EEET 201: ELECTRONIC CIRCUITS

PHYS 162: SOLID STATE DEVICES

CHAPTER 3 : SPECIAL-PURPOSE DIODES HW 3

DUE DATE: SATURDAY, 29 OCTOBER, **201**1

The Zener Diode

3.1 When the reverse current in a particular zener diode increases from 20mA to 30mA,

the zener voltage changes from 5.6V to 5.7V. What is the impedance of this device?

3.2 A certain zener diode has V Z =6.5 at 25°C and TC=0.05%/°C. Determine the zener

voltage at 70°C. (HINT – Refer Example 3-2 Page 111).

3.3 What are the two zener diode models? What is the difference between the two?

Which one is more suitable for circuit analysis?

Zener Diode Application

3.4 Determine the minimum and maximum input voltage required for the zener voltage

regulation for the circuit in Figure 1. Assume I ZK =1.5mA, Z Z =15Ω and V Z =14V at

30mA. (HINT – Refer to Example 3.5 Page 116).

Figure 1

3.5 To what value must R be adjusted in Figure 2 to make I Z = 35mA? Assume

V Z = 12V at 30mA and Z Z = 25Ω.

Figure 2

3.6 A loaded zener regulator shown in Figure 3 has V Z =5.1V at I Z =49mA, I ZK =1mA,

Z Z =7Ω. And I ZM =70mA. Determine the minimum and maximum load current I L

passing through R L .

Figure 3

3.7 A loaded zener regulator shown in Figure 4 has V Z =5.6V at I Z =45mA, I ZK =1mA,

Z Z =7Ω. (HINT – Refer to Example 3-7

Page 118)

(a) Determine V OUT at I ZK and I ZM

(P D(max) =1W).

(b) Calculate the value of R that should be

used.

(c) Determine the minimum value of R L

that can be used.

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Figure 4

3.8 Describe the basic operation of a varactor diode.

3.9 Describe the basic operation of light emitting diode (LED).

3.10 Determine the approximate value of the forward current I F for a green coloured

LED when the forward voltage is 3V. (HINT – Refer to V-I characteristic curve of

LED’s)

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The Zener Diode

3.1

Z Z = ΔV Z

ΔI Z

HW 3 SOLUTION

Z Z =

5.6V − 5.65V

30mA − 20mA = 0.1V

10mA = 10Ω

3.2 ΔV Z = V Z × TC × ΔT

TC = 0.05%/°C = 0.0005/°C

ΔV Z = 6.5V × 0.0005 × 70°C − 25°C = 0.153V

V Z + ΔV Z = 6.5 + 0.153 = 6. 653V

3.3 Zener diodes are modeled in two approximations.

o Ideal Zener Diode

o Practical Zener Diode

The Ideal model ignores the AC resistance of the zener diode while the Practical

model takes the AC resistance into account.

Anyone of them can be used for circuit analysis. It depends upon the type of

application. If high accuracy voltage regulation is required then practical model

should be used otherwise ideal model is suitable.

Zener Diode Application

3.4 At I ZK =1.5mA, the output voltage is

V OUT = 14V − ΔV Z = 14V − I Z − I ZK Z Z = 14V − 30mA − 1.5mA (15Ω)

V OUT = 14V − 28.5mA 15Ω = 13. 57V

V IN min = I ZK R + V OUT = 1.5mA 560Ω + 13.57V = 14. 41V

For V IN(max) , we need to find maximum zener current I ZM . So

I ZM = P D max

= 1W = 71. 4mA

V Z 14V

So at I ZM , the output voltage is

V OUT = 14V + ΔV Z = 14V + I ZM − I Z Z Z = 14V + 71.4mA − 30mA (15Ω)

V OUT = 14V + 41.4mA 15Ω = 14. 62V

V IN max = I ZM R + V OUT = 71.4mA 560Ω + 14.62V = 54. 6V

3.5 First find out V out at I Z =35mA.

V OUT = V Z + ΔV Z

V OUT = 12V + ΔV Z = 12V + 35mA − 30mA 25Ω

V OUT = 12V + 0.125V = 12. 125V

R = V IN − V OUT

I Z

=

18V − 12.125V

35mA

= 168Ω

3.6 I L is maximum when I Z is minimum. So find V Z at I ZK

V Z min = V Z − ΔV Z = 5.1V − 49mA − 1mA 7Ω = 4. 76V

V R = V IN − V Z min = 8V − 4.76V = 3.24V

Now find total current in the circuit

I T = V R

R = 3.24V

22Ω = 147mA

I L max = I T − I ZK = 147mA − 1mA = 146mA

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I L is minimum when I Z is maxmum. So find V Z at I ZK

V Z max = V Z + ΔV Z = 5.1V + 70mA − 49mA 7Ω = 5. 25V

V R = V IN − V Z max = 8V − 5.25V = 2.75V

Now find total current in the circuit

I T = V R

R = 2.25V

22Ω = 125mA

I L min = I T − I ZM = 125mA − 70mA = 55mA

3.7

(a) V OUT at I ZK :

V OUT = 5.6V − ΔV Z = 5.6V − I Z − I ZK Z Z = 5.6V − 45mA − 1mA (7Ω)

V OUT = 5.6V − 44mA 7Ω = 5. 2V

V OUT at I ZM :

V OUT = 5.6V + ΔV Z = 5.6V + I ZM − I Z Z Z

I ZM = P D max

= 1W = 178. 6mA

V Z 5.6V

V OUT = 5.6V + 178.6mA − 45mA 7Ω = 6. 5V

(b)

(c)

Calculate R for maximum zener diode current that occurs when there is no load as

shown in Figure 1(a)

R = V IN − V Z 8V − 6.5V

= = 8. 4Ω

I ZM 178.6mA

The minimum load resistance (maximum load current) is at minimum zener current

(I ZK =1mA) as shown in Figure 1(b). So we first need to find maximum load current

I L(max) .

I L max = I T − I ZK

I T = V IN − V OUT(min )

R

=

8V − 5.2V

8.4Ω

= 333. 3mA

I L max = 333.3mA − 1mA = 332. 3mA

R L min = V OUT(min )

I L max

= 5.2V

332.3mA = 3Ω

3.8

Figure 5 Solution Circuit for Q3.7

- The capacitance of a material can be determined by plate area A, dielectric constant

ε and plate separation d and is expressed as

C = Aε

d

- As the reverse bias increases, the depletion region becomes wide and increases the

plate separation d. This decreases the capacitance.

- As the reverse bias decreases, the depletion region becomes narrow and decreases

the plate separation d. This increases the capacitance.

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3.9

- We know that the free electrons in the n-type have high energy.

- When the LED is forward biased, electrons cross the pn junction and recombines

with the holes in the p-type.

- When these high energy electrons recombine with holes, they release energy in the

form of photons.

- The emission of these photons is called electroluminescence.

- The doping determines the wavelength of the emitted photons.

3.10 I F =55mA @ V F =3V

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