 0% and %ΔV CE = 0%.

(b) Problem 17: E Th = 3.13 V, R Th = 6.78 kΩ

ETH

−VBE

3.13 V − 0.7 V 2.43 V

I B =

= =

R + ( β + 1) R 6.78 k Ω+ (180 + 1)1 kΩ 187.78 kΩ

Th

E

= 12.94 μA

I C = βI B = (180)(12.94 μA) = 2.33 mA

V CE = V CC − I C (R C + R E ) = 18 V − (2.33 mA)(3.3 kΩ + 1 kΩ)

= 7.98 V

2.33 mA − 2.28 mA

(c) %ΔI C = × 100% = 2.19%

2.28 mA

%ΔV CE = 7.98 V − 8.2 V × 100% = 2.68%

8.2 V

For situations where βR E > 10R 2 the change in I C and/or V CE due to significant change in

β will be relatively small.

(d) %ΔI C = 2.19% vs. 49.83% for problem 11.

%ΔV CE = 2.68% vs. 48.70% for problem 11.

(e) Voltage-divider configuration considerably less sensitive.

II.

The resulting %ΔI C and %ΔV CE will be quite small.

39

VCC

−VBE

16 V − 0.7 V

22. (a) I B =

=

RB + β ( RC + RE) 470 k Ω + (120)(3.6 kΩ+ 0.51 k Ω)

= 15.88 μA

(b) I C = βI B = (120)(15.88 μA)

= 1.91 mA

(c) V C = V CC − I C R C

= 16 V − (1.91 mA)(3.6 kΩ)

= 9.12 V

23. (a) I B =

VCC

−VBE

30 V − 0.7 V

=

R + β ( R + R ) 6.90 kΩ+ 100(6.2 kΩ+ 1.5 k Ω )

B C E

= 20.07 μA

I C = βI B = (100)(20.07 μA) = 2.01 mA

(b) V C = V CC − I C R C

= 30 V − (2.01 mA)(6.2 kΩ) = 30 V − 12.462 V = 17.54 V

(c) V E = I E R E ≅ I C R E = (2.01 mA)(1.5 kΩ) = 3.02 V

(d) V CE = V CC − I C (R C + R E ) = 30 V − (2.01 mA)(6.2 kΩ + 1.5 kΩ)

= 14.52 V

VCC

−VBE

22 V − 0.7 V

24. (a) I B =

=

RB + β ( RC + RE) 470 k Ω+ (90)(9.1 kΩ+ 9.1 k Ω)

= 10.09 μA

I C = βI B = (90)(10.09 μA) = 0.91 mA

V CE = V CC − I C (R C + R E ) = 22 V − (0.91 mA)(9.1 kΩ + 9.1 kΩ)

= 5.44 V

VCC

−VBE

22 V − 0.7 V

(b) β = 135, I B =

=

RB + β ( RC + RE) 470 k Ω+ (135)(9.1 kΩ+ 9.1 k Ω)

= 7.28 μA

I C = βI B = (135)(7.28 μA) = 0.983 mA

V CE = V CC − I C (R C + R E ) = 22 V − (0.983 mA)(9.1 kΩ + 9.1 kΩ)

= 4.11 V

(c)

0.983 mA − 0.91 mA

% Δ I C

= × 100% = 8.02%

0.91 mA

4.11 V − 5.44 V

% Δ V CE

= × 100% = 24.45%

5.44 V

(d) The results for the collector feedback configuration are closer to the voltage-divider

configuration than to the other two. However, the voltage-divider configuration

continues to have the least sensitivities to change in β.

40

25. 1 MΩ = 0 Ω, R B = 150 kΩ

VCC

−VBE

12 V − 0.7 V

I B =

=

RB + β ( RC + RE) 150 k Ω+ (180)(4.7 kΩ+ 3.3 k Ω)

= 7.11 μA

I C = βI B = (180)(7.11 μA) = 1.28 mA

V C = V CC −I C R C = 12 V − (1.28 mA)(4.7 kΩ)

= 5.98 V

Full 1 MΩ: R B = 1,000 kΩ + 150 kΩ = 1,150 kΩ = 1.15 MΩ

VCC

−VBE

12 V − 0.7 V

I B =

=

R + β ( R + R ) 1.15 M Ω+ (180)(4.7 kΩ+ 3.3 k Ω)

B C E

= 4.36 μA

I C = βI B = (180)(4.36 μA) = 0.785 mA

V C = V CC − I C R C = 12 V − (0.785 mA)(4.7 kΩ)

= 8.31 V

V C ranges from 5.98 V to 8.31 V

26. (a) V E = V B − V BE = 4 V − 0.7 V = 3.3 V

VE

3.3 V

(b) I C ≅ I E =

R = = 2.75 mA

E

1.2 kΩ (c) V C = V CC − I C R C = 18 V − (2.75 mA)(2.2 kΩ)

= 11.95 V

(d) V CE = V C − V E = 11.95 V − 3.3 V = 8.65 V

VR V 11.95 V 4 V

B C

−VB

−

(e) I B = = =

= 24.09 μA

RB

RB

330 kΩ

IC

2.75 mA

(f) β = = = 114.16

I 24.09 μA

B

VCC + VEE −VBE

6 V + 6 V − 0.7 V

27. (a) I B =

=

RB

+ ( β + 1) RE

330 k Ω+ (121)(1.2 k Ω)

= 23.78 μA

I E = (β + 1)I B = (121)(23.78 μA)

= 2.88 mA

−V EE + I E R E − V E = 0

V E = −V EE + I E R E = −6 V + (2.88 mA)(1.2 kΩ)

= −2.54 V

VEE

−VBE

12 V − 0.7 V

28. (a) I B =

=

RB

+ ( β + 1) RE

9.1 k Ω+ (120 + 1)15 kΩ

= 6.2 μA

(b) I C = βI B = (120)(6.2 μA) = 0.744 mA

(c) V CE = V CC + V EE − I C (R C + R E )

= 16 V + 12 V − (0.744 mA)(27 kΩ)

= 7.91 V

(d) V C = V CC − I C R C = 16 V − (0.744 mA)(12 kΩ) = 7.07 V

41

29. (a) I E =

8 V − 0.7 V 7.3 V

=

2.2 kΩ

2.2 kΩ = 3.32 mA

(b) V C = 10 V − (3.32 mA)(1.8 kΩ) = 10 V − 5.976

= 4.02 V

(c) V CE = 10 V + 8 V − (3.32 mA)(2.2 kΩ + 1.8 kΩ)

= 18 V − 13.28 V

= 4.72 V

30. (a) βR E > 10R 2 not satisfied ∴Use exact approach:

Network redrawn to determine the Thevenin equivalent:

(b) I C = βI B = (130)(13.95 μA) = 1.81 mA

R Th = 510 k Ω = 255 kΩ

2

18 V + 18 V

I =

= 35.29 μA

510 kΩ + 510 kΩ E Th = −18 V + (35.29 μA)(510 kΩ)

= 0 V

18 V − 0.7 V

I B =

255 k Ω + (130 + 1)(7.5 k Ω)

= 13.95 μA

(c) V E = −18 V + (1.81 mA)(7.5 kΩ)

= −18 V + 13.58 V

= −4.42 V

(d) V CE = 18 V + 18 V − (1.81 mA)(9.1 kΩ + 7.5 kΩ)

= 36 V − 30.05 V = 5.95 V

31. (a) I B =

VR V 8 V 0.7 V

B C

−VBE

−

= =

R R 560 kΩ

B

B

= 13.04 μA

VCC

−V

(b) I C =

R

C

C

=

18 V − 8 V 10 V

=

3.9 kΩ

3.9 kΩ = 2.56 mA

(c) β =

I

I

C

B

= 2.56 mA

13.04 μA

= 196.32

(d) V CE = V C = 8 V

42

I 2.5 mA

32. I B = C

β = 80

= 31.25 μA

VR V 12 V 0.7 V

B CC

−VBE

−

R B = = = = 361.6 kΩ

IB

IB

31.25 μA

VR V 12 V 6 V 6 V

C CC

−V

VCC

−V

C

CEQ

−

R C = = = = =

I I I 2.5 mA 2.5 mA

C C CQ

= 2.4 kΩ

Standard values:

R B = 360 kΩ

R C = 2.4 kΩ

33.

VCC

I

C sat

=

R + R

C

20 V

R + R

4 E E

E

= 10 mA

= 10 mA ⇒ 20 V

5R

E

= 10 mA ⇒ 5R E =

20 V

10 mA = 2 kΩ

R E = 2 k 5Ω = 400 Ω

R C = 4R E = 1.6 kΩ

I 5 mA

I B = C

β = 120

= 41.67 μA

20 V −0.7 V −5 mA(0.4 k Ω) 19.3 −2 V

R B = V RB /I B = =

41.67 μA 41.67 μA

= 415.17 kΩ

Standard values: R E = 390 Ω, R C = 1.6 kΩ, R B = 430 kΩ

VE

VE

3 V

34. R E = ≅ = = 0.75 kΩ

IE

IC

4 mA

V ( )

R V

C CC

−V

VCC − V

C

CE

+ V

Q E

R C = = =

IC IC IC

24 V −(8 V + 3 V) 24 V −11 V 13 V

= = = = 3.25 kΩ

4 mA 4 mA 4 mA

V B = V E + V BE = 3 V + 0.7 V = 3.7 V

RV

2 CC

R2

(24 V) ⎫

V B = ⇒ 3.7 V = ⎬ 2 unknowns!

R2 + R1 R2 + R1

⎭

∴ use βR E ≥ 10R 2 for increased stability

(110)(0.75 kΩ) = 10R 2

R 2 = 8.25 kΩ

Choose R 2 = 7.5 kΩ

43

Substituting in the above equation:

7.5 k Ω(24 V)

3.7 V =

7.5 kΩ+

R1

R 1 = 41.15 kΩ

Standard values:

R E = 0.75 kΩ, R C = 3.3 kΩ, R 2 = 7.5 kΩ, R 1 = 43 kΩ

35. V E = 1 V = 1 (28 V) = 5.6 V

5 CC

5

VE

5.6 V

R E = = = 1.12 kΩ (use 1.1 kΩ)

I

E

5 mA

VCC

28 V

V C = + VE

= + 5.6 V = 14 V + 5.6 V = 19.6 V

2 2

V

R C

= V CC − V C = 28 V − 19.6 V = 8.4 V

VR 8.4 V

R C =

C

= = 1.68 kΩ (use 1.6 kΩ)

IC

5 mA

V B = V BE + V E = 0.7 V + 5.6 V = 6.3 V

RV

2

V B = CC

R2

(28 V)

⇒ 6.3 V = (2 unknowns)

R2 + R1

R2 + R1

IC

5 mA

β = = = 135.14

I

B

37 μA

βR E = 10R 2

(135.14)(1.12 kΩ) = 10(R 2 )

R 2 = 15.14 kΩ (use 15 kΩ)

(15.14 k Ω)(28 V)

Substituting: 6.3 V =

15.14 kΩ+

R1

Solving, R 1 = 52.15 kΩ (use 51 kΩ)

Standard values:

R E = 1.1 kΩ

R C = 1.6 kΩ

R 1 = 51 kΩ

R 2 = 15 kΩ

36. I 2 kΩ =

18 V − 0.7 V

2 kΩ

= 8.65 mA ≅ I

37. For current mirror:

I(3 kΩ) = I(2.4 kΩ) = I = 2 mA

38.

I

DQ

= I = 6 mA

DSS

44

4.3 kΩ

39. V B ≅

( −18 V) = −9 V

4.3 kΩ+ 4.3 kΩ

V E = −9 V − 0.7 V = −9.7 V

−18 V −( −9.7 V)

I E =

= 4.6 mA = I

1.8 kΩ

40. I E =

V

Z

−V

R

E

BE

5.1 V − 0.7 V

=

1.2 kΩ

= 3.67 mA

41.

VCC

10 V

IC

= = = 4.167 mA

sat

RC

2.4 kΩ From characteristics I

B max

≅ 31 μA

Vi

−VBE

10 V − 0.7 V

I B = =

= 51.67 μA

R 180 kΩ

B

51.67 μA 31 μA, well saturated

V o = 10 V − (0.1 mA)(2.4 kΩ)

= 10 V − 0.24 V

= 9.76 V

42.

I

C sat

= 8 mA = 5 V RC

5 V

R C = = 0.625 kΩ

8 mA

I C 8 mA

sat

I

B max

= = = 80 μA

β 100

Use 1.2 (80 μA) = 96 μA

R B = 5 V − 0.7 V = 44.79 kΩ

96 μA

Standard values:

R B = 43 kΩ

R C = 0.62 kΩ

45

43. (a) From Fig. 3.23c:

I C = 2 mA: t f = 38 ns, t r = 48 ns, t d = 120 ns, t s = 110 ns

t on = t r + t d = 48 ns + 120 ns = 168 ns

t off = t s + t f = 110 ns + 38 ns = 148 ns

(b) I C = 10 mA: t f = 12 ns, t r = 15 ns, t d = 22 ns, t s = 120 ns

t on = t r + t d = 15 ns + 22 ns = 37 ns

t off = t s + t f = 120 ns + 12 ns = 132 ns

The turn-on time has dropped dramatically

168 ns:37 ns = 4.54:1

while the turn-off time is only slightly smaller

148 ns:132 ns = 1.12:1

44. (a) Open-circuit in the base circuit

Bad connection of emitter terminal

Damaged transistor

(b) Shorted base-emitter junction

Open at collector terminal

(c) Open-circuit in base circuit

Open transistor

45. (a) The base voltage of 9.4 V reveals that the 18 kΩ resistor is not making contact with the

base terminal of the transistor.

If operating properly:

V B ≅

18 k Ω(16 V)

18 kΩ+ 91 kΩ = 2.64 V vs. 9.4 V

As an emitter feedback bias circuit:

VCC

−VBE

16 V − 0.7 V

I B =

=

R1

+ ( β + 1) RE

91 k Ω+ (100 + 1)1.2 kΩ

= 72.1 μA

V B = V CC − I B (R 1 ) = 16 V − (72.1 μA)(91 kΩ)

= 9.4 V

46

(b) Since V E > V B the transistor should be “off”

18 k Ω(16 V)

With I B = 0 μA, V B =

18 kΩ + 91 kΩ = 2.64 V

∴ Assume base circuit “open”

The 4 V at the emitter is the voltage that would exist if the transistor were shorted

collector to emitter.

1.2 k Ω(16 V)

V E =

1.2 kΩ+ 3.6 kΩ = 4 V

46. (a) R B ↑, I B ↓, I C ↓, V C ↑

(b) β↓, I C ↓

(c) Unchanged, I

C sat

not a function of β

(d) V CC ↓, I B ↓, I C ↓

(e) β↓, I C ↓, V ↓ ,

R C

VR E

↓ , V CE ↑

ETh −VBE ETh −VBE

47. (a) I B =

≅

RTh + ( β + 1) RE RTh + βRE

⎡ ETh −V ⎤

BE

ETh −VBE

I C = βI B = β ⎢ ⎥ =

⎣R

R

Th

+ β RE

⎦ Th

+ RE

β

RTh

As β↑,

β ↓, I C↑,

V C = V CC −

and V C ↓

V

R C

VR C

(b) R 2 = open, I B ↑, I C ↑


and V CE ↓

(c) V CC ↓, V B ↓, V E ↓, I E ↓, I C ↓

(d) I B = 0 μA, I C = I CEO and I C (R C + R E ) negligible

with V CE ≅ V CC = 20 V

↑

(e) Base-emitter junction = short I B ↑ but transistor action lost and I C = 0 mA with

V CE = V CC = 20 V

48. (a) R B open, I B = 0 μA, I C = I CEO ≅ 0 mA

and V C ≅ V CC = 18 V

(b) β↑, I C ↑, V ↑,

V ↑,

V CE ↓

R C

R E

(c) R C ↓, I B ↑, I C ↑, V E ↑

(d) Drop to a relatively low voltage ≅ 0.06 V

(e) Open in the base circuit

47

V

49. I B =

CC

−V

R

B

BE

12 V − 0.7 V 11.3 V

= = = 22.16 μA

510 kΩ

510 kΩ I C = βI B = (100)(22.16 μA) = 2.216 mA

V C = −V CC + I C R C = −12 V + (2.216 mA)(3.3 kΩ)

= −4.69 V

V CE = V C = −4.69 V

50. βR E ≥ 10R 2

(220)(0.75 kΩ) ≥ 10(16 kΩ)

165 kΩ ≥ 160 kΩ (checks)

Use approximate approach:

16 k Ω− ( 22 V)

V B ≅

16 k Ω + 82 kΩ = −3.59 V

V E = V B + 0.7 V = −3.59 V + 0.7 V = −2.89 V

I C ≅ I E = V E /R E = 2.89/0.75 kΩ = 3.85 mA

I 3.85 mA

I B = C

= = 17.5 μA

β 220

V C = −V CC + I C R C

= −22 V + (3.85 mA)(2.2 kΩ)

= −13.53 V

V −V

51. I E =

R

E

BE

8 V − 0.7 V 7.3 V

= = = 2.212 mA

3.3 kΩ

3.3 kΩ V C = −V CC + I C R C = −12 V + (2.212 mA)(3.9 kΩ)

= −3.37 V

52. (a) S(I CO ) = β + 1 = 91

(b) S(V BE ) =

I C

−β

−90

=

470 kΩ = −1.92 ×10−4 S

R B

2.93 mA

1

(c) S(β) =

β = 90

= 32.56 × 10−6 A

1

(d) ΔI C = S(I CO )ΔI CO + S(V BE )ΔV BE + S(β)Δβ

= (91)(10 μA − 0.2 μA) + (−1.92 × 10 −4 S)(0.5 V − 0.7 V) + (32.56 × 10 −6 A)(112.5 − 90)

= (91)(9.8 μA) + (1.92 × 10 −4 S)(0.2 V) + (32.56 × 10 −6 A)(22.5)

= 8.92 × 10 −4 A + 0.384 × 10 −4 A + 7.326 × 10 −4 A

= 16.63 × 10 −4 A

≅ 1.66 mA

48

53. For the emitter-bias:

(1 + RB

/ RE) (1 + 510 k Ω/1.5 k Ω)

(a) S(I CO ) = (β + 1)

= (100 + 1)

( β + 1) + RB

/ RE

(100 + 1) + 510 k Ω/1.5 kΩ

= 78.1

(b) S(V BE ) =

R

B

−β

−100

=

+ ( β + 1) R 510 k Ω+ (100 + 1)1.5 kΩ

E

= −1.512 × 10 −4 S

IC (1 + R / ) 2.92 mA(1 + 340)

1 B

RE

(c) S(β) = =

β (1 + β + R / R ) 100(1 + 125 + 340)

1 2

B

= 21.37 × 10 −6 A

E

(d) ΔI C = S(I CO )ΔI CO + S(V BE )ΔV BE + S(β)Δβ

= (78.1)(9.8 μA) + (−1.512 × 10 −14 S)(−0.2 V) + (21.37 × 10 −6 A)(25)

= 0.7654 mA + 0.0302 mA + 0.5343 mA

= 1.33 mA

54. (a) R Th = 62 kΩ || 9.1 kΩ = 7.94 kΩ

1 + RTh

/ RE

(1 + 7.94 k Ω/ 0.68 k Ω)

S(I CO ) = (β + 1)

= (80 + 1)

( β + 1) + RTh

/ RE

(80+ 1) + 7.94 k Ω/0.68 kΩ

(81)(1 + 11.68)

= = 11.08

81+

11.68

−β

−80

(b) S(V BE ) =

=

RTh

+ ( β + 1) RE

7.94 k Ω+ (81)(0.68 k Ω)

−80

=

7.94 kΩ+ 55.08 kΩ = −1.27 × 10−3 S

IC (1 + R / ) 1.71 mA(1 + 7.94 k / 0.68 k )

1 Th

RE

Ω Ω

(c) S(β) = =

β (1 + β + R / R ) 80(1 + 100 + 7.94 k Ω/ 0.68 k Ω)

=

1 2

Th

1.71 mA(12.68)

80(112.68)

E

= 2.41 × 10 −6 A

(d) ΔI C = S(I CO )ΔI CO + S(V BE ) ΔV BE + S(β)Δβ

= (11.08)(10 μA − 0.2 μA) + (−1.27 × 10 −3 S)(0.5 V − 0.7 V) + (2.41 × 10 −6 A)(100 − 80)

= (11.08)(9.8 μA) + (−1.27 × 10 −3 S)(−0.2 V) + (2.41 × 10 −6 A)(20)

= 1.09 × 10 −4 A + 2.54 × 10 −4 A + 0.482 × 10 −4 A

= 4.11 × 10 −4 A = 0.411 mA

49

55. For collector-feedback bias:

(1 + RB

/ RC) (1 + 560 k Ω/3.9 k Ω)

(a) S(I CO ) = (β + 1)

= (196.32 + 1)

( β + 1) + RB

/ RC

(196.32 + 1) + 560 k Ω/3.9 kΩ

1+

143.59

= (197.32)

(197.32 + 143.59)

= 83.69

(b) S(V BE ) =

R

B

−β

−196.32

=

+ ( β + 1) R 560 k Ω+ (196.32 + 1)3.9 kΩ

C

= −1.477 × 10 −4 S

IC ( R ) 2.56 mA(560 k 3.9 k )

1 B

+ RC

Ω+ Ω

(c) S(β) = =

β ( R + R ( β + 1)) 196.32(560 kΩ+ 3.9 k Ω (245.4 + 1))

1 B C 2

= 4.83 × 10 −6 A

(d) ΔI C = S(I CO )ΔI CO + S(V BE ) ΔV BE + S(β)Δβ

= (83.69)(9.8 μA) + (−1.477 × 10 −4 S)(−0.2 V) + (4.83 × 10 −6 A)(49.1)

= 8.20 × 10 −4 A + 0.295 × 10 −4 A + 2.372 × 10 −4 A

= 10.867 × 10 −4 A = 1.087 mA

56. Type S(I CO ) S(V BE ) S(β)

Collector feedback 83.69 −1.477 × 10 −4 S 4.83 × 10 −6 A

Emitter-bias 78.1 −1.512 × 10 −4 S 21.37 × 10 −6 A

Voltage-divider 11.08 −12.7 × 10 −4 S 2.41 × 10 −6 A

Fixed-bias 91 −1.92 × 10 −4 S 32.56 × 10 −6 A

S(I CO ): Considerably less for the voltage-divider configuration compared to the other three.

S(V BE ): The voltage-divider configuration is more sensitive than the other three (which have

similar levels of sensitivity).

S(β): The voltage-divider configuration is the least sensitive with the fixed-bias

configuration very sensitive.

In general, the voltage-divider configuration is the least sensitive with the fixed-bias the most

sensitive.

57. (a) Fixed-bias:

S(I CO ) = 91, ΔI C = 0.892 mA

S(V BE ) = −1.92 × 10 −4 S, ΔI C = 0.0384 mA

S(β) = 32.56 × 10 −6 A, ΔI C = 0.7326 mA

(b) Voltage-divider bias:

S(I CO ) = 11.08, ΔI C = 0.1090 mA

S(V BE ) = −1.27 × 10 −3 S, ΔI C = 0.2540 mA

S(β) = 2.41 × 10 −6 A, ΔI C = 0.0482 mA

50

(c) For the fixed-bias configuration there is a strong sensitivity to changes in I CO and β and

less to changes in V BE .

For the voltage-divider configuration the opposite occurs with a high sensitivity to

changes in V BE and less to changes in I CO and β.

In total the voltage-divider configuration is considerably more stable than the fixed-bias

configuration.

51

Chapter 5

1. (a) If the dc power supply is set to zero volts, the amplification will be zero.

2. −

(b) Too low a dc level will result in a clipped output waveform.

(c) P o = I 2 R = (5 mA) 2 2.2 kΩ = 55 mW

P i = V CC I = (18 V)(3.8 mA) = 68.4 mW

Po

(ac) 55 mW

η = = = 0.804 ⇒ 80.4%

P (dc) 68.4 mW

i

1 1

3. x C = = = 15.92 Ω

2π fC 2 π(1 kHz)(10μF)

f = 100 kHz: x C = 0.159 Ω

Yes, better at 100 kHz

4. −

Vi

10 mV

5. (a) Z i =

I =

i

0.5 mA

= 20 Ω (=r e )

(b) V o = I c R L

= αI c R L

= (0.98)(0.5 mA)(1.2 kΩ)

= 0.588 V

Vo

(c) A v =

V =

i

= 58.8

(d) Z o = ∞ Ω

0.588 V

10 mV

Io

(e) A i =

I

i

α I

I

=

e

e

= α = 0.98

(f)

I b = I e − I c

= 0.5 mA − 0.49 mA

= 10 μA

52

6. (a) r e =

Vi

I

i

48 mV

= = 15 Ω

3.2 mA

(b) Z i = r e = 15 Ω

(c) I C = αI e = (0.99)(3.2 mA) = 3.168 mA

(d) V o = I C R L = (3.168 mA)(2.2 kΩ) = 6.97 V

(e) A v =

Vo

6.97 V

V = 48 mV

= 145.21

i

(f)

I b = (1 − α)I e = (1 − 0.99)I e = (0.01)(3.2 mA)

= 32 μA

7. (a) r e =

(b) I b =

26 mV 26 mV

= = 13 Ω

I E

(dc) 2 mA

Z i = βr e = (80)(13 Ω)

= 1.04 kΩ

IC α I β

e

Ie Ie

= = ⋅ =

β β β + 1 β β + 1

= 2 mA

81

= 24.69 μA

Io

I

L

(c) A i = =

Ii

Ib

ro( β Ib)

I L =

ro

+ RL

ro

⋅ β Ib

ro + RL ro

A i =

= ⋅β

I r + R

(d) A v =

b o L

40 kΩ

=

(80)

40 kΩ+ 1.2 kΩ

= 77.67

RL

ro

1.2 kΩ

40 kΩ

− =−

r 13 Ω

1.165 kΩ

= −

13 Ω

= −89.6

e

53

8. (a) Z i = βr e = (140)r e = 1200

r e = 1200

140 = 8.571 Ω

Vi

30 mV

(b) I b =

Z = = 25 μA

1.2 kΩ i

(c) I c = βI b = (140)(25 μA) = 3.5 mA

(d) I L =

A i =

rI

o c

(50 k Ω)(3.5 mA)

=

r + R 50 kΩ+ 2.7 kΩ o

I

I

L

i

L

= 3.321 mA

25 μA

= 132.84

= 3.321 mA

Vo −AR

i L

(e) A v = = =

V Z

i

i

(2.7 k Ω)

−(132.84) 1.2 k Ω

= −298.89

VCC

−VBE

12 V − 0.7 V

9. (a) r e : I B = =

= 51.36 μA

RB

220 kΩ

I E = (β + 1)I B = (60 + 1)(51.36 μA)

= 3.13 mA

26 mV 26 mV

r e = = = 8.31 Ω

3.13 mA

I E

Z i = R B || βr e = 220 kΩ || (60)(8.31 Ω) = 220 kΩ || 498.6 Ω

= 497.47 Ω

r o ≥ 10R C ∴ Z o = R C = 2.2 kΩ

(b) A v =

RC

−2.2 kΩ

− =

r 8.31 Ω = −264.74

e

(c) Z i = 497.47 Ω (the same)

Z o = r o || R C = 20 kΩ || 2.2 kΩ

= 1.98 kΩ

(d) A v =

−RC

ro

−1.98 kΩ

=

r 8.31 Ω = −238.27

e

A i = −A v Z i /R C

= −(−238.27)(497.47 Ω)/2.2 kΩ

= 53.88

54

RC

10. A v = − ⇒ r e =

r

r e =

V

11. (a) I B =

e

RC

4.7 kΩ

− =− = 23.5 Ω

A ( − 200)

v

26 mV 26 mV 26 mV

⇒ I E = = = 1.106 mA

I E

r e

23.5 Ω

I 1.106 mA

I B = E

= = 12.15 μA

β + 1 91

VCC

−VBE

I B = ⇒ V CC = I B R B + V BE

RB

= (12.15 μA)(1 MΩ) + 0.7 V

= 12.15 V + 0.7 V

= 12.85 V

CC

−V

R

B

BE

10 V − 0.7 V

=

= 23.85 μA

390 kΩ

I E = (β + 1)I B = (101)(23.85 μA) = 2.41 mA

26 mV 26 mV

r e = = = 10.79 Ω

I E

2.41 mA

I C = βI B = (100)(23.85 μA) = 2.38 mA

(b) Z i = R B || βr e = 390 kΩ || (100)(10.79 Ω) = 390 kΩ || 1.08 kΩ

= 1.08 kΩ

r o ≥ 10R C ∴Z o = R C = 4.3 kΩ

(c) A v =

(d) A v =

RC

−4.3 kΩ

− =

r 10.79 Ω = −398.52

e

RC

ro

(4.3 k Ω) (30 k Ω) 3.76 kΩ

− =− =−

r 10.79 Ω 10.79 Ω = −348.47

e

12. (a) Test βR E ≥ 10R 2

?

(100)(1.2 kΩ) ≥ 10(4.7 kΩ)

120 kΩ > 47 kΩ (satisfied)


RV

2

4.7 k (16 V)

V B = CC

Ω

=

R + R 39 kΩ+ 4.7 kΩ = 1.721 V

1 2

V E = V B − V BE = 1.721 V − 0.7 V = 1.021 V

VE

1.021 V

I E =

R = = 0.8507 mA

1.2 kΩ r e =

E

26 mV 26 mV

= = 30.56 Ω

0.8507 mA

I E

55

(b) Z i = R 1 || R 2 || β r e

= 4.7 kΩ || 39 kΩ || (100)(30.56 Ω)

= 1.768 kΩ

r o ≥ 10R C ∴ Z o ≅ R C = 3.9 kΩ

(c) A v =

RC

3.9 kΩ

− =−

r 30.56 Ω = −127.6

e

(d) r o = 25 kΩ

(b) Z i (unchanged) = 1.768 kΩ

Z o = R C || r o = 3.9 kΩ || 25 kΩ = 3.37 kΩ

(c) A v =

( RC

ro) (3.9 k Ω) (25 k Ω) 3.37 kΩ

− =− =−

r

30.56 Ω 30.56 Ω

e

= −110.28 (vs. −127.6)

?

13. βR E ≥ 10R 2

(100)(1 kΩ) ≥ 10(5.6 kΩ)

100 kΩ > 56 kΩ (checks!) & r o ≥ 10R C


RC

RC

3.3 kΩ

A v = − ⇒ re

=− =− = 20.625 Ω

r A − 160

e

v

26 mV 26 mV 26 mV

r e = ⇒ I

E

= = = 1.261 mA

IE

re

20.625 Ω

VE

I E =

R

⇒ V E = I E R E = (1.261 mA)(1 kΩ) = 1.261 V

E

V B = V BE + V E = 0.7 V + 1.261 V = 1.961 V

5.6 kΩ V

V B =

CC

5.6 kΩ+ 82 kΩ = 1.961 V

5.6 kΩ V CC = (1.961 V)(87.6 kΩ)

V CC = 30.68 V

14. Test βR E ≥ 10R 2

?

(180)(2.2 kΩ) ≥ 10(56 kΩ)

396 kΩ < 560 kΩ (not satisfied)

Use exact analysis:

(a) R Th = 56 kΩ || 220 kΩ = 44.64 kΩ

56 k Ω(20 V)

E Th =

220 kΩ + 56 kΩ = 4.058 V

ETh

−VBE

4.058 V − 0.7 V

I B =

=

R + ( β + 1) R 44.64 k Ω+ (181)(2.2 k Ω)

Th

E

56

= 7.58 μA

I E = (β + 1)I B = (181)(7.58 μA)

= 1.372 mA

26 mV 26 mV

r e = = = 18.95 Ω

1.372 mA

I E

(b) V E = I E R E = (1.372 mA)(2.2 kΩ) = 3.02 V

V B = V E + V BE = 3.02 V + 0.7 V

= 3.72 V

V C = V CC − I C R C

= 20 V − βI B R C = 20 V − (180)(7.58 μA)(6.8 kΩ)

= 10.72 V

(c) Z i = R 1 || R 2 || βr e

= 56 kΩ || 220 kΩ || (180)(18.95 kΩ)

= 44.64 kΩ || 3.41 kΩ

= 3.17 kΩ

RC

ro

r o < 10R C ∴ A v = −

re

(6.8 k Ω) (50 k Ω)

= −

18.95 Ω

= −315.88

VCC

−VBE

20 V − 0.7 V

15. (a) I B =

=

RB

+ ( β + 1) RE

390 k Ω+ (141)(1.2 k Ω)

19.3 V

= = 34.51 μA

559.2 kΩ

I E = (β + 1)I B = (140 + 1)(34.51 μA) = 4.866 mA

26 mV 26 mV

r e = = = 5.34 Ω

4.866 mA

I E

(b) Z b = βr e + (β + 1)R E

= (140)(5.34 kΩ) + (140 + 1)(1.2 kΩ) = 747.6 Ω + 169.9 kΩ

= 169.95 kΩ

Z i = R B || Z b = 390 kΩ || 169.95 kΩ = 118.37 kΩ

Z o = R C = 2.2 kΩ

(c) A v =

β R

Z

C

− =

b

(140)(2.2 k Ω)

−

169.95 kΩ = −1.81

(d) Z b = βr e +

⎡ ( β + 1) + RC

/ ro

⎢

⎣1 + ( R + R )/ r

C E o

⎤

⎥R

⎦

⎡(141) + 2.2 k Ω/ 20 kΩ⎤

= 747.6 Ω ⎢

1 + (3.4 k Ω)/ 20 kΩ

⎥1.2 kΩ

⎣

⎦

E

57

= 747.6 Ω + 144.72 kΩ

= 145.47 kΩ

Z i = R B || Z b = 390 kΩ || 145.47 kΩ = 105.95 kΩ

Z o = R C = 2.2 kΩ (any level of r o )

β R ⎡

C

r ⎤

e

RC

− ⎢1

+ ⎥+

V Z

o

b

ro ro

A v = =

⎣ ⎦

V

R

i

C

1+

r

o

−(140)(2.2 k Ω) ⎡ 5.34 Ω⎤

2.2 kΩ

1+ +

145.47 kΩ ⎢ 20 k 20 k

=

⎣ Ω⎥

⎦ Ω

2.2 kΩ

1+

20 kΩ

− 2.117 + 0.11

=

= −1.81

1.11

16. Even though the condition r o ≥ 10R C is not met it is sufficiently close to permit the use of the

approximate approach.

β RC β RC RC

A v = − =− =− = −10

Zb β RE RE

RC

8.2 kΩ

∴ R E = = = 0.82 kΩ

10 10

26 mV 26 mV

I E = = = 6.842 mA

3.8 Ω

r e

V E = I E R E = (6.842 mA)(0.82 kΩ) = 5.61 V

V B = V E + V BE = 5.61 V + 0.7 V = 6.31 V

I 6.842 mA

I B = E

= = 56.55 μA

( β + 1) 121

VR V 20 V 6.31 V

B CC

−VB

−

and R B = = = = 242.09 kΩ

I I 56.55 μA

B

17. (a) dc analysis the same

∴ r e = 5.34 Ω (as in #15)

B

(b) Z i = R B || Z b = R B || βr e = 390 kΩ || (140)(5.34 Ω) = 746.17 Ω vs. 118.37 kΩ in #15

Z o = R C = 2.2 kΩ (as in #15)

(c) A v =

−RC

−2.2 kΩ

=

r 5.34 Ω

e

= −411.99 vs −1.81 in #15

(d) Z i = 746.17 Ω vs. 105.95 kΩ for #15

Z o = R C || r o = 2.2 kΩ || 20 kΩ = 1.98 kΩ vs. 2.2 kΩ in #15

58

18. (a) I B =

=

RC

ro

1.98 kΩ

A v = − =− = −370.79 vs. −1.81 in #15

re

5.34 Ω

Significant difference in the results for A v .

VCC

−VBE

R + ( β + 1) R

B

E

22 V − 0.7 V 21.3 V

=

330 k Ω+ (81)(1.2 kΩ+ 0.47 k Ω) 465.27 kΩ

= 45.78 μA

I E = (β + 1)I B = (81)(45.78 μA) = 3.71 mA

26 mV 26 mV

r e = = = 7 Ω

3.71 mA

I E

(b) r o < 10(R C + R E )

⎡ ( β + 1) + RC

/ ro

∴Z b = βr e + ⎢

⎣1 + ( R + R )/ r

C E o

⎤

⎥R

⎦

E

= (80)(7 Ω) +

⎡(81) + 5.6 k Ω/ 40 kΩ⎤

⎢ 1+ 6.8 k Ω/ 40 kΩ

⎥1.2 kΩ

⎣

⎦

= 560 Ω +

⎡81+

0.14⎤

⎢ 1+

0.17 ⎥1.2 kΩ

⎣ ⎦

(note that (β + 1) = 81 R C /r o = 0.14)

= 560 Ω + [81.14 /1.17]1.2 kΩ = 560 Ω + 83.22 kΩ

= 83.78 kΩ

Z i = R B || Z b = 330 kΩ || 83.78 kΩ = 66.82 kΩ

−β R ⎛

C

r ⎞

e

RC

⎜1+ ⎟+

Zb ro ro

A v =

⎝ ⎠

RC

1+

ro

−(80)(5.6 k Ω) ⎛ 7 Ω ⎞ 5.6 kΩ

⎜1+ ⎟+

83.78 kΩ ⎝ 40 kΩ⎠

40 kΩ

=

1+ 5.6 k Ω/40 kΩ

− (5.35) + 0.14

=

1+

0.14

= −4.57

19. (a) I B =

VCC

−VBE

16 V − 0.7 V 15.3 V

= =

R + ( β + 1) R 270 k Ω+ (111)(2.7 k Ω) 569.7 kΩ

B

E

59

= 26.86 μA

I E = (β + 1)I B = (110 + 1)(26.86 μA)

= 2.98 mA

26 mV 26 mV

r e = = = 8.72 Ω

I E

2.98 mA

βr e = (110)(8.72 Ω) = 959.2 Ω

(b) Z b = βr e + (β + 1)R E

= 959.2 Ω + (111)(2.7 kΩ)

= 300.66 kΩ

Z i = R B || Z b = 270 kΩ || 300.66 kΩ

= 142.25 kΩ

Z o = R E || r e = 2.7 kΩ || 8.72 Ω = 8.69 Ω

(c) A v =

RE

2.7 kΩ

=

R + r 2.7 kΩ+ 8.69 Ω ≅ 0.997

E

e

20. (a) I B =

VCE

−VBE

8 V − 0.7 V

=

R + ( β + 1) R 390 k Ω+ (121)5.6 kΩ B

E

I E = (β + 1)I B = (121)(6.84 μA) = 0.828 mA

26 mV 26 mV

r e = = = 31.4 Ω

I E

0.828 mA

r o < 10R E :

Z b = βr e + ( β + 1) RE

1 + RE

/ ro

(121)(5.6 k Ω)

= (120)(31.4 Ω) +

1+ 5.6 k Ω/40 kΩ

= 3.77 kΩ + 594.39 kΩ

= 598.16 kΩ

Z i = R B || Z b = 390 kΩ || 598.16 kΩ

= 236.1 kΩ

Z o ≅ R E || r e

= 5.6 kΩ || 31.4 Ω

= 31.2 Ω

= 6.84 μA

(b) A v = ( β + 1) R / E

Z

1 + R / r

E

o

(121)(5.6 k Ω) / 598.16 kΩ

=

1+ 5.6 k Ω/ 40 kΩ

= 0.994

V0

(c) A v =

V = 0.994

i

b

V o = A v V i = (0.994)(1 mV) = 0.994 mV

60

21. (a)

?

Test βR E ≥ 10R 2

(200)(2 kΩ) ≥ 10(8.2 kΩ)

400 kΩ ≥ 82 kΩ (checks)!


8.2 k Ω(20 V)

V B =

8.2 kΩ+ 56 kΩ = 2.5545 V

V E = V B − V BE = 2.5545 V − 0.7 V ≅ 1.855 V

VE

1.855 V

I E =

R = = 0.927 mA

E

2 kΩ I 0.927 mA

I B = E

= = 4.61 μA

( β + 1) (200 + 1)

I C = βI B = (200)(4.61 μA) = 0.922 mA

(b) r e =

26 mV 26 mV

= = 28.05 Ω

0.927 mA

I E

(c) Z b = βr e + (β + 1)R E

= (200)(28.05 Ω) + (200 + 1)2 kΩ

= 5.61 kΩ + 402 kΩ = 407.61 kΩ

Z i = 56 kΩ || 8.2 kΩ || 407.61 kΩ

= 7.15 kΩ || 407.61 kΩ

= 7.03 kΩ

Z o = R E || r e = 2 kΩ || 28.05 Ω = 27.66 Ω

(d) A v =

22. (a) I E =

r e =

RE

2 kΩ

=

R + r 2 kΩ+ 28.05 Ω = 0.986

V

E

EE

E

e

−V

R

BE

6 V − 0.7 V

=

= 0.779 mA

6.8 kΩ

26 mV 26 mV

= = 33.38 Ω

0.779 mA

I E

(b) Z i = R E || r e = 6.8 kΩ || 33.38 Ω

= 33.22 Ω

Z o = R C = 4.7 kΩ

α RC

(0.998)(4.7 k Ω)

(c) A v = =

re

33.38 Ω

= 140.52

23. α =

β 75

= = 0.9868

β + 1 76

61

VEE

−VBE

5 V − 0.7 V 4.3 V

I E = = = = 1.1 mA

RE

3.9 kΩ

3.9 kΩ 26 mV 26 mV

r e = = = 23.58 Ω

I E

1.1 mA

RC

(0.9868)(3.9 k Ω)

A v = α =

= 163.2

r 23.58 Ω

e

VCC

−VBE

12 V − 0.7 V

24. (a) I B = =

RF

+ β RC

220 kΩ+ 120(3.9 k Ω)

= 16.42 μA

I E = (β + 1)I B = (120 + 1)(16.42 μA)

= 1.987 mA

r e =

26 mV 26 mV

= = 13.08 Ω

1.987 mA

I E

(b) Z i = βr e ||

R

A

F

v

Need A v !

