Unit 1-Clippers and Clampers

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Unit I:- CLIPPERS AND CLAMPERS

AEC unit 1 Notes

Staff :- KRS TE Dept PESIT Session :- Aug-Dec 2012

Diode Clippers

A clipperis a circuit that is used to eliminate a portion of an input signal. There are two basic typesof clippers: seriesclippers and shunt/parallelclippers. As shown in Figure 4-1, the series clippercontains a diode that is in series with the load. The shunt clipper contains a diode that is in parallelwith the load.

FIGURE 4-1 Basic clippers.

The series clipper is a familiar circuit. The half-wave rectifier is nothing more than a series clipper. When thediode in the series clipper is conducting, the load waveform follows the input waveform. When the diode is

not conducting, the output is approximately 0 V or fixed dc voltage which is connected in parallel. (Figure4.2). The direction of the diode determines the polarity of the output waveform. If the diode symbol (in theschematic diagram) points toward the source, the circuit is apositive series clipper, meaning that it clips thepositive alternation of the input. If the diode symbol points toward the load, the circuit is a negative seriesclipper, meaning that it clips the negative alternation of the input (Figure 4.11). With this di

Ideally, a series clipper has an output of when the diode is conducting (ignoring the voltage across

the diode). When the diode is not conducting, the input voltage is dropped across the diode, and .

Unlike a series clipper, a shunt clipperprovides an output when the diode is not conducting. For example,refer to Figure 4-1. When the diode is off(not conducting), the component acts as an open. When this is the

case, and form a voltage divider, and the output from the circuit is found using

When the diode in the circuit is on (conducting), it shorts out the load. In this case, the circuit ideallyhas an

output of . Again, this relationship ignores the voltage across the diode. In practice, the output from

the circuit is generally assumed to equal 0.7 V, depending upon whether the circuit is apositive shuntclipper or a negative shunt clipper. The direction of the diode determines whether the circuit is a positive or

negative shunt clipper. The series current-limiting resistor( ) is included to prevent the conducting diodefrom shorting out the source.

A biased clipper is a shunt clipper that uses a dc voltage source to bias the diode. A biased clipper is

shown in Figure 4-2. (Several more are shown in Figures 4.9 and 4.10). The biasing voltage ( )determines the voltage at which the diode begins conducting. The diode in the biased clipper turns on when

the load voltage reaches a value of . In practice, the dc biasing voltage is usually set using apotentiometer and a dc supply voltage, as shown in Figure 4.10.


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FIGURE 4-2 A biased clipper.

Clippers are used in a variety of systems, most commonly to perform one of two functions:

1. Altering the shape of a waveform2. Protecting circuits from transients

The first application is apparent in the operation of half-wave rectifiers. As you know, these circuits areseries clippers that change an alternating voltage into a pulsating dc waveform. A transient is an abruptcurrent or voltage spike of extremely short duration. Left unprotected, many circuits can be damaged bytransients. Clippers can be used to protect sensitive circuits from the effects of transients, as illustrated inFigure 4.12.

Various clipper circuits and their waveforms are given below. The output waveforms are taken using Pspicesimulation with Vm =+5V, Vbe =0.7V and Vdc =2V .

1. Series positive clipper

+ive cycle :- anode is at ground potentialand cathode sees variable +ive voltagefrom 0 to +Vm

For comlpete, cycle, diode becomereverse biased and hence Vo =0V

-ive cycle :- anode is at ground potential

and cathode sees variable -ive voltagefrom 0 to Vm. So in complete cycle, thediode is forward biased and

Vo= Vin + Vd and

At negative peak, Vo= -Vm+ Vd

= -4.3V


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2.Series negative clipper

+ive cycle :- cathode is at ground potentialand anode sees variable +ive voltage from0 to +Vm. So in complete cycle, the diodeis forward biased and

Vo= Vin - Vd and

At positive peak, Vo= Vm- Vd = 4.3V

-ive cycle :- cathode is at ground potentialand anode sees variable -ive voltage from0 to -Vm.

When magnitude of /Vin/ >Vd diodebecome reverse biased and hence Vo =0V

3.Shunt positive clipper

+ive cycle :- cathode is at ground potentialand anode sees variable +ive voltage from0 to +Vm

When Vin >Vd diode become forwardbiased and hence Vo =Vd =+0.7V

-ive cycle :- cathode is at ground potential

and anode sees variable -ive voltage from0 to Vm. So complete cycle, the diode isreverse biased and

Vo= Vin and at negative peak, Vo= -Vm=-5V


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4.Shunt negative clipper

+ive cycle :- anode is at ground potentialand cathode sees variable +ive voltagefrom 0 to +Vm

So complete cycle, the diode is reversebiased and Vo =Vin

At positvive peak Vo=+5V

-ive cycle :- anode is at ground potentialand cathode sees variable -ive vols from 0to Vm.

