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In the following circuit I have to find the output voltage, Vo, related to the input voltage, Vin. We assume that both diodes, D1 and D2, are the same and have a 0.7 V drop when they conduct and op-amps are ideal.

I am thinking that at start, before any input is applied, both D1 and D2 are reversed biased and no current flows in the circuit. When a positive input is applied (Vin > 0), op-amp 1 will immediately saturate at the negative supply (V -> V+), which means that V' will be negative and D1 will conduct when D2 will be reversed biased.

On the other hand, when a negative input is applied (Vin < 0) opamp 1 will immediately saturate at the positive supply (V- < V+) which means that V' will be positive and D2 will conduct when D1 will be reversed biased. Then, in both cases, we will have a circuit with closed loop op-amps with negative feedbacks. Thus, by using Kirchhoff's law I can define the output voltage related to the input voltage knowing that no currents flow in or out of the inputs of the op-amps.

Is this process right?

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Yeah, you are on the right path.

When Vin > 0, D1 = forward biased, D2 = reverse biased

We have two cascaded inverting amplifiers now.

So Vout will be:

$$V_{out} = V_{in}*(-R/R)*(-R/R) = V_{in}$$

When Vin < 0, D1 = reverse biased, D2 = forward biased

This means the non-inverting terminal of the second op-amp will be at some voltage V', and the same voltage would be there at the inverting terminal too.

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This is in contrast with the first case, where both terminals of the second op-amp were at '0'. So we were able to find the Vo straightforward.

Let us try to find V in this case now.

At node A, since voltage = 0, we can write a KCL equation:

$$\frac{V'}{2R} + \frac{V_{in}}{R }+ \frac{V'}{R} = 0 $$ $$\implies V' = -\frac{2}{3}V_{in}$$

The second op-amp is now nothing but a non-inverting configuration.

$$\therefore V_{out} = V'(1+\frac{R}{2R}) = -V_{in} $$

So the conclusion is this circuit acts as a precision full-wave rectifier. The analysis remains the same even if 0.7 V drops are considered, because the diode drops get compensated at the outputs of the op-amp. The same output is still obtained. Hence the term "precision" full-wave rectifier, which acts as a rectifier with ideal diodes.

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  • \$\begingroup\$ Thank you!Is my thought about the saturation at start right? I mean, does the opamp 1 saturates immediately when an input(positive or negative) is applied and then comes at linear region so that V+=V-? \$\endgroup\$ – MJ13 Apr 30 '18 at 8:19
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    \$\begingroup\$ Yea cz it is open loop at first. Then -ve feedback is established by diode later on. \$\endgroup\$ – Mitu Raj Apr 30 '18 at 11:20

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