Chapter 2 C apte DIODE Part ７ Sinusoidal Inputs: Half-Wave ... - FKE

Chapter C apte 2 

DIODE 

Part 7 

Sinusoidal Inputs: 

Half-Wave Rectification 

Chapter 2 DIODE, Part 7 Half-Wave Rectification 1

AC INPUTS (SINUSOIDAL INPUTS) ; HALF-WAVE RECTIFICATION 

The diode analysis to include time-varying functions such as Sinusoidal and Square waveform 

Example of sinusoidal input 

rectifier 

T : period p ( (one full cycle) y ) 

Half-wave rectifier 

Vm : amplitude 

t : time 

vi : input 

vo : output t t 


t = 0 T/2 

The polarity of the applied voltage vi is such as to “establish” pressure in the direction 

indicated and “turn on” the diode diode. 

short circuit:Ideal diode 

Conduction region(0 T/2) 

t = T/2 T 

Th The polarity l it of f th the applied li d voltage lt vi iis such h as t to “t “turn off” ff” the th di diode. d 

Nonconduction region(T/2 T) 

open circuit:Ideal diode 

v 0 = iR = (0)R = 0 V 


Sketch of input,v i and output,v o for half-wave rectifier 

Ideal Diode (no ( effect of Si diode) ) 

The output signal v0 has a net +ve 

area above the axis over the a full 

period and an average value determined 

by 

The process of removing one-half the 

input signal to establish a dc level is 

called half-wave rectification. 

Eq.(1) 


Effect of Si Diode 

Si Diode 

The applied signal must have at least 0.7V (V T) before the diode can “turn on”. 

Nonconducting region: For v i less than 00.7V, 7V the diode is in an “open open circuit circuit” state and v 0 =0V 0V. 

Conducting region: v0 = vi - VT (VT =0.7V). 

The net effect is a reduction in area above the axis, which naturally reduces the resulting 

ddc voltage l llevel. l 

an average value (V m >> V T) Eq. (2) 

If V m is sufficiently larger than V T, eq.(1) is often applied as approximation for V dc. 


Example (1): 

(a) Sketch the output v 0 and determine the dc level of the output. Use ideal diode model. 

Solution: In this situation the diode will conduct during the –ve part of input. 

- 6.36 V 

V Vd dc =-00.318 318 V Vm =-00.318 318 (20V) =-636V 6.36 V 

The –ve sign indicates that the polarity of the output is 

opposite to the defined polarity. 


(b) Replace the ideal diode with a silicon diode. 

Solution 

V dc = - 0.318 (V m - V T )= - 0.318 (20 – 0.7 )= -0.318 (19.3 V) = - 6.14 V 

The resulting drop in dc level 

is 0.22 V (6.36 - 6.14 V) 

or about 3.5%. 

(c) Vm is increased to 200 V and compare solutions using Eq.(1) and Eq.(2) 

Solution 

E Eq.(1) (1) V Vdc = - 00.318 318 V Vm = - 00.318 318 (200 ) )= - 63 63.6 6 V 

Eq.(2) Vdc = - 0.318 (Vm - VT )= - 0.318 (200 – 0.7 )= -0.318 (199.3 V) = - 63.38 V 

The difference is very small that can be ignored. 


Peak Inverse Voltage (PIV) : maximum reverse-bias before entering Zener avalanche region. 

Requirement for the diode to behave as rectifier 

V m ≤ 

PIV 

If V , there will a short circuited diode 

m ≥ PIV 

or the diode will enter zener avalanche region. 

This will make the diode to loose its function 

as rectifier. 


AC INPUTS (SINUSOIDAL INPUTS) ; FULL-WAVE RECTIFICATION 

(a) Positive Region 

4 diodes t =0 = 0 T/2 

Full-Wave Bridge rectifier 

Ideal diode 

Polarity y of R 

Conduction path for +ve region of v i 

D 2 & D 3 : ON ; D 1 & D 4 : OFF 

Chapter 2 DIODE, Part 7 Full-Wave Rectification 9

(b) Negative Region 

Ideal diode 

Polarity of R 

Conduction path for -ve region of vi D D2 &D & D3 :OFF : OFF ; ;D D1 &D & D4 :ON : ON 

The important result is that the polarity across the load resistor R is the same as 

+ve region, g establishing g a second ppositive ppulse. 