When magnitude of in put volatge i.e / Vin/>Vd, the diode become forward biasedand hence Vo =-Vd =0.7V

1.1.Series positve clipper with +ive biasvoltage (connected parallel to the load)

+ive cycle :- anode is at +Vdc andcathode sees variable +ive voltage from 0to +Vm

Vin< Vd+Vdc , the diode is forward biasedand hence Vo = Vin+Vd

When Vin >Vd+Vdc, the diode becomereverse biased and hence Vo =+Vdc =2V

-ive cycle :- anode is at +Vdc potentialand cathode sees variable -ive voltagefrom 0 to Vm. So complete cycle, thediode is forward biased and

Vo= Vin + Vd and

at negative peak, Vo= -Vm+ Vd = - 4.3V

1.2.Series positve clipper with -ive bias

voltage (connected parallel to the load)

+ive cycle :- anode is at -Vdc and cathodesees variable +ive voltage from 0 to +Vm

So complete cycle, the diode is reversebiased and Vo=-Vd =-2V

-ive cycle :- anode is at -Vdc potential and


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cathode sees variable -ive voltage from 0to Vm.

/Vin/< Vd+Vdc , the diode is reversebiased and hence Vo = -Vd

When /Vin />Vd+Vdc, the diode become

forward biased and hence Vo =+Vdc =2VVo= Vin + Vd and

at negative peak, Vo= -Vm+ Vd = - 4.3V

1.3. Series positve clipper with +ive biasvoltage in series

+ive cycle :- anode is at ground potentaland cathode sees variable +ive voltage of(Vin+Vdc) i.e +Vdc to Vm+Vdc

So complete cycle, the diode is reversebiased and hence

Vo = 0V

-ive cycle :- anode is at ground potentialand cathode sees variable voltage from+Vdc to (Vm-Vdc).

When the magnitude of Vin i.e /Vin/Vdc+Vd, the diode becomesforward biased and Vo= -Vin + Vdc+Vd = -(Vin-Vdc-Vd) and at negative peak,

Vo= -Vm+ Vdc+Vd =-5+2+0.7V =-2.3V

1.4.Series positve clipper with -ive biasvoltage in series

+ive cycle :- anode is at ground potentaland cathode sees variable voltage of (Vin-Vdc) i.e Vdc to Vm-Vdc

Vin< Vdc-Vd , the diode is forward biasedand hence

Vo = Vin Vdc +Vd = Vin-(Vdc-Vd)

When Vin >Vd+Vdc, the diode becomesreverse biased and hence Vo =0V

-ive cycle :- anode is at ground potential


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and cathode sees variable ive voltage ofVdc to (Vm+Vdc). So in complete cycle,the diode is forward biased and Vo = Vin Vdc +Vd = Vin-(Vdc-Vd)

At negative peak, Vo= -Vm-Vdc+Vd =-6.3V

2.1.Series negative clipper with +ive biasvoltage (Connected in parallel)

+ive cycle :- Cathode is at +Vdc andanoode sees variable +ive voltage from 0to +Vm

Vin< Vd+Vdc , the diode is reverse biasedand hence Vo= +2Vdc

When Vin >Vd+Vdc, the diode becomeforward biased and hence Vo =Vin-Vd .

At positive peak, Vo=5-0.7 =4.3V

-ive cycle :- cathode is at +Vdc potentialand anode sees variable -ive voltage from0 to Vm. So complete cycle, the diode isreverse biased and Vo=+vd =2v

2.2.Series negative clipper with -ive biasvoltage (connected in parallel)

+ive cycle :- Cathode is at -Vdc andcathode sees variable +ive voltage from 0to +Vm

Vin< Vd+Vdc , the diode is forward biasedand hence

Vo = Vin+Vd

When Vin >Vd+Vdc, the diode becomereverse biased and hence Vo =+Vdc =2V

-ive cycle :- anode is at +vdc potential andcathode sees variable -ive voltage from 0to Vm. So complete cycle, the diode isforward biased and

Vo= Vin + Vd and

at negative peak, Vo= -Vm+ Vd =-4.3V


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2.3.Series negative clipper with +ive biasvoltage in series

+ive cycle :- cathode is at -2Vdc potentaland anode sees variable voltage of 0 toVm

In complete cycle, the diode is forwardbiased and hence

Vo = Vin +Vdc-Vd = Vin+(Vdc-Vd). So atpositive peak Vo =Vm+Vdc-Vd =6.3V

When Vin >Vd+Vdc, the diode becomesreverse biased and hence Vo =0V

-ive cycle :- cathode is at -2Vdc potentaland anode sees variable voltage of 0 to