−RC

−3.9 kΩ

A v = =

re

13.08 Ω = −298

Z i = (120)(13.08 Ω) || 220 k Ω

298

= 1.5696 kΩ || 738 Ω

= 501.98 Ω

Z o = R C || R F = 3.9 kΩ || 220 kΩ

= 3.83 kΩ

(c) From above, A v = −298

−RC

25. A v =

r

e

= −160

R C = 160(r e ) = 160(10 Ω) = 1.6 kΩ

A i =

β RF

R + β R

F

C

= 19 ⇒ 19 =

200RF

R + 200(1.6 k Ω)

F

19R F + 3800R C = 200R F

R F = 3800 R

C

3800(1.6 k Ω)

=

181 181

= 33.59 kΩ

VCC −VBE

I B =

R + β R

F

I B (R F + βR C ) = V CC − V BE

C

62

and V CC = V BE + I B (R F + βR C )

26 mV 26 mV

with I E = = = 2.6 mA

r e

10 Ω

I 2.6 mA

I B = E

= = 12.94 μA

β + 1 200 + 1

∴V CC = V BE + I B (R F + βR C )

= 0.7 V + (12.94 μA)(33.59 kΩ + (200)(1.6 kΩ))

= 5.28 V

26.

(a) A v : V i = I b βr e + (β + 1)I b R E

I o + I′ = I C = βI b

but I i = I′ + I b

and I′ = I i − I b

Substituting, I o + (I i − I b ) = βI b

and I o = (β + 1)I b − I i

Assuming (β + 1)I b I i

I o ≅ (β + 1)I b

and V o = −I o R C = −(β + 1)I b R C

Vo − ( β + 1) IbRC

Therefore, =

V I βr + ( β + 1) I R

β IR

b C

≅

β Ir

b e+

β IR

b E

Vo RC RC

and A v = ≅− ≅−

V r + R R

(b) V i ≅ βI b (r e + R E )

For r e R E

V i ≅ βI b R E

i b e b E

i e E E

Now I i = I′ + I b

V −V

R

=

i o

F

+ I

b

63

Since V o V i

Vo

I i = − + I

R

Vo

or I b = I i +

RF

and V i = βI b R E

Vo

V i = βR E I i + β R

E

RF

but V o = A v V i

β AVR

v i E

and V i = βR E I i +

R

Avβ

REVi

or V i −

R

F

F

b

F

= βR E I i

⎡ Avβ

R ⎤

E

V i ⎢1 − ⎥ = [ β RE

] Ii

⎣ RF

⎦

Vi β RE β RERF

so Z i = = =

I Avβ

RE

1−

R + β ( −A ) R

R

Vi

Z i =

I

with Z i =

Z i ≅

Z o : Set V i = 0

i F v E

i

F

= x || y where x = βR E and y = R F /|A v |

x⋅

y ( β RE )( RF / Av

)

=

x + y β R + R / A

β RERF

β R A + R

E v F

E F v

V i = I b βr e + (β + 1)I b R E

V i ≅ βI b (r e + R E ) = 0

since β, r e + R E ≠ 0 I b = 0 and βI b = 0

64

Vo

V ⎡ 1 1 ⎤

o

∴ I o = + = Vo

⎢ + ⎥

RC RF ⎣RC RF

⎦

Vo 1 RCRF

and Z o = = =

I 1 1

+

R + R

R R

o C F

C

F

= R C || R F

RC

2.2 kΩ

(c) A v ≅ − =−

RE

1.2 kΩ = −1.83

β RR

E F

(90)(1.2 k Ω)(120 k Ω)

Z i ≅

=

β R A + R (90)(1.2 k Ω )(1.83) + 120 kΩ

E v F

= 40.8 kΩ

Z o ≅ R C || R F

= 2.2 kΩ || 120 kΩ

= 2.16 kΩ

VCC

−VBE

9 V − 0.7 V

27. (a) I B = =

RF

+ β RC

(39 kΩ+ 22 k Ω ) + (80)(1.8 k Ω)

8.3 V 8.3 V

=

= = 40.49 μA

61 kΩ+ 144 kΩ 205 kΩ I E = (β + 1)I B = (80 + 1)(40.49 μA) = 3.28 mA

26 mV 26 mV

r e = = = 7.93 Ω

I E

3.28 mA

Z i = R

F 1

|| β re

= 39 kΩ || (80)(7.93 Ω) = 39 kΩ || 634.4 Ω = 0.62 kΩ

Z o = R C || R = 1.8 kΩ || 22 kΩ = 1.66 kΩ

F 2

−R′

−RC

RF

1.8 kΩ 22 kΩ

2

(b) A v = = =−

r r

7.93 Ω

=

28. A i ≅ β = 60

29. A i ≅ β = 100

e

e

−1.664 kΩ

7.93 Ω = −209.82

30. A i = −A v Z i /R C = −(−127.6)(1.768 kΩ)/3.9 kΩ = 57.85

31. (c) A i =

β RB

(140)(390 k Ω)

=

R + Z 390 kΩ+ 0.746 kΩ = 139.73

B

b

(d) A i =

Z

i

− Av

= −(−370.79)(746.17 Ω)/2.2 kΩ

R

C

= 125.76

65

32. A i = −A v Z i /R E = −(0.986)(7.03 kΩ)/2 kΩ = −3.47

Io α Ie

33. A i = = = α = 0.9868 ≅ 1

I I

i

e

34. A i = −A v Z i /R C = −(−298)(501.98 Ω)/3.9 kΩ = 38.37

35. A i =

Zi

−− ( 209.82)(0.62 k Ω)

− Av

= R 1.8 kΩ

C

= 72.27

VCC

−VBE

18 V − 0.7 V

36. (a) I B = =

= 25.44 μA

RB

680 kΩ

I E = (β + 1)I B = (100 + 1)(25.44 μA)

= 2.57 mA

26 mV

r e =

2.57 mA = 10.116 Ω

RC

3.3 kΩ

Av

=− =−

NL

r 10.116 Ω = −326.22

(b) −

e

Z i = R B || βr e = 680 kΩ || (100)(10.116 Ω)

= 680 kΩ || 1,011.6 Ω

= 1.01 kΩ

Z o = R C = 3.3 kΩ

(c)

(d)

(e)

A

v L

=

RL

4.7 kΩ

Av

=

NL

R + R 4.7 kΩ+ 3.3 kΩ (−326.22)

L

o

= −191.65

Zi

(1.01 k Ω)

A

i L

= − Av

=−( −191.65)

L

R

L

4.7 kΩ

= 41.18

A

v L

=

V −β

Ib( RC R )

o

L −100(1.939 k Ω)

= =

V I ( β r ) 100(10.116 Ω)

i b e

= −191.98

Z i = R B || βr e = 1.01 kΩ

RC( β Ib)

I L = = 41.25I b

RC

+ RL

RBIi

I b = = 0.9985I i

RB

+ βre

Io

IL

IL

Ib

A

i L

= = = ⋅ = (41.25)(0.9985)

Ii Ii Ib Ii

= 41.19

Z o = R C = 3.3 kΩ

66

37. (a)

A

v NL

= −326.22

RL

A

v L

= Av

R + R

L

R L = 4.7 kΩ:

R L = 2.2 kΩ:

R L = 0.5 kΩ:

As R L ↓,

o

NL

A

v L

=

A

v L

=

A

v L

=

Av L

↓

4.7 kΩ

( −326.22)

4.7 kΩ + 3.3 kΩ

2.2 kΩ

( −326.22)

2.2 kΩ+ 3.3 kΩ

0.5 kΩ

( −326.22)

0.5 kΩ+ 2.3 kΩ

= −191.65

= −130.49

= −42.92

(b) No change for Z i , Z o , and

A

v NL

!

VCC

−VBE

12 V − 0.7 V

38. (a) I B = =

= 11.3 μA

RB

1 MΩ

I E = (β + 1)I B = (181)(11.3 μA) = 2.045 mA

26 mV 26 mV

r e = = = 12.71 Ω

I E

2.045 mA

RC

3 kΩ

A

V NL

= − =−

r 12.71 Ω = −236

(b) −

e

Z i = R B || βr e = 1 MΩ || (180)(12.71 Ω) = 1 MΩ || 2.288 kΩ

= 2.283 kΩ

Z o = R C = 3 kΩ

(c) No-load: A v =

A

v NL

= −236

(d)

A

vs

Zi

2.283 k Ω− ( 236)

= Av

=

NL

Z + R 2.283 kΩ+ 0.6 kΩ

i

s

= −186.9

(e) V o = −I o R C = −βI b R C

V i = I b βr e

Vo β IbRC RC

3 kΩ

A v = =− =− =−

Vi β Ibre re

12.71 Ω = −236

Vo Vo Vi

Av

= = ⋅

s

Vs Vi Vs

(1 M Ω β re) Vs 2.288 k Ω( Vs

)

V i =

=

(1 M Ω β re) + Rs

2.288 kΩ+ 0.6 kΩ = 0.792 V s

A

v s

= (−236)(0.792)

= −186.9 (same results)

67

(f)

(g)

No change!

A

vs

R s ↑,

Zi

2.283 k Ω− ( 236)

= ( Av

) =

NL

Z + R 2.283 kΩ+ 1 kΩ = −164.1

i

A ↓

v s

s

(h) No change!

VCC

−VBE

24 V − 0.7 V

39. (a) I B = =

= 41.61 μA

RB

500 kΩ

I E = (β + 1)I B = (80 + 1)(41.61 μA) = 3.37 mA

26 mV 26 mV

r e = = = 7.715 Ω

I E

3.37 mA

RL

4.3 kΩ

A

v NL

= − =−

re

7.715 Ω = −557.36

Z i = R B || βr e = 560 kΩ || (80)(7.715 Ω)

= 560 kΩ || 617.2 Ω

= 616.52 Ω

Z o = R C = 4.3 kΩ

(b) −

(c)

A

v L

=

Vo

RL

2.7 k Ω− ( 557.36)

= Av

=

NL

V R + R 2.7 kΩ+ 4.3 kΩ

i L o

= −214.98

Vo Vo Vi

Av

= = ⋅

s

Vs Vi Vs

ZV

i s

616.52 Ω Vs

V i = =

Zi

+ Rs

616.52 Ω+ 1 kΩ = 0.381 V s

A

v s

= ( − 214.98)(0.381)

= −81.91

(d)

A

is

⎛ Rs

+ Z ⎞

i

⎛1 kΩ + 616.52 Ω⎞

=− Av

( 81.91)

s

⎜ ⎟=− − ⎜ ⎟

⎝ RL

⎠

⎝ 2.7 kΩ

⎠

= 49.04

(e)

A

v L

=

Vo

RL

5.6 k Ω− ( 557.36)

= Av

=

NL

V R + R 5.6 kΩ+ 4.3 kΩ = −315.27

i L o

Vi

the same = 0.381

Vs

Vo

Vi

A = vs

V

⋅ i

V

= (−315.27)(0.381) = −120.12

s

As R L ↑, A ↑

v s

68

(f)

A

v L

the same = −214.98

Vi

Zi

616.52 Ω

= =

Vs Zi + Rs

616.52 Ω+ 0.5 kΩ = 0.552

Vo

Vi

A = vs

V

⋅ i

V

= (−214.98)(0.552) = −118.67

s

As R s ↓, A ↑

v s

(g) No change!

40. (a) Exact analysis:

R2

16 k Ω(16 V)

E Th = VCC

=

R + R 68 kΩ+ 16 kΩ = 3.048 V

(b) −

1 2

R Th = R 1 || R 2 = 68 kΩ || 16 kΩ = 12.95 kΩ

I B =

ETh

−VBE

3.048 V − 0.7 V

=

R + ( β + 1) R 12.95 k Ω+ (101)(0.75 k Ω)

Th

= 26.47 μA

I E = (β + 1)I B = (101)(26.47 μA)

= 2.673 mA

26 mV 26 mV

r e = = = 9.726 Ω

I E

2.673 mA

−RC

2.2 kΩ

Av

= =−

NL

re

9.726 Ω = −226.2

Z i = 68 kΩ || 16 kΩ || βr e

= 12.95 kΩ || (100)(9.726 Ω)

= 12.95 kΩ || 972.6 Ω

= 904.66 Ω

Z o = R C = 2.2 kΩ

E

(c)

(d)

A

v L

=

A

i L

=

RL

5.6 k Ω− ( 226.2)

( Av

) =

NL

R + Z 5.6 kΩ+ 2.2 kΩ = −162.4

L

− A

vL

Z

R

o

i

L

= −(−162.4) (904.66 Ω )

5.6 kΩ

= 26.24

69

(e)

A

v L

=

−RC

Re

−2.2 kΩ 5.6 kΩ

=

r 9.726 Ω

e

= −162.4

Z i = 68 kΩ || 16 kΩ || 972.6 Ω

41. (a)

= 904.66 Ω

Zi

A

i L

= − Av

L

R

L

βr e

= ( − 162.4)(904.66 Ω )

5.6 kΩ

= 26.24

Z o = R C = 2.2 kΩ

Same results!

A

v L

=

RL

R + Z

L

R L = 4.7 kΩ:

R L = 2.2 kΩ:

R L = 0.5 kΩ:

R L ↓,

Av L

↓

o

A

vNL

A

v L

=

A

v L

=

A

v L

=

4.7 kΩ

( −226.4)

4.7 kΩ+ 2.2 kΩ

2.2 kΩ

( −226.4)

2.2 kΩ+ 2.2 kΩ

0.5 kΩ

( −226.4)

0.5 kΩ+ 2.2 kΩ

= −154.2

= −113.2

= −41.93

(b) Unaffected!

VCC

−VBE

18 V − 0.7 V

42. (a) I B =

=

RB

+ ( β + 1) RE

680 k Ω+ (111)(0.82 k Ω)

= 22.44 μA

I E = (β + 1)I B = (110 + 1)(22.44 μA)

= 2.49 mA

26 mV 26 mV

r e = = = 10.44 Ω

I E

2.49 mA

RC

3 kΩ

Av

=− =−

NL

re

+ RE

10.44 Ω+ 0.82 kΩ

= −3.61

Z i ≅ R B || Z b = 680 kΩ || (βr e + (β + 1)R E )

= 680 kΩ || (610)(10.44 Ω) + (110 + 1)(0.82 kΩ)

= 680 kΩ || 92.17 kΩ

= 81.17 kΩ

Z o ≅ R C = 3 kΩ

(b) −

70

(c)

A

v L

=

Vo

RL

4.7 k Ω( −3.61)

= Av

=

NL

V R + R 4.7 kΩ+ 3 kΩ

i L o

= −2.2

Vo Vo Vi

Av

= = ⋅

s

Vs Vi Vs

ZV

i s

81.17 k Ω ( Vs)

V i = =

Zi

+ Rs

81.17 kΩ+ 0.6 kΩ = 0.992 V s

A

v s

= (−2.2)(0.992)

= −2.18

(d) None!

(e)

A

v L

– none!

Vi

Zi

81.17 kΩ

= =

Vs Zi + Rs

81.17 kΩ+ 1 kΩ = 0.988

A

v s

= (−2.2)(0.988)

= −2.17

R s ↑, Av s

↓, (but only slightly for moderate changes in R s since Z i is typically much larger

than R s )

43. Using the exact approach:

ETh

−VBE

I B =

RTh

+ ( β + 1) RE

2.33 V − 0.7 V

=

10.6 k Ω+ (121)(1.2 k Ω)

R2

E Th = V

R + R

1 2

CC

12 kΩ

=

(20 V) = 2.33 V

91 kΩ+ 12 kΩ

= 10.46 μA R Th = R 1 || R 2 = 91 kΩ || 12 kΩ = 10.6 kΩ

I E = (β + 1)I B = (121)(10.46 μA)

= 1.266 mA

26 mV 26 mV

r e = = = 20.54 Ω

1.266 mA

I E

(a)

A

vNL

RE

1.2 kΩ

≅ =

r + R 20.54 Ω+ 1.2 kΩ = 0.983

e

E

Z i = R 1 || R 2 || (βr e + (β +1)R E )

= 91 kΩ || 12 kΩ || ((120)(20.54 Ω) + (120 + 1)(1.2 kΩ))

= 10.6 kΩ || (2.46 kΩ + 145.2 kΩ)

= 10.6 kΩ || 147.66 kΩ

= 9.89 kΩ

Z o = R E || r e = 1.2 kΩ || 20.54 Ω

= 20.19 Ω

71

(b) −

(c)

(d)

RL

2.7 k Ω(0.983)

A

v L

= Av

=

NL

RL

+ Zo

2.7 kΩ+ 20.19 Ω

= 0.976

Zi

9.89 k Ω(0.976)

Av

= A

s

v

=

L

Zi

+ Rs

9.89 kΩ+ 0.6 kΩ

= 0.92

A

v L

= 0.976 (unaffected by change in R s )

Zi

9.89 k Ω(0.976)

Av

= A

s

v

=

L

Zi

+ Rs

9.89 kΩ+ 1 kΩ

= 0.886 (vs. 0.92 with R s = 0.6 kΩ)

As R s ↑, A ↓

v s

(e) Changing R s will have no effect on

A

v NL

, Z i , or Z o .

(f)

A = L

( A ) v

v L

R

5.6 k Ω(0.983)

=

NL

R + Z 5.6 kΩ+ 20.19 Ω

L

o

= 0.979 (vs. 0.976 with R L = 2.7 kΩ)

Zi

9.89 k Ω(0.979)

Av

= ( A )

s

v

=

L

Z + R 9.89 kΩ+ 0.6 kΩ

i

s

= 0.923 (vs. 0.92 with R L = 2.7 kΩ)

As R L ↑, A ↑, A ↑

v L

v s

VEE

−VBE

6 V − 0.7 V

44. (a) I E = =

RE

2.2 kΩ

= 2.41 mA

26 mV 26 mV

r e = = = 10.79 Ω

I E

2.41 mA

RC

4.7 kΩ

A

v NL

= =

r 10.79 Ω = 435.59

(b) −

e

Z i = R E || r e = 2.2 kΩ || 10.79 Ω = 10.74 Ω

Z o = R C = 4.7 kΩ

(c)

RL

5.6 k Ω(435.59)

A

v L

= Av

=

NL

RL

+ Ro

5.6 kΩ+ 4.7 kΩ = 236.83

Zi

10.74 Ω ( Vs)

V i = Vs

=

Zi

+ Rs

10.74 Ω+ 100 Ω = 0.097 V s

Vo

Vi

A = vs

V

⋅ i

V

= (236.83)(0.097)

s

= 22.97

72

(d) V i = I e ⋅ r e

V o = −I o R L

−4.7 k Ω( I )

I o =

e

4.7 kΩ+ 5.6 kΩ = −0.4563I e

Vo

A

v L

=

V = + (0.4563 Ie) RL

0.4563(5.6 k Ω)

=

i

Ie

⋅re

10.79 Ω

= 236.82 (vs. 236.83 for part c)

A

v s

: 2.2 kΩ || 10.79 Ω = 10.74 Ω

Zi

10.74 Ω ( Vs)

V i = ⋅ Vs

=

Zi

+ Rs

10.74 Ω+ 100 Ω = 0.097 V s

Vo

Vi

A = vs

V

⋅ i

V

= (236.82)(0.097)

s

= 22.97 (same results)

(e)

RL

2.2 kΩ

A

v L

= Av

=

(435.59)

NL

RL

+ Ro

2.2 kΩ+ 4.7 kΩ

= 138.88

V0 Vi Vi Zi

10.74 Ω

Av

= ⋅ , = =

s

Vi Vs Vs Zi + Rs

10.74 Ω+ 500 Ω = 0.021

A = (138.88)(0.021) = 2.92

v s

A

v s

very sensitive to increase in R s due to relatively small Z i ; R s ↑,

A

v L

sensitive to R L ; R L ↓,

Av L

↓

Av s

↓

(f) Z o = R C = 4.7 kΩ unaffected by value of R s !

(g) Z i = R E || r e = 10.74 Ω unaffected by value of R L !

45. (a)

(b)

A

A

v1

v2

A

v L

=

A

vs

RA

L v NL

1 k Ω− ( 420)

= =

RL

+ Ro

1 kΩ+ 3.3 kΩ = −97.67

RA

L v NL

2.7 k Ω− ( 420)

= =

R + R 2.7 kΩ+ 3.3 kΩ = −189

A

L

o

⋅ A = (−97.67)(−189) = 18.46 × 10 3

v1 v2

V V V V

V V V V

o o o1 i1

= = ⋅ ⋅

s i2 i1

s

V i =

=

V

⋅ ⋅

i

Av

A

2 v1

V

s

ZV

i s

1 k Ω( Vs)

=

Z + R 1 kΩ+ 0.6 kΩ = 0.625

i

A

v s

= (−189)(−97.67)(0.625)

= 11.54 × 10 3

s

73

(c)

(d)

AZ

v i

−− ( 97.67)(1 k Ω)

Ai

=− =

= 97.67

1

R 1 kΩ

A

i2

L

L

−AZ

v i

−− ( 189)(1 k Ω)

= =

= 70

R 2.7 kΩ

A = A ⋅ A = (97.67)(70) = 6.84 × 10 3

iL

i1 i2

(e) No effect!

(f)

No effect!

(g) In phase

46. (a)

Zi

1.2 kΩ

2

Av

= A

(1)

1 v

=

1NL

Zi

+ Z 1.2 k 20

2 o

Ω+ Ω

1

= 0.984

RL

2.2 kΩ

A

v 2

= Av

= ( −640)

2 NL

RL

+ Zo

2.2 kΩ+ 4.6 kΩ

2

= −207.06

(b)

A = A ⋅ A = (0.984)(−207.06)

A

vL

v1 v2

vs

= −203.74

Zi

= Av

Z + R

i

s

L

50 kΩ

=

( −203.74)

50 kΩ+ 1 kΩ

= −199.75

Zi

1

(c) Ai

=− A

1 v1

Z

i2

(50 k Ω)

= −(0.984) 1.2 k Ω

= −41

Zi

2

Ai

=− A

2 v2

R

L

(1.2 k Ω)

= −− ( 207.06) 2.2 k Ω

= 112.94

Zi

1

(d) Ai

=− A

L vL

R

L

= −(−203.74) (50 k Ω )

2.2 kΩ

= 4.63 × 10 3

74

(e) A load on an emitter-follower configuration will contribute to the emitter resistance (in

fact, lower the value) and therefore affect Z i (reduce its magnitude).

(f) The fact that the second stage is a CE amplifier will isolate Z o from the first stage and R s .

(g) The emitter-follower has zero phase shift while the common-emitter amplifier has a

180° phase shift. The system, therefore, has a total phase shift of 180° as noted by the

negative sign in front of the gain for A in part b.

47. For each stage:

6.2 kΩ

V B =

(15 V) = 3.08 V

24 kΩ+ 6.2 kΩ V E = V B − 0.7 V = 3.08 V − 0.7 V = 2.38 V

VE

2.38 V

I E ≅ I C =

R = = 1.59 mA

1.5 kΩ 48. r e =

E

V C = V CC − I C R C = 15 V − (1.59 mA)(5.1 kΩ)

= 6.89 V

26 mV 26 mV

= = 16.35 Ω

1.59 mA

I E

R

i 2

= R 1 || R 2 || βr e = 6.2 kΩ || 24 kΩ || (150)(16.35 Ω)

= 1.64 kΩ

RC

Ri

5.1 k Ω 1.64 kΩ

2

Av

=− =

= −75.8

1

re

16.35 Ω

RC

−5.1 kΩ

Av

=− =

2

re

16.35 Ω = −311.9

A v = A A = (−75.8)(−311.9) = 23,642

v1 v2

v T

49.

V B 1

V B 2

V

I

3.9 kΩ

=

(20 V) = 4.4 V

3.9 kΩ+ 6.2 kΩ+ 7.5 kΩ

6.2 kΩ+ 3.9 kΩ

=

(20 V) = 11.48 V

3.9 kΩ+ 6.2 kΩ+ 7.5 kΩ

= V − 0.7 V = 4.4 V − 0.7 V = 3.7 V

E1 B1

VE

3.7 V

1

≅ I = =

R 1 kΩ = 3.7 mA ≅ IE

≅ I

2 C2

C1 E1

E

V = V CC − I C R C = 20 V − (3.7 mA)(1.5 kΩ)

C 2

= 14.45 V

75

26 mV 26 mV

50. r e = = = 7 Ω

I E

3.7 mA

re

Av

=− = −1

1

re

RE

1.5 kΩ

Av

= =

2

r 7 Ω ≅ 214

e

Av = A

T v

A

1 v

= (−1)(214) = −214

2

V o = AV = (−214)(10 mV) = −2.14 V

vT

i

51. R o = R D = 1.5 kΩ (V o (from problem 50) = −2.14 V)

RL

10 kΩ

V o (load) = ( Vo

) = ( −2.14 V)

R + R 10 kΩ+ 1.5 kΩ

52. I B =

o

L

= −1.86 V

VCC

−VBE

(16 V −1.6 V)

=

β R + R (6000)(510 Ω ) + 2.4 MΩ

D E B

14.4 V

= = 2.64 μA

5.46 MΩ

I C ≅ I E = β D I B = 6000(2.64 μA) = 15.8 mA

V E = I E R E = (15.8 mA)(510 Ω) = 8.06 V

53. From problem 69, I E = 15.8 mA

r e = 26 = 26 V = 1.65 Ω

I E

15.8 mA

RE

510 Ω

A v = =

r + R 1.65 Ω+ 510 Ω = 0.997 ≈ 1

e

E

54. dc:

I

I

r

B

VCC

−VBE

16 V −1.6 V

≅ =

R + β R 2.4 M Ω+ (6000)(510 Ω )

B D E

= β I = (6000)(2.64 μA)

= 15.84 mA

C D B

e2

26 mV 26 mV

= = = 1.64 Ω

I 15.84 mA

E2

= 2.64 μA

ac:

Z ≅ β r = (6000)(1.64 Ω ) = 9.84 kΩ

i

D e2

Vi

I

b

=

1

9.84 kΩ

⎛ V ⎞

i

Vo = ( − βDIb )( R ) (6000) (200 )

1 C

=− ⎜ ⎟ Ω

⎝9.84 kΩ

⎠

= −121.95

Vi

and

A

V

o

v

= ≅ −121.95

Vi

76

VCC

−VEB

16 V − 0.7 V

1

55. I B = =

R + ββ R 1.5 M Ω+ (160)(200)(100 Ω)

B

1 2

E

= 3.255 μA

I C ≅ β 1 β 2 I B = (160)(200)(3.255μA) ≅ 104.2 mA

V

C 2

= V CC − I C R C = 16 V − (104.2 mA)(100 Ω) = 5.58 V

V

B 1

= I B R B = (3.255 μA)(1.5 MΩ) = 4.48 V

56. From problem 55: I

E 1

= 0.521 mA

26 mV 26 mV

re

= = = 49.9 Ω

1

I

E

(mA) 0.521 mA

Ri

= β r

1 e

= 160(49.9 Ω) = 7.98 kΩ

i

β1β2RC

(160)(200)(100 Ω)

A v =

=

ββ R + R (160)(200)(100 Ω ) + 7.98 kΩ

57. r e =

58. −

59. −

60. −

61. −

1 2

C

i1

= 0.9925

V o = A v V i = 0.9975 (120 mV)

= 119.7 mV

26 mV 26 mV

= = 21.67 Ω

1.2 mA

I E (dc)

βr e = (120)(21.67 Ω) = 2.6 kΩ

Vo

62. (a) A v =

V

i

= −160

V o = −160 V i

Vi −hreVo Vi −hreAV v i

Vi(1 −hreAv)

(b) I b = = =

hie hie hie

−4

V i ( 1 − (2×

10 )(160))

=

1 kΩ

I b = 9.68 × 10 −4 V i

(c) I b = 1 k

i

V

Ω = 1 × 10−3 V i

77

−

1× 10 Vi

− 9.68×

10

(d) % Difference =

−3

1×

10 Vi

= 3.2 %

3 −4

V

i

× 100%

(e) Valid first approximation

63. % difference in total load =

R − R 1/ h

L L oe

2.2 k Ω−(2.2 kΩ 50 k Ω)

= × 100%

2.2 kΩ

2.2 kΩ−2.1073 kΩ

= × 100%

2.2 kΩ

= 4.2 %

R

L

× 100%

In this case the effect of 1/h oe can be ignored.

64. (a) V o = −180V i (h ie = 4 kΩ, h re = 4.05 × 10 −4 )

−4

Vi

− (4.05×

10 )(180 Vi)

(b) I b =

4 kΩ

= 2.32 × 10 −4 V i

(c) I b =

Vi

Vi

h = 4 kΩ = 2.5 × 10−4 V i

ie

(d) % Difference =

2.5× 10 V − 2.32×

10

−4 −4

i

−4

2.5×

10 Vi

V

i

× 100% = 7.2%

(e) Yes, less than 10%

65. From Fig. 5.18

min max

h oe : 1 μS 30 μS

Avg = (1 + 30) μ S

2

= 15.5 μS

66. (a) h fe = β = 120

h ie ≅ βr e = (120)(4.5 Ω) = 540 Ω

h oe = 1 1

r = = 25 μS

40 kΩ h ie

o

1 kΩ

(b) r e ≅ = = 11.11 Ω

β 90

β = h fe = 90

1 1

r o = =

20 μS

hoe

= 50 kΩ

78

67. (a) r e = 8.31 Ω (from problem 9)

(b)

h

fe

= β = 60

h

ie

= βr e = (60)(8.31 Ω) = 498.6 Ω

(c) Z i = R B || h ie = 220 kΩ || 498.6 Ω = 497.47 Ω

Z o = R C = 2.2 kΩ

(d) A v =

−hfeRC

−(60)(2.2 k Ω)

=

h 498.6 Ω

ie

A i ≅ h fe = 60

= −264.74

(e) Z i = 497.47 Ω (the same)

1

Z o = r o || R C , r o = = 40 kΩ

25 μ S

= 40 kΩ || 2.2 kΩ

= 2.09 kΩ

−hfe( ro RC

) −(60)(2.085 k Ω)

(f) A v =

=

= −250.90

h

498.6 Ω

ie

A i = −A v Z i /R C = −(−250.90)(497.47 Ω)/2.2 kΩ = 56.73

68. (a) 68 kΩ || 12 kΩ = 10.2 kΩ

Z i = 10.2 kΩ || h ie = 10.2 kΩ || 2.75 kΩ

= 2.166 kΩ

Z o = R C || r o

= 2.2 kΩ || 40 kΩ

= 2.085 kΩ

(b) A v =

=

−hfeR′

C

R′

C

= R C || r o = 2.085 kΩ

hie

−(180)(2.085 k Ω)

= −136.5

2.75 kΩ

Io Io I′

i

A i = = ⋅

I I′

I

i i i

⎛ hf 10.2 k

=

e

⎞⎛

Ω ⎞

⎜

1+ hoeR

⎟⎜

L

10.2 kΩ+ 2.68 kΩ

⎟

⎝ ⎠⎝ ⎠

⎛ 180 ⎞

= ⎜

⎟(0.792)

⎝1 + (25 μS)(2.2 k Ω)

⎠

= 135.13

79

69. (a) Z i = R E || h ib

= 1.2 kΩ || 9.45 Ω

= 9.38 Ω

1

Z o = R C || = 2.7 kΩ ||

h ob

×

1

A

−6

1 10 V

= 2.7 kΩ || 1 MΩ ≅ 2.7 kΩ

−hfb ( RC 1/ hob

) −− ( 0.992)( ≅2.7 k Ω)

(b) A v =

=

hib

9.45 Ω

= 284.43

A i ≅ −1

(c) α = −h fb = −(−0.992) = 0.992

α 0.992

β =

1 α = = 124

− 1−0.992

r e = h ib = 9.45 Ω

1 1

r o = = = 1 MΩ

1 μA/V

hob

70. (a)

(b)

(c)

−4

hfehreRL

(180)(2 10 )(2.2 k )

Z′ × Ω

i

= hie

− = 2.75 kΩ−

1 + hoeRL

(1+ 25μS)(2.2 k Ω)

= 2.68 kΩ

Z i = 10.2 kΩ || Z ′

i

= 2.12 kΩ

1 1

Z ′

o

= =

−4

hoe −( hfehre / hie) 25 μS − (180)(2× 10 ) / 2.75 kΩ

= 83.75 kΩ

Z o = 2.2 kΩ || 83.75 kΩ = 2.14 kΩ

−hfeRL

−(180)(2.2 k Ω)

Av

= =

h + ( h h −h h ) R 2.75 k Ω + (2.75 k Ω)(25μS) − (180)(2× 10 ) 2.2 kΩ

ie ie oe fe re L

= −140.3

hfe

(180)

A′ i

= = = 170.62

1+ h R 1 + (25μS)(2.2 k Ω)

A i =

oe

L

Io Io I′ i ⎛ 10.2 kΩ

⎞

= ⋅ = (170.62) ⎜ ⎟

Ii

I ′ I

i i ⎝ 10.2 k Ω + 2.68 k Ω⎠

= 135.13

−4

( )

80

−hfehreRL

71. (a) Z i = h ie = 1 + h R

oe L

−4

(140)(1.5× 10 )(2.2 k Ω)

= 0.86 kΩ −

1 + (25 μS)(2.2 k Ω)

= 0.86 kΩ − 43.79 Ω

= 816.21 Ω

Z ′

i

= R B || Z i

= 470 kΩ || 816.21 Ω

= 814.8 Ω

−hfeRL

(b) A v =

hie + ( hie hoe −hfehre)

RL

−(140)(2.2 k Ω)

=

−4

0.86 k Ω+ ((0.86 k Ω)(25 μS) − (140)(1.5× 10 ))2.2 kΩ

= −357.68

I h

o

fe

140

(c) A i = = =

Ii 1+ hoeRL

1 + (25 μS)(2.2 k Ω)

= 132.70

I ⎛

o

I ⎞⎛

o

I ⎞

i

A′ 470 k Ω I ′

i

i

= =⎜ ⎟ I i =

I′ I

i ⎝ i ⎜ I′

⎟

⎠⎝ i ⎠

470 kΩ + 0.816 kΩ

= (132.70)(0.998)

Ii

I ′

i

= 0.998

= 132.43

(d) Z o =

1

( /( ))

h − h h h + R

oe fe re ie s

=

=

1

25 μS − (140)(1.5× 10 ) /(0.86 kΩ+ 1 k Ω)

1

13.71 μS

−4

( )

≅ 72.9 kΩ

72. (a)

(b)

hfbhrbRL

Z′ − × Ω

i

= hib

− = 9.45 Ω−

1+ hobRL

1 + (0.5μA/V)(2.2 k Ω)

= 9.67 Ω

Z i = 1.2 kΩ || Z ′

i

= 1.2 kΩ || 9.67 Ω = 9.59 Ω

−4

( 0.997)(1 10 )(2.2 k )

hfb

0.997

A′ −

i

= = =−0.996

1+ h R 1 + (0.5 μA/V)(2.2 k Ω)

ob

L

Io

I′ i ⎛ 1.2 kΩ

⎞

Ai

= ⋅ = ( −0.996) I I

⎜

i i

1.2 k Ω + 9.67 k Ω

⎟

′ ⎝ ⎠

≅ −0.988

81

(c)

(d)

−hfbRL

Av

=

hib + ( hibhob −hfbhrb ) RL

−− ( 0.997)(2.2 k Ω)

=

−4

9.45 Ω+ ((9.45 Ω)(0.5 μA/V) −( − 0.997)(1× 10 ))(2.2 k Ω)

= 226.61

1

Z ′

o

=

hob − ⎡ ⎣ hfbhrb / h ⎤

ib ⎦

1

=

−4

0.5 μA/V −⎡

⎣

( − 0.997)(1× 10 ) /9.45 Ω⎤

⎦

= 90.5 kΩ

Z o = 2.2 kΩ ||

Z ′

o

= 2.15 kΩ

73. −

74. (a) h fe (0.2 mA) ≅ 0.6 (normalized)

h fe (1 mA) = 1.0

% change =

h

fe

(0.2 mA) − h (1mA)

h

fe

fe

(0.2 mA)

= 0.6 − 1 × 100%

0.6

= 66.7%

× 100%

(b) h fe (1 mA) = 1.0

h fe (5 mA) ≅ 1.5

hfe

(1 mA) − hfe

(5 mA)

% change =

h (1 mA)

= 1 − 1.5

1

= 50%

fe

× 100%

× 100%

75. Log-log scale!

(a) I c = 0.2 mA, h ie = 4 (normalized)

I c = 1 mA, h ie = 1(normalized)

% change = 4 − 1 × 100% = 75%

4

(b) I e = 5 mA, h ie = 0.3 (normalized)

% change = 1 − 0.3 × 100% = 70%

1

82

76. (a) h oe = 20 μS @ 1 mA

I c = 0.2 mA, h oe = 0.2(h oe @ 1 mA)

= 0.2(20 μS)

= 4 μS

(b) r o =

1 1

= = 250 kΩ 6.8 kΩ

4 μS

hoe

Ignore 1/h oe

77. (a) I c = 10 mA, h oe = 10(20 μS) = 200 μS

1 1

(b) r o = = = 5 kΩ vs. 8.6 kΩ

hoe

200μS

Not a good approximation

78. (a) h re (0.1 mA) = 4(h re (1 mA))

= 4(2 × 10 −4 )

= 8 × 10 −4

(b) h re V ce = h re A v ⋅ V i

= (8 × 10 −4 )(210)V i

= 0.168 V i

In this case h re V ce is too large a factor to be ignored.

79. (a) h fe

(b) h oe

(c) h oe ≅ 30 (normalized) to

h oe ≅ 0.1 (normalized) at low levels of I c

(d) mid-region

80. (a) h ie is the most temperature-sensitive parameter of Fig. 5.33.

(b) h oe exhibited the smallest change.

(c) Normalized: h fe(max) = 1.5, h fe(min) = 0.5

For h fe = 100 the range would extend from 50 to 150—certainly significant.

(d) On a normalized basis r e increased from 0.3 at −65°C to 3 at 200°C—a significant

change.

(e) The parameters show the least change in the region 0° → 100°C.

83

81. (a) Test:

βR E ≥ 10R 2

70(1.5 kΩ) ≥ 10(39 kΩ)

?

105 kΩ ≥ 390 kΩ

No!

R Th = 39 kΩ || 150 kΩ = 30.95 kΩ

39 k Ω(14 V)

E Th =

39 kΩ+ 150 kΩ = 2.89 V

ETh

−VBE

2.89 V − 0.7 V

I B =

=

R + ( β + 1) R 30.95 k Ω+ (71)(1.5 k Ω)

Th

= 15.93 μA

V B = E Th − I B R Th

= 2.89 V − (15.93 μA)(30.95 kΩ)

= 2.397 V

E

V E = 2.397 V − 0.7 V = 1.697 V

VE

1.697 V

and I E =

R = = 1.13 mA

1.5 kΩ E


= 14 V − 1.13 mA(2.2 kΩ + 1.5 kΩ)

= 9.819 V

Biasing OK

(b) R 2 not connected at base:

VCC

− 0 14 V − 0.7 V

I B =

=

= 51.85 μA

R + ( β + 1) R 150 k Ω + (71)(1.5 k Ω )

B

E

V B = V CC − I B R B = 14 V − (51.85 μA)(150 kΩ)

= 6.22 V as noted in Fig. 5.187.

84

Chapter 6

1. −

2. From Fig. 6.11:

V GS = 0 V, I D = 8 mA

V GS = −1 V, I D = 4.5 mA

V GS = −1.5 V, I D = 3.25 mA

V GS = −1.8 V, I D = 2.5 mA

V GS = −4 V, I D = 0 mA

V GS = −6 V, I D = 0 mA

3. (a) V DS ≅ 1.4 V

(b) r d =

V 1.4 V

I = 6 mA

= 233.33 Ω

(c) V DS ≅ 1.6 V

(d) r d =

V 1.6 V

I = 3 mA

= 533.33 Ω

(e) V DS ≅ 1.4 V

(f) r d =

V 1.4 V

I = 1.5 mA

= 933.33 Ω

(g) r o = 233.33 Ω

r

233.33 Ω 233.33 Ω

= =

0.5625

o

r d =

[ 1−V

V ] [ 1 −( −1 V) ( −4 V) ]

GS

= 414.81 Ω

P

2 2

233.33 Ω 233.33 Ω

=

1 −( −2V) ( −4V)

0.25

(h) r d =

[ ]

2

= 933.2 Ω

(i)

533.33 Ω vs. 414.81 Ω

933.33 Ω vs 933.2 Ω

Eq. (6.1) is valid!

4. (a) V GS = 0 V, I D = 8 mA (for V DS > V P )

V GS = −1 V, I D = 4.5 mA

ΔI D = 3.5 mA

(b) V GS = −1 V, I D = 4.5 mA

V GS = −2 V, I D = 2 mA

ΔI D = 2.5 mA

85

(c) V GS = −2 V, I D = 2 mA

V GS = −3 V, I D = 0.5 mA

ΔI D = 1.5 mA

(d) V GS = −3 V, I D = 0.5 mA

V GS = −4 V, I D = 0 mA

ΔI D = 0.5 mA

(e) As V GS becomes more negative, the change in I D gets progressively smaller for the same

change in V GS .

(f)

Non-linear. Even though the change in V GS is fixed at 1 V, the change in I D drops from a

maximum of 3.5 mA to a minimum of 0.5 mA—a 7:1 change in ΔI D .

5. The collector characteristics of a BJT transistor are a plot of output current versus the output

voltage for different levels of input current. The drain characteristics of a JFET transistor are

a plot of the output current versus input voltage. For the BJT transistor increasing levels of

input current result in increasing levels of output current. For JFETs, increasing magnitudes

of input voltage result in lower levels of output current. The spacing between curves for a

BJT are sufficiently similar to permit the use of a single beta (on an approximate basis) to

represent the device for the dc and ac analysis. For JFETs, however, the spacing between the

curves changes quite dramatically with increasing levels of input voltage requiring the use of

Shockley’s equation to define the relationship between I D and V GS . VC sat

and V P define the

region of nonlinearity for each device.

6. (a) The input current I G for a JFET is effectively zero since the JFET gate-source junction is

reverse-biased for linear operation, and a reverse-biased junction has a very high resistance.

(b) The input impedance of the JFET is high due to the reverse-biased junction between the

gate and source.

(c) The terminology is appropriate since it is the electric field established by the applied

gate to source voltage that controls the level of drain current. The term “field” is

appropriate due to the absence of a conductive path between gate and source (or drain).

7. V GS = 0 V, I D = I DSS = 12 mA

V GS = V P = −6 V, I D = 0 mA

Shockley’s equation: V GS = −1 V, I D = 8.33 mA; V GS = −2 V, I D = 5.33 mA; V GS = −3 V,

I D = 3 mA; V GS = −4 V, I D = 1.33 mA; V GS = −5 V, I D = 0.333 mA.

8. For a p-channel JFET, all the voltage polarities in the network are reversed as compared to an

n-channel device. In addition, the drain current has reversed direction.

