So, in this article, we will learn about** precision rectifier. **And we will also see how to design it. Now, the first question which comes to our mind is what is a **precision rectifier**. And how is it different from the **normal rectifier circuit?** So, first of all, let’s understand that.

Now, in the previous Articles, we have learned about the half-wave rectifier circuit. And if we assume this diode as an ideal diode in that case whenever the sine wave is applied to this rectifier circuit then at the output we will get either a positive or negative half cycle. So, here as you can see we will get only positive half cycles. And for the ideal half-wave rectifier, the transfer characteristic will look like this. It means that when the input is negative at that time the output will be zero.

And whenever input is positive at that time the output will follow the input signal. But in the actual rectifier circuit, this diode will conduct only when the applied input signal is more than the forward voltage drop across the diode.

And if you see the actual waveform of the rectifier then it will look like this. So, the peak waveform of the rectifier will be slightly less than the ideal rectifier output.

And if you see the transfer characteristic then it will look like this. So, as you can see over here, the diode will start conducting only when this applied input signal is more than this threshold voltage. Now, this voltage drop across the diode will not be a much problem for the AC to DC conversion.

Because in this case, the applied input signal across the rectifier used to be much larger than the threshold voltage drop across the diode. But whenever this rectifier circuit is used for the small-signal conditioning and signal processing then it will not work.

For example, let’s say the input that is applied to this rectifier is smaller than this diode voltage drop. So, in this case, this rectifier circuit will not be able to rectify the signal. For example, let’s say you want to demodulate the AM signal.

Now, usually, the received signal strength used to be much smaller than the diode voltage drop. So, in this case, suppose if you want to demodulate the envelope from the carrier signal then it is not possible in the case of the normal rectifier circuit.

**precision rectifier**

So, in such a case, the precision rectifier circuit is very useful. So, this circuit is designed using the op-amp and diode. And in this rectifier circuit, the diode acts like an ideal diode.

If You See Wiki Covered This Topic Breifely So, the rectifier circuit behaves as an ideal rectifier circuit. And because of this, this circuit is also known as the super diode. So, if you see the transfer characteristic of the precision rectifier then it will look like this. So, now let’s understand how this circuit works.

So, whenever the applied input signal is positive that means Vin is greater than zero, in this case, the diode will behave like a closed switch. So, if you see the circuit, it is nothing but the buffer circuit.

So, whatever input is applied to this positive terminal of the op-amp, the same input will appear across the output terminal. That means whenever the input is greater than zero, at that time, the output will follow the input signal.

And whenever the input is less than zero, at that time diode will get reversed biased. And simply it will act as an open switch. So, in this case, the output of this rectifier will be equal to zero volts.

Now, in this circuit, this diode behaves like an ideal diode because if you observe this circuit here this diode is connected in the feedback loop. Now, here let’s assume that Vf is the voltage drop across this diode.

And whenever this diode is connected in the closed-loop configuration, then the effective voltage drop will get divided by the open-loop gain of this op-amp. So, for example, let’s say the drop across this diode is equal to 0.7 V.

And usually the open-loop gain of the op-amp used to be in the range of 10 to the power to 10 to the power 6. So, here let’s assume that it is equal to ten to the power 5.

So, now if you see the effective forward voltage drop, it will be approximately equal to zero. So, we can say that in this closed-loop configuration, the effective forward voltage drop across the diode is equal to zero volts.

And that’s why it behaves like an ideal diode. So, now let’s analy=ise this circuit for the different cases of the input signal. Now, first of all, let’s assume that for the given input signal, this diode is conducting. And also assume that the output of this op-amp is equal to Vo’.

So, as this diode is conducting, there will be a drop of 0.7 V across this diode. And here we are assuming that the silicon diode is used in this rectifier circuit. So, the Vout will be equal to this Vo’ – 0.7V. Or in another way, we can say that this Vo’ will be equal to Vo + 0.7VSo, this will be the output of this op-amp.

Now, if you observe this circuit, here as this diode is conducting, this path will get completed. Or we can say that the op-amp is operated in the negative feedback configuration. And we know that in this negative feedback configuration, we can apply the concept of either virtual ground or virtual short.