-Vm

When /Vin/Vdc-Vd, the diode is reversebiased and hence Vo = 0V

2.4.Series negative clipper with -ive biasvoltage in series

+ive cycle :- cathode is at +2Vdc andanode sees variable voltage of 0 to Vm

Vin< Vdc+Vd , the diode is reverse biasedand hence Vo=0

When Von> Vdc+Vd, the diode becomes

forward biased and henceVo = Vin Vdc -Vd = Vin-(Vdc+Vd). So atpositive peak, Vo =2.3V

-ive cycle :- cathode is at +2Vdc andanode sees variable voltage of 0 to -VmSo in complete cycle, the diode is reverse


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biased and Vo = 0V

3.1Shunt positve clipper with +ive shuntbias voltage

+ive cycle :- cathode is at +Vdc andanode sees variable +ive voltage from 0 to+Vm

Vin< Vdc+Vd, the diode is reverse biased

and hence

Vo = Vin

When Vin >Vd+Vdc, the diode becomeforward biased and hence Vo =+Vdc+Vd=2.7V

-ive cycle :- cathode is at +Vdc and anodesees variable -ive voltage from 0 to -Vm

So complete cycle, the diode is reversebiased

Vo= Vin and at negative peak, Vo= -Vm

=-5V

Vdc =2V

3.2 Shunt positve clipper with -ive shuntbias voltage

+ive cycle :- cathode is at -Vdc and anodesees variable +ive voltage from 0 to +Vm

So complete cycle, the diode is forwardbiased and

Vo = -Vdc+Vd =- (Vdc-Vd)= -1.3V

-ive cycle :- cathode is at -Vdc and anodesees variable -ive voltage from 0 to -Vm

So when the magnitude of input volatge

Vdc =2V


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i.e /Vin/< Vdc, the diode is forward biasedand

Vo = - (Vdc-Vd)= -1.3V

So complete cycle, the diode is reversebiased

Vo= Vin and at negative peak, Vo= -Vm= -5V

4.1 Shunt negative clipper with +ive biasvoltage connected in parallel

. +ive cycle :- anode is at +Vdc andcathode sees variable +ive voltage from 0to +Vm

When VinVdc-Vd, the diode is reversebiased and Vo =Vin. At positive peak,V0=+5V

-ive cycle :- anode is at +Vdc and cathodesees variable -ive voltage from 0 to -Vm

So complete cycle, the diode is forwardbiased and

Vo =Vdc-Vd =1.3V

4.2 Shunt negative clipper with -ive biasvoltage (connected in parallel)

+ive cycle :- anode is at -Vdc and cathodesees variable +ive voltage from 0 to +Vm

So in complete cycle , the diode is reversebiased and Vo =Vin At +ive peak, Vo=+Vm=5V

-ive cycle :- anode is at -Vdc and cathodesees variable -ive voltage from 0 to -Vm


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When the magnitude of Vin i.e/Vin/Vdc+Vd , the didoe becomesforward biased and Vo = -Vdc-Vd =-2.7V

5.Two side clipper :- (Combination ofpositive shunt clipper with +ive bias andnegative shunt clipper with negative bias)

+ive cycle :-

Cathode of D1 is at +Vdc1 and anode seesvariable +ive voltage from 0 to +Vm

Anode of D2 is at Vdc2 and cathode sees

variable +ive voltage from 0 to +VmFor complete cycle, diode D2 is reversedbiased

When VinVdc+Vdc1, D1 becomesforward biased and D2 in reverse biasedcondition. Vo =Vdc1+Vd1 =2.7V

-ive cycle :-

Cathode of D1 is at +Vdc1 and anode sees

variable -ive voltage from 0 to -VmAnode of D2 is at Vdc2 and cathode seesvariable -ive voltage from 0 to Vm

For complete cycle, D1 is reverse biased

When /Vin/Vdc2+Vd2, D2 becomesforward and D1 in reverse biasedcondition.

Vo =-Vdc2-Vd2 = -2.7V

In this type, the positive and negativeclipping level can be independently varied


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6.1 Two level clipper where Vdc1> Vdc2(note If Vdc1< Vdc2, both diodes conductand it is not desirable condition) Assuneideal diode