86

9. (b) I DSS = 10 mA, V P = −6 V

10. V GS = 0 V, I D = I DSS = 12 mA

V GS = V P = −4 V, I D = 0 mA

V

P

I

V GS = = −2 V, ID = DSS

= 3 mA

2 4

V GS = 0.3V P = −1.2 V, I D = 6 mA

V GS = −3 V, I D = 0.75 mA (Shockley’s equation)

11. (a) I D = I DSS = 9 mA

(b) I D = I DSS (1 − V GS /V P ) 2

= 9 mA(1 − (−2 V)/(−3.5 V)) 2

= 1.653 mA

(c) V GS = V P = −3.5 V, I D = 0 mA

(d) V GS < V P = −3.5 V, I D = 0 mA

12. V GS = 0 V, I D = 16 mA

V GS = 0.3V P = 0.3(−5 V) = −1.5 V, I D = I DSS /2 = 8 mA

V GS = 0.5V P = 0.5(−5 V) = −2.5 V, I D = I DSS /4 = 4 mA

V GS = V P = −5 V, I D = 0 mA

13. V GS = 0 V, I D = I DSS = 7.5 mA

V GS = 0.3V P = (0.3)(4 V) = 1.2 V, I D = I DSS /2 = 7.5 mA/2 = 3.75 mA

V GS = 0.5V P = (0.5)(4 V) = 2 V, I D = I DSS /4 = 7.5 mA/4 = 1.875 mA

V GS = V P = 4 V, I D = 0 mA

14. (a) I D = I DSS (1 − V GS /V P ) 2 = 6 mA(1 − (−2 V)/(−4.5 V)) 2

= 1.852 mA

I D = I DSS (1 − V GS /V P ) 2 = 6 mA(1 − (−3.6 V)/(−4.5 V)) 2

= 0.24 mA

⎛ I ⎞ ⎛ 3 mA ⎞

D

(b) V GS = VP

1 − = ( −4.5 V) 1−

⎜

I ⎟

⎜

DSS

6 mA ⎟

⎝ ⎠ ⎝ ⎠

= −1.318 V

⎛ I ⎞ ⎛ 5.5 mA ⎞

D

V GS = VP

1 − = ( −4.5 V) 1−

⎜

I ⎟

⎜

DSS

6 mA ⎟

⎝ ⎠ ⎝

⎠

= −0.192 V

15. I D = I DSS (1 − V GS /V P ) 2

3 mA = I DSS (1 − (−3 V)/(−6 V)) 2

3 mA = I DSS (0.25)

I DSS = 12 mA

87

16. From Fig. 6.22:

−0.5 V < V P < −6 V

1 mA < I DSS < 5 mA

For I DSS = 5 mA and V P = −6 V:

V GS = 0 V, I D = 5 mA

V GS = 0.3V P = −1.8 V, I D = 2.5 mA

V GS = V P /2 = −3 V, I D = 1.25 mA

V GS = V P = −6 V, I D = 0 mA

For I DSS = 1 mA and V P = −0.5 V:

V GS = 0 V, I D = 1 mA

V GS = 0.3V P = −0.15 V, I D = 0.5 mA

V GS = V P /2 = −0.25 V, I D = 0.25 mA

V GS = V P = −0.5 V, I D = 0 mA

PD

max

17. V DS = V

DS max

= 25 V, I D =

V

DSmax

120 mW

= = 4.8 mA

25 V

PD

120 mW

max

I D = I DSS = 10 mA, V DS = = = 12 V

I 10 mA

DSS

PD

120 mW

max

I D = 7 mA, V DS = = = 17.14 V

I 7 mA

D

88

18. V GS = −0.5 V, I D = 6.5 mA

2.5 mA

V GS = −1 V, I D = 4 mA

Determine ΔI D above 4 mA line:

2.5 mA x

= ⇒ x = 1.5 mA

0.5 V 0.3 V

I D = 4 mA + 1.5 mA = 5.5 mA corresponding with values determined from a purely

graphical approach.

19. Yes, all knees of V GS curves at or below |V P | = 3 V.

20. From Fig 6.25, I DSS ≅ 9 mA

At V GS = −1 V, I D = 4 mA

I D = I DSS (1 − V GS /V P ) 2

I

D

= 1 − V GS /V P

I

DSS

VGS

V = 1 − I

D

P

I

DSS

VGS

−1 V

V P = =

I

D

4 mA

1−

1−

I 9 mA

DSS

= −3 V (an exact match)

21. I D = I DSS (1 − V GS /V P ) 2

= 9 mA(1 − (−1 V)/(−3 V)) 2

= 4 mA, which compares very well with the level obtained using Fig. 6.25.

22. (a) V DS ≅ 0.7 V @ I D = 4 mA (for V GS = 0 V)

ΔVDS

0.7 V − 0 V

r = =

= 175 Ω

ΔI

4 mA − 0 mA

23. −

D

(b) For V GS = −0.5 V, @ I D = 3 mA, V DS = 0.7 V

r = 0.7 V

3 mA = 233 Ω

ro

175 Ω

(c) r d = =

2 2

(1 −VGS

/ VP

) (1 −( −0.5 V)/( −3 V)

= 252 Ω vs. 233 Ω from part (b)

24. The construction of a depletion-type MOSFET and an enchancement-type MOSFET are

identical except for the doping in the channel region. In the depletion MOSFET the channel is

established by the doping process and exists with no gate-to-source voltage applied. As the

gate-to-source voltage increases in magnitude the channel decreases in size until pinch-off

occurs. The enhancement MOSFET does not have a channel established by the doping

sequence but relies on the gate-to-source voltage to create a channel. The larger the

magnitude of the applied gate-to-source voltage, the larger the available channel.

89

25. −

26. At V GS = 0 V, I D = 6 mA

At V GS = −1 V, I D = 6 mA(1 − (−1 V)/(−3 V)) 2 = 2.66 mA

At V GS = +1 V, I D = 6 mA(1 − (+1 V)/(−3 V)) 2 = 6 mA(1.333) 2 = 10.667 mA

At V GS = +2 V, I D = 6 mA(1 − (+2 V)/(−3 V)) 2 = 6 mA(1.667) 2 = 16.67 mA

V GS

I D

−1 V 2.66 mA

0 6.0 mA

+1 V 10.67 mA

+2 V 16.67 mA

Δ I D

Δ I D

Δ I D

= 3.34 mA

= 4.67 mA

= 6 mA

From −1 V to 0 V, ΔI D = 3.34 mA

while from +1 V to +2 V, ΔI D = 6 mA − almost a 2:1 margin.

In fact, as V GS becomes more and more positive, I D will increase at a faster and faster

rate due to the squared term in Shockley’s equation.

V

P

27. V GS = 0 V, I D = I DSS = 12 mA; V GS = −8 V, I D = 0 mA; V GS = 2

= −4 V, ID = 3 mA;

V GS = 0.3V P = −2.4 V, I D = 6 mA; V GS = −6 V, I D = 0.75 mA

28. From problem 20:

VGS

V P =

I

1−

I

=

D

DSS

1

−0.21395

+ 1 V + 1 V + 1 V

= = =

14 mA 1 − 1.473 1 − 1.21395

1−

9.5 mA

≅ −4.67 V

29. I D = I DSS (1 − V GS /V P ) 2

4 mA

(1 −( −2 V) ( −5 V))

D

I DSS = =

2 2

( )

1−V

I

GS

V

P

= 11.11 mA

30. From problem 14(b):

⎛ I ⎞ ⎛ 20 mA ⎞

D

V GS = VP

1 − = ( −5 V) 1−

⎜

I ⎟

⎜

DSS

2.9 mA ⎟

⎝ ⎠ ⎝

⎠

= (−5 V)(1 − 2.626) = (−5 V)(−1.626)

= 8.13 V

31. From Fig. 6.34,

P

D max

= 200 mW, I D = 8 mA

P = V DS I D

Pmax

and V DS =

I

D

200 mW

= = 25 V

8 mA

90

32. (a) In a depletion-type MOSFET the channel exists in the device and the applied voltage

V GS controls the size of the channel. In an enhancement-type MOSFET the channel is

not established by the construction pattern but induced by the applied control voltage

V GS .

(b) −

(c) Briefly, an applied gate-to-source voltage greater than V T will establish a channel

between drain and source for the flow of charge in the output circuit.

33. (a) I D = k(V GS − V T ) 2 = 0.4 × 10 −3 (V GS − 3.5) 2

V GS

I D

3.5 V 0

4 V 0.1 mA

5 V 0.9 mA

6 V 2.5 mA

7 V 4.9 mA

8 V 8.1 mA

(b) I D = 0.8 × 10 −3 (V GS − 3.5) 2

V GS

I D

3.5 V 0

4 V 0.2 mA

5 V 1.8 mA

6 V 5.0 mA

7 V 9.8 mA

8 V 16.2 mA

For same levels of V GS , I D attains

twice the current level as part (a).

Transfer curve has steeper slope.

For both curves, I D = 0 mA for

V GS < 3.5 V.

D(on)

34. (a) k =

( VGS

(on)

−VT

)

I

4mA

=

(6 V − 4 V)

2 2

I D = k(V GS − V T ) 2 = 1 × 10 −3 (V GS − 4 V) 2

= 1 mA/V 2

(b) V GS I D For V GS < V T = 4 V, I D = 0 mA

4 V 0 mA

5 V 1 mA

6 V 4 mA

7 V 9 mA

8 V 16 mA

(c) V GS I D (V GS < V T )

2 V 0 mA

5 V 1 mA

10 V 36 mA

91

35. From Fig. 6.58, V T = 2.0 V

At I D = 6.5 mA, V GS = 5.5 V: I D = k(V GS − V T ) 2

6.5 mA = k(5.5 V − 2 V) 2

k = 5.31 × 10 −4

I D = 5.31 × 10 −4 (V GS − 2) 2

36. I D = kV ( − V) 2

GS (on)

and ( V V ) 2

V T =

V

GS (on)

GS (on)

T

I

D

−

T

=

k

I

D

− VT

=

k

D

V

GS (on)

− I k

3mA

= 4 V −

3

0.4×

10 −

= 4 V − 2.739 V

= 1.261 V

= 4 V − 7.5 V

37. I D = k(V GS − V T ) 2

I D

k = (V GS − V T ) 2

I D

= V GS − V T

k

V GS = V T +

= 27.36 V

I D

k

30 mA

= 5 V +

3

0.06×

10 −

38. Enhancement-type MOSFET:

2

I D = kV (

GS

− VT

)

dI

⎡

D

d ⎤

= 2 kV (

GS

−VT ) ⎢ ( VGS −VT

) ⎥

dVGS

⎣dVGS

1

⎦

dI

D

= 2k(V GS − V T )

dV

GS

92

Depletion-type MOSFET:

I D = I DSS (1 − V GS /V P ) 2

2

dI

D

d ⎛ V ⎞

GS

= I

DSS ⎜1−

⎟

dVGS

dVGS

⎝ VP

⎠

⎡ V ⎤

GS

d ⎡ V ⎤

GS

= IDSS

21 ⎢ − ⎥ ⎢0−

⎥

⎣ VP ⎦dVGS ⎣ VP

⎦

dI

dV

D

GS

=

=

=

−

1

V P

⎛ V ⎞⎛

GS

1 ⎞

2I

DSS ⎜1−

⎟⎜−

⎟

⎝ VP

⎠⎝ VP

⎠

2I

⎛

DSS

V ⎞

GS

− ⎜1−

⎟

VP

⎝ VP

⎠

2I

⎛

DSS

V ⎞⎛

P

V ⎞

GS

− ⎜ ⎟⎜1−

⎟

VP ⎝VP ⎠⎝ VP

⎠

2I

DSS

( VGS

− VP

)

V

=

2

P

dI

D

For both devices = k 1 (V GS − K 2 )

dVGS

revealing that the drain current of each will increase at about the same rate.

39. I D = k(V GS − V T ) 2 = 0.45 × 10 −3 (V GS − (−5 V)) 2

= 0.45 × 10 −3 (V GS + 5 V) 2

V GS = −5 V, I D = 0 mA; V GS = −6 V, I D = 0.45 mA; V GS = −7 V, I D = 1.8 mA;

V GS = −8 V, I D = 4.05 mA; V GS = −9 V, I D = 7.2 mA; V GS = −10 V, I D = 11.25 mA

41. −

42. (a) −

43. −

V 0.1 V

(b) For the “on” transistor: R = = = 25 ohms

I 4 mA

V 4.9 V

For the “off” transistor: R =

I

= 0.5 μA

= 9.8 MΩ

Absolutely, the high resistance of the “off” resistance will ensure V o is very close to 5 V.

93

Chapter 7

1. (a) V GS = 0 V, I D = I DSS = 12 mA

V GS = V P = −4 V, I D = 0 mA

V GS = V P /2 = −2 V, I D = I DSS /4 = 3 mA

V GS = 0.3V P = −1.2 V, I D = I DSS /2 = 6 mA

(b)

(c)

(d)

I

D Q

≅ 4.7 mA

V

DS Q

= V DD −

= 6.36 V

I

DQ

R = 12 V − (4.7 mA)(1.2 kΩ)

D

I

D Q

= I DSS (1 − V GS /V P ) 2 = 12 mA(1 − (−1.5 V)/(−4 V)) 2

= 4.69 mA

V

DS Q

= V DD − ID

R

Q D

= 12 V − (4.69 mA)(1.2 kΩ)

= 6.37 V

excellent comparison

2. (a)

(b)

2

I = I DSS (1 − V / V )

D Q

GS

P

= 10 mA ( 1 −( −3 V) ( − 4.5 V) ) 2

= 10 mA(0.333) 2

I = 1.11 mA

D Q

V

GS Q

= −3 V

(c) V DS = V DD − I D (R D + R S )

= 16 V − (1.11 mA)(2.2 kΩ)

= 16 V − 2.444 V

= 13.56 V

V D = V DS = 13.56 V

V G = V = −3 V

GS Q

V S = 0 V

94

3. (a)

I

DQ

VDD

−VD

14 V − 9 V

= =

R 1.6 kΩ

D

= 3.125 mA

(b) V DS = V D − V S = 9 V − 0 V = 9 V

⎛

(c) I D = I DSS (1 − V GS /V P ) 2 I

D

⇒ V GS = VP

1−

⎜

⎝ I

⎛

V GS = (−4 V)

⎜

1−

⎝

= −1.5 V

∴V GG = 1.5 V

3.125 mA ⎞

8 mA ⎟

⎠

DSS

⎞

⎟

⎠

4.

V

GS Q

= 0 V, I D = I DSS = 5 mA

V D = V DD − I D R D

= 20 V − (5 mA)(2.2 kΩ)

= 20 V − 11 V

= 9 V

5. V GS = V P = −4 V

∴ I

D Q

= 0 mA

and V D = V DD − I

= 18 V

DQ

R = 18 V − (0)(2 kΩ)

6. (a)(b) V GS = 0 V, I D = 10 mA

V GS = V P = −4 V, I D = 0 mA

V

P

V GS = = −2 V, ID = 2.5 mA

2

D

V GS = 0.3V P = −1.2 V, I D = 5 mA

V GS = −I D R S

I D = 5 mA:

V GS = −(5 mA)(0.75 kΩ)

= −3.75 V

(c)

I

D Q

≅ 2.7 mA

V

GS Q

≅ −1.9 V

(d) V DS = V DD − I D (R D + R S )

= 18 V − (2.7 mA)(1.5 kΩ + 0.75 kΩ)

= 11.93 V


= 18 V − (2.7 mA)(1.5 kΩ)

= 13.95 V

V G = 0 V

V S = I S R S = I D R S

= (2.7 mA)(0.75 kΩ)

= 2.03 V

95

7. I D = I DSS (1 − V GS /V P ) 2 ⎛ 2I R

= I

DSS ⎜

⎝

2

⎛ IDSS R ⎞

S 2

⎛2IDSS R ⎞

S

⎜ I 1 I I

2 ⎟ + ⎜ − ⎟ +

⎝ VP

⎠ ⎝ VP

⎠

I R

2 2

D S D S

1+ +

2

VP

VP

D D DSS

= 0

Substituting: 351.56 I 2 − 4.75I D

D + 10 mA = 0

2

− b±

b −4ac

I D =

= 10.91 mA, 2.60 mA

2a

I

D Q

= 2.6 mA (exact match #6)

V GS = −I D R S = −(2.60 mA)(0.75 kΩ)

= −1.95 V vs. −2 V (#6)

8. V GS = 0 V, I D = I DSS = 6 mA

V GS = V P = −6 V, I D = 0 mA

V

P

V GS = = −3 V, ID = 1.5 mA

2

⎞

⎟

⎠

V GS = 0.3V P = −1.8 V, I D = 3 mA

V GS = −I D R S

I D = 2 mA:

V GS = −(2 mA)(1.6 kΩ)

= −3.2 V

(a)

I

D Q

= 1.7 mA

V

GS Q

= −2.8 V

(b) V DS = V DD − I D (R D + R S )

= 12 V − (1.7 mA)(2.2 kΩ + 1.6 kΩ)

= 5.54 V


= 12 V − (1.7 mA)(2.2 kΩ)

= 8.26 V

V G = 0 V


= (1.7 mA)(1.6 kΩ)

= 2.72 V (vs. 2.8 V from V S = ( V

GS Q

))

9. (a)

I

D Q

= I S =

VS

1.7 V

R = 0.51 kΩ S

= 3.33 mA

(b)

V =− I R = −(3.33 mA)(0.51 kΩ)

GSQ

DQ

S

≅ −1.7 V

96

(c) I D = I DSS (1 − V GS /V P ) 2

3.33 mA = I DSS (1 − (−1.7 V)/(−4 V)) 2

3.33 mA = I DSS (0.331)

I DSS = 10.06 mA

(d) V D = V DD −

I

DQ

R

D

= 18 V − (3.33 mA)(2 kΩ) = 18 V − 6.66 V

= 11.34 V

(e) V DS = V D − V S = 11.34 V − 1.7 V

= 9.64 V

10. (a) V GS = 0 V

∴I D = I DSS = 4.5 mA

(b) V DS = V DD − I D (R D + R S )

= 20 V − (4.5 mA)(2.2 kΩ + 0.68 kΩ)

= 20 V − 12.96

= 7.04 V

(c) V D = V DD − I D R D

= 20 V − (4.5 mA)(2.2 kΩ)

= 10.1 V

(d) V S = I S R S = I D R S

= (4.5 mA)(0.68 kΩ)

= 3.06 V

11. Network redrawn:

V GS = 0 V, I D = I DSS = 6 mA

V GS = V P = −6 V, I D = 0 mA

V

P

V GS = = −3 V, ID = 1.5 mA

2

V GS = 0.3V P = −1.8 V, I D = 3 mA

V GS = −I D R S = −I D (0.39 kΩ)

For I D = 5 mA, V GS = −1.95 V

97

From graph

I

D Q

≅ 3.55 mA

V

GS Q

≅ −1.4 V

V S = − ( V GSQ ) = −(−1.4 V)

= +1.4 V

R2

110 k Ω(20 V)

12. (a) V G = VDD

=

R1 + R2

910 kΩ + 110 kΩ

= 2.16 V

V GS = 0 V, I D = I DSS = 10 mA

V GS = V P = −3.5 V, I D = 0 mA

V

P

V GS = = −1.75 V, ID = 2.5 mA

2

V GS = 0.3V P = −1.05 V, I D = 5 mA

V

GS Q

= V G − I D R S

V

GS Q

= 2.16 − I D (1.1 kΩ)

I D = 0: V

GS Q

= V G = 2.16 V

V

GS Q

= 0 V, I D = 2.16 V = 1.96 mA

1.1 kΩ

(b)

I

D Q

≅ 3.3 mA

V

GS Q

≅ −1.5 V

(c) V D = V DD − ID

R

Q D

= 20 V − (3.3 mA)(2.2 kΩ)

= 12.74 V


= (3.3 mA)(1.1 kΩ)

= 3.63 V

(d)

V = V DD − I ( R + R )

DS Q

DQ

D S

= 20 V − (3.3 mA)(2.2 k Ω + 1.1 kΩ)

= 20 V − 10.89 V

= 9.11 V

98

13. (a) I D = I DSS = 10 mA, V P = −3.5 V

V GS = 0 V, I D = I DSS = 10 mA

V GS = V P = −3.5 V, I D = 0 mA

VP

−3.5 V

V GS =

2 2

V GS = 0.3V P = −1.05 V, I D = 5 mA

V G

= = −1.75 V, I D = 2.5 mA

I

D Q

≅ 5.8 mA vs. 3.3 mA (#12)

V

GS Q

≅ −0.85 V vs. −1.5 V (#12)

110 k Ω(20 V)

=

110 kΩ + 910 kΩ

= 2.16 V

(b) As R S decreases, the intersection on the vertical axis increases. The maximum occurs at

I D = I DSS = 10 mA.

VG

2.16 V

∴ RS

min

= = = 216 Ω

I 10 mA

DSS

14. (a) I D =

VR V 18 V 9 V 9 V

D DD

−VD

−

= = =

R R 2 kΩ

2 kΩ D

D

= 4.5 mA

(b) V S = I S R S = I D R S = (4.5 mA)(0.68 kΩ)

= 3.06 V

V DS = V DD − I D (R D + R S )

= 18 V − (4.5 mA)(2 kΩ + 0.68 kΩ)

= 18 V − 12.06 V

= 5.94 V

R2

91 k Ω(18 V)

(c) V G = VDD

=

R + R 750 kΩ+ 91 kΩ = 1.95 V

1 2

V GS = V G − V S = 1.95 V − 3.06 V = −1.11 V

VGS

−1.11 V

(d) V P = = = − 4.44 V

I

D

4.5 mA

1−

1−

I 8 mA

DSS

= −1.48 V

99

15. (a) V GS = 0 V, I D = I DSS = 6 mA

V GS = V P = −6 V, I D = 0 mA

V GS = V P /2 = −3 V, I D = 1.5 mA

V GS = 0.3V P = −1.8 V, I D = 3 mA

V GS = V SS − I D R S

V GS = 4 V − I D (2.2 kΩ)

4 V

V GS = 0 V, I D =

2.2 kΩ

I D = 0 mA, V GS = 4 V

= 1.818 mA

I

D Q

≅ 2.7 mA

V

GS Q

≅ −2 V

(b) V DS = V DD + V SS − I D (R D + R S )

= 16 V + 4 V − (2.7 mA)(4.4 kΩ)

= 8.12 V

V S = −V SS + I D R S = −4 V + (2.7 mA)(2.2 kΩ)

= 1.94 V

or V S = − ( V GSQ ) = −(−2 V) = +2 V

16. (a) I D =

V VDD + VSS −VDS

12 V+ 3 V − 4 V 11 V

= = =

R R + R 3 kΩ+ 2 kΩ 5 kΩ D

S

= 2.2 mA

(b) V D = V DD − I D R D = 12 V − (2.2 mA)(3 kΩ)

= 5.4 V

V S = I S R S + V SS = I D R S + V SS

= (2.2 mA)(2 kΩ) + (−3 V)

= 4.4 V − 3 V

= 1.4 V

(c) V GS = V G − V S

= 0 V − 1.4 V

= −1.4 V

17. (a) I

D Q

= 4 mA

(b) V

D Q

= 12 V −4 mA(1.8 k Ω ) = 12 V − 7.2 V = 4.8 V

DS Q

V = 4.8 V

(c) P s = (12 V)(4 mA) = 48 mW

P d = (4.8 V)(4 mA) = 19.2 mW

100

18. V GS = 0 V, I D = I DSS = 6 mA

V GS = V P = −4 V, I D = 0 mA

V GS = V P /2 = −2 V, I D = I DSS /4 = 1.5 mA

V GS = 0.3V P = −1.2 V, I D = I DSS /2 = 3 mA

V GS = −I D R S = −I D (0.43 kΩ)

I D = 4 mA, V GS = −1.72 V

(b) V DS = V DD − I D (R D + R S )

= 14 V − 2.9 mA(1.2 kΩ + 0.43 kΩ)

= 9.27 V


= 14 V − (2.9 mA)(1.2 kΩ)

= 10.52 V

19. (a) V GS = 0 V, I D = I DSS = 8 mA

V GS = V P = −8 V, I D = 0 mA

V

P

V GS = = −4 V, ID = 2 mA

2

V GS = 0.3V P = −2.4 V, I D = 4 mA

V GS = +1 V, I D = 10.125 mA

V GS = +2 V, I D = 12.5 mA

V GS = −V SS − I D R S

= −(−4 V) − I D (0.39 kΩ)

V GS = 4 − I D (0.39 kΩ)

I D = 0: V GS = +4 V

4

V GS = 0: I D = = 10.26 mA

0.39 kΩ

I

D Q

≅ 9 mA

V

GS Q

≅ +0.5 V

101

(b) V DS = V DD − I D (R D + R S ) + V SS

= 18 V − 9 mA(1.2 kΩ + 0.39 kΩ) + 4 V

= 22 V − 14.31 V

= 7.69 V

20. I D = k(V GS − V T ) 2

V S = − ( V GSQ ) = −0.5 V

V DS = V DD − I D (R D + R S )

I

D(on)

= 5 mA =

5 mA

VDD

V

2

2 2 DS = 0 V; I D =

(7 V − 4 V) 9 V

RD

+ RS

=

K = 0.556 × 10 −3 A/V 2

and I D = 0.556 × 10 −3 (V GS − 4 V) 2

k =

( VGS

(on)

−VTh

)

I D = 0 mA, V DS = V DD

= 22 V

22 V

1.2 kΩ + 0.51 kΩ

= 12.87 mA


= 22 V − (8.25 mA)(1.2 kΩ)

= 12.1 V


= (8.25 mA)(0.51 kΩ)

= 4.21 V

(d) V DS = V D − V S

= 12.1 V − 4.21 V

= 7.89 V

vs. 7.9 V obtained graphically

102

R2

6.8 MΩ

21. (a) V G = VDD

=

(24 V)

R + R 10 MΩ+ 6.8 MΩ

1 2

= 9.71 V

V GS = V G − I D R S

V GS = 9.71 − I D (0.75 kΩ)

At I D = 0 mA, V GS = 9.71 V

9.71 V

At V GS = 0 V, I D =

0.75 kΩ

D(on)

k =

( VGS

(on)

−VGS ( Th)

)

I

= 0.556 × 10 −3 A/V 2

∴I D = 0.556 × 10 −3 (V GS − 3 V) 2

= 12.95 mA

5 mA 5 mA

= =

(6 V − 3 V) (3 V)

2 2 2

V GS

I D

3 V 0 mA

4 V 0.556 mA

5 V 2.22 mA

6 V 5 mA

7 V 8.9 mA

I

D Q

≅ 5 mA

V

GS Q

≅ 6 V

103

(b) V D = V DD − I D R D = 24 V − (5 mA)(2.2 kΩ)

= 13 V


= (5 mA)(0.75 kΩ)

= 3.75 V

R2

18 kΩ

22. (a) V G = VCC

=

(20 V)

R1 + R2

91 kΩ+ 18 kΩ

= 3.3 V

(b) V GS = 0 V, I D = I DSS = 6 mA

V GS = V P = −6 V, I D = 0 mA

V

P

V GS = = −3 V, ID = 1.5 mA

2

V GS = V P = −1.8 V, I D = 3 mA

I

D Q

≅ 3.75 mA

V

GS Q

≅ −1.25 V

(c) I E = I D = 3.75 mA

(d) I B =

I C

3.75 mA

= = 23.44 μA

β 160

(e) V D = V E = V B − V BE = V CC − I B R B − V BE = 20 V − (23.44 μA)(330 kΩ) − 0.7 V

= 11.56 V

(f)

V C = V CC − I C R C = 20 V − (3.75 mA)(1.1 kΩ)

= 15.88 V

23. Testing:

βR E ≥ 10R 2

(100)(1.2 kΩ) ≥ 10(10 kΩ)

120 kΩ > 100 kΩ (satisfied)

RV DD

2

10 k Ω(16 V)

(a) V B = V G = =

R1 + R2

40 kΩ+ 10 kΩ

= 3.2 V

(b) V E = V B − V BE = 3.2 V − 0.7 V = 2.5 V

VE

2.5 V

(c) I E =

R = = 2.08 mA

E

1.2 kΩ I C ≅ I E = 2.08 mA

I D = I C = 2.08 mA

104

(d) I B =

I C

2.08 mA

= = 20.8 μA

β 100

(e) V C = V G − V GS

⎛ I

D

V GS = VP

1−

⎜

⎝ I

DSS

⎞

⎟

⎠

⎛ 2.08 mA ⎞

= ( −6 V) ⎜

1−

6 mA ⎟

⎝

⎠

= −2.47 V

V C = 3.2 − (−2.47 V)

= 5.67 V

V S = V C = 5.67 V


= 16 V − (2.08 mA)(2.2 kΩ)

= 11.42 V

(f)

V CE = V C − V E = 5.67 V − 2.5 V

= 3.17 V

(g) V DS = V D − V S = 11.42 V − 5.67 V

= 5.75 V

⎛ I

D

24. V GS = VP

1−

⎜

⎝ I

= −1.75 V

DSS

V GS = −I D R S : R S =

⎞ ⎛

= (−6 V) 1

⎟

⎜

−

⎠ ⎝

4 mA ⎞

8 mA ⎟

⎠

VGS

−− ( 1.75 V)

− = = 0.44 kΩ

I 4 mA

R D = 3R S = 3(0.44 kΩ) = 1.32 kΩ

Standard values: R S = 0.43 kΩ

R D = 1.3 kΩ

D

⎛ I ⎞ ⎛

D

2.5 mA ⎞

25. V GS = VP

1−

⎜

= (−4 V)

I ⎟

⎜

1−

⎝ DSS ⎠

10 mA ⎟

⎝

⎠

= −2 V

V GS = V G − V S

and V S = V G − V GS = 4 V − (−2 V)

= 6 V

VS

6 V

R S = = = 2.4 kΩ (a standard value)

I

D

2.5 mA

R D = 2.5R S = 2.5(2.4 kΩ) = 6 kΩ ⇒ use 6.2 kΩ

RV

2

V G = DD

R2

(24 V)

⇒ 4 V = ⇒ 88 MΩ + 4R 2 = 24R 2

R1 + R2

22 MΩ+R2

20R 2 = 88 MΩ

R 2 = 4.4 MΩ

Use R 2 = 4.3 MΩ

105

26. I D = k(V GS − V T ) 2

I D

k = (V GS − V T ) 2

I D

= V GS − V T

k

D

and V GS = V T +

I 6mA

= 4 V + = 7.46 V

−3 2

k

0.5×

10 A/V

VR V 16 V 7.46 V 8.54 V

D DD

−VDS VDD −VGS

−

R D = = = = =

ID ID ID

6 mA 6 mA

= 1.42 kΩ

Standard value: R D = 0.75 kΩ

R G = 10 MΩ

27. (a) I D = I S =

VS

4 V

R = 1 kΩ = 4 mA

S

V DS = V DD − I D (R D + R S ) = 12 V − (4 mA)(2 kΩ + 1 kΩ)

= 12 V − (4 mA)(3 kΩ)

= 12 V − 12 V

= 0 V

JFET in saturation!

(b) V S = 0 V reveals that the JFET is nonconducting and the JFET is either defective or an

open-circuit exists in the output circuit. V S is at the same potential as the grounded side

of the 1 kΩ resistor.

(c) Typically, the voltage across the 1 MΩ resistor is ≅ 0 V. The fact that the voltage across

the 1 MΩ resistor is equal to V DD suggests that there is a short-circuit connection from

gate to drain with I D = 0 mA. Either the JFET is defective or an improper circuit

connection was made.

75 k Ω(20 V)

28. V G =

= 3.7 V (seems correct!)

75 kΩ+ 330 kΩ V GS = 3.7 V − 6.25 V = −2.55 V (possibly okay)

I D = I DSS (1 − V GS /V P ) 2

= 10 mA(1 − (−2.55 V)/(−6 V)) 2

= 3.3 mA (reasonable)

VS

6.25 V

However, I S =

R = = 6.25 mA ≠ 3.3 mA

S

1 kΩ V

R D

= I D R D = I S R D = (6.25 mA)(2.2 kΩ)

= 13.75 V

and VR

+ V

S R

= 6.25 V + 13.75 V

D

= 20 V = V DD

∴V DS = 0 V

1. Possible short-circuit from D-S.

2. Actual I DSS and/or V P may be larger in magnitude than specified.

106

VS

6.25 V

29. I D = I S =

R = = 6.25 mA

S

1 kΩ V DS = V DD − I D (R D + R S )

= 20 V − (6.25 mA)(2.2 kΩ + 1 kΩ)

= 20 V − 20 V

= 0 V (saturation condition)

RV DD

2

75 k Ω(20 V)

V G = =

= 3.7 V (as it should be)

R1 + R2

330 kΩ+ 75 kΩ V GS = V G − V S = 3.7 V − 6.25 V = −2.55 V

2

⎛ V ⎞

GS

I D = I

DSS ⎜1−

⎟ = 10 mA(1 − (−2.55 V)/(6 V)) 2

⎝ VP

⎠

= 3.3 mA ≠ 6.25 mA

In all probability, an open-circuit exists between the voltage divider network and the gate

terminal of the JFET with the transistor exhibiting saturation conditions.

30. (a) V GS = 0 V, I D = I DSS = 8 mA

V GS = V P = +4 V, I D = 0 mA

V

P

V GS = = +2 V, ID = 2 mA

2

V GS = 0.3V P = 1.2 V, I D = 4 mA

V GS = I D R S

I D = 4 mA;

V GS = (4 mA)(0.51 kΩ)

= 2.04 V

I = 3 mA, V = 1.55 V

D Q

GS Q

(b) V DS = V DD + I D (R D + R S )

= −18 V + (3 mA)(2.71 kΩ)

= −9.87 V


= −18 V − (3 mA)(2.2 kΩ)

= −11.4 V

107

I

D

(on)

31. k =

( V −V

) ( −7 V −( −3 V) )

GS (on) GS ( Th)

= 0.25 × 10 −3 A/V 2

I D = 0.25 × 10 −3 (V GS + 3 V) 2

4 mA 4 mA

= =

( − 4 V)

2 2 2

V GS

−3 V

−4 V

−5 V

−6 V

−7 V

−8 V

I D

0 mA

0.25 mA

1 mA

2.25 mA

4 mA

6.25 mA

V GS = V DS = V DD + I D R D

At I D = 0 mA, V GS = V DD = −16 V

VDD

16 V

At V GS = 0 V, I D =

R = 2 kΩ = 8 mA

D

(b) V DS = V GS = −7.25 V

(c) V D = V DS = −7.25 V

or V DS = V DD + I D R D

= −16 V + (4.4 mA)(2 kΩ)

= −16 V + 8.8 V

V DS = −7.2 V = V D

32.

VGS

−1.5 V

= = −0.375

VP

4 V

Find −0.375 on the horizontal axis.

Then move vertically to the I D = I DSS (1 − V GS /V P ) 2 curve.

Finally, move horizontally from the intersection with the curve to the left to the I D /I DSS axis.

I

D

= 0.39

IDSS

and I D = 0.39(12 mA) = 4.68 mA vs. 4.69 mA (#1)

V = V DD − I D R D = 12 V − (4.68 mA)(1.2 kΩ)

DS Q

= 6.38 V vs. 6.37 V (#1)

108

VP

4 V

33. m =

IDSS

R = S

(10 mA)(0.75 k Ω)

= 0.533

VGG

0.533(0)

M = m

V = 4 V

P

= 0

Draw a straight line from M = 0 through m = 0.533 until it crosses the normalized curve of I D

2

⎛ V ⎞

GS

= I

DSS ⎜1−

⎟ . At the intersection with the curve drop a line down to determine

⎝ VP

⎠

VGS

V = −0.49

P

so that V

GS Q

= −0.49V P = −0.49(4 V)

= −1.96 V (vs. −1.9 V #6)

If a horizontal line is drawn from the intersection to the left vertical axis we find

I

D

= 0.27

IDSS

and I D = 0.27(I DSS ) = 0.27(10 mA) = 2.7 mA

(vs. 2.7 mA from #6)

(a)

V

GS Q

= −1.96 V,

I

D Q

= 2.7 mA

(b) −

(c) −

(d) V DS = V DD − I D (R D + R S ) = 11.93 V (like #6)

V D = V DD − I D R D = 13.95 V (like #6)

V G = 0 V, V S = I D R S = 2.03 V (like #6)

RV DD

2

110 k Ω(20 V)

34. V GG = =

R1 + R2

110 kΩ+ 910 kΩ = 2.16 V

VP

3.5 V

m =

IDSS

R = S

(10 mA)(1.1 k Ω )

= 0.318

VGG

(2.16 V)

M = m × 0.318

V = 3.5

= 0.196

P

Find 0.196 on the vertical axis labeled M and mark the location. Move horizontally to the

vertical axis labeled m and then add m = 0.318 to the vertical height (≅ 1.318 in total)—mark

the spot. Draw a straight line through the two points located above, as shown below.

109

Continue the line until it intersects the I D = I DSS (1 − V GS /V P ) 2 curve. At the intersection move

horizontally to obtain the I D /I DSS ratio and move down vertically to obtain the V / V ratio.

GS

p

I

I

D

DSS

= 0.33 and

I

D Q

= 0.33(10 mA) = 3.3 mA

vs. 3.3 mA (#12)

VGS

V = −0.425 and V

GS

= −0.425(3.5 V)

Q

P

= −1.49 V

vs. 1.5 V (#12)

VP

6 V

35. m =

IDSS

R = S

(6 mA)(2.2 k Ω)

= 0.4545

VGG

(4 V)

M = m = 0.4545 V (6 V)

P

= 0.303

Find 0.303 on the vertical M axis.

Draw a horizontal line from M = 0.303 to the vertical m axis.

Add 0.4545 to the vertical location on the m axis defined by the horizontal line.

Draw a straight line between M = 0.303 and the point on the m axis resulting from the

addition of m = 0.4545.

Continue the straight line as shown below until it crosses the normalized

I D = I DSS (1 − V GS /V P ) 2 curve:

110

At the intersection drop a vertical line to determine

VGS

V = −0.34

P

and V

GS Q

= −0.34(6 V)

= −2.04 V (vs. −2 V from problem 15)

At the intersection draw a horizontal line to the I D /I DSS axis to determine

I

D

= 0.46

I

DSS

and I

D Q

= 0.46(6 mA)

= 2.76 mA (vs. 2.7 mA from problem 15)

(a)

I

D Q

= 2.76 mA, V

GS Q

= −2.04 V

(b) V DS = V DD + V SS − I D (R D + R S )

= 16 V + 4 V − (2.76 mA)(4.4 kΩ)

= 7.86 V (vs. 8.12 V from problem 15)

V S = −V SS + I D R S = −4 V + (2.76 mA)(2.2 kΩ)

= −4 V + 6.07 V

= 2.07 V (vs. 1.94 V from problem 15)

111

Chapter 8

1. g m0 =

2I

DSS

2(15 mA)

=

V −5 V

P

= 6 mS

2. g m0 =

2IDSS

2IDSS

2(12 mA)

⇒ VP

= = = 2.4 V

V

g 10 mS

P

V P = −2.4 V

m0

3. g m0 = 2 I

DSS

V

P

⇒ I DSS = ( g )( ) mO

VP

5 mS(3.5 V)

= = 8.75 mA

2 2

⎛ VGS 2(12 mA) 1 V

4. g m =

0

1

Q

⎞ ⎛ − ⎞

gm

− = 1−

⎜ V ⎟ ⎜ ⎟

⎝ P ⎠ −3 V ⎝ −3 V ⎠

= 5.3 mS

5. g m = 2 I ⎛ V

DSS

GS

1

Q

⎞

−

V ⎜

P

V ⎟

⎝ P ⎠

6 mS = 2 I

1

1 V

DSS ⎛ − ⎞

⎜ − ⎟

2.5 V ⎝ −2.5 V ⎠

I DSS = 12.5 mA

6. g m = g

0

m

I

D

2 IDSS

IDSS

/4 2(10 mA) 1

= =

I V I 5 V 4

DSS P DSS

20 mA ⎛1

⎞

= ⎜ ⎟

5 V ⎝2

⎠ = 2 mS

7. g m0 =

2I

DSS

2(8 mA)

= = 3.2 mS

V 5 V

P

⎛ VGS /4

g m =

0

1

Q

⎞ ⎛ V ⎞

P

gm

− = 3.2mS 1−

⎜

V ⎟ ⎜ ⎟

⎝ P ⎠ ⎝ VP

⎠

= 2.4 mS

1 3

= 3.2 mS

⎛ ⎜1− ⎞ ⎟=

3.2 mS

⎛ ⎜ ⎞

⎟

⎝ 4⎠ ⎝4⎠

8. (a) g m = y fs = 4.5 mS

(b) r d =

1 = 1

25μS

= 40 kΩ

yos

9. g m = y fs = 4.5 mS

1 1

r d = = 25μS

= 40 kΩ

yos

Z o = r d = 40 kΩ

A v (FET) = −g m r d = −(4.5 mS)(40 kΩ) = −180

112

10. A v = −g m r d ⇒ g m =

−Av

r

d

( −200)

=−

(100 k Ω )

= 2 mS

11. (a) g m0 =

2I

DSS

2(10 mA)

= = 4 mS

V 5 V

P

(b) g m =

ΔI

ΔV

D

GS

6.4 mA − 3.6 mA

=

2 V −1 V

= 2.8 mS

⎛ VGS 1.5 V

(c) Eq. 8.6: g m =

0

1

Q

⎞ ⎛ − ⎞

gm

− = 4 mS 1−

⎜ V ⎟ ⎜ ⎟

⎝ P ⎠ ⎝ −5 V ⎠

= 2.8 mS

(d) g m =

ΔI

ΔV

D

GS

3.6 mA −1.6 mA

=

3 V −2 V

= 2 mS

⎛ VGS 2.5 V

(e) g m =

0

1

Q

⎞ ⎛ − ⎞

gm

− = 4 mS 1−

⎜ V ⎟ ⎜ ⎟

⎝ P ⎠ ⎝ −5 V ⎠

= 2 mS

12. (a) r d =

ΔV

ΔI

DS

D

VGS

constant

(15 V − 5 V) 10 V

= =

(9.1 mA −8.8 mA) 0.3 mA

= 33.33 kΩ

(b) At V DS = 10 V, I D = 9 mA on V GS = 0 V curve

∴g m0 = 2 I

DSS

2(9 mA)

= = 4.5 mS

V 4 V

P

13. From 2N4220 data:

g m = y fs = 750 μS = 0.75 mS

1 1

r d = = = 100 kΩ

10μS

yos

2I

DSS

2(8 mA)

14. (a) g m (@ V GS = −6 V) = 0, g m (@ V GS = 0 V) = g m0 = = = 2.67 mS

VP

6 V

(b) g m (@ I D = 0 mA) = 0, g m (@ I D = I DSS = 8 mA) = g m0 = 2.67 mS

15. g m = y fs = 5.6 mS, r d =

1 1

= = 66.67 kΩ

15 μS

yos

16. g m =

r d =

2I

⎛ VGS 2(10 mA) 2 V

1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 4 V ⎝ −4 V ⎠

1 1

= = 40 kΩ

25 μS

yos

= 2.5 mS

113

17. Graphically, V

GS Q

= −1.5 V

2I

⎛ VGS 2(10 mA) 1.5 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 4 V ⎝ −4 V ⎠

Z i = R G = 1 MΩ

Z o = R D || r d = 1.8 kΩ || 40 kΩ = 1.72 kΩ

A v = −g m (R D || r d ) = −(3.125 mS)(1.72 kΩ)

= −5.375

= 3.125 mS

18.

