# Clamper circuits – Positive clamper, Negative clamper, Types, Video Explained

## What is Clamper Circuit?

A clamper circuit changes the DC level of the signal to the desired level. And at the same time, it does not change the shape of the waveform. So, as you can see over here, in this diagram, this clamper circuit shifts the entire waveform upwards. Or we can say that this clamper circuit shifts the entire waveform towards the positive side.

So, this type of clamper circuit is known as the positive clamper circuit. Similarly, if the clamper circuit shifts the DC level towards the negative side, then such a clamper circuit is known as the negative clamper circuit. And this clamper circuit can be designed using the diode, resistor, and capacitor.

And many times to shift the DC level to the desired voltage, the reference level or the biasing voltage is also added with this clamper circuit.

## Types of Clamper Circuits

So, this clamper circuit can be classified broadly into three categories.

• The positive clamper circuit
• The negative clamper circuit
• The clamper circuit with the biasing voltage

The positive clamper circuit, the negative clamper circuit, and the clamper circuit with the biasing voltage. So, in this article, one by one we will see all the three types of clamper circuits, but first of all, let’s talk about the positive clamper circuit.

## Positive Camper Circuit (with circuit analysis)

So, this circuit which is shown over here is an example of the positive clamper circuit. Now, in general, any type of waveform can be applied to this clamper circuit but here we will analyze the circuit by applying a sine wave to this circuit.

So, let’s say, a sine wave with a peak value of Vm is applied to this clamper circuit.

So, at the output, this entire waveform will get shifted by a DC value of VM. So, let’s analyze this circuit, and let’s find out how we will get this type of waveform. But before we do that here we are assuming that the RC time constant of the circuit is very large.

So, if T is the time period of the signal, then the RC time constant of the circuit should be much larger than this time period. Or in-general, it should be at least 10 times more than the time period of the signal. Alright so considering this assumption, let’s analyze this circuit.

### positive half cycle

So, during this positive half cycle, this diode will act as an open circuit. Because the voltage at the cathode is more than the anode. So, considering the ideal diode, it will act as an open circuit.

So, during the positive half cycle, this capacitor will get charge through this path. But considering the RC time constant is very large than the time period, during the positive half cycle, it will not be able to charge to the peak value.

And we can assume that during the positive half cycle, the output will follow the input signal. So, during the positive half cycle, you will get this type of waveform.

### negative half cycle

Now, during the negative half cycle, the polarity of the input voltage will get reversed. So, now this diode will act as a short circuit. And effectively this resistor RL will also get short-circuited. So, during this negative half cycle, this capacitor will get charged through this path. And it will get charged to the peak value of VM.

And during the negative half cycle, if you see the output then up to this point it will be equal to zero. And during that time, the capacitor will get charged to the peak value of VM. Now, after this point, if you see the voltage across the diode, it will be equal to Vin + VM. So, this is the voltage that will appear between these two terminals.

Now, after this point, as Vin is going from-VM to 0 volts, so this voltage will be always greater than 0. So, due to that now the cathode of the diode is more positive than the anode. That means after this point, this diode will act as an open circuit. And then after the voltage, V out can be given as Vin + VM.

So, whenever this Vin is equal to -VM, at that time, the output will be equal to 0. And similarly, whenever the Vin is equal to0, at that time, the output voltage will be equal to VM. So, during this time, the output will look like this. which means this portion of the waveform will get DC shifted by VM volt.

### Next Positive Half Cycle

Now, during the next positive half cycle, this diode will still remain reversed biased. Because the voltage across the diode is equal to Vin + VM. And as Vin is greater than zero, the voltage at the cathode is more than the anode.

So, this diode will still remain reverse biased. And during the positive half cycle, the output voltage Vout can be given as Vin + VM. So, due to that, during the next positive half cycle, we will get this type of waveform. That means the entire waveform will get DCshifted by Vm volt.

And during the next negative half cycle also, this diode will still remain reverse biased. So, due to that, we will get this type of waveform. So, in a steady-state condition, if you seethe an output waveform, then it will look like this. That means the entire waveform will get shifted upwards by Vm volt.

And this is the simulation result of the positive clamper circuit. So, as you can see, after a couple of half-cycles, the output will reach the steady-state condition. So, here this blue waveform is the input waveform and this green waveform is the output waveform. And here as the diode is non-ideal so there will be a voltage drop across the diode also. And that is why this output is going below the zero volts.

Alright so, this is all about the positive clamper circuit. And in this circuit, just by reversing the direction of the diode, it can be converted into the negative clamper circuit.

Also Read: Peak Detector Circuit Explained

## Negative Clamper Circuit (with circuit analysis)

So, here we will analyze this circuit by considering the square wave as an input waveform.