Input and output of full wave rectifier for one full cycle. 

Si Since th the area above b th the axis i ffor one ffull ll cycle l iis now ttwice, i th the ddc llevel l hhas also l bbeen 

doubled. (1)) 

Effect of Si diode 

Conduction path 

KVL of conduction path 


Peak Inverse Voltage (PIV) : maximum reverse-bias before entering Zener avalanche region. 

Requirement for the diode to behave as rectifier 

V m ≤ 

PIV 

If V , there will a short circuited diode 

m ≥ PIV 

or the diode will enter zener avalanche region. 

This will make the diode to loose its function 

as rectifier. 


CLIPPERS 

Clipper has the ability to “clip” clip off a portion of the input signal without distorting the 

remaining part of the alternating waveform. Half-wave rectifier is an example of the 

simplest form of diode clipper ; one resistor and one diode. 

Depending on the orientation of the diode, the +ve and –ve region of the input signal is 

“clipped” off. 

SSeries i clipper li 

Example of clipped input 

Chapter 2 DIODE, Part 7 Clipper 13

Procedure for analyzing Networks 

Additional dc supply 

(1) Make a mental sketch of the response of the network based on the direction of the 

diode and the applied voltage levels 

The direction of the diode suggests that the signal vi must be +ve to turn it on. 

The voltage vi must be greater than V volts to turn the diode “ON”. 

The –ve ve region of the input signal is “pressuring” pressuring the diode into the “OFF” OFF state supported 

further by dc supply. 

Here, we can be quite sure that the diode is an open circuit for –ve region of the input signal. 


(2) Determine the applied voltage (transition voltage) that will cause a change in state 

for the diode. 

For the ideal diode the transition between the states will occur at the point of 

v d = 0 V and i d = 0 A. 

The level of v i that will cause a 

transition in state is 

v i = V 

(3) Be continually aware of the defined terminals and polarity of v 0 

Example: Short-circuit state 

The output v0 can be determined by KVL in 

clockwise direction 

An input greater than V volts the diode is in 

the short short-circuit circuit state state. 

An input less than V volts the diode is in 

the open-circuit state 


(4) It can be helpful to sketch the input signal above the output and determine the output 

at instantaneous values of the input. 

Determining v 0 when v i = V m 

Keep in mind that at an instantaneous value of v i the input 

can be treated as a dc supply of that value and the 

corresponding dc value (the instantaneous value) of the output 

determined. 

Example: vi = Vm V Vm > V ; the diode is in short short-circuit circuit, then v 0 = V Vm – VV. 

vi = V ; the diode change state 

vi = - Vm, v0 = 0 V 

Sketching v v0 Chapter 2 DIODE, Part 7 Clipper 16

Example (2): Determine the output waveform. 

Substituting ideal diode. 

The transition level (v d=0 & i d=0) 

v o = v i + 5V 

Here, v o = 0V 

v i = -5 V 

For v i more –ve than – 5V the diode 

will enter its open-circuit state, 

For v i more +ve than – 5V the diode 

is in the short-circuit state. 

Series clipper 

The diode will be in the “ON” state for 

the positive region of v i. Note here 

that V=5V also will aid to this effect. 

v o = v i + 5V 



Circuit with square-wave input is easier to analyze than with 

(b) v i = -10V (T/2 T) 

Th The di diode d iis iin 

opencircuit 

state. 

v 0 = iRR R = (0)R ( ) = 0V 

Output Voltage g 

sinusoidal wave input. 

(a) v vi = 20V (0T/2) (0T/2) 

The diode is in shortcircuit 

state. 

v 0 = 20V + 5V = 25V 

The clipper not only clipped 

off the input signal but raised 

the h ddc llevel l of f the h signal i l bby 

5V. 


Parallel clipper 

Example of response to a parallel clipper 



Short-circuit 

Short circuit 

The polarity of the dc supply and the 

Direction of the diode strongly suggest 

That the diode will be in “ON” state for 

-ve region of the input signal. 

Use ideal diode (v (vd=0 =0 & i id=0) =0) 

The transition level is 

vo = V = 4V 

vi = V = 4V 

Since the dc supply is “pressuring” the 

diode to stay in short-circuit state, 

th the iinput t voltage lt must t be b greater t than th 

4V for the diode to be in the “OFF” 

state. 