V

GS Q

= −1.5 V

2I

⎛ VGS 2(12 mA) 1.5 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 6 V ⎝ −6 V ⎠

Z i = R G = 1 MΩ

1 1

Z o = R D || r d , r d =

yos

= = 25 kΩ

40 μS

= 1.8 kΩ || 25 kΩ

= 1.68 kΩ

A v = −g m (R D || r d ) = −(3 mS)(1.68 kΩ) = −5.04

= 3 mS

19. g m = y fs = 3000 μS = 3 mS

1 1

r d = = 50 μS

= 20 kΩ

yos

Z i = R G = 10 MΩ

Z o = r d || R D = 20 kΩ|| 3.3 kΩ = 2.83 kΩ

A v = −g m (r d || R D )

= −(3 mS)(2.83 kΩ)

= −8.49

20.

V

GS Q

= 0 V, g m = g m0 =

2I

DSS

2(6 mA)

= = 2 mS, r d =

V 6 V

Z i = 1 MΩ

Z o = r d || R D = 25 kΩ || 2 kΩ = 1.852 kΩ

A v = −g m (r d || R D ) = −(2 mS)(1.852 kΩ) ≅ −3.7

P

1 1

= = 25 kΩ

40 μS

yos

114

21. g m = 3 mS, r d = 20 kΩ

Z i = 10 MΩ

RD

3.3 kΩ

Z o =

=

RD

+ RS

3.3 kΩ+ 1.1 kΩ

1+ gmRS

+ 1 + (3 mS)(1.1 k Ω ) +

r

20 kΩ

3.3 kΩ

3.3 kΩ

=

= = 730 Ω

1+ 3.3 + 0.22 4.52

d

−gmRD

−(3 mS)(3.3 k Ω)

A v =

=

RD

+ RS

3.3 kΩ+ 1.1 kΩ

1+ gmRS

+ 1 + (3 mS)(1.1 k Ω ) +

rd

20 kΩ

−9.9 9.9

=

=− = −2.19

1+ 3.3 + 0.22 4.52

22. g m = y fs = 3000 μS = 3 mS

1 1

r d = = = 100 kΩ

10 μS

yos

Z i = R G = 10 MΩ (the same)

Z o = r d || R D = 100 kΩ || 3.3 kΩ = 3.195 kΩ (higher)

A v = −g m (r d || R D )

= −(3 mS)(3.195 kΩ)

= −9.59 (higher)

23.

V

GS Q

= −0.95 V

g m = 2 I ⎛ V

DSS

GS

1

Q

⎞

−

V ⎜

P

V ⎟

⎝ P ⎠

2(12 mA) ⎛ −0.95 V

= ⎜1−

3 V ⎝ −3 V

= 5.47 mS

⎞

⎟

⎠

Z i = 82 MΩ || 11 MΩ = 9.7 MΩ

Z o = r d || R D = 100 kΩ || 2 kΩ = 1.96 kΩ

A v = −g m (r d || R D ) = −(5.47 mS)(1.96 kΩ) = −10.72

V o = A v V i = (−10.72)(20 mV) = −214.4 mV

24.

V

GS Q

= −0.95 V (as before), g m = 5.47 mS (as before)

Z i = 9.7 MΩ as before

RD

Z o =

RD

+ R

1+ gmRS

+

r

but r d ≥ 10(R D + R S )

d

S

115

RD

2 kΩ 2 kΩ 2 kΩ

∴Z o = = = =

1+ gmRS

1 + (5.47 mS)(0.61 k Ω ) 1+

3.337 4.337

= 461.1 Ω

−gmRD

A v = since r d ≥ 10(R D + R S )

1 + gmRS

−(5.47 mS)(2 k Ω ) 10.94

=

=− = −2.52 (a big reduction)

4.337 (from above) 4.337

V o = A v V i = (−2.52)(20 mV) = −50.40 mV (compared to −214.4 mV earlier)

25.

26.

V

GS Q

= −0.95 V, g m (problem 23) = 5.47 mS

Z i (the same) = 9.7 MΩ

Z o (reduced) = r d || R D = 20 kΩ || 2 kΩ = 1.82 kΩ

A v (reduced) = −g m (r d || R D ) = −(5.47 mS)(1.82 kΩ) = −9.94

V o (reduced) = A v V i = (−9.94)(20 mV) = −198.8 mV

V

GS Q

= −0.95 V (as before), g m = 5.47 mS (as before)

Z i = 9.7 MΩ as before

RD

Z o =

RD

+ R

1+ gmRS

+

r

d

S

since r d < 10(R D + R S )

2 kΩ

=

2 kΩ + 0.61 kΩ

1 + (5.47 mS)(0.61 k Ω ) +

20 kΩ

2 kΩ

2 kΩ

=

=

1+ 3.33 + 0.13 4.46

= 448.4 Ω (slightly less than 461.1 Ω obtained in problem 24)

−gmRD

A v =

R

1+ g R +

m

S

D

+ R

r

d

S

−(5.47 mS)(2 k Ω)

=

2 kΩ + 0.61 kΩ

1 + (5.47 mS)(0.61 k Ω ) +

20 kΩ

−10.94 −10.94

=

= = −2.45 slightly less than −2.52 obtained in problem 24)

1+ 3.33 + 0.13 4.46

27.

2I

⎛ VGS 2(9 mA) 2.85 V

V

GS Q

= −2.85 V, g m = 1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 4.5 V ⎝ −4.5 V ⎠

Z i = R G = 10 MΩ

Z o = r d || R S || 1/g m = 40 kΩ || 2.2 kΩ || 1/1.47 mS = 512.9 Ω

680.27 Ω

gm( rd RS) (1.47 mS)(40 kΩ

2.2 k Ω) 3.065

A v =

= =

1 + g ( r R ) 1 + (1.47 mS)(40 kΩ 2.2 k Ω ) 1+

3.065

= 0.754

m d S

= 1.47 mS

116

28.

V

GS Q

= −2.85 V, g m = 1.47 mS

Z i = 10 MΩ (as in problem 27)

Z o = r d

|| R S || 1/g m = 20 kΩ || 2.2 kΩ || 680.27 Ω = 506.4 Ω < 512.9 Ω (#27)

1.982 kΩ

gm( rd RS) 1.47 mS(20 kΩ

2.2 k Ω) 2.914

A v =

= =

1 + g ( r R ) 1+ 1.47 mS(20 kΩ 2.2 k Ω ) 1+

2.914

m d S

= 0.745 < 0.754 (#27)

29.

V

GS Q

= −3.8 V

2I

⎛ VGS 2(6 mA) 3.8 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 6 V ⎝ −6 V ⎠

= 0.733 mS

The network now has the format examined in the text and

Z i = R G = 10 MΩ

Z o = r′ d

|| R S || 1/ g′

m

=

= 33.3 kΩ || 3.3 kΩ || 1/0.66 mS

= 3 kΩ || 1.52 kΩ

≅ 1 kΩ

r′

d

= r d + R D = 30 kΩ + 3.3 kΩ = 33.3 kΩ

gmrd

(0.733 mS)(30 k Ω) 21.99

g′ m

= = =

r + R 30 kΩ+ 3.3 kΩ 33.3 kΩ d

D

= 0.66 mS

g′ m( r′ d

RS) 0.66 mS(3 k Ω) 1.98 1.98

A v =

= = =

1 + g′ ( r′

R ) 1+ 0.66 mS(3 k Ω ) 1+

1.98 2.98

= 0.66

m d S

30.

GS Q

V = −1.75 V, g m = 2.14 mS

r d ≥ 10R D , ∴Z i ≅ R S || 1/g m = 1.5 kΩ || 1/2.14 mS

= 1.5 kΩ || 467.29 Ω

= 356.3 Ω

r d ≥ 10R D , ∴Z o ≅ R D = 3.3 kΩ

r d ≥ 10R D , ∴ A v ≅ g m R D = (2.14 mS)(3.3 kΩ) = 7.06

V o = A v V i = (7.06)(0.1 mV) = 0.706 mV

117

31.

32.

2I

⎛ VGS 2(8 mA) 1.75 V

V

GS Q

= −1.75 V, g m = 1

Q

⎞

DSS

⎛ − ⎞

− = ⎜1−

⎟ = 2.14 mS

V ⎜

P

V ⎟

⎝ P ⎠ 2.8 V ⎝ −2.8 V ⎠

⎡rd

+ R ⎤

D

⎡ 25 kΩ + 3.3 kΩ ⎤ 28.3 kΩ

Z i = R S || ⎢ ⎥= 1.5 kΩ 1.5 k

1 gmr

⎢ = Ω

d

1 (2.14 mS)(25 k )

⎥

⎣ + ⎦ ⎣ + Ω ⎦

54.5

= 1.5 kΩ || 0.52 kΩ = 386.1 Ω

Z o = R D || r d = 3.3 kΩ || 25 kΩ = 2.92 kΩ

gmRD + RD / rd

(2.14 mS)(3.3 k Ω ) + 3.3 k Ω/ 25 kΩ

A v =

=

1 + R / r

1+ 3.3 k Ω/ 25 kΩ

D

d

7.062 + 0.132 7.194

= = = 6.36

1+

0.132 1.132

V o = A v V i = (6.36)(0.1 mV) = 0.636 mV

V

GS Q

≅ −1.2 V, g m = 2.63 mS

r d ≥ 10R D , ∴ Z i ≅ R S || 1/g m = 1 kΩ || 1/2.63 mS = 1 kΩ || 380.2 Ω = 275.5 Ω

Z o ≅ R D = 2.2 kΩ

A v ≅ g m R D = (2.63 mS)(2.2 kΩ) = 5.79

1 1

33. r d =

y

= os

20 μS

= 50 kΩ, V

GS Q

= 0 V

g m = g m0 = 2 I

DSS

2(8 mA)

= = 5.33 mS

VP

3

A v = −g m R D = −(5.33 mS)(1.1 kΩ) = −5.863

V o = A v V i = (−5.863)(2 mV) = 11.73 mV

34.

V

GS Q

= −0.75 V, g m = 5.4 mS

Z i = 10 MΩ

r o ≥ 10R D , ∴

r o ≥ 10R D , ∴

Z o ≅ R D = 1.8 kΩ

A v ≅ −g m R D = −(5.4 mS)(1.8 kΩ)

= −9.72

35. Z i = 10 MΩ

Z o = r d || R D = 25 kΩ || 1.8 kΩ = 1.68 kΩ

A v = −g m (r d || R D )

2I

⎛ VGS 2(12 mA) 0.75 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = 1−

V ⎜ P

V ⎟ ⎜ ⎟

⎝ P ⎠ 3.5 V ⎝ −3.5 V ⎠

A v = −(5.4 mS)(1.68 kΩ)

= −9.07

= 5.4 mS

118

36. g m = y fs = 6000 μS = 6 mS

1 1

r d = = 35 μS

= 28.57 kΩ

yos

r d ≤ 10R D , ∴A v = −g m (r d || R D )

= −(6 mS)(28.57 kΩ || 6.8 kΩ)

= −32.94

V o = A v V i = (−32.94)(4 mV)

= −131.76 mV

5.49 kΩ

37. Z i = 10 MΩ || 91 MΩ ≅ 9 MΩ

2I

⎛ VGS 2(12 mA) 1.45 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = ⎜1−

⎟ = 4.13 mS

V ⎜

P

V ⎟

⎝ P ⎠ 3 V ⎝ −3 V ⎠

Z o = r d || R S || 1/g m = 45 kΩ || 1.1 kΩ || 1/4.13 mS

= 1.074 kΩ || 242.1 Ω

= 197.6 Ω

gm( rd RS) (4.13 mS)(45 kΩ

1.1 k Ω)

A v =

=

1 + gm( rd RS) 1 + (4.13 mS)(45 kΩ 1.1 k Ω)

(4.13 mS)(1.074 k Ω) 4.436

= =

1 + (4.13 mS)(1.074 k Ω ) 1+

4.436

= 0.816

38. g m = 2kV

( GS

− V

( ))

Q GS Th

= 2(0.3 × 10 −3 )(8 V − 3 V)

= 3 mS

39.

V

GS Q

= 6.7 V

g m = 2kV

( )

GS

− V

Q T

= 2(0.3 × 10 −3 )(6.7 V − 3 V) = 2.22 mS

RF + rd RD

10 MΩ+ 100 kΩ 2.2 kΩ

Z i =

=

1 + gm( rd RD) 1 + (2.22 mS)(100 kΩ 2.2 k Ω)

10 MΩ+ 2.15 kΩ

=

≅ 1.73 MΩ

1+ 2.22 mS(2.15 k Ω)

Z o = R F || r d || R D = 10 MΩ || 100 kΩ || 2.2 kΩ = 2.15 kΩ

A v = −g m (R F || r d || R D ) = −2.22 mS(2.15 kΩ) = −4.77

119

40. g m = 2kV

( GS

V)

Q T

− = 2(0.2 × 10 −3 )(6.7 V − 3 V)

= 1.48 mS

RF + rd RD

10 MΩ+ 100 kΩ 2.2 kΩ

Z i =

=

1 + g ( r R ) 1 + (1.48 mS)(100 kΩ 2.2 k Ω)

m d D

10 MΩ+ 2.15 kΩ

=

= 2.39 MΩ > 1.73 MΩ (#39)

1 + (1.48 mS)(2.15 k Ω )

41.

Z o = R F || r d || R D = 2.15 kΩ = 2.15 kΩ (#39)

A v = −g m (R F || r d || R D ) = −(1.48 mS)(2.15 kΩ)

= −3.182 < −4.77 (#39)

V = 5.7 V, g GS m = ( )

Q

2kVGS

V

Q T

− = 2(0.3 × 10 −3 )(5.7 V − 3.5 V)

= 1.32 mS

1

r d = = 33.33 kΩ

30 μ S

A v = −g m (R F || r d || R D ) = −1.32 mS(22 MΩ || 33.33 kΩ || 10 kΩ)

= −10.15

V o = A v V i = (−10.15)(20 mV) = −203 mV

42. I D = k(V GS − V T ) 2

I

D(on)

4 mA

∴k =

= = 0.444 × 10 −3

2

2

( VGS

(on)

−VT

) (7 V − 4 V)

g m = 2 kV (

GS

− VGS ( Th)

) = 2(0.444 × 10 −3 )(7 V − 4 V)

Q

= 2.66 mS

43.

A v = −g m (R F || r d || R D ) = −(2.66 mS)(22 MΩ || 50 kΩ || 10 kΩ) = −22.16

≅ 8.33 kΩ

V o = A v V i = (−22.16)(4 mV) = −88.64 mV

V

GS Q

= 4.8 V, g m = 2kV

( GS

V )

Q GS( Th)

8.33 kΩ

− = 2(0.4 × 10 −3 )(4.8 V − 3 V) = 1.44 mS

A v = −g m (r d || R D ) = −(1.44 mS)(40 kΩ || 3.3 kΩ) = −4.39

V o = A v V i = (−4.39)(0.8 mV) = −3.51 mV

120

1 1

44. r d =

y

= os

25 μS

= 40 kΩ

2I

DSS

2(8 mA)

V

GS Q

= 0 V, ∴g m = g m0 = = = 6.4 mS

VP

2.5 V

A

v

= g m (r d || R D )

8 = (6.4 mS)(40 kΩ || R D )

8

40 kΩ ⋅ RD

= 1.25 kΩ =

6.4 mS 40 kΩ+

RD

and R D = 1.29 kΩ

Use R D = 1.3 kΩ

45.

1 1

VGS

= V

Q P

= ( − 3 V) = −1 V

3 3

2

2

⎛ VGS

⎞

Q ⎛ −1 V ⎞

ID

= I 1 12 mA 1

Q DSS

− = ⎜ − ⎟ = 5.33 mA

⎜ V ⎟

⎝ P ⎠ ⎝ −3 V ⎠

VS

1 V

R S = = = 187.62 Ω ∴Use R S = 180 Ω

I

D

5.33 mA

Q

2I

⎛ VGS 2(12 mA) 1 V

g m = 1

Q

⎞

DSS

⎛ − ⎞

− = ⎜1−

⎟ = 5.33 mS

V ⎜

P

V ⎟

⎝ P ⎠ 3 V ⎝ −3 V ⎠

A v = −g m (R D || r d ) = −10

−10

or R D || 40 kΩ = = 1.876 kΩ

5.33 mS

RD

⋅40 kΩ

= 1.876 kΩ

RD

+ 40 kΩ 40 kΩR D = 1.876 kΩR D + 75.04 kΩ 2

38.124R D = 75.04 kΩ

R D = 1.97 kΩ ⇒ R D = 2 kΩ

121

Chapter 9

1. (a) 3, 1.699, −1.151

(b) 6.908, 3.912, −0.347

(c) results differ by magnitude of 2.3

2. (a) log 10 2.2 × 10 3 = 3.3424

(b) log e (2.2 × 10 3 ) = 2.3 log 10 (2.2 × 10 3 ) = 7.6962

(c) log e (2.2 × 10 3 ) = 7.6962

3. (a) same 13.98

(b) same −13.01

(c) same 0.699

4.

Po

(a) dB = 10 log 10

P = 10 log 100 W

10

i

5 W = 10 log 10 20 = 10(1.301)

= 13.01 dB

(b) dB = 10 log 10

100 mW

5 mW = 10 log 1020 = 10(1.301)

= 13.01 dB

(c) dB = 10 log 100 μ W

10

20 μW

= 6.9897 dB

= 10 log 10 5 = 10(0.6987)

5.

dBm

P2

25 W

G = 10 log 10 = 10 log10

1 mW

600 Ω

1 mW

= 43.98 dBm

1

600 Ω

V2

6. G dB = 20 log 10

V = 20 log 100 V

10

25 V = 20 log 104 = 20(0.6021)

= 12.04 dB

V2

25 V

7. G dB = 20 log 10 20 log10

V = = 20 log 10 2500

1 10 mV

= 20(3.398) = 67.96 dB

8. (a) Gain of stage 1 = A dB

Gain of stage 2 = 2 A dB

Gain of stage 3 = 2.7 A dB

A + 2A +2.7A = 120

A = 21.05 dB

122

Vo

1

(b) Stage 1: A

v 1

= 21.05 dB = 20 log 10

Vi

1

21.05

Vo

1

= 1.0526 = log 10

20

Vi

1

V

10 1.0526 o1

=

Vi

1

Vo

1

and

V = 11.288

i1

Vo

2

Stage 2: A

v 2

= 42.1 dB = 20 log 10

Vi

2

Vo

2

2.105 = log 10

Vi

2

V

10 2.105 o2

=

Vi

2

Vo

2

and

V = 127.35

i2

Vo

3

Stage 3: : A

v 3

= 56.835 dB = 20 log 10

Vi

3

Vo

3

2.8418 = log 10

Vi

3

V

10 2.8418 o3

=

Vi

3

Vo

3

and

V = 694.624

i3

Av = A

T v

⋅A 1 v

⋅ A

2 V

= (11.288)(127.35)(694.624) = 99,8541.1

3

?

A T = 120 dB = 20 log 10 99,8541.1

120 dB ≅ 119.99 dB (difference due to level of accuracy carried through calculations)

P2

9. (a) G dB = 20 log 10

P

1

48 W

= 10 log10

= 69.83 dB

5 μ W

V

PR

o

o o 20 log

10

(48 W)(40 k Ω)

(b) G v = 20 log 10 = 20 log10

=

Vi

Vi

100 mV

= 82.83 dB

123

(c) R i =

(d) P o =

V i

P

(100 mV)

= = 2 kΩ

5 μW

2 2

2

Vo

R ⇒ V o = PR

o o

= (48 W)(40 k Ω ) = 1385.64 V

o

10. (a) Same shape except A v = 190 is now level of 1. In fact, all levels of A v are divided by 190

to obtain normalized plot.

0.707(190) = 134.33 defining cutoff frequencies

at low end f 1 ≅ 230 Hz (remember this is a log scale)

at high end f 2 ≅ 160 kHz

(b)

11. (a)

A

v

Vo

1

= =

Vi

1 + ( f / f )

1

2

1 1

f 1 = =

2π RC 2 π(1.2 k Ω)(0.068 μF)

= 1950.43 Hz

A v

=

1

⎛1950.43 Hz ⎞

1+ ⎜ ⎟

⎝ f ⎠

2

(b)

A

V dB

100 Hz: A

v

= 0.051 −25.8

1 kHz: A

v

= 0.456 −6.81

2 kHz: A

v

= 0.716 −2.90

5 kHz: A

v

= 0.932 −0.615

10 kHz: A

v

= 0.982 −0.162

(c) f 1 ≅ 1950 Hz

(d)(e)

124

1

12. (a) f 1 =

2π

RC

θ = tan −1 f1

f

= 1.95 kHz

−1

= tan

1.95 kHz

f

(b)

f

θ = tan −1 1.95 kHz

f

100 Hz 87.06°

1 kHz 62.85°

2 kHz 44.27°

5 kHz 21.3°

10 kHz 11.03°

(c) f 1 =

1

2π

RC

= 1.95 kHz

(d) First find θ = 45° at f 1 = 1.95 kHz. Then sketch an approach to 90° at low frequencies and

0° at high frequencies. Use an expected shape for the curve noting that the greatest

change in θ occurs near f 1. The resulting curve should be quite close to that plotted above.

13. (a) 10 kHz

(b) 1 kHz

(c) 20 kHz → 10 kHz → 5 kHz

(d) 1 kHz → 10 kHz → 100 kHz

125

14. From example 9.9, r e = 15.76 Ω

−RC RL ro

4 k 2.2 k 40 k

A v =

− Ω Ω Ω

=

r

15.76 Ω

e

= −86.97 (vs. −90 for Ex. 9.9)

f

L s

f

L C

=

: r o does not affect R i ∴

f

L s

=

1

2 π ( R + R ) C

s i S

1 1

=

2 π( Ro + RL) CC 2 π( RC ro + RL)

CC

R C || r o = 4 kΩ || 40 kΩ = 5.636 kΩ

1

=

2 π (5.636 kΩ+ 2 k Ω)(1 μF)

= 28.23 Hz (vs. 25.68 Hz for Ex. 9.9)

f LC

the same ≅ 6.86 Hz

f

L E

: R e not affected by r o , therefore,

f

L E

1

= ≅ 327 Hz is the same.

2π R C

In total, the effect of r o on the frequency response was to slightly reduce the mid-band

gain.

e

E

15. (a) βR E ≥ 10R 2

(120)(1.2 kΩ) ≥ 10(10 kΩ)

144 kΩ ≥ 100 kΩ (checks!)

10 k Ω(14 V)

V B =

10 kΩ+ 68 kΩ = 1.795 V

V E = V B − V BE = 1.795 V − 0.7 V

= 1.095 V

VE

1.095 V

I E =

R = = 0.913 mA

1.2 kΩ r e =

E

26 mV 26 mV

= = 28.48 Ω

0.913 mA

I E

(b)

A

Vmid

( RL

RC) −(3.3 kΩ 5.6 k Ω)

=− =

r 28.48 Ω

e

= −72.91

(c) Z i = R 1 || R 2 || βr e

= 68 kΩ || 10 kΩ || (120)(28.48 Ω)

= 2.455 kΩ

3.418 kΩ

126

(d)

(e)

Vo Vo Vi

Av

= = ⋅

s

Vs Vi Vs

Vi

Zi

2.455 kΩ

= =

Vs Zi + Rs

2.455 kΩ+ 0.82 kΩ

= 0.75

A

v s

= (−72.91)(0.75)

= −54.68

1 1

fL S

= =

2 π ( Rs + Ri) Cs

2 π(0.82 kΩ+ 2.455 k Ω)(0.47 μF)

= 103.4 Hz

1 1

f

L C

=

=

2 π ( Ro + RL) CC

2 π(5.6 kΩ+ 3.3 k Ω)(0.47 μF)

= 38.05 Hz

1

⎛R′

s

⎞

f

L E

= : R e = R E || ⎜ + re

⎟

2π ReCE

⎝ β ⎠

R′

s

= R s || R 1 || R 2 = 0.82 kΩ || 68 kΩ || 10 kΩ

= 749.51 Ω

⎛749.51

Ω ⎞

R e = 1.2 kΩ || ⎜ + 28.48 Ω⎟

⎝ 120

⎠

= 1.2 kΩ || 34.73 Ω

= 33.75 Ω

f

L E

= 1 1

2π RC

=

e E

2 π(33.75 Ω)(20 μF)

= 235.79 Hz

(f)

f 1 ≅

f

L E

= 235.79 Hz

(g)(h)

127

VCC

−VBE

20 V − 0.7 V 19.3 V

16. (a) I B =

= =

RB

+ ( β + 1) RE

470 k Ω+ (111)(0.91 k Ω) 470 kΩ+ 101.01 kΩ

= 33.8 μA

I E = (β + 1)I B = (111)(33.8 μA)

= 3.752 mA

26 mV

r e =

3.752 mA = 6.93 Ω

(b)

A

vmid

Vo −( RC RL) −(3 kΩ

4.7 k Ω) −1.831 kΩ

= = = =

V r

6.93 Ω 6.93 Ω

i

= −264.24

e

(c) Z i = R B || βr e = 470 kΩ || (110)(6.93 Ω) = 470 kΩ || 762.3 Ω

= 761.07 Ω

(d)

(e)

(f)

A

Z

761.07 Ω

= A = ( −264.24)

+ 761.07 Ω+ 0.6 kΩ

= −147.76

i

vs

( mid) vmid

Zi

Rs

1 1

f

L S

=

=

2 π ( RS + Zi) CS

2 π(600 Ω+ 761.07 Ω)(1 μF)

= 116.93 Hz

1 1

f

L C

=

=

2 π ( Ro + RL) CC

2 π(3 kΩ+ 4.7 k Ω)(1 μF)

= 20.67 Hz

1

⎛R′

s ⎞

f Re = RE ⎜ + r

L E

=

e ⎟

2π R ⎝ β

eCE

⎠

1

⎛Rs

R

=

B ⎞

= 0.91 kΩ ⎜ + re

2 π (12.21 Ω)(6.8 μF)

⎟

⎝ β ⎠

= 1.917 kHz

⎛0.6 kΩ

470 kΩ

⎞

= 0.91 kΩ ⎜

+ 6.93 Ω⎟

⎝ 110

f

⎠

1 ≅ f

L E

= 1.917 kHz

= 910 Ω12.38

Ω

= 12.21 Ω

(g, h)

128

17. (a) βR E ≥ 10R 2

(100)(2.2 kΩ) ≥ 10(30 kΩ)

220 kΩ ≥ 300 kΩ (No!)

R Th = R 1 || R 2 = 120 kΩ || 30 kΩ = 24 kΩ

30 k Ω(14 V)

E Th =

30 kΩ+ 120 kΩ = 2.8 V

ETh

−VBE

2.8 V − 0.7 V

I B =

=

R + ( β + 1) R 24 kΩ+ 222.2 kΩ

Th

= 8.53 μA

I E = (β + 1)I B = (101)(8.53 μA)

= 0.86 mA

26 mV 26 mV

r e = = = 30.23 Ω

0.86 mA

I E

E

(b)

A

v mid

=

RE

RL

r + R R

e E L

2.2 kΩ

8.2 kΩ

=

30.23 Ω+ 2.2 kΩ 8.2 kΩ

= 0.983

(c) Z i = R 1 || R 2 || β(r e + R′

E

) R′

E

= R E || R L = 2.2 kΩ || 8.2 kΩ = 1.735 kΩ

= 120 kΩ || 30 kΩ || (100)(30.23 Ω + 1.735 kΩ)

= 21.13 kΩ

(d)

Vo Vo Vi

A

v s

= = ⋅

V V V

s i s

Vi

Zi

21.13 kΩ

= =

V Z + R 21.13 kΩ+ 1 kΩ = 0.955

s i s

(e)

f

f

L S

L C

1

=

2 π ( Rs + Ri)

Cs

1

=

2 π (1 kΩ+ 21.13 k Ω)(0.1 μF)

= 71.92 Hz

1

=

2 π ( R + R ) C

o L C

⎛R′

s

⎞

R o = R E || ⎜ + re

⎟

⎝ β ⎠

⎛0.96 kΩ

⎞

= (2.2 kΩ) || ⎜ + 30.23 Ω⎟

⎝ 100

⎠

= 39.12 Ω

f LC

1

=

2 π (39.12 Ω+ 8.2 k Ω)(0.1 μF)

= 193.16 Hz

R′ = R R R

s

s

1 2

= 1 kΩ120 kΩ30 kΩ

= 0.96 kΩ

129

(f)

f

1 low

≅ 193.16 Hz

(g)(h)

18. (a) I E =

r e =

V

EE

−V

R

E

EB

4 V − 0.7 V

=

= 2.75 mA

1.2 kΩ

26 mV 26 mV

= = 9.45 Ω

2.75 mA

I E

(b)

RC

R

A

v mid

=

re

= 205.1

L

3.3 kΩ

4.7 kΩ

=

9.45 Ω

(c) Z i = R E || r e = 1.2 kΩ || 9.45 Ω

= 9.38 Ω

(d)

(e)

Zi

9.38 Ω(205.1)

A

v s (mid)

= Av

=

mid

Zi

+ Rs

9.38 Ω+ 100 Ω

= 17.59

1 1

f

L s

=

=

2 π ( Rs + Zi) Cs

2 π(100 Ω+ 9.38 Ω)(10 μF)

= 145.5 Hz

1 1

f

L C

=

=

2 π ( Ro + RL) CE

2 π(3.3 kΩ+ 4.7 k Ω)(10 μF)

= 1.989 Hz

(f) f =

f

L s

≅ 145.5 Hz

(g, h)

130

19. (a) V GS = −I D R S

I D =

I

DSS

⎛

⎜

⎝

V

−

V

1

GS

P

2

⎞

⎟

⎠

V

GSQ

I

DQ

≅ −2.45 V

≅ 2.1 mA

(b) g m0 = 2 I

V

DSS

P

2(6 mA)

= = 2 mS

6 V

⎛ VGS ( 2.45 V)

g m =

0

1

Q

⎞ ⎛ − ⎞

gm

− = 2 mS 1−

⎜ V ⎟ ⎜ ⎟

⎝ P ⎠ ⎝ ( − 6 V) ⎠

= 1.18 mS

(c)

A

v mid

= −g m (R D || R L )

= −1.18 mS(3 kΩ || 3.9 kΩ) = −1.18 mS(1.6956 kΩ)

= −2

(d) Z i = R G = 1 MΩ

(e)

A

v s

= A v = −2

(f)

f

L G

= 1 1

2 π ( R + R ) C

= 2 π(1 kΩ+ 1 M Ω)(0.1 μF)

sig

= 1.59 Hz

i

G

f

f

L C

1

=

2 π ( Ro + RL)

CC

1

=

2 π (3 kΩ+ 3.9 k Ω)(4.7 μF)

= 4.91 Hz

1

L S

= Req

2π ReqCS

1

=

2 π (496.69 Ω)(10 μF)

= 32.04 Hz

1 1

= RS

= 1.2 kΩ = 1.2 kΩ || 847.46 Ω

g

1.18 mS

m

= 496.69 Ω

(g) f 1 ≅

f

L S

≅ 32 Hz

(h, i)

131

20. (a) same as problem 19

V ≅ −2.45 V,

GS Q

I

D Q

≅ 2.1 mA

(b) g m0 = 2 mS, g m = 1.18 mS (r d has no effect!)

(c)

A

v mid

= −g m (R D || R L || r d )

= −1.18 mS(3 kΩ || 3.9 kΩ || 100 kΩ)

= −1.18 mS (1.67 kΩ)

= −1.971 (vs. −2 for problem 19)

(d) Z i = R G = 1 MΩ (the same)

(e)

(f)

A

v s (mid)

=

Z

1 MΩ

( A )

Z

= −1.969 vs. −2 for problem 19

i

v

=

mid

i

+ Rsig

1 MΩ+ 1 kΩ (−1.971)

f

L G

= 1.59 Hz (no effect)

f

L C

: R o = R D || r d = 3 kΩ || 100 kΩ = 2.91 kΩ

1 1

f

L C

=

=

2 π ( Ro + RL) CC

2 π(2.91 kΩ+ 3.9 k Ω)(4.7 μF)

= 4.97 Hz vs. 4.91 Hz for problem 19

RS

f

L S

: R eq = 1 + RS (1 + gmrd )/( rd + ( RD RL

))

1.2 kΩ

=

1 + (1.2 k Ω )(1 + (1.18 mS)(100 k Ω)) /(100 kΩ+ 3 kΩ

3.9 k Ω)

= 1.2 k Ω

1+

1.404

≅ 499.2 Ω

1 1

f

L S

: = =

2π ReqC S

2 π(499.2 Ω)(10 μF)

= 31.88 Hz vs. 32.04 for problem 19.

Effect of r d = 100 kΩ insignificant!

21. (a) V G =

68 k Ω(20 V)

68 kΩ+ 220 kΩ = 4.72 V

V GS = V G − I D R S

V GS = 4.72 V − I D (2.2 kΩ) V

I D = I DSS (1 − V GS /V P ) 2

(b) g m0 = 2 I

V

DSS

P

GSQ

I

DQ

2(10 mA)

= = 3.33 mS

6 V

≅ −2.55 V

≅ 3.3 mA

⎛ V ⎞

GS

⎛ ( −2.55 V) ⎞

g m = gm0

⎜1− ⎟= 3.33 mS⎜1−

⎟

⎝ VP

⎠ ⎝ −6 V ⎠

= 1.91 mS

132

(c)

A

v mid

= −g m (R D || R L )

= −(1.91 mS)(3.9 kΩ || 5.6 kΩ)

= −4.39

(d) Z i = 68 kΩ || 220 kΩ = 51.94 kΩ

(e)

(f)

Vo Vi

A

v s (mid)

=

V

⋅ V

V

i

s

Z

Ω

+ 51.94 kΩ+ 1.5 kΩ = 0.972

i

V = i

51.94 k

=

s

Zi

Rs

A = (−4.39)(0.972) = −4.27

f

v s (mid)

L G

= 1 1

2 π ( R + R ) C

= 2 π(1.5 kΩ+ 51.94 k Ω)(1 μF)

sig

= 2.98 Hz

i

G

f

L C

= 1 1

2 π ( Ro + RL) C

=

C

2 π(3.9 kΩ+ 5.6 k Ω)(6.8 μF)

= 2.46 Hz

f

L S

1

=

2π R C

eq

S

1

=

2 π (388.1 Ω)(10 μF)

= 41 Hz

R eq = R S ||

1

g m

1

= 1.5 kΩ ||

1.91 mS

= 1.5 kΩ || 523.56 Ω

= 388.1 Ω

(g) f 1 ≅

f

L S

= 41 Hz

(h, i)

133

22. (a)

f

H i

=

1

π R C

2

Th1

i

1

=

2 π (614.56 Ω)(931.92 pF)

= 277.89 kHz

f

H o

=

1

π R C

2

Th2

o

1

=

2 π (2.08 k Ω)(28 pF)

= 2.73 MHz

R = R R R R

Th1 s 1 2 i

= 0.82 kΩ68 kΩ10 kΩ3.418 kΩ

= 0.81 kΩ 2.547 kΩ

= 614.56 Ω

C i = CW + C (1 )

1 be

+ Cbc − Av

= 5 pF + 40 pF + 12 pF(1 − (−72.91))

= 931.92 pF ↑Prob. 15

R

Th 2

= R C || R L = 5.6 kΩ || 3.3 kΩ

= 2.08 kΩ

C o = CW + C

o ce

+ CMo

= 8 pF + 8 pF + 12 pF

= 28 pF

1

(b) f β ≅

=

2 πβ r ( C + C )

(c)

mid e be bc

1

2 π (120)(28.48 Ω)(40 pF + 12 pF)

= 895.56 kHz ↑ Prob. 15

f T = βf β = (120)(895.56 kHz)

= 107.47 MHz

23. (a)

f

H i

1

=

2π R C

Th1

i

R

Th 1

= R s || R B || R i

VCC

−VBE

20 V − 0.7 V

R i : I B =

=

R + ( β + 1) R 470 k Ω+ (111)(0.91 k Ω)

B

= 33.8 μA

E

I E = (β + 1)I B = (110 + 1)(33.8 μA)

= 3.75 mA

26 mV 26 mV

r e = = = 6.93 Ω

I E

3.75 mA

R i = βr e = (110)(6.93 Ω)

= 762.3 Ω

R

Th 1

= R s || R B || R i = 0.6 kΩ || 470 kΩ || 762.3 Ω

= 335.50 Ω

134

f

H i

=

C i : C i = C

A v :

1

2 π (335.50 Ω)( )

Wi

A

v mid

=

be

C i

+ C + (1 − A v )C bc

−( RL

RC) −(4.7 kΩ

3 k Ω)

=

r

6.93 Ω

e

= −264.2

C i = 7 pF + 20 pF + (1 − (−264.2)6 pF

= 1.62 nF

f

H i

=

1

2 π (335.50 Ω)(1.62 nF)

≅ 293 kHz

1

f

H o

=

2π RTh C

2 o

R = R C || R L = 3 kΩ || 4.7 kΩ = 1.831 kΩ

Th 2

C o =

C + C + C

Wo

ce Mo

≅ C f = C bc

= 11 pF + 10 pF + 6 pF

= 27 pF

1

f

H o

=

(b) f β =

(c)

2 π (1.831 k Ω)(27 pF)

= 3.22 MHz

1

2 πβ r ( C + C )

mid e be bc

1

=

2 π (110)(6.93 Ω)(20 pF + 6 pF)

= 8.03 MHz

f T = β mid f β = (110)(8.03 MHz)

= 883.3 MHz

135

24. (a)

f

H i

=

1

π R C

2

Th1

i

1

=

2 π (955 Ω)(58 pF)

= 2.87 MHz

f

H o

=

1

π R C

2

Th2

o

1

=

2 π (38.94 Ω)(32 pF)

= 127.72 MHz

R = R R R Z

Th1 s 1 2 b

Z b = βr e + (β + 1)(R E || R L )

= (100)(30.23 Ω) + (101)(2.2 kΩ || 8.2 kΩ)

= 3.023 kΩ + 175.2 kΩ

= 178.2 kΩ

R

Th 1

= 1 kΩ || 120 kΩ || 30 kΩ || 178.2 kΩ

= 955 Ω

C i = CW + C

i be

+ Cbc

(No Miller effect)

= 8 pF + 30 pF + 20 pF

= 58 pF

24 kΩ

⎛ R1R2 R3

⎞

R

Th 2

= R E || R L || ⎜re

+ ⎟

⎝ β ⎠

⎛ 24 kΩ1 kΩ⎞

= 2.2 kΩ || 8.2 kΩ || ⎜30.23

Ω+

⎟

⎝

100 ⎠

= 1.735 kΩ || (30.23 Ω + 9.6 Ω)

= 1.735 kΩ || 39.83 Ω

= 38.94 Ω

C o = CW

+ C

o ce

= 10 pF + 12 pF

= 32 pF

(b) f β =

(c)

1

2 πβ r ( C + C )

mid e be bc

1

=

2 π (100)(30.23 Ω)(30 pF + 20 pF)

= 1.05 MHz

f T = β mid f β = 100(1.05 MHz) = 105 MHz

136

25. (a)

f

H i

=

1

π R C

2

Th1

i

R

Th 1

= R s || R E || R i

VEE

−VBE

4 V − 0.7 V

R i : I E = =

= 2.75 mA

RE

1.2 kΩ

26 mV 26 mV

r e = = = 9.45 Ω

I E

2.75 mA

R i = R E || r e = 1.2 kΩ || 9.45 Ω

= 9.38 Ω

C i : C i = C

Wi

+ C (no Miller cap-noninverting!)

be

= 8 pF + 24 pF

= 32 pF

R i = 0.1 kΩ || 1.2 kΩ || 9.38 Ω = 8.52 Ω

f

H i

=

f

H o

=

(b) f β =

(c)

C o = C

1

≅ 584 MHz

2 π (8.52 Ω)(32 pF)

1

R

Th

= R C || R L = 3.3 kΩ || 4.7 kΩ = 1.94 kΩ

R C

2

2π Th2

o

Wo

+ C + (no Miller)

bc

= 10 pF + 18 pF

= 28 pF

1

f

H o

=

2 π (1.94 k Ω)(28 pF)

= 2.93 MHz

1

2 πβ r ( C + C )

mid e be bc

1

=

2 π (80)(9.45 Ω)(24 pF + 18 pF)

= 5.01 MHz

f T = β mid f β = (80)(5.01 MHz)

= 400.8 MHz

137

26. (a) From problem 19 g m0 = 2 mS, g m = 1.18 mS

(b) From problem 19

A

≅ A = −2

vmid v s (mid)

(c)

(d)

f

H i

f Hi

1

=

2π R C

Th1

1

=

2 π (999 Ω)(21 pF)

= 7.59 MHz

f

H o

=

1

π R C

2

Th2

o

i

1

=

2 π (1.696 k Ω)(12 pF)

= 7.82 MHz

R

Th 1

= R sig || R G

= 1 kΩ || 1 MΩ

= 999 Ω

C i = C + C + C

Wi

gs Mi

C

M i

= (1 − A v )C gd

= (1 − (−2)4 pF

= 12 pF

C i = 3 pF + 6 pF + 12 pF

= 21 pF

R

Th 2

= R D || R L

= 3 kΩ || 3.9 kΩ

= 1.696 kΩ

C o = C + C + C

Wo

ds Mo

⎛ 1 ⎞

C

M o

= ⎜1−

⎟4 pF

⎝ −2

⎠

= (1.5)(4 pF)

= 6 pF

C o = 5 pF + 1 pF + 6 pF

= 12 pF

2I

DSS

2(10 mA)

27. (a) g m0 = = = 3.33 mS

VP

6 V

From problem #21 V

GS Q

≅ −2.55 V, I

D Q

≅ 3.3 mA

⎛ VGS g m =

0

1

Q

⎞ ⎛ −2.55 V ⎞

gm

−

⎜

= 3.33 mS 1

V ⎟ ⎜ − ⎟ = 1.91 mS

⎝ P ⎠ ⎝ −6 V ⎠

138

(b)

A

v mid

= −g m (R D || R L )

= −(1.91 mS)(3.9 kΩ || 5.6 kΩ)

= −4.39

Z i = 68 kΩ || 220 kΩ = 51.94 kΩ

Vi

Zi

51.94 kΩ

= =

Vs

Zi

+ Rsig

51.94 kΩ+ 1.5 kΩ = 0.972

A

v s

= (−4.39)(0.972)

(mid)

= −4.27

(c)

f

H i

=

1

Th

2π RTh C

1

1 i

R = R sig || R 1 || R 2

= 1.5 kΩ || 51.94 kΩ

= 1.46 kΩ

C i = C + C + (1 − A ) C

Wi

gs v gd

= 4 pF + 12 pF + (1 − (−4.39))8 pF

= 59.12 pF

1

f

H i

=

2 π (1.46 k Ω)(59.12 pF)

= 1.84 MHz

f

H o

=

1

Th2

2π RTh C

2 o

⎛ 1 ⎞

C o = C + C + 1 C

o

⎜ − ⎟

⎝ Av

⎠

W ds gd

⎛ 1 ⎞

= 6 pF + 3 pF + ⎜1−

⎟8 pF

⎝ ( −4.39)

⎠

= 18.82 pF

1

f

H o

=

2 π (2.3 k Ω)(18.82 pF)

= 3.68 MHz

R = R D || R L = 3.9 kΩ || 5.6 kΩ

= 2.3 kΩ

(d)

139

28.