So, to this circuit, if we apply a square wave which is varying from Vm to -Vm, the nat the output we will get this type of waveform.

That means the entire waveform will get shifted downwards by Vm volt. So, let’s analyze this circuit, and let’s see how we will get this type of waveform. So, during the positive half cycle, the input to this circuit is equal to Vm volt.

And due to that, this diode will get forward biased. And considering the ideal diode, it can be represented as a short circuit. And due to that, effectively this resistor will also get short-circuited.

Also read: What is Precision Rectifier

### positive half cycle

So, due to that during the positive half cycle, this capacitor will get charged through this path. And it will get charged to the peak value of the voltage VM.

And during the positive half cycle, the output voltage V-out will be equal to zero.

### negative half cycle

Now, after that during the negative half cycle, the polarity of the input voltage will get reversed. And now if you observe the voltage which is appearing across the diode is equal to -2Vm. That means the summation of these two voltages.

And due to that this diode will get reverse biased. And we can replace it with an open circuit. So, during the negative half cycle, the output voltage out will be equal to -Vm -Vm = -2Vm. So, during the negative half cycle, the output waveform will look like this.

### Next Positive Half cycle

Now, once again during the next positive half cycle, this diode will act as a short circuit. Because now, the voltage which is appearing across the diode will be positive. Because during the previous cycle, the capacitor has lost some finite charge. And due to that, the voltage across the capacitor will be slightly less than VM.

And due to that, the voltage which is appearing across the diode will be positive. So, during this positive half cycle, this diode will act as a short circuit. And due to that once again this capacitor will get charged to the peak value of voltage VM. And during this cycle, the output voltage V out will be equal to 0. And once again during the negative half cycle, we will get an output voltage out that is equal to -2Vm.

So, if you see the overall output waveform, then it will look like this. That means the entire input waveform, has been Dc shifted by a voltage of -VM.

Now, even though this clamper circuit changes the DC level of the input waveform, but the peak to peak voltage of the output waveform is the same as the input waveform. So, here if you see, the input waveform is varying from VM to -VM.

That means the peak to peak voltage of the waveform is equal to 2Vm. While at the output the waveform is varying from 0 to -2Vm. That means here also the peak to peak voltage is equal to 2Vm.

So, in this way, using this negative clamper circuit the DC level of the input waveform can be shifted towards the negative side. Now, as I said earlier, to shift the reference level or the DC level of the waveform, sometimes the biasing voltage is also used with this camper circuit.

## Negative Clamper Circuit with Biasing Voltage (with circuit analysis)

So, now let’s see the clamper circuit with the biasing voltage. So, this circuit which is shown over here is the negative clamper circuit with the biasing voltage.

So, let’s analyze this circuit by applying a square wave as an input.

### Positive Half cycle

So, now during the positive half cycle, this diode will act as a short circuit. Because here, this voltage Vm is greater than a volt.

So, during the positive half cycle, this capacitor will start charging through this path. And if you apply the KVL, then the voltage across the capacitor will be equal to Vm – V volt. Ane during this positive half cycle, if you see the output voltage, then it will be equal to V volt.

### negative half cycle

Now, during the negative half cycle, if you see the voltage at the anode of this diode, it will be equal to -2Vm +V volt. that means during this negative half cycle, this diode will act as an open circuit. And during this negative half cycle, the output voltage Vout will be equal to -2Vm + V volt.

So, during the negative half cycle, if you see the output waveform, then it will look like this. So, the peak value of the waveform will be equal to -2Vm +V volt. And then after once again, we will get this type of repetitive waveform.

So, if you see the overall output waveform, then it can be represented by this pink waveform. That means this waveform is varying from V volt to -2Vm + V volt. Now, without this biasing voltage, the output waveform would have been varied from zero to -2Vm volt. But due to this biasing voltage, the entire waveform has been DC shifted by V volt.

## Positive Clamper Circuit with Biasing Voltage (with circuit analysis)

So, this circuit, which we have discussed is the negative clamper circuit with the biasing voltage. And similarly, we can also design a positive clamper circuit with the biasing voltage. And here, for the analysis, we will assume a sine wave as an input. Now, here to easily analyze this circuit, let’s take some numbers.

So, let’s say, here the biasing voltage is equal to 3V and the input waveform is varying from 10 to -10 V. Now, in this circuit, if the biasing voltage is not there, then the output waveform would look like this. that means the output waveform would get DC shifted by 10 V. But due to this additional 3V, this waveform will get further shifted by 3V.