So, let’s quickly prove that even if this diode is connected in the feedback path, then also it is possible to apply the concept of virtual ground or virtual short. Now, for the op-amp, we know that the output voltage Vo can be given as the open-loop gain, multiplied by the difference between the two input signals.

So, let’s say the input that is applied at the positive terminal is equal to V+. And the input which is applied at the negative input terminal is equal to V-. So, the output will be equal to the open-loop gain times the difference between the two input signals. Now, this Vo’ can be written as Vo +0.7V. And that will get divided by the open-loop gain.

Now, as we have just seen, the open-loop gain op-amp used to be very large. So, this quantity will be approximately equal to zero. And that will be equal to the difference between the two input signals.

So, we can say that whatever input is applied to the positive terminal, the same input will appear at the negative terminal. And if you observe over here, this V- is connected to the output.

So, at output also, we will get the input voltage. That means Vout will be equal to Vin. So, whenever this diode is conducting at that time the output voltage will follow the input voltage. And this diode will conduct only when the applied input voltage is greater than zero. Now, similarly, let’s analyze the circuit for the negative input voltage.

Now, for a moment, here let’s assume that this diode is non-conducting. So, if this diode is non-conducting in that case, the op-amp will operate in the open-loop configuration. And if you observe at this node, the output at this node will be equal to 0.

Because this Vout is connected to the ground through this load resistance. And as this output terminal is also connected to the negative input terminal of the op-amp, at this terminal also the output will be equal to zero volts. Now, if you observe over here, the diode is reversed biased.

And one end of the diode is connected to the zero volts. It means that at this terminal the voltage should be negative. And that is only possible whenever the applied input voltage to the diode is negative.

So, now let’s see the output waveform of this precision rectifier. So, let’s say, the sine wave is applied as an input to this rectifier circuit. And if you see the output then it will look like this. So, whenever the input is greater than zero, at that time the output will follow the input signal.

And whenever the input is less than zero, at that time the output will be equal to zero. Now, in this precision rectifier circuit, there is one limitation. So, let’s see that. So, let’s assume that the input that is applied to the rectifier is less than zero. And we have seen that in that case, the output of the op-amp will be equal to the negative saturation voltage.

And in this condition, the op-amp will also operate in the open-loop condition. Now, whenever this input becomes positive at that time the first of all the op-amp needs to come out of this negative saturation voltage.

So, it will take some time for the op-amp to come out from this negative saturation voltage to some finite positive voltage. Now, this will not be a problem for the op-amp whenever it is operated at low frequencies. Let’s say, it is operated at 50 to 100 Hz. But whenever it is operated at much higher frequencies.

Let’s say, from 50 to 100 kHz, at that time the output of the op-amp will get distorted. Because the op-amp will take some time to come out from this negative saturation voltage to some finite voltage. And this will limit the maximum operating frequency for the rectifier circuit. So, suppose if you want to increase the operating frequency range then somehow we need to prevent this op-amp from going into the negative saturation.

## Modified Precision Rectifier

**precision rectifier circuit**

And that can be avoided by using this modified circuit. So, using this circuit we can prevent the -amp from going into the negative saturation. So, let’s understand how this circuit works. So, in this circuit, whenever the input is greater than zero, at that time the output of the op-amp will be equal to zero.

Because in this case, the diode D1 will conduct and this diode D2 will be off.

And the output will be equal to 0 volts. And whenever this input is less than zero, at that time this diode D1 will be off and this diode D2 will conduct. And the output of the rectifier circuit can be given as -R2/R1 times the input voltage. so, if you see the output of the circuit, it is never going into saturation. So, in this way, we can prevent this circuit from going into negative saturation.

So, now let’s analyze the circuit and let’s understand how this circuit works. Now, in this circuit, for a moment let’s assume that this diode D1 is conducting. And whenever this diode D1 will conduct at that time, this path will get completed. And because of the completion of the path, we can say that there is a negative feedback from the output to the input side. It means that the op-amp is operated in the linear region.

And we can apply the concept of virtual ground. And we can say that this node over here will be at ground potential. Because this positive input terminal is also at ground potential. So, as this diode D1 is conducting. so there will be a 0.7V of the voltage drop across this diode.