A

v T

= Av ⋅Av ⋅Av ⋅ Av

4

= A v

= (20) 4

= 16 × 10 4

1 2 3 4

29.

2

f ′ = ( 1/ n

2 − 1 ) f2

= (

1/4

− )

2 1 (2.5MHz)

1.18

= 0.435(2.5 MHz)

= 1.09 MHz

f1

40 Hz

30. f

1′= =

1/ n

1/4

2 −1 2 −1

40 Hz

=

0.435

= 91.96 Hz

31. (a) v = 4 π V ⎡ 1 1

m

⎢sin 2π fst+ sin 2 π(3 fs) t+

sin 2 π(5 fs)

t

⎣ 3 5

1 1

⎤

+ sin 2 π(7 fs) t+ sin 2 π(9 fs)

t+ …

7 9

⎥⎦

= 12.73 × 10 −3 1

(sin 2 π(100 × 10 3 ) t+ sin 2 π(300×

10 3 ) t

3

1 3 1 3 1

3

+ sin 2 π(500× 10 ) t+ sin 2 π(700× 10 ) t+ sin 2 π(900×

10 ) t)

5 7 9

(b) BW ≅ 0.35

t r

0.35

≅

0.7 μs

≅ 500 kHz

At 90% or 81 mV, t ≅ 0.75 μs

At 10% or 9 mV, t ≅ 0.05 μs

t r ≅ 0.75 μs − 0.05 μs = 0.7 μs

V −V′

90 mV −80 mV

(c) P = = = 0.111

V 90 mV

P (0.111)(100 kHz)

f = L

f

o s

π

= π

≅ 3.53 kHz

140

Chapter 10

RF

250 k Ω

1. V o = − V

1

=− (1.5 V)

R 20 kΩ

1

= −18.75 V

V

2. A v =

V

o

i

R

=−

R

For R 1 = 10 kΩ:

500 kΩ

A v = −

10 kΩ

= −50

For R 1 = 20 kΩ:

500 kΩ

A v = −

20 kΩ = −25

R

f ⎛1 MΩ

⎞

3. V o = − V

1

=− ⎜ ⎟

R ⎝20 kΩ

⎠ V 1 = 2 V

1

F

1

V 1 = 2 V −50

= −40 mV

RF

200 kΩ

4. V o = − V

1

=− V

1

R 20 kΩ

1

= −10 V 1

For V 1 = 0.1 V:

V o = −10(0.1 V) = −1 V

For V 1 = 0.5 V:

V o = −10(0.5 V) = −5 V

V o ranges

from

−1 V to −5 V

⎛ R ⎞

F ⎛ 360 kΩ

⎞

5. V o = ⎜1+ ⎟ V

1

= ⎜1 + ⎟( −0.3 V)

⎝ R1

⎠ ⎝ 12 kΩ

⎠

= 31(−0.3 V) = −9.3 V

⎛ R ⎞

F ⎛ 360 kΩ

⎞

6. V o = ⎜1+ ⎟ V

1

= ⎜1+

⎟ V

1

⎝ R1

⎠ ⎝ 12 kΩ

⎠

V 1 = 2.4 V = 77.42 mV

31

= 2.4 V

141

⎛ R ⎞

F

7. V o = ⎜1+

⎟ V

1

⎝ R1

⎠

For R 1 = 10 kΩ:

⎛ 200 kΩ

⎞

V o = ⎜1 + ⎟(0.5 V) = 21(0.5 V) = 10.5 V

⎝ 10 kΩ

⎠

For R 1 = 20 kΩ:

⎛ 200 kΩ

⎞

V o = ⎜1+

⎟(0.5 V) = 11(0.5 V) = 5.5 V

⎝ 20 kΩ

⎠

V o ranges from 5.5 V to 10.5 V.

⎡Rf Rf Rf

⎤

8. V o = − ⎢ V1 + V2 + V3⎥

⎣ R1 R2 R3

⎦

⎡330 k 330 k 330 k

=

Ω (0.2 V) + Ω ( 0.5 V)+ Ω ⎤

−⎢

−

(0.8 V)

33 kΩ 22 kΩ 12 kΩ

⎥

⎣

⎦

= −[10(0.2 V) + 15(−0.5 V) + 27.5(0.8 V)]

= −[2 V + (−7.5 V) + 2.2 V]

= −[24 V − 7.5 V] = −16.5 V

⎡RF RF R ⎤

F

9. V o = − ⎢ V1 + V2 + V3⎥

⎣ R1 R2 R3

⎦

⎡68 kΩ 68 kΩ 68 kΩ

⎤

= −⎢

(0.2 V)+ ( − 0.5 V) + ( + 0.8 V)

33 kΩ 22 kΩ 12 kΩ

⎥

⎣

⎦

= −[0.41 V − 1.55 V + 4.53 V]

= −3.39 V

1

10. v o (t) = − v1

() t dt

RC

∫

1

= −

1.5 dt

(200 k Ω)(0.1 μF)

∫

= −50(1.5t) = −75t

11. V o = V 1 = +0.5 V

R

f 100 kΩ

12. V o = − V

1

=− (1.5 V)

R 20 kΩ 1

= −5(1.5 V) = −7.5 V

13. V 2 =

V 3 =

⎡200 kΩ

⎤

− ⎢ (0.2 V)

20 kΩ

⎥ = −2 V

⎣ ⎦

⎛ 200 kΩ

⎞

⎜1 + ⎟(0.2 V) = +4.2 V

⎝ 10 kΩ

⎠

142

⎛ 400 kΩ⎞ ⎛−100 kΩ⎞ ⎛ 100 kΩ⎞

14. V o = ⎜1 + ⎟(0.1 V) ⋅ ⎜ ⎟+ ⎜−

⎟(0.1 V)

⎝ 20 kΩ ⎠ ⎝ 20 kΩ ⎠ ⎝ 10 kΩ

⎠

= (2.1 V)(−5) + (−10)(0.1 V)

= −10.5 V − 1 V = −11.5 V

⎡600 k 600 k 300 k

15. V o = − Ω (25 mV) + Ω ⎤⎛

( 20 mV)

Ω ⎞

⎢ −

15 k 30 k ⎥⎜ − ⎟

⎣ Ω Ω ⎦⎝ 30 kΩ

⎠

⎡ ⎛300 kΩ

⎞ ⎤

+ ⎢−⎜

⎟( −20 mV)

15 kΩ

⎥

⎣ ⎝ ⎠ ⎦

= −[40(25 mV) + (20)(−20 mV)](−10) + (−20)(−20 mV)

= −[1 V − 0.4 V](−10) + 0.4 V

= 6 V + 0.4 V = 6.4 V

⎛ R

f ⎞

16. V o = ⎜1+ ⎟ V

Io

+ I

Io R

f

⎝ R1

⎠

⎛ 200 kΩ

⎞

= ⎜1 + ⎟(6 mV) + (120 nA)(200 k Ω)

⎝ 2 kΩ

⎠

= 101(6 mV) + 24 mV

= 606 mV + 24 mV = 630 mV

17.

I

Io

4 nA

I + IB

= I 20 nA +

IB + + = = 22 nA

2 2

I

Io

4 nA

I

− IB

= I 20 nA

IB − − = − = 18 nA

2 2

18. f 1 = 800 kHz

f1

800 kHz

f c =

3

A = 150×

10

v2

= 5.3 Hz

19. A CL =

SR 2.4 V/ μs

=

ΔV

/ Δt

0.3 V/10 μs

i

= 80

R f

200 kΩ

20. A CL = =

R1

2 kΩ = 100

K = A CL V i = 100(50 mV) = 5 V

SR 0.4 V / μs

w s ≤ = = 80 × 10 3 rad/s

K 5 V

3

w 80 10

f s = s

×

= = 12.73 kHz

2π

2π

143

21. V Io = 1 mV, typical

I Io = 20 nA, typical

R

VIo

⎛ f ⎞

V o (offset) = ⎜1+

⎟ + I Io R f

⎝ R1

⎠

⎛ 200 kΩ

⎞

= ⎜1 + ⎟(1 mV) + (200 kΩ)(20 nA)

⎝ 20 kΩ

⎠

= 101(1 mV) + 4000 × 10 −6

= 101 mV + 4 mV = 105 mV

22. Typical characteristics for 741

R o = 25 Ω, A = 200 K

R f

200 kΩ

(a) A CL = − =−

R1

2 kΩ

= −100

(b) Z i = R 1 = 2 kΩ

R 25

(c) Z o = o

Ω

=

1+

β A 1

1 + (200,000)

100

= 25 Ω = 0.0125 Ω

2001

Vo

120 mV

23. A d =

V = d

1 mV

= 120

Vo

20 μV

A c = = = 20 × 10 −3

Vc

1 mV

Ad

120

Gain (dB) = 20 log = 20log 20 10

−3

Ac

×

= 20 log(6 × 10 3 ) = 75.56 dB

144

24. V d = V i1 − V i2 = 200 μV − 140 μV = 60 μV

Vi1 + Vi2 (200 μV + 140 μV)

V c = = = 170 μV

2 2

(a) CMRR =

A

d

A = 200

c

A

d

6000

A c = = = 30

200 200

Ad

5

(b) CMRR = 10

A =

c

A

d

6000

A c =

5 =

5 = 0.06 = 60 × 10

−3

10 10

⎡ 1 V ⎤

c

Using V o = A d V d ⎢1 + ⎥

⎣ CMRR Vd

⎦

⎡ 1 170 V

(a) V o = 6000(60 μV) 1 μ ⎤

⎢ + 200 60 μ V ⎥ = 365.1 mV

⎣

⎦

⎡ 1 170 V

(b) V o = 6000(60 μV) 1 μ ⎤

⎢ + 10 5 60 μ V ⎥ = 360.01 mV

⎣

⎦

145

Chapter 11

RF

180 k Ω

1. V o = − V

1

=− (3.5 mV)

R 3.6 kΩ

1

= −175 mV

⎛ R ⎞

F ⎛ 750 kΩ

⎞

2. V o = ⎜1+ ⎟ V

1

= ⎜1+

⎟(150 mV, rms)

⎝ R1

⎠ ⎝ 36 kΩ

⎠

= 3.275 V, rms ∠0°

⎛ 510 kΩ⎞ ⎡ 680 kΩ⎤⎡ 750 kΩ⎤

3. V o = ⎜1 + ⎟(20 μV)

− −

18 kΩ ⎢ 22 kΩ ⎥⎢ 33 kΩ

⎥

⎝ ⎠ ⎣ ⎦⎣ ⎦

= (29.33)(−30.91)(−22.73)(20 μV)

= 412 mV

4.

⎛ 420 kΩ

⎞

420 kΩ

420 kΩ ⎜1+

⎟ = +15 − = −22

= −30

⎝ R1

⎠

R2

R2

R 1 = 420 k Ω

R 2 = 420 k Ω

R 3 = 420 k Ω

14

22

30

R 1 = 71.4 kΩ R 2 = 19.1 kΩ R 3 = 14 kΩ

5.

V o = (+15)(−22)(−30)V 1 = 9000(80 μV) = 792 mV

= 0.792 V

RF1

150 kΩ

V o1 = − V

1

=− V

1

R1 R1

Vo

150 kΩ

1

= Av

=− 15 =−

1

V

R

1 1

R 1 =

150 kΩ = 10 kΩ

15

RF

2

150 kΩ

V o2 = − V

1

=− V

1

R2 R2

Vo

2

150 kΩ

= Av

=− 30 =−

2

V

R

1 2

R 2 =

150 kΩ = 5 kΩ

30

146

⎡RF

R ⎤

F ⎡470 kΩ 470 kΩ

⎤

6. V o = − ⎢ V1 + V2⎥=−

(40 mV) + (20 mV)

R1 R ⎢

2

47 kΩ

12 kΩ

⎥

⎣

⎦ ⎣

⎦

= −[400 mV + 783.3 mV] = −1.18 V

⎛

7. V o = 10 kΩ ⎞⎛ 150 kΩ+ 300 kΩ ⎞

⎜ ⎟⎜ ⎟V

300 k

1

−

Ω V2

⎝10 kΩ+ 10 kΩ⎠⎝ 150 kΩ ⎠ 150 kΩ

= 0.5(3)(1 V) − 2(2 V) = 1.5 V − 4 V = −2.5 V

⎧⎡330 kΩ ⎤⎛470 kΩ⎞

470 kΩ

⎫

8. V o = − ⎨⎢ (12 mV) (18 mV)

33 k ⎥⎜ ⎟+

⎬

⎩⎣ Ω ⎦⎝ 47 kΩ ⎠ 47 kΩ

⎭

= −[(−120 mV)(10) + 180 mV] = −[−1.2 V + 0.18 V]

= +1.02 V

9.

10.

V1

12 V

11. I L =

R = 2 kΩ = 6 mA

1

12. V o = −I 1 R 1 = −(2.5 mA)(10 kΩ) = −25 V

13.

Io

R ⎛

F

1 ⎞

= ⎜ ⎟

V1 R1

⎝ Rs

⎠

100 kΩ ⎛ 1 ⎞

I o

= ⎜ ⎟(10 mV)

200 kΩ⎝10

Ω⎠

= 0.5 mA

⎛ 2R

⎞

14. V o =

1+

[V

⎜ R ⎟ 2 − V 1 ]

⎝ p ⎠

⎛ 2(5000) ⎞

= ⎜1 + ⎟[1 V − 3 V] = −22 V

⎝ 1000 ⎠

147

15. f OH =

16. f OL =

17. f OL =

f OH =

1 1

=

2π RC

1 1

2 π(2.2 k Ω)(0.05 μF)

= 1.45 kHz

1 1

=

2π RC

1 1

2 π(20 k Ω)(0.02 μF)

= 397.9 Hz

1 1

=

= 318.3 Hz

2π RC

1 1

2 π(10 k Ω)(0.05 μF)

1 1

=

2π RC

2 2

2 π(20 k Ω)(0.02 μF)

= 397.9 Hz

148

Chapter 12

1.

I

I

BQ

CQ

VCC

−VBE

18 V − 0.7 V

= =

= 14.42 mA

RB

1.2 kΩ

= β I = 40(14.42 mA) = 576.67 mA

BQ

P i = V CC I dc ≅

VCC I

C Q

= (18 V)(576.67 mA)

≅ 10.4 W

I C (rms) = βI B (rms)

= 40(5 mA) = 200 mA

P o = I

2 (rms) R = (200 mA) 2 (16 Ω) = 640 mW

C

C

2.

I

B Q

=

I

C Q

=

V

CC

−V

R

I BQ

B

BE

18 V − 0.7 V

=

= 11.5 mA

1.5 kΩ

β = 40(11.5 mA) = 460 mA

P i (dc) = V CC I dc = V ( )

CC

IC + I

Q BQ

= 18 V(460 mA + 11.5 mA)

= 8.5 W

⎡P ≈ V I

⎣

i CC C Q

= 18 V(460 mA) = 8.3 W⎤

⎦

3. From problem 2:

I

C Q

= 460 mA, P i = 8.3 W.

For maximum efficiency of 25%:

Po

%η = 100% ×

P = P o

× 100% = 25%

i

8.3 W

P o = 0.25(8.3 W) = 21 W

[If dc bias condition also is considered:

V C = V CC −

I

CQ

R = 18 V − (460 mA)(16 Ω) = 10.64 V

C

collector may vary ±7.36 V about Q-point, resulting in maximum output power:

2 2

VCE

( P) (7.36 V)

P o = = =1.69 W

2R

2(16)

C

149

4. Assuming maximum efficiency of 25%

with P o (max) = 1.5 W

Po

%η =

P × 100%

i

P i = 1.5 W

0.25 = 6 W

Assuming dc bias at mid-point, V C = 9 V

VCC

−VC

18 V − 9 V

IC

= =

= 0.5625 A

Q

RC

16 Ω

P i (dc) = VCC I

C Q

= (18 V)(0.5625 A)

= 10.38 W

at this input:

Po

%η =

P × 100% = 1.5 W

× 100% = 14.45%

10.38 W

i

5. R p =

2

⎛ N ⎞

1

⎜ ⎟

⎝ N2

⎠

R

s

=

2

⎛25

⎞

⎜ ⎟ (4 Ω)

⎝ 1 ⎠

= 2.5 kΩ

6. R 2 = a 2 R 1

a 2 R2

8 kΩ

= =

R 8 Ω = 1000

1

a = 1000 = 31.6

7. R 2 = a 2 R 1

8 kΩ = a 2 (4 Ω)

a 2 = 8 k Ω

4 Ω = 2000

a = 2000 = 44.7

8. (a) P pri = P L = 2 W

2

VL

(b) P L =

RL

V L = PR

L L

= (2 W)(16 Ω )

= 32 = 5.66 V

(c) R 2 = a 2 R 1 = (3.87) 2 (16 Ω) = 239.6 Ω

2

Vpri

P pri =

Rpri

= 2 W

2

V

pri

= (2 W)(239.6 Ω)

V pri = 479.2 = 21.89 V

[or, V pri = aV L = (3.87)(5.66 V) = 21.9 V]

150

(d) P L =

I R

2

L

I L =

L

PL

2 W

R = 16 Ω

L

= 353.55 mA

2

P pri = 2 W = IpriR pri

= (239.6 Ω) I

2 pri

2 W

I pri =

= 91.36 mA

239.6 Ω

I 353.55 mA

or, I pri = L

= = 91.36 mA

a 3.87

9. I dc = I

C Q

= 150 mA

P i = VCCI C Q

= (36 V)(150 mA) = 5.4 W

Po

2 W

%η = 100% 100%

P × = 5.4 W

× = 37%

10.

i

11.

151

12. (a) P i = V CC I dc = (25 V)(1.75 A) = 43.77 W

Where, I dc = 2 2 Vp

I

2 22 V

p

π = π R

= π

⋅ 8 Ω

= 1.75 A

L

(b) P o =

V

2 2

p (22 V)

2R = 2(8 Ω )

= 30.25 W

L

Po

(c) %η =

P

i

× 100% =

30.75 W

43.77 W

× 100% = 69%

13. (a) max P i = V CC I dc

⎛ 2 V ⎞

CC ⎡ 2 25 V⎤

= V CC ⋅ ⎜ ⋅ ⎟= (25 V)

π R ⎢ ⋅

L

π 8 Ω ⎥

⎝ ⎠ ⎣ ⎦

= 49.74 W

2 2

VCC

(25 V)

(b) max P o =

2R = 2(8 Ω )

= 39.06 W

L

(c) max %η = max P

100%

39.06 W

o

max P × = i

49.74 W

× 100%

= 78.5%

14. (a)

V

L (peak)

= 20 V

⎡2 V ⎤

L

P i = V CC I dc = VCC

⎢ ⋅ ⎥

⎣π

RL

⎦

⎡2 20 V⎤

= (22 V) ⎢ ⋅

π 4 Ω ⎥

⎣ ⎦ = 70 W

2 2

VL

(20 V)

P o =

2R = L

2(4 Ω )

= 50 W

Po

50 W

%η = × 100% = × 100% = 71.4%

P 70 W

i

⎡ 2 4 V⎤

(b) P i = (22 V) ⎢ ⋅

π 4 Ω ⎥

⎣ ⎦ = 14 W

2

(4)

P o =

2(4) = 2 W

Po

%η =

P × 100% = 2 W × 100% = 14.3%

14 W

i

152

15.

16. (a) max P o (ac) for V

L peak

= 30 V:

2 2

VL

(30 V)

max P o (ac) =

2R = 2(8 Ω )

= 56.25 W

L

(b) max P i (dc) = V CC I dc = V

CC

⎡2 V ⎤

o ⎡ 2 30 V⎤

⎢ ⋅ ⎥ = VCC

π R ⎢ ⋅

L

π 8 Ω ⎥

⎣ ⎦ ⎣ ⎦

= 71.62 W

(c) max %η = max Po

max P

i

56.25 W

× 100% =

71.62 W × 100%

= 78.54%

(d) max P

ZQ

= 2 V 2 (30)

π

⋅ = ⋅ = 22.8 W

2 2

CC

2 2

RL

π 8

17. (a) P i (dc) = V CC I dc = V CC ⋅ 2 ⎛V

⎞

o

⎜ ⎟

π ⎝RL

⎠

= 30 V ⋅ 2 ⎡ 2 ⋅ 8 ⎤

⎢ ⎥

π ⎣ 8 ⎦

= 27 W

(b) P o (ac) =

V

(rms) (8 V)

=

R 8 Ω

2 2

L

L

= 8 W

(c) %η =

Po

8 W

× 100% = × 100% = 29.6%

P 27 W

i

(d) P 2Q = P i − P o = 27 W − 8 W = 19 W

153

18. (a) P o (ac) =

V (rms) (18 V)

=

R 8 Ω

2 2

L

L

= 40.5 W

⎡2 VL (b) P i (dc) = V CC I dc =

peak

⎤

VCC

⎢ ⋅ ⎥

⎢⎣π

RL

⎥⎦

⎡2 18 2 V⎤

= (40 V) ⎢ ⋅ ⎥ = 81 W

⎣π

8 Ω ⎦

Po

(c) %η =

P

i

× 100% =

40.5 W

81 W

× 100% = 50%

(d)

P

2 Q

= P i − P o = 81 W − 40.5 W = 40.5 W

A2

19. %D 2 =

A

1

A3

%D 3 =

A

1

A4

%D 4 =

A

1

0.3 V

× 100% =

2.1 V

0.1 V

× 100% =

2.1 V

0.05 V

× 100% =

2.1 V

× 100% ≅ 14.3%

× 100% ≅ 4.8%

× 100% ≅ 2.4%

20. %THD =

D + D + D × 100%

2 2 2

2 3 4

2 2 2

= (0.143) + (0.048) + (0.024) × 100%

= 15.3%

21. D 2 =

( VCE

+ VCE

)

1

2

V

max

CEmax

−V

min

CEmin

× 100%

=

1 (20 V + 2.4 V) − 10 V

2

20 V − 2.4 V

× 100%

= 1.2 V

17.6 V

× 100% = 6.8%

22. THD = D2 + D3 + D4 = (0.15) + (0.01) + (0.05)

≅ 0.16

2 2

I1 RC

(3.3 A) (4 Ω)

P 1 = = = 21.8 W

2 2

P = (1 + THD 2 )P 1 = [1 + (0.16) 2 )]21.8 W

= 22.36 W

2 2 2 2 2 2

154

23. P D (150°C) = P D (25°C) − (T 150 − T 25 ) (Derating Factor)

= 100 W − (150°C − 25°C)(0.6 W/°C)

= 100 W − 125(0.6) = 100 − 75

= 25 W

TJ

− TA

200° C − 80°

C

24. P D =

=

θJC + θCS + θSA

0.5 ° C/W + 0.8 ° C/W + 1.5 ° C/W

120°

C

= = 42.9 W

2.8 ° C/W

TJ − TA

25. P D =

θ

JA

200° C − 80° C 120°

C

= =

(40C/W) ° 40C/W °

= 3 W

155

Chapter 13

1.

2.

3.

4.

5.

156

6.

7. Circuit operates as a window detector.

9.1 kΩ

Output goes low for input above

( + 12 V) = 7.1 V

9.1 k Ω + 6.2 kΩ

1 kΩ

Output goes low for input below

( + 12 V) = 1.7 V

1 k Ω + 6.2 kΩ

Output is high for input between +1.7 V and +7.1 V.

8.

11010 26

9. (16 V) = (16 V)

5

2 32

= 13 V

VREF

10 V 10 V

10. Resolution = = = = 2.4 mV/count

n 12

2 2 4096

11. See section 13.3.

12. Maximum number of count steps = 2 12 = 4096

13. 2 12 = 4096 steps at T = 1 = 1 = 50 ns/count

f 20 MHz

Period = 4096 counts ×

ns

50 count

= 204.8 μs

157

14.

f =

1.44

( R + 2 R ) C

A

B

= 350 kHz

C =

1.44

7.5 k Ω + 2(7.5 k Ω)(350 kHz)

≅ 183 pF

15.

T = 1.1 R A C

20 μs = 1.1(7.5 kΩ)C

−6

20×

10

C =

3

1.1(7.5×

10 )

= 2.4 × 10 −9

= 2400 × 10 −12

= 2400 pF

16. T = 1 = 1 = 100 μs

f 10 kHz

T = 1.1 R A C = 1.1(5.1 kΩ)(5 nF) = 28 μs

158

+

2 ⎛V

−V

⎞

C

17. f o = ⎜ + ⎟

RC

1 1⎝

V ⎠

V + = 12 V

R3

+ 11 kΩ

V C = ( V ) = ( + 12 V)

R + R 1.8 k Ω + 11 kΩ

2 3

= 10.3 V

2 ⎡12 V −10.3 V ⎤

f o =

(4.7 k Ω)(0.001 μF) ⎢ 12 V ⎥

⎣ ⎦

= 60.3 × 10 3 ≅ 60 kHz

18. With potentiometer set at top:

R3 + R4

+ 5 kΩ+ 18 kΩ

V C = V =

(12 V) = 11.74 V

R2 + R3 + R4

510 Ω+ 5 kΩ+ 18 kΩ

resulting in a lower cutoff frequency of

+

2 ⎛V −V 2 12 V 11.74 V

f o = C

⎞

⎛ − ⎞

⎜ + ⎟=

3 ⎜ ⎟

RC

1 1⎝

V ⎠ (10 × 10 )(0.001 μF) ⎝ 12 V ⎠

= 4.3 kHz

With potentiometer set at bottom:

R4

+ 18 kΩ

V C = V =

(12 V)

R2 + R3 + R4

510 Ω+ 5 kΩ+ 18 kΩ

= 9.19 V

resulting in a higher cutoff frequency of

+

2 ⎛V −V 2 12 V 9.19 V

f o =

C

⎞

⎡ − ⎤

⎜ + ⎟=

RC

1 1

V (10 k Ω)(0.001 μF) ⎢ 12 V ⎥

⎝ ⎠

⎣ ⎦

= 61.2 kHz

19. V + = 12 V

R3

+ 10 kΩ

V C = V =

(12 V) = 10.4 V

R2 + R3

1.5 k Ω + 10 kΩ

+

2 ⎛V −V 2 12 V 10.4 V

f o =

C

⎞

⎛ − ⎞

⎜ + ⎟=

⎜ ⎟

RC

1 1⎝

V ⎠ 10 k Ω( C1) ⎝ 12 V ⎠

= 200 kHz

2

C 1 =

(0.133)

10 k Ω(200 kHz)

= 133 × 10 −12 = 133 pF

20. f o =

21. C 1 =

0.3 0.3

=

RC

1 1

(4.7 k Ω)(0.001 μF)

= 63.8 kHz

0.3 0.3

Rf =

= 300 pF

(10 k Ω)(100 kHz)

1

159

22. f L = ± 8 fo

V

3

8(63.8×

10 ) ⎡ 0.3 0.3 ⎤

= ±

⎢f o

= =

⎥

6 V ⎣ RC

1 1

4.7 k Ω(0.001 μF)

⎦

= 85.1 kHz = 63.8 kHz

23. For current loop: mark = 20 mA

space = 0 mA

For RS − 232 C:

mark = −12 V

space = +12 V

24. A line (or lines) onto which data bits are connected.

25. Open-collector is active-LOW only.

Tri-state is active-HIGH or active-LOW.

160

Chapter 14

1. A f =

A −2000 −2000

= =

1+ β A ⎛ 1 ⎞

201

1 + ⎜− ⎟( −2000)

⎝ 10 ⎠

= −9.95

2.

dA

A

f

f

1 dA 1

= =

β A A ⎛ 1 ⎞

⎜−

⎟

⎝ 20 ⎠

( −1000)

(10%)

= 0.2%

A −300 −300

3. A f = = = = −14.3

1+ β A ⎛ 1 ⎞ 21

1 + ⎜− ⎟( −300)

⎝ 15 ⎠

R if = (1 + βA)R i = 21(1.5 kΩ) = 31.5 kΩ

R 50 k

R of = o

Ω

= = 2.4 kΩ

1+

β A 21

RoRD

4. R L =

R + R

o

D

= 40 kΩ || 8 kΩ = 6.7 kΩ

A = −g m R L = −(5000 × 10 −6 )(6.7 × 10 3 ) = −33.5

−R2

−200 kΩ

β = =

R + R 200 kΩ+ 800 kΩ = −0.2

1 2

A −33.5 −33.5

A f = = =

1+ β A 1 + ( −0.2)( −33.5) 7.7

= −4.4

5. DC bias:

I B =

VCC

−VBE

16 V − 0.7 V

=

R + ( β + 1) R 600 kΩ+ 76(1.2 k Ω)

B

E

=

15.3 V

691.2 kΩ

= 22.1 μA

I E = (1 + β)I B

= 76(22.1 μA) = 1.68 mA

[V CE = V CC − I C (R C + R E ) = 16 V − 1.68 mA(4.7 kΩ + 1.2 kΩ) ≅ 6.1 V]

r e =

26 mV 26 mV

= ≅ 15.5 Ω

I E

(mA) 1.68 mA

h ie = (1 + β)r e = 76(15.5 Ω) = 1.18 kΩ = Z i

Z o = R C = 4.7 kΩ

161

−hfe

−75

A v = =

= −31.5 × 10−3

hie

+ RE

1.18 kΩ+ 1.2 kΩ β = R E = −1.2 × 10 3

(1 + βA) = 1 + (−1.2 × 10 3 )(−31.5 × 10 −3 )

= 38.8

−3

Av

− 31.5×

10

A f = =

= 811.86 × 10 −6

1+

β Av

38.8

A

v f

= −A f R C = −(811.86 × 10 −6 )(4.7 × 10 3 ) = −3.82

Z

i f

= (1 + βA v )Z i = (38.8)(1.18 kΩ) = 45.8 kΩ

Z

o f

= (1 + βA v )Z o = (38.8)(4.7 kΩ) = 182.4 kΩ

without feedback (R E bypassed):

−RC

4.7 kΩ

A v = =−

r 15.5 Ω = −303.2

e

1 1

6. C = =

π π

= 2.6 × 10 −9 = 2600 pF = 0.0026 μF

3 3

2 Rf 6 2 (10 × 10 )(2.5×

10 ) 6

1 1

7. f o = ⋅

2π

RC ⎛Rc

⎞

6+ 4 ⎜

R ⎟

⎝ ⎠

1 1

= ⋅

3 −12 2 π (6× 10 )(1500× 10 )

3 3

6+ 4(18× 10 /6×

10 )

= 4.17 kHz ≅ 4.2 kHz

1 1

8. f o = =

3 −12

2πRC

2 π(10× 10 )(2400×

10 )

= 6.6 kHz

CC

1 2

9. C eq =

C + C =

f o =

1 2

(750 pF)(2000 pF)

750 pF + 2000 pF

1 1

=

= 577 pF

−6 −12

2π

LCeq

2π

40× 10 )(577 × 10 )

= 1.05 MHz

10. f o =

1

2π

LC

eq

,

1

=

2 π (100μH)(3300 pF)

= 277 kHz

CC

1 2

where C eq =

C + C

1 2

(0.005 μF)(0.01 μF)

=

0.005 μF + 0.01 μF

= 3300 pF

162

11. f o =

12. f o =

1

,

2π

L C

eq

=

π

= 159.2 kHz

1

−3 −12

2 (4× 10 )(250×

10 )

1

2π

LC

eq

,

1

=

2 π (1800μH)(150 pF)

= 306.3 kHz

L eq = L 1 + L 2 + 2 M

= 1.5 mH + 1.5 mH + 2(0.5 mH)

= 4 mH

where L eq = L 1 + L 2 + 2 M

= 750 μH + 750 μH + 2(150 μH)

= 1800 μH

13. See Fig. 14.33a and Fig. 14.34.

14. f o =

1

RC ln(1/(1 −η))

T

T

for η = 0.5:

1.5

f o ≅

R C

T

T

(a) Using R T = 1 kΩ

1.5 1.5

C T = =

R f (1 k Ω)(1 kHz)

T

o

= 1.5 μF

(b) Using R T = 10 kΩ

1.5 1.5

C T = =

= 1000 pF

R f (10 k Ω)(150 kHz)

T

o

163

Chapter 15

1. ripple factor =

V r

VNL −VFL

2. %VR =

V

FL

(rms) 2 V/ 2

= = 0.028

V 50 V

dc

× 100% =

28 V − 25 V

25 V

× 100% = 12%

3. V dc = 0.318V m

Vdc 20 V

V m = = = 62.89 V

0.318 0.318

V r = 0.385V m = 0.385(62.89 V) = 24.2 V

4. V dc = 0.636V m

Vdc 8 V

V m = = = 12.6 V

0.636 0.636

V r = 0.308V m = 0.308(12.6 V) = 3.88 V

5. %r =

V r

(rms)

V

dc

× 100%

V r (rms) = rV dc = 8.5 × 14.5 V = 1.2 V

100

6. V NL = V m = 18 V

V FL = 17 V

VNL −VFL

18 V −17 V

%VR = × 100% =

VFL

17 V

= 5.88%

× 100%

7. V m = 18 V

C = 400 μF

I L = 100 mA

2.4Idc

2.4(100)

V r = = = 0.6 V, rms

C 400

4.17Idc

V dc = V m −

C

= 18 V − 4.17(100) = 16.96 V

400

≅ 17 V

2.4Idc

2.4(120)

8. V r =

C = 200

= 1.44 V

9. C = 100 μF

V dc = 12 V

R L = 2.4 kΩ

I

dc

V

12 V

2.4 kΩ

dc

= = =

R

L

5mA

V r (rms) =

2.4I

2.4(5)

dc

C = 100

= 0.12 V

164

2.4Idc

2.4(150)

10. C = = = 100 μF

rV (0.15)(24)

dc

11. C = 500 μF

I dc = 200 mA

R = 8% = 0.08

2.4Idc

Using r =

CV

V m = V dc +

V dc =

4.17Idc

C

dc

2.4I

2.4(200)

dc

rC = 0.08(500)

= 12 V

= 12 V + (200)(4.17)

500

= 12 V + 1.7 V = 13.7 V

2.4Idc

2.4(200)

12. C = = = 6857 μF

V (0.07)

r

13. C = 120 μF

I dc = 80 mA

V m = 25 V

4.17Idc

V dc = V m −

C

= 25 V − 4.17(80)

120

= 22.2 V

2.4Idc

%r =

CV

dc

2.4(80)

× 100% =

(120)(22.2) × 100%

= 7.2%

rV ⋅ ′

dc

2(80)

14. V ′

r

= = = 1.6 V, rms

100 100

15. V r = 2 V

V dc = 24 V

R = 33 Ω, C = 120 μF

X C =

1.3 1.3

C = 120

= 10.8 Ω %r =

V r

V

dc

× 100% = 2 V

24 V × 100%

= 8.3%

X

C

10.8

V′ r

= Vr

= (2 V) = 0.65 V

R 33

V ′

dc

= V dc − I dc R = 24 V − 33 Ω (100 mA)

= 20.7 V

V

%r′ = r

′

× 100% = 0.65 V × 100% = 3.1%

V ′

20.7 V

dc

165

16.

RL

V′ dc

= V

R+

R

L

dc

500

= (40 V)

50 + 500

= 36.4 V

V ′

dc

36.4 V

I dc = = = 72.8 mA

R 500 Ω

L

17.

1.3 1.3

X C =

C = 100

= 13 Ω

X

C

13

V′ r

= Vr

= (2.5 V)

R 100

= 0.325 V, rms

18. V NL = 60 V

RL

1 kΩ

V FL = Vdc

=

(50 V) = 45.46 V

R+ RL

100 Ω+ 1 kΩ VNL −VFL

50 V − 45.46 V

%VR = × 100% = × 100%

VFL

45.46 V

= 10 %

19. V o = V Z − V BE = 8.3 V − 0.7 V = 7.6 V

V CE = V i − V o = 15 V − 7.6 V = 7.4 V

Vi

−VZ

15 V − 8.3 V

I R = =

= 3.7 mA

R 1.8 kΩ

Vo

7.6 V

I L =

R = = 3.8 mA

L

2 kΩ I 3.8 mA

I B = C

= = 38 μA

β 100

I Z = I R − I B = 3.7 mA − 38 μA = 3.66 mA

166

R1 + R2

V

z

+ V

BE2

R2

20. V o = ( )

33 kΩ+ 22 kΩ

= (10 V + 0.7 V)

22 kΩ

= 26.75 V

⎛ R ⎞

1 ⎛ 12 kΩ

⎞

21. V o = ⎜1+ ⎟ VZ = ⎜1+

⎟10 V

⎝ R2

⎠ ⎝ 8.2 kΩ

⎠

= 24.6 V

22. V o = V L = 10 V + 0.7 V = 10.7 V

23.

24. I L = 250 mA

V m = V r (rms) ⋅ 2 = 2 (20 V) = 28.3 V

⎛2.4Idc

⎞

V

r peak

= 3 V r (rms) = 3⎜

C

⎟

⎝ ⎠

⎛2.4(250)

⎞

= 3⎜

⎟

⎝ 500 ⎠ = 2.1 V

V dc = V m − V = 28.3 V − 2.1 V = 26.2 V

r peak

V i (low) = V dc −

V

r peak

= 26.2 V − 2.1 V = 24.1 V

25. To maintain V i (min) ≥ 7.3 V (see Table 15.1)

V ≤ V m − V i (min) = 12 V − 7.3 V = 4.7 V

r peak

so that

V r peak

4.7 V

V r (rms) = = = 2.7 V

3 1.73

The maximum value of load current is then

Vr

(rms) C (2.7 V)(200)

I dc = = = 225 mA

2.4 2.4

167

⎛

2

26. V o = V ref 1 R ⎞

⎜ + ⎟ + I adj R L

⎝ R 1 ⎠

⎛ 1.8 kΩ

⎞

= 1.25 V ⎜1+

⎟ + 100 μA(2.4 kΩ)

⎝ 240 Ω ⎠

= 1.25 V(8.5) + 0.24 V

= 10.87 V

⎛

2

27. V o = V ref 1 R ⎞

⎜ + ⎟ + I adj R 2

⎝ R 1 ⎠

⎛ 1.5 kΩ

⎞

= 1.25 V ⎜1+

⎟ + 100 μA(1.5 kΩ)

⎝ 220 Ω ⎠

= 9.9 V

168

Chapter 16

1. (a) The Schottky Barrier diode is constructed using an n-type semiconductor material and a

metal contact to form the diode junction, while the conventional p-n junction diode uses

both p- and n-type semiconductor materials to form the junction.

(b) −

2. (a) In the forward-biased region the dynamic resistance is about the same as that for a p-n

junction diode. Note that the slope of the curves in the forward-biased region is about

the same at different levels of diode current.

(b)

In the reverse-biased region the reverse saturation current is larger in magnitude than

for a p-n junction diode, and the Zener breakdown voltage is lower for the Schottky

diode than for the conventional p-n junction diode.

3.

ΔI R

100 μA −0.5 μA = = 1.33 μ A/ ° C

Δ° C 75 ° C

Δ I = (1.33 μA/ ° C) Δ C = (1.33 μA/ ° C)(25° C) = 33.25μA

R

I = 0.5μA+33.25μA

=33.75μ

A

R

4.

X C

= 1 1

2π

fC

= 2 π(1 MHz)(7 pF)

=22.7 kΩ

R

DC

VF

400 mV

= = = 40 Ω

I 10 mA

F

5. Temperature on linear scale

T(1/2 power level of 100 mW) ≅ 95°C

6. V F a linear scale V F (25°C) ≅ 380 mV = 0.38 V

At 125°C, V F ≅ 280 mV

ΔVF

100 mV

=

Δ 100°

C

T

= 1 mV/°C

∴ At 100°C V F = 280 mV + (1 mV/°C)(25°C)

= 280 mV + 25 mV

= 305 mV

Increase temperature and V F drops.

169

7. (a) C T (V R ) =

n

1/3

( )

=

C(0) 80 pF

=

1+

V

4.2 V

R

VT

⎛ ⎞

⎜1+

⎟

⎝ 0.7 V ⎠

80 pF

= 41.85 pF

1.912

(b) k = C T (V T + V R ) n

= 41.85 pF(0.7 V + 4.2 V) 1/3

≅ 71 × 10 −12

1.698

8. (a) At −3 V, C = 40 pF

At −12 V, C = 20 pF

ΔC = 40 pF − 20 pF = 20 pF

Δ C 40 pF

(b) At −8 V, = = 2 pF/V

ΔVR

20 V

Δ C 60 pF

At −2 V, = = 6.67 pF/V

ΔVR

9 V

ΔC

increases at less negative values of V R .

ΔV

R

9. Ratio =

Ct

( −1 V) 92pF

=

C ( −8 V) 5.5 pF

t

= 16.73

Ct

( −1.25 V)

C ( −7 V)

t

= 13

10. C t ≅ 15 pF

1 1

Q = =

2π

fRsCt

2 π(10 MHz)(3 Ω)(15 pF)

= 354.61 vs 350 on chart

11.