Any y input p less than 4V will 

result in a short-circuited diode. 


Determining v 0 for open state. 

open-circuit 

Input & Output 

Voltage 

v 0 = v i 


Effect of Si diode (V T = 0.7V) in parallel clipper 

The diode in the “ON” state 

The transition voltage is determined at i d = 0A & v d = 0.7V 

KVL in the clockwise direction 

For input greater than 3.3V the diode will be an open 

circuit. v0 = vi For input less than 3.3V the diode will be in the 

“ON” state. t t 

The only effect 

of f was to drop d the h 

transition level to 

3.3 V from 4V. 


SUMMARY : CLIPPER 



Center-Tapped Transformer 

This full-wave rectifier appears with two diodes but requiring a center-tapped (CT) 

transformer to establish the input signal across each section of the secondary of the 

transformer. 

(a) Consider the +ve portion of v i applied to primary of the transformer. 

D 1 assumes the short-circuit equivalent and D 2 the open-circuit equivalent, as 

determined by the secondary voltages and the resulting current directions. 

Chapter 2 DIODE, Part 7 Full-Wave Rectification,CT 25

(b) Consider the –ve portion of the input applied to primary of the transformer. 

Now, D1 assumes the open-circuit equivalent and D2 the short-circuit equivalent, as 

determined by the secondary voltages and the resulting current directions directions. 

The voltage across the load resistor R maintain to the same polarity. The net effect is 

the same output with the same dc levels as appearing in 4 diode bridge full-wave rectifier. 

(c) PIV 

Chapter 2 DIODE, Part 7 Full-Wave Rectification,CT 26

Example (5): Determine the output waveform and calculate the output dc level and the 

required PIV of each diode. 

Solution 

(a) +ve region of the input voltage. 

Redrawing the circuit. 

Here, v 0 = ½ v i or v omax= ½ v imax =1/2(10) = 5 V. 


(b) -ve region of the input voltage. 

Th The roles l of f th the di diodes d will ill bbe iinterchanged. t h d 

(c) -ve region of the input voltage. 

Vdc = 0.636 (5 ( V) ) = 3.18 V 

Output waveform for +ve and –ve region of input 

The effect of removing two diodes from the bridge configuration was therefore to reduce 

the available dc level. 


CLAMPERS 

The clamping circuit is one that will “clamp” a signal to a different dc level. 

The circuit must have a capacitor capacitor, a diode and a resistive element and it can also 

have an independent dc supply to introduce an additional shift. 

(a) 0 T/2 (b) T/2 T 

v 0 = 0 V 

KVL The resulting output 

waveform. 

Chapter 2 DIODE, Part 7 Clampers 29

The resulting output 

waveform waveform. 

The output p signal g is clamped p to 0V for the interval 

0 to T/2 but maintains the same total swing (2V) as 

the input. 

For the clamping circuit circuit, the total swing of the output 

Is equal to the total swing of the input signal. 


Example (6): Determine v o for the input indicated. 

C=0.1μF 

Solution: 

Frequency is 1000Hz, resulting in a period of 1 ms (T (T=1 1 / f ) and an interval of 0.5ms 

between levels. 

KVL around the input loop 

t1t2 :the diode is in its 

short circuit state state. 

The output,v0 is across R 

but it is also directly across 

5V battery. y The result is 

v0 = 5V. 

charging circuit 

The capacitor will charge 

up to 25V 


t 2t 3 :the diode is in its open 

circuit state. 

The open circuit equivalent will 

remove the 5 V battery from 

having any effect on v 0 . 

KVL around the outside loop 

The time constant of the discharging 

discharging circuit 

For all practical purposes, the capacitor will fully charge and discharge in five time 

constants. 

The total discharge time is 

5 τ = 5(10 ms) = 50 ms 


Note that the output swing of 30V matches the input swing. 

The resulting output waveform 


Example (7): Repeat example (6) using silicon diode with VT = 0.7V. 

(a) Short Circuit state (t2t3) KVL in the output section 

(b) Open Circuit state (t 2t 3) 

KVL in the input section 

The resulting output waveform 


Example of Clamping Circuit with square wave input (Considering ideal diode) 


Example of Clamping Circuit with sinusoidal wave input (Considering ideal diode)

Chapter 2 C apte DIODE Part ７ Sinusoidal Inputs: Half-Wave ... - FKE

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?