So, at this node, the voltage will be equal to 0 -0.7, which is equal to -0.7V. Now, if you observe the output terminal, itis connected to the ground through this resistor RL. And it is also connected to the ground terminal through this resistor R2.

It means that the output of the circuit will be equal to zero volts. So, no current is flowing through these two resistors.

It means that the cathode of the diode D2will is at zero voltage. And we can say that this diode D2 is non-conducting. Now, as this diode D2 is non-conducting, some forward current will also flow through this diode.

And it should flow in this direction. So, if we apply the KCL at this node, then some finite amount of current should also come into this node. And that is only possible whenever the applied input signal is greater than 0.

So, we can conclude that whenever the input signal is greater than zero, at that time D1 will conduct and D2 will remain off. And the output of the op-amp will be equal to zero volts.

Similarly, now let’s analyze the circuit for the negative input voltage. Now, here let’s assume that the applied input signal is less than zero volts. Now, here for a moment let’s assume that this diode D1 is conducting.

And if it is conducting, then there will be a closed-loop between this output and input terminal. And we can also apply the concept of virtual ground. So, this node will be at ground potential. And as the applied input is negative, the current will flow from zero to the negative input voltage.

And to satisfy the KCL, some finite amount of current should also come into this node. It means that the current that is flowing through the diode should flow in this direction. But if this diode D1 is conducting, the forward current should flow in this direction. It means that this diode D1 is non-conducting.

So simply it will act as an open circuit. And some finite amount of current is entering into this node through this path. It means that this diode D2 is conducting and the current is flowing in this direction.

So, if we assume these conditions, then the equivalent circuit will look like this. So, if you observe over here, now the circuit gets completed through this diode D2 and this resistor R2.

And as there is negative feedback, we can once again apply the concept of virtual ground. So, this node over here will be at ground potential.

So, now if we apply the KCL at this node, then we can write -(Vin)/R1 that is equal to Vout/R2. And fro this we can say that output voltage V out will be equal to (-R2/R1)*Vin So, this is the output voltage whenever the applied input signal is less than zero volts.

### precision Full Wave Rectifier

So, in this modified rectifier circuit, the transfer characteristic will look like this. It means that whenever the input is greater than zero, the output will remain zero. And whenever input is less than zero, at that time the output can be given by the expression -R2/R1 times Vin. So, the slope of this line will be equal to-R2/R1.

Now, let’s see the output waveform of this rectifier circuit. So, if we apply the sinewave to this rectifier, then the output waveform will look like this. So during the positive half cycle, the output is zero. And during the negative half cycle output will be equal to -R2/R1 times the input voltage.

Similarly, you can also show that whenever the directions of this diode D1 and D2 are reversed at that time the transfer characteristics of this rectifier will look like this. So, here whenever the input is less than zero, at that time, the output will be equal to zero. And whenever the input is greater than zero, at that time output will be equal to -R2/R1 times the input voltage.

So, try to get the transfer characteristics for this circuit. Now, so far we have only discussed the half-wave rectifier circuit. But using this circuit, it is also possible to design the precision full-wave rectifier circuit.

So, let’s see using this circuit how we can design the full-wave rectifier circuit. Now, for this circuit, we understood that the output will be equal to zero whenever the input is greater than zero. And the output will be equal to -R2/R1 times the input voltage whenever the input is less than zero.

So, by setting the value of R1 and R2 we can set the gain of the output waveform. Now, here to design this full-wave rectifier circuit, the gain of this half-wave rectifier is set to -2. So, at this node, the output waveform will look like this.

So, whenever the input is equal to -Vm, at that time the output will be equal to 2Vm. So, here this yellow signal is the output waveform of this half-wave rectifier.

And this blue signal is the input that is applied to this half-wave rectifier. Now, whenever we add these two signals then we will get the waveform of the full-wave rectifier. So, as you can see, using the precision rectifier and the adder circuit it is easy to design the precision full-wave rectifier circuit.

So, I hope in this Article, you understood how to design the precision half-wave and full-wave rectifier circuit. So, if you have any questions or suggestions, do let me know in the comment section below. If you like this article, hit the like button and subscribe to the kohikhi for more such Articles.

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