ΔC

Δ C × 100%

TC C =

C ( ) × 100% ⇒ T 1 =

o

T1 − T0

TCC( Co)

+ T o

(0.11 pF)(100)

=

(0.02)(22 pF)

+ 25

= 50°C

12. V R from −2 V to −8 V

C t (−2 V) = 60 pF, C t (−8 V) = 6 pF

Ratio =

Ct

( −2 V) 60 pF

=

C ( −8 V) 6 pF

t

= 10

170

13. Q(−1 V) = 82, Q(−10 V) = 5000

Q( −10 V) 5000

Ratio = = = 60.98

Q( −1 V) 82

6

f 10 10 Hz

BW = o

×

= = 121.95 kHz

Q 82

6

f 10 10 Hz

BW = o

×

= = 2 kHz

Q 5000

14. High-power diodes have a higher forward voltage drop than low-current devices due to larger

IR drops across the bulk and contact resistances of the diode. The higher voltage drops result

in higher power dissipation levels for the diodes, which in turn may require the use of heat

sinks to draw the heat away from the body of the structure.

15. The primary difference between the standard p-n junction diode and the tunnel diode is that

the tunnel diode is doped at a level from 100 to several thousand times the doping level of a

p-n junction diode, thus producing a diode with a “negative resistance” region in its

characteristic curve.

1 1

16. At 1 MHz: X C = =

6 −12

2π

fC 2 π(1× 10 Hz)(5×

10 F)

= 31.83 kΩ

1

At 100 MHz: X C =

6 −12

2 π (100× 10 Hz)(5×

10 F)

= 318.3 Ω

At 1 MHz:

X

L S

= 2πfL = 2π(1 × 10 6 Hz)(6 × 10 −9 H)

= 0.0337 Ω

At 100 MHz: X

L S

= 2π(100 × 10 6 Hz)(6 × 10 −9 H)

= 3.769 Ω

L s effect is negligible!

R and C in parallel:

f = 1 MHz

Z T =

(152 Ω∠ 180 ° )(31.83 kΩ∠ − 90 ° )

−152 Ω− j31.83 kΩ

= −152.05 Ω∠0.27° ≅ −152 Ω∠0°

f = 100 MHz

(152 Ω∠ 180 ° )(318.3∠ − 90 ° )

Z T =

−152 Ω− j318.3

= −137.16 Ω∠25.52 ≠ −152 Ω∠0°

At very high frequencies X C has some impact!

171

17. The heavy doping greatly reduces the width of the depletion region resulting in lower levels

of Zener voltage. Consequently, small levels of reverse voltage can result in a significant

current levels.

18. At V T = 0.1 V,

I F ≅ 5.5 mA

At V T = 0.3 V

I F ≅ 2.3 mA

ΔV

0.3 V −0.1 V

R = =

ΔI

2.3 mA −5.5 mA

0.2 V

= = −62.5 Ω

−3.2 mA

E 2 V

19. I sat =

R = ≅ 5.13 mA

0.39 kΩ From graph: Stable operating points: I T ≅ 5 mA, V T ≅ 60 mV

I T ≅ 2.8 mA, V T = 900 mV

E 0.5 V

20. I sat =

R = = 9.8 mA

51 Ω

Draw load line on characteristics.

⎛ 1 ⎞

21. f s = ⎜ ⎟

⎝2π

LC ⎠

22. W = hf=

2

Rl

C

1−

L

⎛ 2 6

1 ⎞ −

(10 Ω ) (1×

10 F)

=

1−

3 6

−3

⎜ −

−

2 π (5× 10 H)(1×

10 F) ⎟ 5×

10 H

⎝

⎠

= (2250.79 Hz)(0.9899)

≅ 2228 Hz

v

h = λ

−34 °

8

(6.624× 10 J ⋅ s)(3×

10 m/s)

−10

(5000)(10 m)

= 3.97 × 10 −19 J

3.97 × 10 −19 ⎡ 1 eV ⎤

J ⎢

−19

1.6×

10 J ⎥ = 2.48 eV

⎣ ⎦

172

23. (a) Visible spectrum: 3750 A → 7500

(b) Silicon, peak relative response ≅ 8400 A

°

(c) BW = 10,300 A − 6100 A = 4200 A

° ° °

24.

−9 2

4×

10 W/m

= 2,486 f

−12

c

1.609×

10

From the intersection of V A = 30 V and 2,486 f c we find

I λ ≅ 440 μA

25. (a) Silicon

°

(b) 1 A = 10 −10 m,

−7

6×

10 m

−10

°

10 m/A

°

⇒ 6000 A → orange

26. Note that V λ is given and not V.

At the intersection of V λ = 25 V and 3000f c we find I λ ≅ 500 μA and

V R = I λ R = (500 × 10 −6 A)(100 × 10 3 Ω) = 50 V

27. (a) Extending the curve:

0.1 kΩ → 1000f c , 1 kΩ → 25f c

3

ΔR

(1 − 0.1) × 10 Ω

=

= 0.92 Ω/f c ≅ 0.9 Ω/f c

Δf

(1000 −25)

f

c

c

(b) 1 kΩ → 25f c , 10 kΩ → 1.3f c

3

ΔR

(10 − 1) × 10 Ω

=

= 379.75 Ω/f c ≅ 380 Ω/f c

Δf

(25 −1.3)

f

c

c

(c) 10 kΩ → 1.3f c , 100 kΩ → 0.15f c

3

ΔR

(100 − 10) × 10

=

= 78,260.87 Ω/f c ≅ 78 × 10 3 Ω/f c

Δfc

(1.3 −0.15)

fc

The greatest rate of change in resistance occurs in the low illumination region.

28. The “dark current” of a photodiode is the diode current level when no light is striking the

diode. It is essentially the reverse saturation leakage current of the diode, comprised mainly

of minority carriers.

29. 10f c → R ≅ 2 kΩ

V o = 6 V =

V i = 21 V

3

(2× 10 Ω)

V i

3 3

2× 10 Ω+ 5× 10 Ω

173

30. Except for low illumination levels (0.01f c )

the % conductance curves appear above the

100% level for the range of temperature. In

addition, it is interesting to note that for

other than the low illumination levels the %

conductance is higher above and below

room temperature (25°C). In general, the %

conductance level is not adversely affected

by temperature for the illumination levels

examined.

31. (a) (b)

(c) Increased levels of illumination result in reduced rise and decay times.

32. The highest % sensitivity occurs between 5250A and 5750A. Fig 16.20 reveals that the CdS

unit would be most sensitive to yellow. The % sensitivity of the CdS unit of Fig. 16.30 is at

°

°

the 30% level for the range 4800A → 7000A. This range includes green, yellow, and orange

in Fig. 16.20.

33. (a) ≅ 5 mW radiant flux

3.5 mW

(b) ≅ 3.5 mW

−13

1.496×

10 W/lm

= 2.34 × 10 10 lms

174

34. (a) Relative radiant intensity ≅ 0.8.

(b)

Relative radiant vs degrees off vertical

θ >30 ° and relative radiant intensity essentially zero-

Drops off very sharply after 25 ° !

35. At I F = 60 mA, Φ ≅ 4.4 mW

At 5°, relative radiant intensity = 0.8

(0.8)(4.4 mW) = 3.52 mW

36. 6, 7, 8

37. −

38. The LED generates a light source in response to the application of an electric voltage. The

LCD depends on ambient light to utilize the change in either reflectivity or transmissivity

caused by the application of an electric voltage.

39. The LCD display has the advantage of using approximately 1000 times less power than the

LED for the same display, since much of the power in the LED is used to produce the light,

while the LCD utilizes ambient light to see the display. The LCD is usually more visible in

daylight than the LED since the sun’s brightness makes the LCD easier to see. The LCD,

however, requires a light source, either internal or external, and the temperature range of the

LCD is limited to temperatures above freezing.

40. η% =

P

max

2

( A 2 )(100 mW/cm )

cm

Pmax

9% =

2 2

(2 cm )(100 mW/cm )

P max = 18 mW

× 100%

× 100%

41. The greatest rate of increase in power will occur at low illumination levels. At higher

illumination levels, the change in V OC drops to nearly zero, while the current continues to rise

linearly. At low illumination levels the voltage increases logarithmically with the linear

increase in current.

42. (a) Fig. 16.48 ⇒ 79 mW/cm 2

(b) It is the maximum power density available at sea level.

(c) Fig. 16.48 ≅ 12.7 mA

(b)

175

43. (a) (b)

(c) The curve of I o vs P density is quite linear while the curve of V o vs P density is only linear in

the region near the optimum power locus (Fig 16.48).

44. Since log scales are present, the differentials must be as small as possible.

ΔR

(7000 −1000) Ω 6000 Ω

= =

ΔT

(40 − 0) ° 40°

= 150 Ω/°C

ΔR

(3 −1) Ω 2 Ω

= =

Δ T 40° 40° = 0.05 Ω/°C

From the above 150 Ω/°C: 0.05 Ω/°C = 3000:1

Therefore, the highest rate of change occurs at lower temperatures such as 20°C.

45. No. 1 Fenwall Electronics Thermistor material.

Specific resistance ≅ 10 4 = 10,000 Ω cm

ρ

R =

A

2x ∴R = 2 × (10,000 Ω) = 20 kΩ

twice

46. (a) ≅ 10 −5 A = 10 μA

(b) Power ≅ 0.1 mW, R ≅ 10 7 Ω = 10 MΩ

(c)

Log scale ≅ 0.3 mW

47. V = IR + IR unk + V m

V = I(R + R unk ) + 0 V

R unk = V R

I −

= 0.2 V

2 mA − 10 Ω

= 100 Ω − 10 Ω

= 90 Ω

176

Chapter 17

1. −

2. −

3. −

4. (a) p-n junction diode

(b) The SCR will not fire once the gate current is reduced to a level that will cause the

forward blocking region to extend beyond the chosen anode-to-cathode voltage. In

general, as I G decreases, the blocking voltage required for conduction increases.

(c) The SCR will fire once the anode-to-cathode voltage is less than the forward blocking

region determined by the gate current chosen.

(d) The holding current increases with decreasing levels of gate current.

5. (a) Yes

(b) No

(c) No. As noted in Fig. 17.8b the minimum gate voltage required to trigger all units is 3 V.

(d) V G = 6 V, I G = 800 mA is a good choice (center of preferred firing area).

V G = 4 V, I G = 1.6 A is less preferable due to higher power dissipation in the gate. Not in

preferred firing area.

6. In the conduction state, the SCR has characteristics very similar to those of a p-n junction

diode (where V T = 0.7 V).

7. The smaller the level of R 1 , the higher the peak value of the gate current. The higher the peak

value of the gate current the sooner the triggering level will be reached and conduction

initiated.

⎛Vsec 8. (a) V P = (rms) ⎞

⎜

2

2

⎟

⎝ ⎠

117 V

= ( 2 ) = 82.78 V

2

V DC = 0.636(82.78 V)

= 52.65 V

(b) V AK = V DC − V Batt = 52.65 V − 11 V = 41.65 V

177

(c) V R = V Z + V GK

= 11 V + 3 V

= 14 V

At 14 V, SCR 2 conducts and stops the charging process.

(d) At least 3 V to turn on SCR 2 .

9. −

(e) V 2 ≅ 1 V = 1 (82.78 V)

2 P

2

= 41.39 V

10. (a) Charge toward 200 V but will be limited by the development of a negative voltage

= V − V that will eventually turn the GTO off.

V GK ( )

Z C 1

(b) τ = R 3 C 1 = (20 kΩ)(0.1 μF)

= 2 ms

5τ = 10 ms

(c) 5τ ′ = 1 (5 )

2 τ = 5 ms = 5R GTO C 1

5 ms 5 ms

R GTO = =

−6

5C

5(0.1×

10 F)

11. (a) ≅ 0.7 mW/cm 2

(b) 0°C → 0.82 mW/cm 2

100°C → 0.16 mW/cm 2

0.82 − 0.16

× 100% ≅ 80.5%

0.82

1

= 10 kΩ

⎛ 1 ⎞

⎜= (20 k Ω− above) ⎟

⎝ 2

⎠

12. V C = V BR + V GK = 6 V + 3 V = 9 V

V C = 40(1 − e −t/RC ) = 9

40 − 40e −t/RC = 9

40e −t/RC = 31

e −t/RC = 31/40 = 0.775

RC = (10 × 10 3 Ω)(0.2 × 10 −6 F) = 2 × 10 −3 s

log e (e −t/RC ) = log e 0.775

−t/RC = −t/2 × 10 −3 = −0.255

and t = 0.255(2 × 10 −3 ) = 0.51 ms

13. −

14.

V

= V ± 10% V

BR 2

= 6.4 V ± 0.64 V ⇒ 5.76 V → 7.04 V

BR1 BR2

15. −

178

16.

V −VP

I

P

> R 1

40 V −[0.6(40 V) + 0.7 V]

6

10×

10 − = 1.53 MΩ > R 1

V −VV

< R 1 ⇒ 40 V − 1 V

I

8 mA

= 4.875 kΩ < R 1

V

∴1.53 MΩ > R 1 > 4.875 kΩ

RB

1

17. (a) η =

R + R

B1 B2

I E=

0

⇒ 0.65 =

2 kΩ

R

B

= 1.08 kΩ

2

2 kΩ+ RB

2

(b) R BB = ( RB

RB

)

+ = 2 kΩ + 1.08 kΩ = 3.08 kΩ

1 2

IE=

0

(c)

V

R B1

= ηV BB = 0.65(20 V) = 13 V

(d) V P = ηV BB + V D = 13 V + 0.7 V = 13.7 V

RB

1

18. (a) η =

R

BB IE

= 0

R B

1

0.55 =

10 kΩ

R

B !

= 5.5 kΩ

R BB = RB

+ R

! B2

10 kΩ = 5.5 kΩ +

R

B 2

= 4.5 kΩ

R

B 2

(b) V P = ηV BB + V D = (0.55)(20 V) + 0.7 V = 11.7 V

V −V

(c) R 1 <

I

p

ok: 68 kΩ < 166 kΩ

P

20 V −11.7 V

= = 166 kΩ

50 μA

179

V −V

(d) t 1 = R 1 C log e

V −V

(e)

P

t 2 = ( R + R ) C log e

B1 2

= 0.546 ms

T = t 1 + t 2 = 6.106 ms

f = 1 = 1 = 163.77 Hz

T 6.106 ms

V

P

= (68 × 10 3 )(0.1 × 10 −6 18.8

) log e = 5.56 ms

8.3

V

V = (0.2 kΩ + 2.2 kΩ)(0.1 × 11.7

10−6 ) log e

1.2

V

(f)

V

V

R2

R2

RV

2

2.2 k Ω(20 V)

= =

R2

+ RBB

2.2 kΩ+ 10 kΩ

= 3.61 V

R2

( VP

− 0.7 V)

≅

R + R

2

B1

2.2 k Ω(11.7 V −0.7 V)

=

2.2 kΩ+ 0.2 kΩ

= 10.08 V

1 1

(g) f ≅

=

= 184.16 Hz

RC log (1/(1 −η)) (6.8 k Ω)(0.1 μF)log 2.22

1

e

difference in frequency levels is partly due to the fact that t 2 ≅ 10% of t 1 .

19. I B = 25 μA

I C = h I = (40)(25 μA) = 1 mA

fe

B

e

180

20.

ΔI

21. (a) D F =

ΔT

0.95 − 0 0.95

= =

25 −( −50) 75

= 1.26%/°C

(b) Yes, curve flattens after 25°C.

22. (a) At 25°C, I CEO ≅ 2 nA

At 50°C, I CEO ≅ 30 nA

−9

ΔI CEO

(30 − 2) × 10 A 28 nA

= = = 1.12 nA/°C

ΔT

(50 − 25) ° C 25°

C

I CEO (35°C) = I CEO (25°C) + (1.12 nA/°C)(35°C − 25°C)

= 2 nA + 11.2 nA

= 13.2 nA

From Fig. 17.55 I CEO (35°C) ≅ 4 nA

Derating factors, therefore, cannot be defined for large regions of non-linear curves.

Although the curve of Fig. 17.55 appears to be linear, the fact that the vertical axis is a

log scale reveals that I CEO and T( °C) have a non-linear relationship.

23.

Io

IC

20 mA

= = = 0.44

Ii

IF

≅ 45 mA

Yes, relatively efficient.

181

24. (a) P D = V CE I C = 200 mW

PD

200 mW

I C = = = 6.67 mA @ V CE = 30 V

V 30 V

V CE =

CEmax

P

I

D

C

200 mW

= = 20 V @ I C = 10 mA

10 mA

PD

200 mW

I C =

V = 25 V

= 8.0 mA @ V CE = 25 V

CE

Almost the entire area of Fig. 17.57 falls within the power limits.

(b) β dc =

I

I

C

F

4 mA

IC

= = 0.4, Fig. 17.56

10 mA

I

F

≅ 4 mA

10 mA = 0.4

25. (a) I C ≥ 3 mA

The fact that the I F characteristics of Fig. 17.57 are fairly horizontal reveals that the level

of I C is somewhat unaffected by the level of V CE except for very low or high values.

Therefore, a plot of I C vs. I F as shown in Fig. 17.56 can be provided without any

reference to the value of V CE . As noted above, the results are essentially the same.

(b) At I C = 6 mA; R L = 1 kΩ, t = 8.6 μs

R L = 100 Ω; t = 2 μs

1 kΩ:100 Ω = 10:1

8.6 μs:2 μs = 4.3:1

ΔR:Δt ≅ 2.3:1

3RB

3

2

26. η =

3R + R = 4

B2 B2

= 0.75, V G = ηV BB = 0.75(20 V) = 15 V

27. V P = 8.7 V, I P = 100 μA Z P =

V V = 1 V, I V = 5.5 mA Z V =

87 kΩ: 181.8 Ω = 478.55:1 ≅ 500:1

V

I

P

P

V

I

V

V

= 8.7 V

100 μA

= 87 kΩ (≅ open)

1 V

= = 181.8 Ω (relatively low)

5.5 mA

182

⎛ V ⎞ ⎛

BB

V ⎞

BB

28. Eq. 17.23: T = RC log e ⎜ ⎟=

RC loge

⎜ ⎟

⎝VBB −VP ⎠ ⎝VBB − ( ηVBB + VD

) ⎠

⎛

⎜

⎛ V ⎞

BB

Assuming ηV BB V D , T = RC log e ⎜ ⎟ = RC log e (1/1− η) = RC log

⎜ 1

e

⎝VBB

(1 −η)

⎠

⎜ RB

1

1−

⎜

⎝

RB

+ R

⎛ RB + R ⎞ ⎛

1 B

R ⎞

2 B1

= RC log e = RC loge

1+

Eq. 17.24

⎜ R ⎟ ⎜

B

R ⎟

⎝ 2 ⎠ ⎝ B2

⎠

1 B2

⎞

⎟

⎟

⎟

⎟

⎠

29. (a) Minimum V BB :

VBB −VP

R max =

I

P

P

≥ 20 kΩ

VBB − ( ηVBB + VD

)

= 20 kΩ

I

V BB − ηV BB − V D = I P 20 kΩ

V BB (1 − η) = I P 20 kΩ + V D

IP20 kΩ + VD

V BB =

1−η

(100 μA)(20 k Ω ) + 0.7 V

=

1−

0.67

= 8.18 V

10 V OK

V

(b) R <

BB

−V

I

V

R < 2 kΩ

V

12 V −1 V

= = 2 kΩ

5.5 mA

⎛ R ⎞

B1

(c) T ≅ RC log e 1+

⎜ R ⎟

⎝ B2

⎠

2 × 10 −3 = R(1 × 10 −6 ⎛ 10 kΩ

⎞

)log e ⎜1+

⎟

⎝ 5 kΩ

⎠

log e 3 = 1.0986

−3

2×

10

R =

−6

(1×

10 )(1.0986)

R = 1.82 kΩ

183

Solutions for Laboratory Manual

to accompany

Electronic Devices and Circuit Theory

Tenth Edition

Prepared by

Franz J. Monssen

185

186

EXPERIMENT 1: OSCILLOSCOPE AND FUNCTION GENERATOR OPERATIONS

Part 1: The Oscilloscope

a. it focuses the beam on the screen

b. adjusts the brightness of the beam on the screen

c. allows the moving of trace in either screen direction

d. selects volts/screen division on y-axis

e. selects unit of time/screen division on x-axis

g. allows for ac or dc coupling of signal to scope and at

GND position; establishes ground reference on screen

h. locates the trace if it is off screen

i. provide for the adjustment of scope from external

reference source

k. determines mode of triggering of the sweep voltage

m. the input impedance of many scopes consists of the parallel combination of a 1 Meg

resistance and a 30pf capacitor

n. measuring device which reduces loading of scope on a circuit and effectively

increases input impedance of scope by a factor of 10.

Part 2: The Function Generator

d. T = l/f = 1/1000 Hz = l ms

e. (calculated): 1 ms*[l cm/.2 ms] = 5cm

(measured): 5 cm = same

f. (calculated): l ms*[cm/.5ms] =2 cm

(measured): 2 cm = same

g. (calculated): 1 ms*[cm/1ms] = l cm

(measured): l cm = same

h. .2 ms/cm takes 5 boxes to display total wave

.5 ms/cm takes 2 boxes to display total wave

1 ms/cm takes 1 box to display total wave

i. 1. adjust timebase to obtain one cycle of the wave

2. count the number of cm's occupied by the wave

3. note the timebase setting

4. multiply timebase setting by number of cm's occupied

by wave. This is equal to the period of the wave.

5. obtain its reciprocal; that's the frequency.

187

j. (calculated): 2cm * [2V/cm] = 4Vp-p

k. 8 * [.5V/cm] = 4Vp-p

1. the signal occupied full screen; the

peak amplitude did not change with a

change in the setting of the vertical sensitivity

m. no: there is no voltmeter built into function

generator

Part 3: Exercises

a. chosen sensitivities: Vert. Sens. = l V/cm

Hor. Sens. = 50 μs/cm

T(calculated): 4cm*[50 μs/cm)= 200 μs

Fig 1.1

b. chosen sensitivities: Vert. Sens. = .l V/cm

Hor. Sens. = 1 ms/cm

T(calculated):5 cm*[l ms/cm] = 5 ms

Fig 1.2

188

c. chosen sensitivities: Vert. Sens. = l V/cm

Hor. Sens. = l μs/cm

T(calculated):10 cm*[1μs/cm]=10 μs

Fig 1.3

Part 4: Effect of DC Levels

a. V (rms) (calculated) = 4V * 1/2 * .707 = 1.41 Volts

b. V (rms) (measured) = 1.35 Volts

c. [(1.41 − 1.35)/1.41) * 100 = 4.74%

d. no trace on screen

e. signal is restored, adjust zero level

f. no shift observed; the shift is proportional to dc

value of waveform

g. (measured) dc level: 1.45 Volts

h.

Fig 1.5

i. Switch AC-GND-DC switch, make copy of waveform above.

The vertical shift of the waveform was equal to the battery voltage.

189

The shape of the sinusoidal waveform was not affected by changing the positions of

the AC-GND-DC coupling switch.

j. The signal shifted downward by an amount equal to

the voltage of the battery.

Fig 1.6

Part 5: Problems

1. b. f = 2000/(2*3.14) = 318Hz

c. T = l/f =1/318 = 3.14ms

d. by inspection: V(peak) = 20V

e. V(peak-peak) = 2*Vpeak = 40V

f. V(rms) =.707 * 20 = 14.1V

g. by inspection: Vdc = 0V

2.

a. f = 2 * 3.14 * 4000/(2 * 3.14*) = 4 KHz

c. T = l/f =1/4 Khz = 250 μs

d. by inspection:Vpeak)= 8 mV

e. V(peak-peak) = 2 * V(peak) = 16 mV

f. V(rms) = .707 * 8 mV = 5.66 mV

g. by inspection: Vdc = 0V

3. V(t) = 1.7 sin (2.51 Kt) volts

Part 6: Computer Exercise

PSpice Simulation 1-1

See Probe Plot page 191.

190

191

EXPERIMENT 2: DIODE CHARACTERISTICS

Part 1: Diode Test

diode testing scale

Test

Forward

Reverse

Both diodes are in good working order.

Part 2. Forward-bias Diode characteristics

Table 2.1

Si (mV)

535

OL

Ge (mV)

252

OL

b.

V R (V)

V D (mV)

I D (mA)

.1

453

.1

.2

481

.2

Table 2.3

.3 .4

498 512

.3 .4

.5

528

.5

.6

532

.6

.7

539

.7

.8

546

.8

V R (V) .9 1 2 3 4 5 6 7 8 9 10

V D (mV) 551 559 580 610 620 630 640 650 650 660 660

I D (mA) .9 1 2 3 4 5 6 7 8 9 10

d.

V R (V)

V D (mV)

I D (mA)

.1

156

.1

.2

187

.2

Table 2.4

.3 .4

206 217

.3 .4

.5

229

.5

.6

239

.6

.7

247

.7

.8

254

.8

V R (V) .9 1 2 3 4 5 6 7 8 9 10

V D (mV) 260 266 300 330 340 360 370 380 390 400 400

I D (mA) .9 1 2 3 4 5 6 7 8 9 10

e. Fig 2.5

192

f. Their shapes are similar, but for a given I D , the potential V D is greater for the silicon diode

compared to the germanium diode. Also, the Si has a higher firing potential than the

germanium diode.

Part 3: Reverse Bias

b. R m = 9.9 Mohms

V R (measured) = 9.1 mV

I S (calculated) = 8.21 nA

c. V R (measured) = 5.07 mV

I S (calculated) = 4.58 μA

d. The I S level of the germanium diode is approximately 500 times as large

as that of the silicon diode.

e. R DC (Si) = 2.44*10 9 ohms

R DC (Ge) = 3.28 M*10 6 ohms

Part 4: DC Resistance

These values are effective open-circuits when compared to resistors in the kilohm range.

a.

b.

Table 2.5

I D (mA) V D (mV) R DC (ohms)

.2 350 1750

1.0 559 559

5.0 630 126

10.0 660 66

Table 2.6

I D (mA) V D (mV) R DC (ohms)

.2 80 400

1.0 180 180

5.0 340 68

10.0 400 40

Part 5: AC Resistance

a. (calculated)r ac = 3.4 ohms

b. (calculated)r ac = 2.9 ohms

c. (calculated)r ac = 27.0 ohms

d. (calculated)r ac = 26.0 ohms

Part 6: Firing Potential

V T (silicon) = 540 mV

V T (germanium) = 260 mV

193

Part 7: Temperature Effects

c. For an increase in temperature, the forward diode current will increase while the voltage

V D across the diode will decline. Since R D = V D /I D , therefore, the resistance of a diode

declines with increasing temperature.

d. As the temperature across a diode increases, so does the current. Therefore, relative to the

diode current, the diode has a positive temperature coefficient.

Part 9: Computer Exercises


1. See Probe plot page 195.

2. R D 600mV = 658 Ω

R D 700 mV = 105 Ω

4. R D 600 mV = 257 Ω

5. See Probe Plot V(D1) versus I(D1)

7. Silicon




194

195

196

EXPERIMENT 3: SERIES AND PARALLEL DIODE CONFIGURATIONS

Part 1: Threshold Voltage V T

Fig 3.2

Firing voltage: Silicon: 595 mV

Germanium: 310 mV

Part 2: Series Configuration

b. V D = .59 V

V O (calculated) = 5 − .595 = 4.41 V

I D = 4.41/2.2 K = 2 mA

c. V D (measured) = .59 V

V O (measured) = 4.4 V

I D (from measured) = 2 mA

e. V D = 595 mV

V O (calculated) = (5 − .595) 1 K/(1 K + 2.2 K) = 1.33 V

I D = 1.36 mA

f. V D = .57 V

V O = 1.36 V

I D (from measured) = 1.36 V/1 K = 1.36 mA

g. V D (measured) = 5 V h. V D (measured) = 5 V

V O (measured) = 0 V


I D (measured) = 0 A

I D (measured) = 0 A

j. V 1 (calculated) = .905 V

V O (calculated) = 4.1 V

I D (calculated) = 1.86 mA



1. 638.0 mV

197

EXPERIMENT 4: HALF-WAVE AND FULL-WAVE RECTIFICATION

Part 1: Threshold Voltage

V T = .64 V

Part 2: Half-wave Rectification

b. Vertical sensitivity = 1 V/cm

Horizontal sensitivity = .2 ms/cm

c.

Fig 4.4

d. Both waveforms are in essential agreement.

e. V dc = (4 − .64)/3.14 = 1.07 V

f. V dc (measured) = .979 V

% difference = (1.07 − .979)/1.07*100 = 8.5%

g. For an ac voltage with a dc value, shifting the coupling switch from its DC to AC

position will make the waveform shift down in proportion to the dc value of the

waveform.

h. Fig 4.6

198

i. V dc (calculated) = −1.07 V

V dc (measured) = −.970 V

Part 3: Half-Wave Rectification (continued)

b.

Fig 4.8

c.

Fig 4.9

The results are in reasonable agreement.

d. The significant difference is in the respective reversal of the two voltage waveforms.

While in the former case the voltage peaked to a positive 3.4 volts, in the latter case, the

voltage peaked negatively to the same voltage.

e. V DC = (.318)*3.4 = 1.08 Volts

f. Difference = [1.08 − .979]/1.08*100 = 9.35%

199

Part 4: Half-Wave Rectification (continued)

b.

Fig 4.11

c.

Fig 4.12

There was a computed 2.1% difference between the two waveforms.

d.

Fig 4.13

200

We observe a reversal of the polarities of the two waveforms caused by the reversal of

the diode in the circuit.

Part 5: Full-Wave Rectification (Bridge Configuration)

a. V (secondary)rms = 14 V

This value differs by 1.4 V rms from the rated voltage of the secondary of the

transformer.

b. V (peak) = 1.41*14 = 20 V

c.

Fig 4.15

Vertical sensitivity: 5 V/cm

Horizontal sensitivity: 2 ms/cm

d.

Fig 4.16

201

Again, the difference between expected and actual was very slight.

e. V dc (calculated) = (.6326)*(20) = 12.7 V

V dc (measured) = 11.36 V

% Difference = −10.6%

g. Vertical sensitivity = 5 V/cm

Horizontal sensitivity = 2 ms/cm

Fig 4.17

i. V dc (calculated) = (.636)*(12) = 7.63 V

j


% Difference = −7.6%

k. The effect was a reduction in the dc level of the output voltage.

Part 6: Full-Wave Center-tapped Configuration

a. V rms (measured) = 6.93 V

V rms (measured) = 6.97 V

As is shown from the data, the difference for both halves of the center-tapped windings

from the rated voltage is .6 volts.

b. Vertical sensitivity = 5 V/cm

Horizontal sensitivity = 2 ms/cm

202

c.

Fig 4.21

d. V dc (calculated) = 3.5 V




1. V p = 8.47 V; relative phase shift is equal to 180°

2. PIV = 2 Vp

3. 180° out of phase


Its amplitude is 7.89 V

5. Yes

6. Reasonable agreement.

203

204

EXPERIMENT 5: CLIPPING CIRCUITS


V T (Si) = .618 V

V T (Ge) = .299 V

Part 2 Parallel Clippers

b. V O (calculated) = 4 V

c. V O (calculated) = −1.5 − .618 = −2.2 V

d.

Fig 5.2

Vertical sensitivity = 1 V/cm


e.

Fig 5.3

No measured differences appeared between expected and observed waveforms.

f. V O (calculated) = 4 V

g. V O (calculated) = .62 V

205

Part 3: Parallel Clippers (continued)

b. V O (calculated) = .61 V

c. V O (calculated) = .34 V

d.

Fig 5.7



e.

Fig 5.8

The waveforms agree.

Part 4: Parallel Clippers (Sinusoidal Input)

b. V O (calculated) = 4 V when V i = 4 V

V O (calculated) = −2 V when V i = −4 V

V O (calculated) = 0 V when V i = 0 V

206

Fig 5.9

c. Waveforms agree within 6.5%.

Part 5: Series Clippers

b. V O (calculated) = 2.5 V when V i = 4 V

c. V O (calculated) = 0 V when V i = −4 V

d.

Fig 5.12

e. agree within 5.1%



f. V O (calculated) = 5.5 V when V i = 4 V

g. V O (calculated) = 0 V when V i = −4 V

207

h.

Fig 5.14

i. no major differences



Part 6: Series Clippers (Sinusoidal Input)


V O (calculated) = 0 V when V i = −4 V

V O (calculated) = 0 V when V i = 0 V

Fig 5.16



208




2. V OUT = 4 V

3. No

4. V OUT = −2.067 V

5. Yes, V OUT (ideal) = −1.5 V

6. Reasonable agreement

7. No significant discrepancies




2. In close agreement

3. No

4. For V 1 = 4 V; V out = V 1 − V D1 − 1.5 V = 4 V − .6 − 1.5 V = 1.9 V

For V 1 = −4 V; I (D1) = 0 A, ∴ V out = 0 V





9. Forward bias voltage of about 600 mV when “ON”.

Reverse diode voltage of diode is −4 V − 1.5 V = −5.5 V

209

210

211

212

213

EXPERIMENT 6: CLAMPING CIRCUITS


V T = .62 V

Part 2: Clampers (R, C, Diode Combination)

b. V C (calculated) = 4 − 0.62 = 3.38 V

V O (calculated) = 0.62 V

c. V O (calculated) = −4 − 3.38 V = −7.38 V

d. Fig 6.2



214

e. Fig 6.3

f. V C (calculated) = −3.38 V

V O (calculated) = −0.62 V

g. V O (calculated) = 7.38 V

h. Fig 6.4

i. Fig 6.5



215

Part 3: Clampers with a DC battery

b. V C (calculated) = 1.88 V

V O (calculated) = 0.62 V + 1.5 V = 2.12 V

c. V O (calculated) = −1.88 V − 4 V = −5.88 V

d. Fig 6.7



e. Fig 6.8

f. V C (calculated) = 4.88 V

V O (calculated) = 1.5 V − 0.62 V = 0.88 V

g. V O (calculated) = 4 V + 4.88 V = 8.88 V

216

h. Fig 6.9



Part 4: Clampers (Sinusoidal Input)


V O (calculated) = −2 V when V i = −3.6 V

V O (calculated) = −1.6 V when V i = 0 V

Fig 6.11



217

Part 5: Clampers (Effect of R)

a. Tau(calculated) = R*C = 103 ms

b. T(calculated) = 1/f = 1 ms

T/2(calculated) = 1 ms/2 = .5 ms

c. 5Tau(calculated) = 5*103 ms = 515 ms

d. otherwise the capacitor voltage will not remain constant

e. 5Tau(calculated) = 5 ms

f. 5 ms/.5 ms = 10

g.

Fig 6.13



i. 5Tau = .5 ms

j. .5 ms/.5ms = 1

k.

Fig 6.14



m. 5Tau = 2.5 T or Tau = 1/2 T

218



1. See Probe Plot page 220.

2. They are the same.

3. V O (calculated) is close to V(2) of Probe plot.


5. V(1, 2) remains at 2 V during the cycle of V(1)

6. It rises exponentially toward its final value of 2 V.


219

220

221

222

EXPERIMENT 7: LIGHT-EMITTING AND ZENER DIODES

Part 1: LED Characteristics

b. V D (measured) = 1.6 V

V R (measured) = 49.1 mV

I D (calculated) = 49.1 mV/101.4 ohms = 484 μA

c. V D (measured) = 1.9 V

V R (measured) = 1.55 V

I D (calculated) = 1.55 V/101.4 ohms = 15.3 mA

d.

E(v) 0 1 2 3 4 5 6

VD (V) 0 1 1.71 1.84 1.93 2.01 2.08

VR (V) 0 0 .34 1.2 2.2 3.1 3.9

ID = VR/R (mA) 0 0 3.3 11.8 21.4 30.6 38.5

e. Fig 7.2

h. The reversed biased Si diode prevents any current from flowing through the circuit,

hence, the LED will not light.

k. V R (V) = 3.48 V, therefore I D (mA) = 1.6 mA and LED is in the “good brightness” region.

Part 2: Zener Diode Characteristics

b. and c.

Table 7.2

E (V) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

V Z (V) 0 1 2 3 4 5 6 7 8 9 10 10 10.1 10.2 10.3 10.4

V R (V) 0 0 0 0 0 0 0 0 0 .1 .97 1.9 2.8 3.7 4.6 4.6

I Z (mA) 0 0 0 0 0 0 0 0 0 .99 9.6 18.7 27.6 36.5 45.4 45.4

223

d.

Fig 7.5

e. V Z (V) (approximated) = (10.4 + 9)/2 = 9.7 V

f. r av (ohms) = (10.4 − 9)/(.045 − .0099) = 39.9 ohms

g. R Z (ohms) = 39.9 ohms

V Z (V) = 9.7 V

Part 3: Zener Diode Regulation

a. R (meas) = 979 ohms

R L (meas) = 986 ohms

V Z (V) = 10.2 V

b. V L (V) = 986*15/(979 + 986) = 7.53 V

V R (V) = 979*15/(979 + 986) = 7.47 V

I R (mA) = 7.47/979 = 7.64 mA

I L (mA) = 7.53/986 = 7.63 mA

I Z (mA) = I R − I L = 10 μA

c. V L (measured) = 7.5 V

V R (measured) 7.49 V

I R (calculated) = 7.65 mA

I L (calculated) = 7.60 mA

I Z (calculated) = 50 μA

d. V L (calculated) = 11.5 V

V R (calculated) 3.54 V



I Z (calculated) = .14 mA

224

e. V L (measured) = 9.82 V

V R (measured) = 3.54 V



I Z (calculated) = .56 mA

The difference is expressed as a percent with calculated value as the standard of

reference.

percent change of: V L = −14.6%

V R = 0.%

I R = −2.21%

I L = −14.4%

I Z = 30.0%

f. R min /(R min + 979)*15 = 9.82 V

R L (calculated) = 1.86 Kohms

g. Since 2.2 Kohms > R min = 1.86 Kohms, therefore, diode is in the “on” state.

Part 4: LED-Zener diode combination

b. V D = 1.86 V

I D = 15.8 mA

V Z = 10.07 V

V ab (calculated) = 11.9 V

c. V L (calculated) = 11.9 V


e. E (calculated) = V R + V L = 6.93 + 11.9 = 18.9 V

f. E (measured) = 19.1 V

The two values are in agreement within 1.06% using E (calculated) as reference.

225



1. – 8. See Circuit diagram

9. Yes

226

EXPERIMENT 8: BIPOLAR JUNCTION TRANSISTOR (BJT) CHARACTERISTICS

Part 2: The Collector Characteristics

d., f., g., h. Table 8.3

i.

Fig 8.3

227

Part 3: Variation of Alpha and Beta

b. The variations for Alpha and Beta for the tested transistor are not really significant,

resulting in an almost ideal current source which is independent of the voltage V CE .

c. The highest Beta’s are found for relatively large values of I C and V CE . This is a generally

well known factor.

d. Beta did increase with increasing levels of I C .

e. Beta did increase with increasing levels of V CE .

Part 5: Exercises

1. Beta(average) = 141

The arithmetic average occurred in the center of Fig 8.3.

2. V BE(average) = .678 V

Given that .7 V differs by only 3.14% from .678, and given that resistive circuit component

can vary by as much as 20%, the assumption of a constant .7 V is entirely reasonable.

3. The Beta of the transistor is increasing. Table 8.3 does substantiate that conclusion.

Beta would be a constant anywhere along that line.



1. See Circuit diagram.

2. Experimental PSpice

α .99 .99

β 150 208

228

EXPERIMENT 9: FIXED- AND VOLTAGE-DIVIDER BIAS OF BJTs

Part 1: Determining β

b. V BE (measured) = .67 V

V RC (measured) = 4.9 V

c. I B

B = (VCC − V BE )/R

BB = (20 − .67)/1.108 M = 17.4 μA

I C = V RC /R C = 4.9/2.73 K = 1.79 mA

d. Beta I C /I B

B = 1.79 mA/17.4 μA = 105

Part 2: Fixed-bias configuration

a. I B (calculated) B = 17 μA

I C (calculated) = 1.79 mA

b. V B (calculated) B = VCC − I B * R

BB = .67 V

V C (calculated) = V CC − I C * R C = 13.4 V

V E (calculated) = 0 V(emitter is at ground)

V CE (calculated) = V C − V E = 13.4 V

c. V B (measured) B = .67 V

V C (measured) = 13.4 V

V E (measured) = 0 V

V CE (measured) = 13.34 V

The difference between measured and calculated values in every case is less

than 10%. It’s almost too good to be true.

d. V BE (measured) = .68 V


I B (from B measured) = 17.4 μA

I C (from measured) = 6.12 mA

Beta(calculated) = 352

Table 9.1

Transistor Type V CE (V) I C (mA) I B (μA) β

2N3904 13.34 1.79 17.4 105

2N4401 3.2 6.12 17.4 352

e.

Table 9.2

% Δ β % Δ I C % Δ V CE % Δ I

BB

242 242 −76.0 0

229

Part 3: Voltage-divider configuration

b.

B

Table 9.3

2N3904 V B (V) V E (V) V C (V) V CE (V)

(calculated) 3.52 2.82 12.47 9.7

(measured) 3.3 2.6 12.9 10.1

B

2N3904 I E (mA) I C (mA) I B μA)

(calculated) 4.07 4.05 30

(measured) 3.76 3.87 36.5

c. The agreement between measured and calculated values fall entirely within reasonable

limits.

d. and e.

Table 9.4

Transistor Type V CE (V) I C (mA) I B (μA) Beta

2N3904 10.1 3.87 36.5 103

2N4401 9.6 4.03 17.2 234

f.

Table 9.5

% Δ β % Δ I C % Δ V CE % Δ I

BB

56 41 4.9 53



1.-3. See circuit diagram.

230

4. See circuit diagram.

5. 8.24%

6. %ΔI B

B = 0.05%

%ΔI C = 8.2%

%ΔI E = 8.15%

7. %ΔV CE = −6.57%

8. S(β) = .995


1.-3. See circuit diagram.

231


B

5. %Δβ = 8.24%

6. %ΔI B = −6.47%

%ΔI C = 1.13%

%ΔI E = 1.10%

7. %ΔV CE = −0.94%

8. S(β) = 1.13%

8.24% = 0.13

9. Circuit with Q2N2222.

10. Same as #9.

11. Same as #9.

Part 5: Problems and Exercises

1. a. I C(sat, fixed bias) = 20/2.73 K = 7.33 mA

b. I C(sat,volt-divider bias) = 20/(1.86 K + 692) = 7.83 mA

c. The saturation currents are not sensitive to the Beta’s in either bias configuration.

2. In the case of the 2N4401 transistor, which had a higher Beta than the 2N3904 transistor, the

Q point of the former shifted higher up the loadline toward saturation. (See data in Table 9.4).

3. a.

Table 9.6

Fixed bias % Δ I C % Δ V CE % Δ I

BB

% Δ β % Δ β % Δ β

1 .314 0

Volt-divider .732 .087 .94

The ideal circuit has Beta independence when the ratio of %Δ I C /%Δ β is equal to 0.

