Op-amps circuit analysis
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 &lt; 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.



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&nbsp;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.