And due to that, we will get this type of waveform. So, let’s analyze this circuit, and let’s find out why we will get this type of waveform. Now, during the positive half cycle, whenever the input is zero, that means at this point, this diode will get forward biased. And due to that, it will act as a short circuit.

So, when Vin is equal to zero, at that time, this capacitor will get charged through this path and it will get charged to the voltage of 3V. So, when Vin is equal to zero, at that time, the voltage across the capacitor will be equal to 3V. Now, after that whenever Vin goes above this0V, at that time, this diode will act as an open circuit. Because if you observe, the voltage at the cathode of the diode is equal to 3V, while the voltage at the anode of the diode is equal to Vin +3 V.

And as Vin is increasing, the voltage at the cathode will be more than the anode. And due to that this diode, will get reverse biased. And we can represent it by an open circuit. So, whenever, this Vin + VC is greater than3V, or in other words, whenever Vin is greater than 0, at that time, the output voltage Vout can be given as Vin + 3V. And during this time, the output waveform will look like this. That means the output waveform will get DC shifted by 3V.

### negative half cycle

Now, as I said earlier, the voltage at the cathode of the diode is equal to Vin + 3V. And during the negative half cycle, as this becomes more and more negative, so this term will be less than 3v. So, due to that, during the negative half cycle, this diode will get forward biased. And we can represent it by a short circuit.

So, during the negative half cycle, up to this point, the output voltage Vout will be equal to 3V. And during that time, this capacitor will get charged to the voltage of Vin + 3V. And up to this point, this capacitor will get charged to the voltage of 3V + 10V. That is equal to 13V.

So, after this point, if you see the output waveform, then it will look like this. Now, after this point, the voltage at the cathode of the diode will be equal to Vin + 13V. And after this point, as Vin is going from-10 to 0V, so this term will be greater than 3V. So, after this point, this diode will act as reverse biased.

And we can represent it by an open circuit. So, after this point, the output voltage Voutcan be given as Vin + 13V. So, whenever Vin is equal to -10V, at that time, the output will be equal to 3V. And when Vin is equal to zero volts, at that time, the output will be equal to 13V. So, during this time, if you see the output waveform, then it will look like this. That means this portion of the waveform will get DC shifted by 13V.

### Positive Half Cycle

Now, during the next positive half cycle, still, this diode will remain reverse biased. Because as I said, the voltage at the cathode is equal to Vin + 13V. And as this Vin is positive, so the voltage at the cathode will be more than the anode.

So, due to that, this diode will still remain reverse biased. And during this time, the output waveform out will be equal to Vin + 13V. So, during the positive half cycle, if you see the output waveform, then it will look like this. And then after the negative half cycle also, this diode will still remain reverse biased.

So, during the negative half cycle also, this entire waveform will get DC shifted by 13V. So, in a steady-state condition, if you see the output waveform, then it will look like this.

That means now, the entire waveform has DC shifted by 13V. And here is the simulated result for a similar type of clamper circuit. So, here this blue waveform is the input waveform, and the green waveform is the output waveform.

as you can see, the output waveform looks similar to the waveform which we have drawn during the analysis. But here, as this diode is non-ideal, so there will be a voltage drop across the diode also. So, in this way, by using the biasing voltage with this clamper circuit, we can shift the DC level of the waveform to the desired level.

## FAQ’S

### How does a clamper circuit work?

A clamper is AN electronic circuit that changes the DC level of a sign to the specified level while not ever-changing the form of the applied signal. In alternative words, the clamper circuit moves the total signal up or right down to set either the positive peak or negative peak of the signal at the specified level.

### What are the different types of clamping circuits?

Negative Clamper.
Negative Clamper with Positive Reference Voltage.
Negative Clamper with Negative Reference Voltage.
Positive Clamper.
Positive Clamper with Positive Reference Voltage.
Positive Clamper with Negative Reference Voltage.

### What are the necessary components of a clamper circuit and why?

The minimum variety of parts of a clamping circuit is 3 – a condenser, a electrical device and a diode. In some cases, DC offer is additionally required to relinquish an extra shift. the character of the wave shape remains alike, however the distinction is within the shifted level. the height to peak price of the wave shape can ne’er amendment

### What is a clipper circuit explain in detail?

In natural philosophy, a clipper could be a circuit designed to forestall a proof from surpassing a preset reference voltage level. … A clipper circuit will take away bound parts of associate degree whimsical undulation close to the positive or negative peaks or each. Clipping changes the form of the undulation and alters its spectral parts.

### What is a positive clamper circuit?

A Positive Clamper circuit is one that consists of a diode, a resistance and a capacitance which shifts the output to the positive portion of the sign. The figure below explains the development of a positive clamper circuit.