Thus, the smaller the ratio, the more Beta independent is the circuit. Using this as a

criterion of stability, it becomes apparent that the voltage divider bias circuit is the more

stable of the two.

232

4. a. I C = β(V CC − .67)/R B

B mA

b. I C = [R 2 /(R 1 + R 2 )*V CC − .7]/[(R 1 || R 2 )/β + R E ] mA

c. In equation 4a, the Beta factor cannot be eliminated by a judicious choice of circuit

components. In 4b however, if the quantity R 1 || R 2 /β is made much smaller than R E , then

I C is no longer dependent upon Beta. In particular:

I C = [R 2 /(R 1 + R 2 )*V CC − .7]/R E mA

In that case, we have achieved Beta independent biasing.

233

EXPERIMENT 10: EMITTER AND COLLECTOR FEEDBACK BIAS OF BJTs

Part 1: Determination of β

b. V B

B (measured) = 5.04 V


c. I B

B (from measured) = (20 − 5.41)/1.1 M = 13.6 μA

I C (from measured) = 4.04/2.2 K = 1.84 mA

d. β = 1.84 mA/13.6 μA = 135

Part 2: Emitter-bias configuration

a. Using KVL:

−20 + I C * (1.01 M/β) + .67 V + I C * (2.23 K) = 0 V

therefore: I C = (20 − .67)/9.1 K = 2.1 mA

I B

B = 2.1 mA/135 = 15 μA

b. and c.

Table 10.1

Calculated Values

Transistor type V B (V) V C (V) V E (V) V BE (V) V CE (V)

2N3904 5.4 15.3 4.7 .70 10.6

2N4401 8.2 12.6 7.4 .8 5.2

B

Transistor type I B (μA) I C (mA)

2N3904 15.0 2.1

2N4401 11.7 3.3

Measured Values

Transistor type V B (V) V C (V) V E (V) V BE (V) V CE (V)

2N3904 4.75 15.9 4.2 .66 11.8

2N4401 8.0 12.5 7.6 .62 4.8

B

Transistor type I B (μA) I C (mA) Beta

2N3904 14.7 2.2 150

2N4401 11.9 3.4 286

d. See Table 10.1.

e. See Table 10.1.

f. In every case, the difference between calculated and measured values were less than 10%

apart.

234

g.

Table 10.3

% Δβ % Δ I C %Δ V CE % Δ I

BB

90.7 54.5 −58.5 −19

Part 3: Collector Feedback Configuration (R E = 0 ohms)

b. Using KVL:

−20 + I C (3.2 K) + I C (395 K)/150 + .7 V = 0 V

from which: I B

B = 21 μA and IC = 3.4 mA

Table 10.4

Calculated Values

B

Transistor type V B (V) V C (V) V CE (V) I B (μA)

I C (mA)

2N3904 .62 9.1 9.1 21.2 3.4

2N4401 .55 6.2 6.2 14.4 4.3

Table 10.5

Measured Values

B

Transistor type V B (V) V C (V) V CE (V) I B (μA)

I C (mA)

2N3904 .68 9.6 9.6 22.4 3.6

2N4401 .63 5.8 5.8 15.1 4.4

Table 10.6

% Δ β % Δ I C % Δ V CE % Δ I

BB

83 22.8 −39.9 −33

Part 4: Collector Feedback Configuration (with R E )

a. For 2N3904:

−20 + I C (3.2 K) + I C (395 K/150) + I C (2.2 K) = 0 V

from which: I B

B = 15 μA and IC = 2.4 mA

for 2N4401:

−20 + I C (3.2 K) + I C (395 K/286) + I C (2.2 K) = 0 V

from which: I B

B = 9.7 μA and IC = 2.8 mA

b. See Table 10.7.

c. See Table 10.8.

d. See Table 10.7.

e. See Table 10.8.

f.

B

Table 10.7

Calculated Values

C (V) V E (V) V CE (V) I C (mA) I E (mA) I B (μA)

B V Transistor V B (V)

2N3904 6.2 12.1 5.4 6.7 2.45 2.5 15

2N4401 6.9 10.8 6.3 4.5 2.8 2.9 9.7

235

B

B

Table 10.8

Measured Values

Transistor V B (V) V C (V) V E (V) V CE (V) I C (mA) I E (mA) I B (μA)

2N3904 5.9 12.6 5.2 7.4 2.3 2.4 19

2N4401 7.0 10.8 6.5 4.3 2.8 2.9 9.2



1−6. See Circuit diagram.

Table 10.9

% Δ β % Δ I C % Δ V CE % Δ I

BB

83.2 23.8 −41.2 −50.3


236

8. %Δβ = 8.87%

9. %ΔI B

B = −2.18%

%ΔI C = 6.50%

%ΔI E = 6.42%

10. %ΔV CE = −7.43%

11. S(β) = .73

12. P(Q2N3904) = 46.41 mW

P(Q2N2222) = 49.40 mW

Yes

13. Yes, see circuit diagram above.

14. Yes, see circuit diagram above.


1−6. See Circuit diagram.


237

8. %Δβ = 8.64%

9. %ΔI B

B = −5.43%

%ΔI C = 2.75%

%ΔI E = 2.69%

10. %ΔV CE = −4.96%

11. S(β) = .32

12-14. See circuit diagrams above.


1. a. I C(sat) 20/(2.2 K + 2.2 K) = 4.55 mA

b. I C(sat) = 20/3 K = 6.67 mA

c. I C(sat) = 20/5.2 K = 3.85 mA

d. Beta does not enter into the calculations.

2. The Q point shifts toward saturation along the loadline.

3. a.

Table 10.10

Emitter bias % Δ I C % Δ V CE % Δ I

BB

% Δ β % Δ β % Δ β

b. The smaller that ratio, the better is the Beta stability of a particular circuit. Looking at the

results, which were computed from measured data, it appears that the collector feedback

circuit with R E = 0 ohms is the most stable. This is counter to expectations.

Theoretically, the most stable of the two collector feedback circuits should be the one

with a finite R E . Since the stability figures of both of those circuits are so small, the

apparent greater stability of the collector feedback circuit without R E is probably the

result of measurement variability.

4. Using KVL:

−V CC + I C /β*R B

B + VBE + I C *R E = 0 V

from this:

I C = (V CC − V BE )/(R B

B/β + RE) mA

This division results in:

I C = β(V CC − V BE )/(R B

B + β*RE) mA

If β*R E >>R B

B then IC = (V CC − V BE )/R E mA

5. Using KVL:

−V CC + I C *R C + I C /β*R B

B + VBE = 0 V

from this:

I C = (V CC − V BE )/(R C + R B /β)

B

if R C >>R B

B/β then IC = (V CC − V BE )/R C mA

6. Using KVL:


B + V

from this:

BE + I C *R E = 0 V

I C = (V CC − V BE )/(R C + R E + R B /β) B mA

if (R C + R E )>>R B

B/β then I

C = (V CC − V BE )/(R C + R E ) mA

238

EXPERIMENT 11: DESIGN OF BJT BIAS CIRCUITS

Part 1: Collector-Feedback Configuration

a. R C = (15 − 7.5)V/5 mA = 1.5 Kohms

R C (commercial) = 1.5 ohms

d. V RC (measured) = 5.14 V

V CEQ (measured) = 7.7 V

I CQ (from measured) = 3.4 mA

β(calculated) = 104

e. The most critical values for proper operation of this design is the voltage V CEQ measured

at 7.7 V. It being within 2.7% of the design makes this a workable design.

f. R B /(β*R

B

C) = 214 K/(104*1.5 K) = 1.37

g. R F1 + R F2 = 189 K

R B (commercial) B

+ 214 K

h. No, the value of R B

B is fixed both by VCC and V BE , neither of which changed.

i. V RC (measured) = 5.64 V


I CQ (from measured) = 3.76 mA

β(calculated) = 3.73 mA/([9.27 − .7)/214 K] = 108

j. The measured voltage V CE is somewhat high due to the measured current I C being below

its design value. In general, the lowest I C which will yield proper V CE is preferable since it

keeps power losses down. For the given specifications, this design, for small signal

operation, will probably work since most likely no clipping will be experienced.

k. R B /(β*R

B

C)(calculated) = 214 K/(108*1.5 K) = 1.4

R B /(β*R

B

C)(calculated) = 1.34 (see above)

The parameters of the circuit do not change significantly with a change of transistor.

Thus, the design is relatively stable in regard to any Beta variation.

l. S(β) = 3.76 mA − 3.4 mA)/3.4 mA = .8

Part 2: Emitter-bias Configuration

a. R C (calculated) = [(V CC − (7.5 + 1.5)]V/5 mA = 1.2 K

R C (commercial) = 1.2 K

b. R E (calculated) = 1.5 V/5 mA = 300 ohms

R E (commercial) = 285 ohms

d. R B (measured) B = R1 + R 2 = 392 K

R B (commercial) B

= 394 K

239

e. V RC (measured) = 6.04 V

V CE (measured) = 7.55 V

I C (from measured) = 4.7 mA


f. All measured values are well within a 10% tolerance of the design parameters. This is

acceptable.

g. R B /(β*R

B

E) = 9.6

h. R B (calculated) B = 950 K

R B (commercial) B

= 1 M

i. Yes, it changed from 214 K to a value of 950 K. The increase in Beta was compensated

for by the increase in R B

B in such a way that ICQ, and consequently V C , V CEQ and V E

remained constant. Hence, so did R C and R E .

j. V RC (measured) = 5.2 V


I CQ (calculated) = 4.2 mA


k. The important voltage V CEQ was measured at 8.61 V while it was specified at 7.5 V. Thus,

it was larger by about 12%. This is probably the largest deviation to be tolerated. If the

design is used for small signal amplification, it is probably OK; however, should the

design be used for Class A, large signal operation, undesirable cut-off clipping may

result.

l. The magnitude of the Beta of a transistor is a property of the device, not of the circuit. All

the circuit design does is to minimize the effect of a changing Beta in a circuit. That the

Betas differed in this case came as no surprise.

m. (calculated)R B /(β*R

B

E) (2N3904) = 10.4

(calculated)R B /(β*R

B

E) (2N4401) = 9.6

n. S(β) = .66

Part 3: Voltage-divider Configuration

a. R C (calculated) = [25 − (1.5 + 7.5)]V/5 mA = 1.2 K

R C (commercial) = 1.25 K

b. R E = 1.5 V/5 mA = 300 ohms

R E (commercial) = 285 ohms

d. R 2 (calculated) = 2.94 K

R 2 (commercial) = 3.2 K

R 1 (calculated) = 17.1 K

R 1 (commercial) = 18.2 VK

240

e. V RC (measured) = 6.47 V




The difference between the calculated and the measured values of I CQ and V CEQ are

insignificant for the operation of this circuit.

f. R 1 || R 2 /(β*R E ) = .066

g. V RC (measured) = 6.98 V




h. The measured values of the previous part show that the circuit design is relatively

independent of Beta.

i. The Betas are about the same.

j. R 1 || R 2 /(β*R E ) (2N4401) = .026

R 1 || R 2 (β*R E ) (2N3904) = .066

k. S(β) = .051


1.

Table 11.1

Configuration I CQ (mA) V CEQ (V)

Collector-feedback 3.4 7.7

Emitter-bias 4.7 7.5

Voltage-divider 5.2 7.1

The critical parameter for this design is the voltage V CEQ . Given that its variation for the

various designs is less than 10%, the results are satisfying.

2.

B

Table 11.2

Configuration

Stability factors

R B /(β RC) S(β)

Collector-feedback 1.4 .8

Emitter-bias 0.6 .66

Voltage-divider .06 .051

The data in adjacent columns is consistent.

The voltage-divider bias configuration was the least sensitive to variations in Beta. This is

expected since the resistor R 2 , while decreasing the current gain of the circuit, stabilized the

circuit in regard to any current changes.

241

3. Using KVL:


B + VBE = 0 V

from which: I C = (V CC − V BE )/(R C + R B )/β) B mA

for stable operation, make: R C >>R B

B/β

4. Using KVL:

−V CC + I C /β*R B

B + IC * R E + V BE = 0 V

from which: I C = (V CC − V BE )/(R E + R B )/β) B mA

for stable operation, make: R E >>R B

B/β

5. Using KVL:

−V BB + I C /Beta*R 1 || R 2 + V BE + I C *R E = 0 V

where: V BB = R 1 /(R 1 + R 2 )*V CC

from which: I C = (V BB − V BE )/(R E + R 1 || R 2 /β) mA

for stable operation: make R E >>R 1 || R 2 /β


PSpice simulation 11-1


2. β = 170.5

3. S = 1.095

4. Yes

5. See Circuit diagram above.

242

PSpice simulation 11-2


2. β = 170.96

3. S = 0.08

4. No


6. Yes

7. Not needed

8. See circuit diagram above.

243

EXPERIMENT 12: JFET CHARACTERISTICS

Part 1: Measurement of the Saturation Current I DSS and Pinch-off Voltage V P

c. V R (measured) = .754 V

d. ID SS = 7.44 mA

e. V p (measured) = −2.53 V

f.

1. ID SS = 8.3 mA, V p = −3.1 V

2. ID SS = 9.1 mA V p = −3.9 V

It is extremely unlikely that two 2N4416 ever have the same saturation current and pinch-off

voltage.

Fig 12.1

Part 2: Drain-Source Characteristics

a. through d.

Table 12.1

V GS (V) 0 −1.0 −2.0

V DS (V) I D (mA) I D (mA) I D (mA)

0.0 0.0 0.0 0.0

1.0 4.63 2.1 .25

2.0 5.61 2.6 .28

3.0 7.32 3.06 .34

4.0 7.40 3.1 .36

5.0 7.43 3.2 .39

6.0 7.5 3.16 .42

7.0 7.5 3.31 .43

8.0 7.5 3.33 .44

9.0 7.3 3.36 .46

10.0 7.3 3.36 .50

11.0 7.1 3.36 .50

12.0 6.81 3.36 .51

13.0 6.76 3.36 .52

14.0 6.71 3.36 .53

244

Fig 12.3

ID SS (Fig 12.3) = 7.5 mA

ID SS (Part 1) = 7.44 mA

V P (Fig 12.3) = −3 V

V P (Part 1) = −2.53 V

Part 3: Transfer Characteristics

a. b.

Table 12.2

V DS (V) 3V 6V 9V 12V

V GS (V) I D (mA) I D (mA) I D (mA) I D (mA)

0 7.32 7.5 7.4 6.81

−1 3.06 3.26 3.36 3.36

−2 .34 .42 .46 .51

d. Given that the various variables in the above Table vary by less than 10%, it is reasonable

that the curves can be replaced on an approximate basis by a single curve defined by

Shockley’s equation if the average values of both ID SS and V GS(off) are used.


1. Shockley’s equation involves four parameters. Given two of them, such as I D and V GS , an

infinite number of curves can be drawn through their interception all of which can satisfy

Shockley’s equation for particular ID SS and V P .

2. V G = V P *[1 − (I D /ID SS ) 1/2 ] V

3. For: ID SS = 10 mA; V P = −5 V; and I D = 4 mA

V GS (calculated) = (−5)*[(.4) 1/2 ] = −3.16 V

a. gm O (calculated) = 2*(7.44 mA)/2.53 = 5.88 ms

b. The slope of the Shockley curve is maximum at V GS = 0 V.

c. gm(calculated) = gm O (1 − V P /V P ) = 0 S when V GS = V P .

The slope of the transfer curve at V GS = V P = 0 S

245

d.

V GS /V P = 1/4 V GS /V P = 1/2 V GS /V P = 3/4

g m 4.41 mS 2.94 mS 1.47 mS

Note: gm 0 = 5.88 mS

e. The slope is a constant value.

f. It is proportional to the derivative of Shockley’s equation.



1-4. See Circuit diagram.


Part A

4. See Probe Plot page 247.

5. I DSS = 16 mA

6. V P = −1.5 V

Part B


5. I DSS = 18.2 mA

V P = −1.4 V

246

247

248

EXPERIMENT 13: JFET BIAS CIRCUITS

Part 1: Fixed-Bias Network

b. ID SS = 12 mA

c. V P (measured) = −6 V

e. I D = 12 −3 (1 − 1/6) 1/2 = 8.33 mA

f. V RD (measured) = 8 V

I DQ (measured) = 8.2 mA

R D (measured) = 976 ohms

Part 2: Self-Bias Network

b. I DQ = 2.64 mA

V GSQ = −3.3 V

c. V GS (calculated) = −3.3 V

V D (calculated) = 12.4 V

V S (calculated) = 3.1 V

V DS (calculated) = 9.3 V

V G (calculated) = 0 V

d. V GS (measured) = −3.4 V

V D (measured) = 12.2 V

V S (measured) = 2.1 V

V DS (measured) = 9.1 V

V G (measured) = 0 V

The percent differences are determined with the calculated values as the reference.

V GS (calculated %) = 3.1%

V D (calculated %) = −1.6%

V S (calculated %) = −.64%

V DS (calculated %) = −2.3%

V G (calculated %) = 0%

Part 3: Voltage Divider-Bias Network

b. For voltage divider-bias-line see Fig. 13.2

c. I DQ (calculated) = 4.8 mA

V GS (calculated) = −2.4 V

d. V D (calculated) = 10.3 V

V S (calculated) = 5.2 V

V DS (calculated ) = 5.1 V

e. V GSQ (measured) = −2.3 V



V DS (measured) = 5.1 V

249

f. The percent differences are determined with calculated values as the reference.

V GS (calculated %) = −4.2 %

V D (calculated %) = .97%

V S (calculated %) = 1.9%

V DS (calculated %) = 1.2%

g. I DQ (measured) = 4.8 mA

I DQ (calculated %) = .4%



1. 928 μA

2. 12.96 V

3. −1.114 V

4. 13.92 mW


6. Negligible due to back bias of gate-source function

7. 12.03 mW

8. No

250


1-8. See circuit diagram.

9. No

251

EXPERIMENT 14: DESIGN OF JFET BIAS CIRCUITS

Part 1: Determining ID SS and V P

b. ID SS (measured) = 10.8 mA

c. V P (measured) = −6 V

Part 2: Self-bias Circuit Design

a. ID Q (calculated) = 5.4 mA

V DSQ (calculated) = 15 V

V DD (calculated) = 30 V

d. R S (calculated) = 1/g m = 333 ohms

R S (commercial) = 330 ohms

e. V RD = V DD − V DSQ − V RS = 30 − 15 − 1.8 = 13.2 V

R D = 2.4 K

f. V DSQ (measured) = 14.7 V


V DSQ (calculated) = 15 V

I DQ (calculated) = 5.4 mA

g. Agreements between calculated and measured values are within 10% of each other and

thus are within acceptable limits.

h. V DSQ (measured) = 13.7 V

I DQ (measured) = 6 mA

ID SS (borrowed JFET) = 9.8 mA

V P (borrowed JFET) = −5.1 V

Even though in our case, the variations between JFETs was relatively small, such may

not be the case in general. Thus, the values of the biasing resistors for the same bias

design but employing different JFETs may differ considerably.

Best is not to use the arithmetic but the geometric average for the range of ID SS and V P .

Thus in our case, the geometric averages would be:

ID SS (geometric average) =

[ID SS(min) * ID SS(max) ] 1/2 = [5 mA*15mA)] 1/2 = 8.66 mA

V P (geometric average) = [1*6] 1/2 = 2.45 V

Statistically, these values are most likely the ones encountered.

Part 3: Voltage-divider Circuit Design

a. V GS (calculated) = −2.6 V

b. R S = (V GG − V GS )/I DQ = (6 − 2.6)V/4 mA = 850 ohms

R S (commercial value) = 820 ohms

V G (calculated) = V GS + I D *R S = 2.6 + 4 mA*820 = 5.85 V

252

c. V RD (calculated) = V DD − V DSQ − V RS = 20 − 8 − 3.28 = 8.72 V

where V RS = I DQ *R S = 4 mA*820 = 3.28 V

R D = [V DD −(V DSQ + V RS )]/I D = [20 − (8 + 3.28)] = 2.18 Kohms

R D (commercial value) = 2 Kohms

d. R 2 = 10*R S = 10*820 = 8.2 Kohms

R 2 (commercial value) = 10 Kohms

Solving equation 14.3 for R 1 we obtain:

R 1 = R 2 *(V DD − V G )/V G = 10 K*(20 − 5.85)/5.85 = 24.2 Kohms


e. V DSQ (measured) = 7.9 V

ID Q (measured) = 4.2 mA

V DSQ (specified) = 8 V

I DQ (specified) = 4 mA

f. %I DQ (calculated) = 5%

%V DSQ (calculated) = −1.25%

Such relative small percent deviations are almost too good to be true.

The voltage divider bias line is parallel to the self-bias line. To shift the Q point in either

direction, it is easiest to adjust the bias voltage V G to bring the circuit parameters within

an acceptable range of the circuit design.

g. In the present case, the percent differences for I DQ and V DSQ were well within the 10%

tolerance allowed. If not, the easiest adjustment would be the moving of the voltagedivider

bias line parallel to itself by means of raising or lowering of V G . This could best

be accomplished by a change of the voltage divider R 2 /(R 1 + R 2 )*V DD . Its value

determines the voltage V G which in turn determines the Q point for the design.

h. V DSQ (measured) = 13.7 V


I DSS (borrowed JFET) = 9.8 mA

V P (borrowed JFET) = −5.1 V


1. R D (commercial value) = 2.7 Kohms


2. R D (commercial value) = 2.4 Kohms


R 1 (commercial value) = 6.8 Kohms


253

3. In the design, use the geometric mean of both the given ranges on IDSS and VP for a

given type JFET.






254


3. See above circuit diagrams.

4. %V DS = 7.37%

5. Yes

255

EXPERIMENT 15: COMPOUND CONFIGURATIONS

Part 1: Determining the BJT(β) amd JFET (I DSS and V P ) Parameters

a. R B (measured) B = 982 Kohms

R C (measured) = 2.6 Kohms

I B

B = (VCC − V BE )/R

BB =

(20 − .7)/982 K = 19.7 μA


I C = V RC /R C = 6.45/2.6 K = 2.48 mA

β(calculated) = 1.48 mA/19.7 μA = 126

Part 2: Capacitive-Coupled Multistage System with Voltage-Divider Bias

b. V B1 = 4.7 K/(4.7 K + 15 K)* 20 = 4.8 V

V E1 = 4.8 − .7 = 4.1 V

I E1 = I C1 = V E1 /R E1 = 4.1/1 K = 4.1 mA

V C1 = V CC − I C1 * R C1 = 20 − 4.1 mA*2.7 K = 9.2 V

V B2 = 2.4 K/(2.4 K + 15 K)*20 = 2.8 V

V E2 = 2.8 − .7 = 2.1 V

I E2 = I C2 = V E2 /R E2 = 2.1/470 = 4.6 mA

V C2 = V CC − I C2 * R C2 = 20 − 4.6 mA * 1.2 K = 14.5 V

Table 15.1

V B1 (V) V C1 (V) V B2 (V) V C2 (V)

Calculated Values 4.8 9.2 2.8 14.5

Measured values 4.7 9.1 2.7 14.2

% Difference −1.1 −1.1 −1.8 −2.1

As can be seen from the above data, the differences between the calculated and measured

values were much less than 10%.

f. We note that the voltages V C1 and V B2 are not the same as they would be if the voltage

across capacitor C C was 0 Volts, indicating a short circuit across that capacitor.

Part 3: DC-Coupled Multistage Systems

Use the same equations to determine the circuit parameters as in Part 2 except that V B2 =V C1 .

b. Table 15.2

V B1 (V) V C1 (V) V B2 (V) V C2 (V)

Calculated Values 4.8 9.2 9.2 13.0

Measured values 4.7 9.1 9.1 12.9

% Difference −1.7 −1.0 −1.0 −.8

Again, the percent differences between calculated and measured values are less than 10%

in every instance.

256

f. The dc collector voltage of stage 1 determines the dc base voltage of stage 2. Note that no

biasing resistors are needed for stage 2.

Part 4: A BJT-JFET Compound Configuration

b. V B

B = 4.7 K/(4.7 k + 15 k) * 30 = 7.2 V

V E = V B

B − .7 V = 6.5 V

I E = I D = 6.5 V/1.2 K = 5.4 mA

V D = V DD *R D = 30 − 5.4 mA*985 = 24.7 V

For the JFET used:

ID SS = 10.1 mA

V P = −3.2 V

determine V GS :

I D /ID SS = [1 − V GS /V P ] 1/2 mA = 5.4 mA/10.1 mA = [1 − V GS /3.2] 1/2 mA

therefore:

[5.4 mA/10.1 mA] 2 = [1 − V GS /3.2]

.286 = [1 − V GS /3.2]

from which: V GS = (1 − .286)*3.2 = −2.28 V

remember: V GS is a negative number:

V C = V B

B − VGS = 7.2 − (−2.28) = 9.5 V

Table 15.3

V B (V) V D (V) V C (V)

Calculated Values 7.2 23.6 9.5

Measured values 7.1 24.4 8.7

% Difference −.56 3.4 −8.4

d. See Table 15.3.

e. Differences were less than 10%.

f. V GS (calculated from measured values) = V B

B − VC = 7.1 − 8.7 = −1.6 V

V GS (measured) = −1.7 V

g. V RD = V DD − V D = 30 − 24.7 = 5.3 V

I D = 5.3 V/985 = 5.4 mA

I D (measured) = 6.4 mA

The percent difference between the measured and the calculated values of I D was 18.5%,

with the calculated value of I D used as the standard of reference.

V E (calculated) = 7.2 − .7 = 6.5 V

I C (calculated) = 6.5 V/1.26 K = 5.2 mA

I C (measured) = 5.06 mA

257

The percent difference between the measured and the calculated values of I C was −2.7%,

with the calculated value of I D used as the standard of reference.


1. a. There will be a change of V B

B and VC for the two stages if the two voltage divider

configurations are interchanged.

b. The voltage divider configuration should make the circuit Beta independent, if it is

well designed. Thus, there should not be much of a change in the voltage and current

levels if the transistors are interchanged.

2. Again, depending on how good the design of the voltage divider bias circuit is, the

changes in the circuit voltages and currents should be kept to a minimum.



1-11. See below.

258


1-11. See below.

259

260

EXPERIMENT 16: MEASUREMENT TECHNIQUES

Part 1: AC and DC Voltage Amplitude Measurements

DC MEASUREMENT

e. V O (calculated) = 2K/(2K + 3.9 K)*12 = 3.86 V

f. V O (measured) = 3.78 V

%Diff. (calculated) = −2%

g. V O (measured shift) = 3.8 V

The shift was down from the center of the screen.

There is almost complete agreement between the two sets of measurements.

The measurement taken with the DMM is the more accurate of the two, especially for a

DMM, since it reads to 1/100 of a volt.

AC MEASUREMENTS

h. V i(rms) (calculated) = 8/2*.707 = 2.82 V

i. V O(rms )(calculated) = [(2 K || 3.9 K + j0)*(2.82 + j0)]/(2.41 K − j1.59 K) = 1.34∠33.4 V

j. V O (measured) = 1.31 V

% diff. (calculated) = −1.51%

k. V O(p − p) (measured) = 3.72 V

l. If we convert the measured rms value of V O to peak value, we obtain 3.78 volts.

Comparing that to the measured peak value of V O which was 3.72 V, we can be satisfied

with the results.

Part 2: Measurements of the Periods and Fundamental Frequencies of Periodic Waveforms

b. Horizontal sensitivity = 100 μs/div

c. number of divisions = 5.6

d. Period(T) = 100 μs/div*5.6 div = 560 μs

e. Frequency(f) = 1/T = 1/560 μs = 1800 Hz

f. f(dial setting) = 1750 Hz

g. The dial setting on the signal generator at best can only give an approximate setting of the

frequency.

h. f(counter) = 1810 Hz

i. Indeed it is, the difference between calculated and measured values is only 10 Hz using

the counter, whereas the difference between signal generator setting and calculated values

was 50 Hz. That measurement which is closest to that of the counter is the better

measurement. In our case, the scope measures better than the signal generator.

261

Part 3: Phase-Shift Measurements

b. V i(rms) (calculated) = 6/2*.707 = 2.12 V

c. V O(rms) = (0 − j1.59 K)*(2.12 + j0)/(1k − j1.59 K) = 1.81∠−31.6 V

V O(p−p)(rms) = 1.81*1.41*2 = 5.1 V

f. A (number of divisions) = .8

g. B (number of divisions) = 10

h. angle θ(calculated) = −31.6 degrees

j. The network is a lag network, i.e., the output voltage V O lags the input voltage by the

angle theta, in our case it lags it by −31.6 degrees.

k. V R(rms) (calculated) = 1.1 V

V R(p−p) (calculated) = 3.1 V

angle theta = 58.4 degrees

The output voltage V O leads the input voltage by 58.4 degrees. Note that an angle of 58.4

degrees is the complement of an angle of 31.6 degrees.

l. V R(p−p) (measured) = 3 V

angle θ = 58 degrees

It’s a lead angle.

Part 4: Loading Effects

c. V O(p−p) (calculated) = 1 K/(1 K + 1 K)*8 = 4 V

d. V O(p−p) (measured) = 3.98 V

f. V O(p−p) (calculated) = 1 M/(1 M + 1 M)*8 = 4 V

V O(p−p) (measured) = 2.7 V

g. The real part of the input impedance of the scope is now in parallel with the R2 resistor

and since for many scopes, that real part is about 1 Mohm, therefore, R scope || R 2 = 500

kohms.

Thus, V O is considerably reduced.

h. R (prime) = 1 M/[Vi/V O − 1] = 1 M/[8/2.7 − 1] = 588 kohms

R (scope) = −R (prime) *R 2 /[R (prime) − R 2 ] = 1.43 Megohms

Most general purpose oscilloscopes have an input impedance consisting of a real part of

1 Megohms in parallel with a 30 pf capacitor. The result obtained for the real part of that

impedance is reasonably close to that.

i. V O(p−p) (calculated) = 1 K/(1 K + 1 M)*8 = 8 mV

j. V O(p−p) (measured) = 7.9 mV

k. The results agree within 1.25 percent.

262


1. No. for the frequency of operation, the capacitor represents an impedance of 1.59k∠−90

ohms. Therefore, in relationship to the existing resistors in the circuit, it cannot be

neglected without making a serious error.

2. It depends upon the waveform. In case of sinusoidal voltages, the advantage is probably

with the DMM. For more complex waveforms, the nod goes to the oscilloscope.

3. For measuring sinusoidal waves, the DMM gives a direct reading of the rms value of the

measured waveform. However, for non-sinusoidal waves, a true rms DMM must be

employed. The oscilloscope only gives peak-peak values, which, if one wants to obtain

the power in an ac circuit, must be converted to rms.

4. T = 5 div*.1 ms/div = .5 ms

f = 1/T = 1/.5 ms = 2 KHz

5. angle theta = 1.5/8*360 = 67.5 degrees

V O /V i = R′/(R′ + R 1 )

therefore: V i /V O = (R′ + R 1 )/R′

solving for R′: R′(V i /V O ) = R′ + R 1

R′(V i /V O − 1) = R 1

Hence:

R′ = R 1 /(V i /V O − 1) ohms







5. 33.74°

6. V out


10 See Probe plot page 266.

11. V in(rms) = 2.84 V

V out(rms) = 1.32 V

12. Yes



15.

R3 2 K

V out = (12 V) = (12 V)

R2 + R3 (212 + 3.9 K)

= 4.067 V

16. Agree

263

264

265

266

267


1. Using VOM, R2 = 100 kΩ

2. Using DMM, R2 = 1 kΩ

3. For R2 = 1 kΩ

4. Both circuits

5. No

268

EXPERIMENT 17: COMMON-EMITTER TRANSISTOR AMPLIFIERS

Part 1: Common-Emitter DC Bias

b. V BB = R 2 /(R 1 + R 2 ) * V CC = 10 K/(10 K + 33 K) * 10 = 2.33 V

V E = V BB − .7 = 1.63 V

V C = V CC − I C * R C = 10 − 1.63 mA * 3 K = 5.1 V

I E = V E /R E = 1.63/1 K = 1.63 mA

r e = 26 mV/I E = 26 mV/1.63 mA = 16 ohms

c. V B (measured) B = 2.25 V

V E (measured) = 1.57 V


I E = V E /R E = 1.57/978 = 1.6 mA

r e = 26 mV/1.6 mA = 16.2 ohms

The two values for r e obtained are within .2 ohms.

This represents a 1.25 percent difference.

Part 2: Common-Emitter AC Voltage Gain

a. A V (no load) = −RC/re = 3.2 K/16 = 198

b. V sig = 8.3 mV(rms)

V O (no load) = 1.47 V (rms)

A V (no load) = 177

The two values of A V agree within 10.6 percent of each other.

Part 3: AC Input Impedance, Z i

Z in = R 1 || R 2 || Beta * r e = 10 K || 33 K || (150 * 16) = 1.8 Kohms

V i (measured) = 12 mV (rms)

V sig = 20 mV(rms)

Z in = [12 mV/(20 mV − 12 mV)] * 1 K = 1.5 Kohms

The two values of the input impedance were within 18.9% of each other. This relatively large

divergence is in part the result of using an assumed value of Beta for our transistor. For a 2N3904

transistor, the geometric average of Beta is closer to 126.

Part 4: Output Impedance

a. Z O (calculated) = R C = 3.2 Kohms

b. V sig(rms) = 10 mV (rms)

V O (no load)(rms) = 1.8 V (rms)

V O (loaded)(rms) = .913 V(rms)

R L = 3.2 Kohms

Z O = [(V O − V L )/V L ] * R L = [(1.8 − .913)/.913] * 3.2 K = 3.1 K

The two values for Z O are within 3.15% of each other.

269

Part 6: Computer Analysis



2. r e = 6.93 Ω



5. 180°

6. As I(B) increases, so does I(C).

As I(C) increases, so does V(RC) and V(RE). Therefore V(C) decreases.

7. Z in (theoretical) = 937.3 Ω



10. Z in (PSpice) = 1.1323 k ≈ Z in (theoretical)

Determining output impedance

1. Z out ≈ RC = 3 k



4. Z out (PSpice) = 2.6392 k ≈ RC

270

271

272

273

EXPERIMENT 18: COMMON-BASE AND EMITTER-FOLLOWER

(COMMON-COLLECTOR TRANSISTOR AMPLIFIERS

Part 1: Common-Base DC Bias

a. V B (calculated) B = 10 K/(10 K + 33 K) * 10 = 2.33 V

V E = V B

B − .7 V = 1.63 V

I E = I C = V E /R E = 1.63 V/1 K = 1.63 mA

V C = 10 − I C * R C = 10 − (1.63 mA) * 3 K = 5.1 V

r e = 26 mV/I E = 26 mV/1.63 mA = 16 ohms

b. V B (measured) B = 2.26 V



I E (from measured values) = V E /R E = 1.57 V/978 = 1.6 mA

r e (from measured values) = 26 mV/I E = 26 mV/1.6 = 16.3 ohms

In every case, the differences between the two sets of values are less than 10% apart.

Such divergence is not excessive given the variability of electronic components.

Part 2: Common-Base AC Voltage Gain

a. A V (calculated) = R C /r e = 3.2 K/16.3 = 197

b. V sig = 50 mV

V O = 2.43 V

A V = 2.43/V sig = 2.43/.05 = 122

The two gains differed by −38 percent with the calculated gain used as the standard of

comparison.

Part 3: CB Input Impedance, Z i

a. Z i = r e = 16.3 ohms

b. V sig = 50 mV

V i = 9.9 mV

R X = 100 ohms

Z i = [V i /(V sig − V i )] * R X = [9.9 mV/(50 mV − 9.9 mV)] * 100 = 23.7 ohms

The two values of the input impedance differed by 45 percent with the theoretical value

of r e (16.3 ohms) used as the standard of comparison.

Part 4: CB Output Impedance, Z O

a. Z O = R C = 3.2 K

b. V sig = 20 mV

V O (measured, no load) = 2.43 V

V L (measured, loaded) = 1.22 V

Z O = [(V O − V L )/V L ] * R L = [(2.43 − 1.22)/1.22] * 3 K = 3.18 Kohms

The agreement between the two values of the output impedance is within less than 1

percent.

274

Part 5: Emitter-Follower DC Bias

a. V B (calculated) B = 2.33 V

V E (calculated) = 1.63 V

I E (calculated) = 1.63 V

V C (calculated) = 10 V

r e (calculated) = 26 mV/I E = 26 mV/1.63 mA = 16 ohms




I E = V E /R E = 1.78 V/1 K = 1.78 mA

r e = 26 mV/1.78 mA = 14.3 ohms

Part 6: Emitter-Follower AC Voltage Gain

a. A V = R E /(R E + r e ) = 1 K/(1 K + 14.3) = .986

b. V sig = 1 V

V O (measured) = .987 V

A V = V O /V sig = .987/1 = .987

The two values of gain are within .1 percent of each other.

Part 7: Emitter Follower (EF) Input Impedance, Z i

a. Z i = R 1 || R 2 || (Beta * (1 K + r e ) = 7.31 Kohms

b. V sig = 2 V

R X = 10 Kohms

f = 1 KHz

V i (measured) = .85 V

Z i = [V i /(V sig − V i )] * R X = [.85/(2 − .85)] * 10 K = 7.39 Kohms

The input impedance calculated from measured values is within 1.1 percent of the

theoretically calculated value of Z i .

Part 8: Emitter Follower (EF) Output Impedance, Z O

a. Z O = r e = 16 ohms

b. V O (measured) = 19.8 mV

V L (measured) = 11.2 mV

Z O = [(V O − V L )/V L ]*R L = [(19.8 mV − 11.2 mV)/11.2 mV] * 100 = 76.8 ohms

In the theoretical formulation, Z O was equated with r e , this is an approximation. A better

expression for the output impedance is: Z O = r e + (R G || R 1 || R 2 )/Beta. Thus it can be seen

that the given formulation was actually a minimum value of the output impedance.

275

Part 9: Computer Analysis


Bias Point Analysis



4. r e = 16.71 Ω

5. A v = 179.53

6. Z in = r e = 16.71 Ω

7. Z out = 3 kΩ

Transient Analysis


2. 38°

4. A v = 141.59, see Probe plot page 278.

Input Impedance

1. Z in = 20.7 Ω, see Probe plot page 279.

Output Impedance

1. Z out = 2.87 kΩ, see Probe plot page 280.

276

277

278

279

280


Bias Point Analysis



4. r e = 16.65 Ω

5. A v = 0.98

6. Z in = 7.31 kΩ

7. Z out ≅ r e = 16.65 kΩ

Transient Data


2. 0.0°

4. 0.981

Input Impedance

1. 7.35 kΩ

Output Impedance

1. 58.63 Ω

281

282

EXPERIMENT 19: DESIGN OF COMMON-EMITTER AMPLIFIERS

Part 2. Computer Analysis



3. β = 139.6

4. V CE = 4.78 V

5. Yes

Transient Analysis

1. A V = 147.9

2. Yes

4. Z in = 2.78 kΩ

5. Yes

8. Z out = 3.893 kΩ

9. Yes

Part 3: Build and Test CE Circuit


V E (measured) = .87 V


I C = I E = V E /R E = .87 V/979 = .89 mA

r e = 26 mV/I E = 26 mV/.89 mA = 29.3 ohms

c. V sig = 10 mV

V L (measured) = .815 V

A V = (R C || R L )/r e = (3.2 K || 10.2 K)/29.3 = 80.7

d. V sig = 20.5 mV

R X = 3.17 Kohms

V i (measured) = 8.8 mV

Z i = (R 1 || R 2 || Beta*r e ) = (100.2 K || 21.6 K || 100 * 29.3 = 2.4 Kohms

e. V O (measured) = 1.08 V

Z O = (V O − V L )/V L *R L = (1.08 − .82)/.82 * 10.2 K = 3.25 Kohms

283

f. Design parameter Measured value

A V 100 min. 80.7

Z i (Kohms) 1 Kmin. 2.38 K

Z O (Kohms) 10 Kmax. 3.35 K

V O max(p−p) 3 Vp−p min. 7.1 Vp−p

The design of the circuit was successful with all parameters, but the gain, meeting and

even exceeding the design specification. The gain is about 20 percent below the expected

value. To increase it, the supply voltage V CC could be increased. This would increase the

quiescent current, lower the dynamic resistance r e and consequently increase the gain of

the amplifier.

284

EXPERIMENT 20: COMMON-SOURCE TRANSISTOR AMPLIFIERS

Part 1: Measurement of I DSS and V P

a. I DS = 9.1 mA

b. V P = −2.9 V

Part 2: DC Bias of Common-Source Circuit

a. V GS = −1.33 V

I D = 2.55 mA

V D = V DD − I D * R D = 20 − 2.55 mA * 2.2 K = 13.8 V

c. V G (measured) = 0 V



V GS (measured) = −1.37 V

I D = V D /R S = 13.8/488 = 2.99 mA

The agreement between calculated and measured values was in most cases within 10

percent of each other, the exception being the 17.3 percent difference between the

calculated and measured value of I D .

Part 3: AC Voltage Gain of Common-Source Amplifier

a. A V = −g m R D

where

g m = I DSS /(2 * | V P |) * (1 − V GS /V P ) 2 = 2 * 9.1 mA/2.9 * (1 − 1.33/2.9) = 3.4 mS

therefore: A V = −3.4 mS * 2.2 K = 7.48

b. V sig = 10 mV

f = 1 KHz

V O (measured) = 758 mV

A V = V O /V sig = 758 mV/100 mV = 7.58

The difference between the theoretical gain and the gain calculated from measured values

was only 1.34 percent.

Part 4: Input and Output Impedance Measurements

a. Z i = R G

Z i (expected) = 1 Megohm

b. Z O = R D

Z O (expected) = 2.25 Kohms

c. V i (measured) = 37.2 mV

Z i (calculated) = V i * R X (V sig − V i ) = 592 Kohms

285

d. V O (measured) = 760 mV

R L (measured) = 9.9 Kohms

V L (measured) = 620 mV

Z O = (V O − V L ) * R L /V L = (760 mV − 620 mV) * 9.9 K/620 mV = 2.24 Kohms

The infinite input impedance of the JFET is predicated upon the assumption of the zero

reverse gate current. Such may not be entirely true. Hence, we observe a 41 percent

difference between the theoretical input impedance and the input impedance calculated from

measured values.

The two values of the output impedance are in far better agreement. They differ only by .44

percent.