+ive cycle :-

Cathode of D1 is at +Vdc1 and anode seesvariable +ive voltage from 0 to +Vm

Anode of D2 is at +Vdc2 and cathode seesvariable +ive voltage from 0 to +Vm

Condition D1 D2 Vo

Vin Vin>Vdc2

OFF OFF

Vin

Vin >Vdc1 ON OF

F

Vdc1 =100V

-ive cycle :-

Cathode of D1 is at +Vdc1 and anode seesvariable -ive voltage from 0 to -Vm

Anode of D2 is at +Vdc2 and cathode seesvariable -ive voltage from 0 to Vm

So complete cycle, D2 is forward iased,&1 is reversed

So Vo = Vdc2

6.2. two side clipper

+ive cycle

Condition D1 D2 Vo

Vin = 0 OFF OFF Vdc2 =25V

Vin = 0 to25V

OFF OFF Vdc2 =25V

>Vin >25V

Vin = 25Vto 137.5V

OFF ON Vo = 25V+ VR2

= 25V+2/3 *Vin

Vin>137.5V

ON ON Vo=Vdc1 =100V

So get Vo =100V dc , Vin = 137.5V


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Negative cycle, both D1 & D2 does notconducts So Vo= 25V

Clampers (DC Restorers)A clamperis a circuit that is designed to shift a waveform above or below a dc reference voltage withoutaltering the shape of the waveform. This results in a change in the dc average of the waveform. Both ofthese statements are illustrated in Figure 4-3. (The clamper has changed the dc average of the inputwaveform from 0 V to +5 V without altering its shape.)

FIGURE 4-3 A clamper with its input and (ideal) output waveforms.

There are two basic types of clampers:

A positive clamper shifts its input waveform in a positive direction, so that it lies above a dcreference voltage. For example, the positive clamper in Figure 4-3 shifts the input waveform so thatit lies above 0 V (the dc reference voltage).

A negative clamper shifts its input waveform in a negative direction, so that it lies below a dcreference voltage.

Both types of clampers, along with their input and output waveforms, are shown in Figure. The direction ofthe diode determines whether the circuit is a positive or negative clamper.

Clamper operation is based on the concept ofswitching time constants. The capacitor charges through thediode and discharges through the load. As a result, the circuit has two time constants:

For the charge cycle, and (where is the resistance of the diode) For the discharge cycle, and (where is the resistance of the load)

Since is normally much greater than , the capacitor charges much more quickly than it discharges.As a result, the input waveform is shifted as illustrated in Figure 4.16.


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A biased clamperallows a waveform to be shifted above (or below) a dc reference other than 0 V. Severalexamples of biased clampers are shown in Figure 4-4.

FIGURE Several biased clampers.

The circuit in Figure (a) uses a dc supply voltage (V) and a potentiometer to set the potential at the cathode

of . By varying the setting of , the dc reference voltage for the circuit can be varied betweenapproximately 0 V and the value of the dc supply voltage.

The zener clamper in Figure (b) uses a zener diode to set the dc reference voltage for the circuit. The dc

reference voltage for this circuit is approximately equal to . Note that zener clampers arelimited to two varieties:

Negative clampers withpositive dc reference voltages Positive clampers with negative dc reference voltages

Positive clamper

Positive clamper

First positive cycle:-

Diode is reverse biased and Vo= Vin .

First negative cycle:-

Diode is forward biased and capacitor is charging with very low time constant. At negative peak, Vc=Vm -Vd After peak diodebecomes reverse biased as Vc>Vin.

Vo = Vin+Vc

Subsequent positive and negative cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In each negative


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cycle, Vc charges to max. value. In both cycles Vo= Vin + Vc

When Vdc =2V

Here Vc capacitor voltage charges to highervoltage

i.e Vc=Vm + Vdc -Vd Other explanation is same asfor the positive clamper.

When Vdc =-2V

Here Vc capacitor voltage charges to lower voltage

i.e Vc=Vm -Vdc -Vd Other explanation is same asfor the positive clamper.

Negative clamper

Negative clamper

First positive cycle:-

Diode is forward biased and capacitor is charging with very low time constant. At positive peak, Vc=Vm.-Vdc After


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peak, diode becomes reverse biased as Vc>Vin.

Vo = Vin-Vc

Subsequent negative and positive cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In eachpositive cycle, Vc charges to max. value. In both cycles

Vo= Vin Vc. (Vin is +ive for positive cycle and ive for negative cycle)

When Vdc =- 2V

Here Vc capacitor voltage charges to highervoltage i.e Vc=Vm + Vdc -Vd Other explanation issame as above

Unit I /d:- Practical Applications -

There are many practical applications of diode . few which we study this semester are

Polarity Insurance:- This circuit is mainly used to protect the expensive system against the reversepolarity voltage . A diode is connected at in put as shown below

Normal condition diode is reverse biased and voltage is applied across the system. When in put isreverse biased, then diode becomes reverse biased and only 0.7 is applied to system thusprotecting the system against the over voltage

Polarity Detector:- This circuit is mainly used to identify the polarity of the unknown voltage Given dcvoltage is to be connected to the Diode & LED circuit as given below

.

If terminal A is +ive and B is -ive then , D1 is forward biased and Led L1 glows.

If terminal A is -ive and B is +ive then , D2 is forward biased and Led L2 glows.

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Unit 1-Clippers and Clampers

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