3. g mo = 21.15 mS; g m = 7.48 mS

4. A v = −17.9

6. A v = −19.47

8. Z in = 954.64 kΩ

11. Z out = 2.34 kΩ

13. A v

14. A R = 4 kΩ = 21.93

v

D

286

EXPERIMENT 21: MULTISTAGE AMPLIFERS: RC COUPLING

Part 1: Measurement of I DSS and V P

I DSS = 10.4 mA

V P = −3.2 V

Part 2: DC Bias of Common-Source Circuit

a. V GS1 (calculated) = −1.36 V

I D1 (calculated) = 3.1 mA

V D1 (calculated) = V DD − I D1 * R D1 = 20 V − 3.1 mA * 2.2 K = 13.2 V

V GS2 (calculated) = −1.38 V

I D2 (calculated) = 3.54 mA

V D2 (calculated) = V DD − I D2 * R D2 = 20 V − 3.54 mA * 2.2 K = 12.2 V

c. V G1 (measured) = 0 V

V S1 (measured) = 1.49

V D1 (measured) = 13.81 V

V GS1 (measured) = −1.04 V

I D1 = V S1 /R S1 = 1.49 V/496 = 3 mA

V G2 (measured) = 0 V

V S2 (measured) = 1.52 V

V D2 (measured) = 11.3 V

V GS2 (measured) = −.8 V

I D2 = V S2 /R S2 = 1.52 V/468 = 3.25 mA

The theoretical and the measured bias values were consistently in close agreement.

Part 3: AC Voltage Gain of Amplifier

a. For stage 2:

A V2 = −g m2 (R D2 || R L ) = (−3.64 mS)(2.2 K || 10 K) = 6.6

For stage 1:

A V1 = −g m1 (R D1 || Z i2 ) = (−3.51 mS)(2.2 K || 1 M) = 7.72

note: Z i2 = R G2 = 1 Megohm

A V = A V1 * A V2 = 6.6 * 7.72 = 50.7

b. V sig (measured) = 20 mV


A V = V L /V sig = 945 mV/20 mV = 47.3

V O1 (measured) = 145 mV

V sig (measured) = 20 mV

A V1 = V O1 /V sig = 145 mV/20 mV = 7.25

A V2 = V L /V O1 = 945 mV/145 mV = 6.52

The voltage gains differed by less than 10 percent from each other.

287

Part 4: Input and Output Impedance Measurements

a. Z i = R G1 = 1 Megohm

b. Z O = R D2 = 2.2 Kohms

c. V i1 (measured) = 7.5 mV

V sig = 20 mV

R X = 1 Megohm

Z i = V i1 * R X /(V sig − V i1 ) = 7.5 mV * 1 M/(20 mV − 7.5 mV) = 600 Kohms

d. V L (measured) = 330 mV


Z O = (V O − V L ) * R L /V L = (410 mV − 330 mV) * 10 K/330 mV = 2.42 Kohms

Again, the input impedance calculated from measured values is about 40 percent below

that which we expected from the assumption that the JFET was ideal and had no reverse

gate current. This seems not to be the case in actuality. There is a reverse leakage current

at the gate which reduces the effective input impedance below that of R G by being in

parallel with it.

The output impedances again are in reasonable agreement, differing by no more than 9

percent from each other.


Pspice Simulation 21-1


3. g mo = 21.15 mS

g mJ1 = 7.48 mS

g mJ2 = 7.48 mS

4. A v1 = 17.95 A v2 = 7.48

6. A v1 = 19.498

288

7. A v2 = 8.275

10. (A v1 )(A v2 ) = 161.35

11. (A v1 )(A v2 ) = 161.35

14. Yes

16. Interchange J1 with J2

17. Z in = 956.89 kΩ

20. Z out = 989.74 Ω

289

EXPERIMENT 22: CMOS CIRCUITS

Part 1: CMOS Inverter Circuit

Table 22.1 Table 22.2

IN OUT IN OUT

0V 5V 0V 5V

5V .3V 5 V .3 V

Part 2: CMOS Gate

Table 22.3

A B OUTPUT

0 V 0 V 5 V

0 V 5 V 0 V

5 V 0 V 0 V

5 V 5 V 0 V

Part 3: CMOS Input-Output Characteristics

a.

IN (V) 0.0 .2 .4 .6 .8 1.0 1.2 1.4 1.8

OUT (V) 5.0 5.0 5.0 5.0 4.9 4.8 4.8 4.7 4.4

IN (V) 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6

OUT (V) 3.9 3.4 1.6 1.1 .75 .6 .4 .3

IN (V) 3.8 4.0 4.2 4.4 4.6 4.8 5.0

OUT (V) .1 .1 .08 .02 .02 .005 0



2. No

VPlot data


290

291

292

EXPERIMENT 23: DARLINGTON AND CASCODE AMPLIFIER CIRCUITS

Part 1: Darlington Emitter-Follower Circuit

a. V B (calculated) B = 2.21 V


A V = R E /(R E + r e ) = 47/(47 + 10) = .83



I B (calculated) B = 199 μA

I E (calculated) = 106 mA

β(calculated) = 106 mA/199 μA = 535

c. V i (measured) = 350 mV


A V = V O /V i = 340 mV/350 mV = .97

Part 2: Darlington Input and Output Impedance

a. Z i (calculated) = 20.6 K || (535 * 47) = 11.3 Kohms

Z O = r e + (R G || R B

B)/β = 9 ohms

b. V sig = 500 mV

V i (measured) = 55.6 mV

Z i = [V i /(V sig − V i ) * Rx = [55.6 mV/(500 mV − 55.6 mV)] * 100 K = 12.5 Kohms

c. V O (measured) = 492 mV


R L = 100 ohms

Z O = [(V O − V L )/V L ] * R L = [(492 mV − 476 mV)/476 mV] * 100 = 4.2 ohms

The two values of the input impedance differed by about 10.6 percent while the two

values of the output impedance differed by 53 percent. It is to be noted however that with

such small values the difference in just one ohm manifests itself as a large percent

change. Given the tolerances of electronic circuit due to their components and that of the

Darlington chip, the results are quite satisfactory.

Part 3: Cascode Circuit

a. V B1 (calculated) = 5.5 V

V E1 (calculated) = 4.8 V

V C1 (calculated) = 11 V

V B2 (calculated) = 12 V

V E2 (calculated) = 11.3 V

V C2 (calculated) = 12.5 V

I E1 = V E1 /R E1 = 4.8 V/1 k = 4.8 mA

I E2 = 11.3/1.8 K = 6.24 mA

r e1 = 26 mV/I E1 = 26 mV/4.8 mA = 5.4 ohms

r e2 = 26 mV/I E2 = 26 mV/6.24 mA = 4.2 ohms

b. V B1 (measured) = 4.69 V

V E1 (measured) = 4.0 V

293

V C1 (measured) = 10.7 V

V B2 (measured) = 12.0 V

V E2 (measured) = 10.5 V


I E1 (calculated) = V E1 /R E1 = 4 V/1 K = 4 mA

I E2 (calculated) = V E2 /R E2 = 10.5/1.8 K = 5.2 mA

r e1 = 26 mV/I E1 = 26 mV/4 mA = 6 ohms

r e2 = 26 mV/I E2 = 26 mV/5.2 mV = 5 ohms

c. A V1 = −1 (as per equation 23.5)

A V2 = R C /r e2 = 1.8 K/5 = 360

d. V sig = 10 mV

V i (measured) = 8 mV

V O2 (measured) = 7.91 mV

V O1 (measured) = 948 mV

A V1 (calculated) = −V O1 /V i = 7.91/8 mV = −.98

A V2 (calculated) = V O2 /V O1 = 948 mV/7.91 mV = 120

A V = V O2 /V i = −948 mV/8 mV = −119

The voltage gains for stage 1 were within 2 percent of each other, while the overall

theoretical gain of 180 differs from the calculated gain from measured values by 34 percent.




2. r e = 249 Ω




6. A V = 0.787

9. Z in = 47.123 kΩ

Z out = 2.04 kΩ

294

295



2. r eQ1

= 5.63 Ω r eQ 2

= 5.6 Ω



8.

RC

1.8 k

For Q 1 ; A V = =

r 5.63 Ω = 319

re

For Q 2 ; A V = =

r

e

e

5.6

5.6

Ω

Ω = 1 296

297

298

EXPERIMENT 24: CURRENT SOURCE AND CURRENT MIRROR CIRCUITS

Part 1: JFET Current Source

a. V DS (measured) = 9.64 V

b. I L = (V DD − V DS )/R L = (10 − 9.64)/51.2 = 7.03 mA

c. Table 24.1

R L (ohms) 20 51 82 100 150

V DS (Volts) 9.88 9.64 9.44 9.34 8.85

I L (mA) 6.1 7.03 6.83 6.60 7.57

Part 2: BJT Current Source

a. I L = 1.9 mA

b. V E (measured) = −.68 V

V C (measured) = .404 V

c. I E (calculated) = 2.13 mA


d.

Table 24.2

R L (kohms) 3.6 4.3 5.1

V E (Volts) −.68 −.67 −.68

V C (Volts) 3.03 1.74 .404

I E (mA) 2.14 2.17 2.13

I L (mA) 1.94 1.92 1.88

Part 3: Current Mirror

a. I X = .9 mA

b. V B1 = .669 V

V C2 = 2.24 V

I X = .89 mA

I L = 1.0 mA

c. I X (calculated) = 1 mA

V B1 (measured) = .669 V


I X = .9 mA

I L = 1.5 mA

299

Part 4: Multiple Current Mirrors

a. I X (calculated) = 1 mA

b. V B1 (measured) = .672 V



I X = 1.01 mA

I L1 = 1.58 mA

I L2 = 1.78 mA

c. I X (calculated) = 1 mA

V B1 (measured) = .672 V



I X = 1.02 mA

I L1 = 1.73 mA

I L2 = 1.44 mA




2. I(R X ) = 933.6 μA

I(R L ) = 1.020 mA

3. Yes

300


5. I(R X ) = 933.6 μA

I(R L ) = 991.3 μA

6. Yes

8. Yes

10. Yes

11. No

12. Yes

301

EXPERIMENT 25: FREQUENCY RESPONSE OF COMMON-EMITTER AMPLIFIERS

Résumé

f L,1 = 1/(2 * 3.24 * 1.39 K * 10 μf) = 11.5 Hz

f L,2 = 1/(2 * 3.24 * 6.1 K * 1 μf) = 26 Hz

f L,E = 1/(2 * 3.14 * 2.2 K * 20 μf) = 3.62 Hz

f H,i = 1/(2 * 3.14 * 1.68 K * 960 μf) = 98.7 KHz

f H,O = 1/(2 * 3.14 * 1.43 K * 45 pf) = 2.43 MHz

Part 1: Low-Frequency Response Calculations

a. C be (specified) = 100 pf

C bc (specified) = 10 pf

C ce (specified) = 15 pf

C W,i (approximated) = 20 pf

C W,o (approximated) = 30 pf

b. β(measured) = 126

c. V B

B (calculated) = 4.08 V



I E (calculated) = 1.54 mA

r e = 26 mV/I E = 26 mV/1.54 mA = 16.9 ohms

d. A V (mid) = (R C || R L )/r e = (3.9 K || 2.2 K)/16.9 = 83.2

e. f L,1 (calculated) = 11.5 Hz

f L,2 (calculated) = 26.2 Hz

f L,E (calculated) = 3.62 Hz

Part 2: Low Frequency Response Measurements

b. V sig (measured) = 30 mV


A V (mid) = 70

Table 25.1

f (Hz) 50 100 200 400 600 800 1 K 2 K 3 K 5 K 10 K

V O(p−p) .4 .5 .9 1.6 1.8 1.9 2.0 2.1 2.1 2.1 2.2

Table 25.2

f (Hz) 50 100 200 400 600 800 1 K 2 K 3 K 5 K 10 K

A V 13.2 16.7 30 53.3 60 63.3 66.7 70 70 70 73.3

Part 3: High Frequency Response Calculations

a. f H,I (calculated) = 98.7 KHz

f H,O (calculated) = 2.47 MHz

302

b. Table 25.3

f (KHz) 10 50 100 300 500 600 700 900 1000 2000

V O(p−p) 2.2 2.2 2.1 1.9 1.6 1.5 1.4 1.3 1.3 .8

Table 25.4

f (KHz) 10 50 100 300 500 600 700 900 1000 2000

A V 73 73 70 63 53 20 46 40 40 27

Part 4: Plotting Bode Plot and Frequency Response

Fig 25.2

from plot: f 1 = 400 Hz

f 2 = 500 Hz




2. r e = 17.2 Ω; A V mid = 81.3

4. See Probe plot 346.

5. Almost identical



8. 20 log(78.028) = 37.84 both gains agree

303

304

305

EXPERIMENT 26: CLASS A AND CLASS B POWER AMPLIFIERS

Part 1: Class A Amplifier: DC Bias

a. V B

B (calculated) = 1.53 V

V E (calculated) = .83 V

I E (calculated) = I C = V E /R E = .83/20 = 41 mA

V C (calculated) = 5.1 V

b. V B

B (measured) = 1.59 V

V E (measured) = .88 V


I E (calculated) = I C = V E /R E = .88/20 = 44 mA

Part 2: Class-A Amplifier: AC Operation

a. P i (calculated) = 400 mW

V O (calculated) = 5.3 Vp−p

P O (calculated) = 29.3 mW

% efficiency (calculated) = 7.3 percent

b. V i (measured) = 65 mV

V o (measured) = 5 Vp−p

c. P i = 400 mW

P O = 26 mW


While the values for the power and the efficiency are fairly consistent between the

theoretical and those calculated from measured values, the low efficiency of the amplifier

is an undesirable feature. In general, Class A amplifiers operate close to a 25 percent

efficiency. This circuit would need to be redesigned to make it a practical circuit.

d. V i (measured) = 32.5 mVp−p

V O (measured) = 3 Vp−p

e. P i (calculated) = 400 mW

P O (calculated) = 9.38 mW


f. P i = 400 mW

P O = 93 mW

% efficiency = 2.3 percent

As stated previously, while the data is consistent, the values of the efficiency makes this

not a practical circuit.

306

Part 3: Class-B Amplifier Operation

a. for V O = 1 V peak

P i (calculated) = 1.59 W

P O (calculated) = 50 mW


for V O = 2 V peak

P i = 1.59 W

P O = 200 mW


b. V i (measured) = 2.9 Vp−p

V O (measured) = 2.7 Vp−p

P i = 890 mW

P O = 91 mW

% efficiency = 10.2%

c. V i (measured) = 5 Vp−p

V O (measured) = 4 Vp−p

I dc (measured) = 159 mA

P i = 1.27 W

P O = 637 mW

% efficiency = 50.2%

Note that the efficiency of the Class B amplifier increases with increasing input signal

and consequent increasing output signal. Also observe that the two stages of the Class B

amplifier shown in Figure 26.2 are in the emitter follower configuration. Thus, the

voltage gain for each stage is near unity. This is what the data of the input and the output

voltages show. Note also, that as the output voltage approaches its maximum value that

the efficiency of the device approaches its theoretical efficiency of about 78 percent.




307

2. Yes.

3. V CE = 5.709 V − 0.719 V = 4.98 V

4. 442.7 mW

5. Q 1


8. Yes

9. No

10. 180°

11. P o (AC) = 4.59 mW

12. %η = 1.04%

13. DC values remain the same

P o (AC) = 1.16 mW

%η = 0.26%

See Probe plot page 310.

1 (10 V)

2

308

309

310



P i (DC) = 372.1 mW

2. Q 1 and Q 2

3. Yes.

4. V(E2) = 4.947 V

1 (10 V)

2

5. V(BE) Q1 = 0.77 V

V(BE) Q2 = −0.81 V

6. Maintain proper bias across Q 1 and Q 2 .

7. 0.7 V


10. V(OUT) p−p = 4.077 V

11. %η = 55.8%

14. V(OUT) p−p = 2.187 V

%η = 16.1%

311

312

EXPERIMENT 27: DIFFERENTIAL AMPLIFIER CIRCUITS

Part 1: DC Bias of BJT Differential Amplifier

a. V B

B (calculated) = 0V

V E (calculated) = −.7 V

V C (calculated) = 5.43 V

I E (calculated) = 457 μA

r e (calculated) = 57 ohms

B

b. Q1 Q2

V B (measured) −.10 V 0 V

V E (measured) −.65 V −.65 V

V C (measured) 5.10 V 4.9 V

I E (calculated) 490 μA 510 μA

r e 53 ohms 51 ohms

Part 2: AC Operation of BJT Differential Amplifier

a. A V,d (calculated) = 179

A V,c (calculated) = .5

b. V O1 (measured) = 1.48 V

V O2 (measured) = 1.43 V

V O,d = (V O,1 + V O,2 )/2 = (1.48 + 1.43)/2 = 1.46 V

A V,d = V O,d /V i = 72.8

c. V O,c (measured) = .55 V

A V,c = V O,c /V i = .55

Part 3: DC Bias of BJT Differential Amplifier with Current Source

a. For either Q1 or Q2:

V B

B (calculated) = 0 V

V E (calculated) = −.7 V


I E (calculated) = .5 mA


For Q3:

V B

B (calculated) = −5 V

V E (calculated) = −5.7 V

V C (calculated) = −.7 V

I E (calculated) = 1 mA


313

b. For Q1, Q2, and Q3:

B

Q 1 Q 2 Q 3

V B (measured) 47 mV 0 mV −4.69 V

V E (measured) −.64 V −.64 V −5.35 V

V C (measured) 7.91 V 2.97 V −.64 V

I E (calculated) 110 μA 612 μA 783 μA

r e (calculated) 236 ohms 42.5 ohms 33.2 ohms

Part 4: AC Operation of Differential Amplifier with Transistor Current Source

a. A V,d = R C /(2 * r e ) = 10 K/(2 * 57.8) = 173

Part 5: JEFT Differential Amplifier

a. For Q1: I DSS = 7.9 mA

V P = −3.1 V

For Q2:

For Q3:

I DSS = 8.1 mA

V P = −3.4 V

I DSS = 11.2 mA

V P = −4.2 V

b. V D,1 (calculated) = 9.84 V

V D,2 (calculated) = 9.84 V

V S,1 (calculated0 = .845 V

c. V G,1 (measured) = 0 V

V D,1 (measured) = 9.71 V

V D,2 (measured) = 9.72 V

V D,3 (measured) = .84 V

d. A V,d = .184

e. V O,1 (measured) = 50 mV

V O,2 (measured) = 46 mV

A V1,d = .5

A V2,d = 4.6

314


Pspice Simulations 27-1


P(DC) VCC = 9.283 mW

P(DC) VEE = 9.356 mW

2. Practically yes

3. VCE ( ) = VCE ( ) = 6 V

Q1 Q2

4. Yes.

5. I(Q 1 ) = 464.2 μA

I(Q 2 ) = 464.2 μA

6. Yes


10. (VOUT1) p−p = (VOUT2) p−p = 3.23 V

phase shift = 180°


A V = 114


15. (VOUT1) p−p = (VOUT2) p−p = 0.98 V

phase shift = 0°


A V = 0

315

316

317

318

319

Pspice Simulations 27-2


P(DC) VCC = 9.853 mW

P(DC) VEE = 14.97 mW

2. VC ( ) Q1

= 5.074 V

VC ( ) Q2

= 5.074 V

Yes

3. I(Q 1 ) = 492.6 μA

I(Q 2 ) = 492.6 μA

I(Q 3 ( = 993.0 μA

IQ ( ) = IQ ( )

4.

1 2

IQ (

3) = 2 IQ (

1) = 2 IQ (

2

)

320


8. Both voltages are 1.7602 V p−p

phase shift = 180°


A V = 125


13. 1.6 mV p−p phase shift = 0°

14. See probe plot page 325.

A V = 0

321

322

323

324

325

EXPERIMENT 28: OP-AMP CHARACTERISTICS

Part 1: Determining the Slew Rate

f. 5 V p-p

g. 12 us

h. 0.41 V/us

Part 2: Determining the Common Mode Rejection Ratio

g. Vout(rms) = 0.263 V Vin(rms)= 8.7 V

h. A(cm) = V(out)/V(in) = 0.0302

i. A(dif)=R1/R2=1000

j. CMR(dB)= 90.4 dB

k. Published values: 90-95 dB


PSpice Simulation: Determining the Slew Rate

b. V(Vout) max= 5 V V(Vout)min= 0 V

c. Time interval = 12 us

d. SR= 0.40 us

e. Published values: 0.3-0.7 us

PSpice Simulation: Determining the Common Mode Rejection Ratio

b. A(cm) = V(out)/V(in) = 0.26 V/8.7 V = 0.03

c. A(dif) = R1/R2 = 1000

d. CMR(dB) = 90.4

e. Published values: 90-95 dB

326

EXPERIMENT 29: LINEAR OP-AMP CIRCUITS

Part 1: Inverting Amplifier

a. V O /V i (calculated) = −R O /R i = 100 K/20 K = −5

b. V O (measured) = −4.87

A V = −V O /V i = 4.87/1 = −4.87

c. V O /V i (calculated) = −R O /R i = −100 K/100 K = −1


A V = −V O /V i = −1.06/1 = −1.06

d.

Fig 29.6

Part 2: Noninverting Amplifier

a. A V (calculated) = (1 + R O /R i ) = (1 + 100 K/20 K) = 6

b. V O (measured) = 5.24 V

A V = V O /V i = 5.25/1 = 5.25

The two gains are within 12.5 percent of agreement.

c. A V (calculated) = (1 + 100 K/100 K) = 2


V O /V i = 2.17

The two gains are within 8.5 percent of agreement.

Part 3: Unity-Gain Follower

a. V i (measured) = 2.06 V


The ratio of the computed gain from measured values is equal to .995, which is

practically identical to the theoretical unity gain.

327

Part 4: Summing Amplifier

a. V O (calculated) = −[100 K/100 K * 1 + 100 K/20 K * 1] = −6 V

b. V O (measured) = −5.02 V

The difference between the two values of V O is equal to 16.3 percent.

c. V O (calculated) = −[100 K/100 K * 1 + 100 K/100 K*1] = −2 V

V O (measured) = −2.01 V

The difference between the two values of V O is equal to .5 percent.




2. (VOUT) peak = 5 V

(VIN) peak = 1 V

3.

Vo

Rout

A V = =− = −5

V R

4.

VOUT 5 V

=− = −5

VIN 1 V

5. Yes

6. 180°

7. Yes

in

in

328

29-1

329



2. (VOUT) peak = 6 V

(VIN) peak = 1 V

3.

V ⎛

o

R ⎞

out ⎛ 100 kΩ

⎞

= ⎜1+ ⎟= ⎜1+

⎟

Vin

⎝ Rin

⎠ ⎝ 20 kΩ

⎠

4.

V

out

6 V = = 6

Vin

1 V

5. Yes

6. 0°

7. Yes

= 6

330

29-2

331

EXPERIMENT 30: ACTIVE FILTER CIRCUITS

Part 1: Low-Pass Active Filter

a. f L (calculated) = 1/(2 * 3.14 * 10 K * .001 μF) = 15.9 KHz

b.

Table 30.1 Low Pass Filter

f (Hz) 100 500 1 K 2 K 5 K 10 K 15 K 20 K 30 K

V O (V) 1.0 1.0 1.0 .99 .95 .85 .74 .59 .52

c. Fig 30.4

d. f L (from graph) = 15 KHz

Part 2: High-Pass Filter

a. f H = 1/(2 * 3.14 * R 2 * C 2 ) = 1/(2 * 3.14 * 10 K*.001 μF) = 15.9 KHz

b. Table 30.2 High-Pass Filter

f (Hz) 1 K 2 K 5 K 10 K 20 K 30 K 50 K 100 K 300 K

V O (V) .06 .13 .31 .54 .78 .94 1.0 1.0 1.0

c.

Fig. 30.5

332

d. f H (from graph) = 15 KHz

Part 3: Band-Pass Active Filter

c.

Table 30.3 Band-Pass Filter

f (Hz) 100 500 1 K 2 K 5 K 10 K 15 K 20 K 30 K

V O (V) .01 .035 .07 .15 .32 .51 .57 .57 .49

f (Hz) 50 K 100 K 200 K 300 K

V O (V) .35 .10

d.

Fig 30.6



1-2. See Probe plot page 334.


5. Slight variance due to PSpice cursor position.

6. f C (calculated) = 15.923 KHz

f C (numeric gain) = 15.937 KHz

f C (log. gain) = 15.848 KHz

333

30-1

334

30-1

335




9.

Numeric

f C (low): 6.5151 KHz

f C (high): 38.826 KHz

Bandwidth: 32.311 KHz

Logarithmic

6.6408 KHz

38.214 KHz

31.573 KHz

10. See tabulation in #9.

336

30-2

337

30-2

338

EXPERIMENT 31: COMPARATOR CIRCUITS OPERATION

Part 1: Comparator with 74IC Used as a Level Detector

a. R 3 = 10 Kohms, V ref = 5 V

R 3 = 20 Kohms, V ref = 6.7 V

c. V ref (measured) = 4.97 V

d. V i (measured) (LED goes on) = 5.01 V

V i (measured) (LED goes off) = 4.98 V

e. V ref (measured) = 6.63 V

V i (measured) (LED goes on) = 6.65 V

V i (mesasured) (LED goes off) = 6.61 V

All values of voltages measured and calculated relative to a particular R 3 are in very close

agreement.

Part 2: Comparator IC Used as a Level Detector

a. R 3 = 10 Kohms V ref (calculated) = 4.98 V

R 3 = 20 Kohms V ref (calculated) = 6.63 V

c. V ref (measured) = 4.97 V (R 3 = 10 Kohms)



e. Replace R 1 with 20 Kohm resistor.

V ref (measured) = 6.67 (R 3 = 20 Kohms)

V i (measured) (LED goes on) = 6.69 V

V i (measured) (LED) goes off) = 6.65 V

f. V i (measured) (LED goes on) = 6.65 V


The agreement between calculated and measured values in every case was near perfect.

Part 3: Window Comparator

a. V + (pin5, calculated) = 7.5 V

V − (pin6, calculated) = 2.5 V

c. V i (pin1, measured) = 7.6 V

V + (pin5, measured) = 7.36 V

V − (pin6, measured) = 2.3 V



339

e. V i (measured) (LED goes on) = 7.46 V


f. V i (measured) (LED goes on) = 7.46 V


Again as in the previous case, the agreement between measured and calculated values

was excellent.




31-1:

2. V in = 6 V; V ref = 5 V

3. Yes. I(D1) = 9.006 mA

340


31-1:

8. V in = 4 V; V ref = 5 V

9. No, I(D1) < 8 mA; I(D1) = 118.8 μA



341

31-1

342

EXPERIMENT 32: OSCILLATOR CIRCUITS 1: THE PHASE-SHIFT OSCILLATOR

Part 1: Determining Vout

d. f(theoretical) = 650 Hz

f. Estimated setting of RPot = 3 kohm

g. Vout (peak-peak) = 29 V

h. Period = 1.54 ms

i. f(experimental) = 649.4 Hz

j. Calculated % difference = 0.15

k. RPot + Rf = 29.5 kohm

l. Open-loop gain = 29.5

m. Calculated % difference = 7.8%

Part 2: PSpice Simulation

b. Vout(peak-peak) = 28.8 V

c. Vout(period) = 1.54 ms

d. Vout(frequency) = 649.4 Hz

e. Vout(peak-peak) = 19.1 V

f. Vout(frequency) = 646.5 Hz

j. P(V(feedback) = -89.9 degrees

P(V(VOUT) = 89.4 degrees

P(V(VOUT) – P(V(feedback) = 180 degrees

343

EXPERIMENT 33: OSCILLATOR CIRCUITS 2

Part 1: Wien Bridge Oscillator

c. T (measured) = 305 μs

d. f = 1/T = 1/305 μs = 3.28 KHz

e. T (measured, C = 0.01 μF) = 3 ms

f (calculated, C = 0.01 μF) = 328 Hz

f. f (calculated, C = .001 μF) = 3.12 KHz

f (calculated, C = .01 μF) = 312 Hz

Again, the agreement between the two sets of values was well within 10 percent.

Part 2: 555 Timer Oscillator

c. T (measured) = 20.1 μs

d. f = 1/T = 49.3 KHz

e. T (measured, C = 0.01 μF) = 203 μs

f = 1/T = 4.93 KHz

f. k = fRC = .48

f = 4.91 KHz

The agreement between the two values differed by only .4 percent.

Part 3: Schmitt-trigger Oscillator

c. T (measured) = 21 μs

d. f = 1/T = 46.9 KHz

e. T (measured, C = 0.01 μF) = 210 μs

f = 1/T = 4.69 KHz

f. f (calculated, C = 0.001 μF) = 46 KHz

f (calculated, C = 0.01 μF) = 4.6 KHz

The measured and calculated values of the frequency for each capacitor were within 2

percent of each other.

344

33-1

345



(VOUT) min = 0 V

(VOUT) max = 10 V

2. Yes.

3. 15.87 μs

4. PW = 8.63 μs

5. f= 63.2 KHz


7. Yes

8. No

9. P = 31.115 μs

10. PW = 23.993 μs

11. f = 41.67 KHz

12. Yes

346

33-1

347

33-1

348

EXPERIMENT 34: VOLTAGE REGULATION—POWER SUPPLIES

Note: The data obtained in this experiment was based on the use of a 10 volt Zener diode.

Part 1: Series Voltage Regulator

a. V L = V Z − V BE = 10 V − .7 V = 9.3 V

b. V O (measured) = 9.3 V

Table 34.1

V i (V) 10 11 12 13 14 15 16

V O (V) 9.25 9.26 9.28 9.30 9.32 9.33 9.35

The voltage regulation of the system was −.54 percent.

Part 2: Improved Series Regulator

a. A = 1 + R 1 /R 2 = 1 + 1 K/2 K = 1.5

V L = A VZ

V L (calculated) = 15 V

b.

Table 34.2

V i (V) 10 12 13 14 16 18 20 22 24

V L (V) 9.44 9.44 9.60 9.64 14.7 14.8 14.9 14.9 14.9

Upon coming near the nominal voltage level, the regulation of the system was −2 percent.

Part 3: Shunt Voltage Regulator

a. V L = (R 1 + R 2 ) * V Z /R 1 = 3 K/2 K * 10 V = 15 V

b. V L (measured) = 14.7 V

Table 34.3

V i (V) 24 26 28 30 32 34 36

V O (V) 14.3 14.4 14.5 14.7 14.7 14.9 15.1

The regulation of this system was 2.7 percent.




2. V in is swept linearly from 2 V to 8 V in 1 V increments.

3. V(V2) = 4.68 V

V(OUT) = 4 V

4. Approx. at V(VIN)) = 6.5 V

5. 0.68 V

6. Yes

7. V L = 4.68 V − 0.68 V = 4 V

349

34-1

350



2. V(IN) increases linearly from 6 V to 16 V in 0.5 V increments.

3. V L = V(OUT) = 1 k Ω+ 1 k Ω

(4.68 V) = 9.36 V

1 kΩ

4. V (OUT) = 9.36 V

theor.

V (OUT) = 8.9197 V

PSpice

5. V(V2) = 4.68 V

V(VOUT) = 8.9197 V

351

34-2

352

EXPERIMENT 35: ANALYSIS OF AND, NAND, AND INVERTER LOGIC GATES

Part 1: The AND Gate: Computer Simulation

a.

Table 35-1

Input terminal 1 Input terminal 2 Output terminal 3

1 1 1

0 1 0

1 0 0

0 0 0

Traces U1A:A and U1A:B are the inputs to the gate.

Trace U1A:Y is the output of the gate.

b. The output is at a logical HIGH if and only if both inputs are HIGH.

c. Over the period investigated, the Off state is the prevalent one.

d.

Terminal 25 ms 125 ms 375 ms

1 1 1 0

2 1 1 0

3 1 0 0

353

Part 2. The AND Gate: Experimental Determination of Logic States

a. Ideally, the same.

b. 10 Hz

c. Should be the same as that for the simulation.

d. The amplitude of the TTL pulses are about 5 volts, that of the Output terminal 3 is

about 3.5 volts.

e. The internal voltage drop of across the gate causes the difference between these voltage

levels.

Part 3: Logic States versus Voltage Levels

b. Example of a calculation: assume: V(V1A:Y) = 3.5 volts, VY = 3.4 volts

3.5V − 3.4V

%deviation = *100 = 2.86 percent

3.5V

c. For this particular example, the calculated percent deviation falls well within the

permissible range.

Part 4: Propagation delay

a. For the current case, the propagation delay at the lagging edge of the applied TTL pulse

should be identical to that at the leading edge of that pulse. Thus, it should measure

about 18 nanoseconds.

b. Ideally, the propagation delays determined by the simulation should be identical to that

determined in the laboratory.

c. From Laboratory data, determine the percent deviation using the same procedure as

before.

354

Part 5: NOT-AND Logic

A. Computer Simulation

a.

Table 35-2

Input1(7408) Input 2(7408) Input1(7404) Output(7404)

1 1 1 0

0 1 0 1

1 0 0 1

0 0 0 1

Traces U1A: A and U1A:B are the inputs to the 7408 gate, U1A:Y its output trace.

Trace U2A:Y is the output of the 7404 gate.

b. The Output of the 7404 gate will be HIGH if and only if the input to both terminals of

the 7408 gate are HIGH, otherwise, the output of the 7404 gate will be LOW.

c. The most prevalent state of the Output terminal of the 7404 gate is HIGH.

d. The PSpice cursor was used to determine the logic states at the requested times. The logic

states are indicated at the left margin.

At t = 25 milliseconds:

A1 = 25.397m, 1

A2 = 0.000, 1

dif = 25.397m

355

At t = 125 milliseconds

A1 = 125.397m, 1

A2 = 0.000, 1

dif = 125.397m


A1 = 375.397m, 0

A2 = 0.000, 1

dif = 375.397m

B. Experimental Determination of Logic States

a. They should be relatively close to each other.

b. They are identical.

c. The output of the 7404 gate is the negation of the output of the 7408 gate.

356

Part 6: The 7400 NAND Gate


Table 35-3

a.


1 1 0

0 1 1

1 0 1

0 0 1

b.


Table 35-4


1 1 0

0 1 1

1 0 1

0 0 1

357

EXPERIMENT 36: ANALYSIS OF OR, NOR AND XOR LOGIC GATES

Part 1: The OR Gate: Computer Simulation

a.

Table 36-1


1 1 1

0 1 1

1 0 1

0 0 0

Traces U1A:A and U1A:b are the inputs to the gate.

Trace U1A:Y is the output of the gate.

b. The output is a logical LOW if and only if both inputs are LOW, otherwise the output is

HIGH.

c. Over the period investigated, the ON, or HIGH, state is the prevalent one. This differs

from that of the AND gate. Its prevalent state was the OFF or LOW state.

358

d. The PSpice cursor was used to determine the logic states at the requested times. The

logic states are indicated at the left margin.

At t = 25 milliseconds:

A1 = 25.397m, 1

A2 = 0.000, 1

dif = 25.397m


A1 = 125.397m, 1

A2 = 0.000, 1

dif = 125.397m


A1 = 375.397m, 0

A2 = 0.000, 1

dif = 375.397m

359

Part 2: The OR Gate: Experimental Determination of Logic States

a. The pulse of 100 milliseconds of the TTL pulse is identical to that of the simulation

pulse.

b. The frequency of 10 Hz of the TTL pulse is identical to that of the simulation pulse.

c. They were determined to be the same at the indicated times.

d. The voltage of the TTL pulse was 5 volts. The voltage at the output terminal was 3.5

volts.

e. The difference in these two voltages is caused by the internal voltage drop across the

7432 gate.

Part 3: Logic States versus Voltage Levels

a. The PSpice simulation produced the identical traces as shown on the PROBE plot for

Figure 36-2.

b. Example of a calculation: assume V(V1A:Y) = 3.6 volts, VY = 3.4 volts

3.6V − 3.4V

%deviation = *100 = 5.56 percent

3.6V

a. It is larger by (5.56-2.86) = 2.7 percent.

Part 4: Combining AND with OR Logic


a.

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Table 36-2

U1A:A U1A:B U1A:Y U2A:A U2A:B U2A:Y U3A:A U3A:B U3A:Y

1 1 1 1 1 1 1 1 1

0 1 0 0 1 0 0 0 0

1 0 0 1 1 1 0 1 1

0 0 0 0 1 0 0 0 0

c.


A1 = 25.397m, 1

A2 = 0.000, 1

dif = 25.397m


A1 = 125.397m, 1

A2 = 0.000, 1

dif = 125.397m

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A1 = 375.397m, 0

A2 = 0.000, 1

dif = 375.397m

b. The output of the 7432 gate, U3A:Y, is evenly divided between the ON state and the OFF

state during the simulation.


a. The logic states of the simulation and those experimentally determined are identical.

b. The logic state of the output terminal U3A:Y is identical to that of the TTL clock.

c. The logic state of the output terminal U3A:Y is identical to that of the output terminal

U2A:Y of the U2A gate.

Part 5: NOR and XOR Logic combined


a.

The output trace of the 7402 NOR gate, U1A:Y and the output trace of the XOR gate,

U2A:Y are both shown in the above plot.

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b.

Table 36-3

U1A:A U1A:B U1A:Y U2A:A U2A:B U2A:Y

1 1 0 1 1 0

0 1 0 0 1 1

1 0 0 1 0 1

0 0 1 0 0 0

c. The output of the 7402 gate, U1A:Y is HIGH if and only if both inputs are LOW,

otherwise the output is LOW.

d. This is a logical inversion of the OR gate.

c. The output of the 7486 gate is HIGH if and only if the two inputs U2A:A and U2A:B

are at opposite logic levels.

f. The logic state of the OR gate is HIGH if both inputs are at opposite logic levels and if

both inputs are HIGH.


a. The experimental data is identical to that obtained from the simulation.

b. Refer to the data in Table 36-3.

c. Refer to the data in Table 36-3.

d. Refer to the data in Table 36-3.

e. The output of the 7486 XOR gate is HIGH if and only if its input terminals have

opposite logic levels, otherwise, its output is at a LOW.

f. For an OR gate, its output is HIGH if both, or at least one input terminal, is HIGH.

Its output will be LOW if both inputs are LOW. For an XOR gate, its output is HIGH if

and only if both input terminals are at opposite logic levels, otherwise, the output will be

LOW.

g. The output of an XOR gate will be HIGH when both input terminals are at opposite

logic levels. Otherwise, its output is at a logical LOW.

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EXPERIMENT 37: ANALYSIS OF INTEGRATED CIRCUITS

Part 1: Positive Edge-Triggered D Flip-Flop


a. The PROBE data shows the flip flop to be in the SET condition.

b. The flip flop goes to RESET at 200 milliseconds because the D input terminal goes

negative. The flip flop goes to SET at 400 milliseconds because both the CLOCK input

and the D input are positive.

c. The importance to note is that the D input can be negative and positive during the time

that the Q output is low.

d. After the initial SET condition of the flip flop, and after a RESET state of 200

milliseconds, the flip flop returns to its SET condition because at 400 milliseconds, both

the CLOCK and the D inputs are positive.

e. Starting from a SET condition, a transition to RESET will occur when the D input is

negative and the CLOCK pulse goes positive. The flip flop will SET again when the D

input is positive and the CLOCK goes positive.

f. The conditions stated in previous answer define a positive edge triggered flip flop as

defined in the first paragraph of Part 1.

g. See above answers.

h.

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i. Let us assume that D is high when a positive CLOCK pulse goes high. This will SET the

flip flop. This SET will be stored, or remembered, until D is negative and the CLOCK

triggers positive again. At that time, the flip flop will RESET. This RESET will be

stored, or remembered, until D is positive and the CLOCK triggers positive again. At

that time the flip flop will SET. Events repeat themselves after this.


a. Both input terminals are held at 5 volts during the experiment.

b. The amplitude of the voltage of the TTL pulse is 5 volts.

c. The amplitude of the output voltage at the Q terminal is 3.5 volts.

d. The difference between the input voltages and the output voltage is caused by the voltage

drop through the flip flop.

e. The experimental and the simulation transition states occur at the same times.

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Part 2: Frequency Division


Answer all questions below with reference to the following PROBE plot.

a. The frequency at the U1A:Q terminal is 5 Hz.

b. The frequency at the U1A:Q terminal is one-half that of the U1A:CLK terminal.

c. The frequency at the U2A:Q terminal is 2.5 Hz.

d. The frequency of the U2A:Q terminal is one-half that of the U2A:CLK terminal.

e. The overall frequency reduction of the output pulse U2A:Q relative to the input pulse

U1A:CLK is one-fourth.

f. Each flip flop reduced its input frequency by a factor of two.

g. It would take four 74107 flip-flops.

B. Experimental Determination of Logic States.

a. The J and CLR terminals of both flip flops are kept at 5 volts during the experiment.

b. The voltage level of the U1A:CLK terminal is 5 volts. The voltage level of the U2A:CLK

terminal is 3.5 volts. The voltage level of the U2A:Q terminal is 3 volts.

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c. Refer to the above PROBE plot.

d.

Pulse

U1A:CLK

U1A:Q

U2A:CLK

U2A:Q

Frequency

10.0 Hz

5.0 Hz

5.0 Hz

2.5 Hz

e. They are identical.

Part 3: An Asynchronous Counter: the 7493A Integrated Circuit


a.

A1 = 22.152m, 0

A2 = 0.000, 0

dif = 22.152m

b. See PROBE plot above.

d. t = 175 milliseconds. There is one clock pulse to the left of the cursor.

A1 = 174.051m, 1

A2 = 0.000, 0

dif = 174.051m

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e. t = 375 milliseconds. There are three clock pulses to the left of the cursor.

A1 = 376.582m, 1

A2 = 0.000, 0

dif = 376.582m

f. t = 575 milliseconds. There are five clock pulses to the left of the cursor.

A1 = 575.949m, 1

A2 = 0.000, 0

dif = 575.949m

g. t = 1.075 seconds. There are ten clock pulses to the left of the cursor.

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A1 = 1.0760, 1

A2 = 0.000, 0

dif = 1.0760

h. At t = 1.075 milliseconds, the output terminals, QA, QB, QC and QD have resumed their

initial states.

i. The MOD 10 counts to ten in binary code after which

it recycles to its original condition.

j. The output terminal QA represents the most significant digit.

k. The indicated propagation delay is about 12.2 nanoseconds.

A1 = 1.0000, 1.7628

A2 = 1.0000, 4.9975

dif = 12.200n, −3.2347


a. The logic states of the output terminals were equal to the number of the TTL pulses.

b. The experimental data is equal to that obtained from the simulation.

c. The propagation delay measured was about 13 nanoseconds.

d. The difference in the experimentally determined propagation delay was 13 nanoseconds

compared to a propagation delay of 12 nanoseconds as obtained from the simulation data.

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Solucionario-Boylestad-10ma-Edicion

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