I'm struggling a bit with op amps again! I'm trying to learn how to use an inverting precision amplifier, and I'm getting very confused about the operation of the op amp in a active rectifier configuration. I'm sure it's a simple answer and I'm overcomplicating it, so here goes.

With a unity gain inverting op amp as shown here, when Vin goes negative, Vout goes positive. Opamps will do whatever is necessary to keep the two inputs the same. Since our non-inverting terminal is connected to GND, the opamp will do whatever it can to keep the inverting terminal GND, giving us our virtual ground. So it raises Vout to match the inverse of Vin to keep our inverting terminal close to GND, yes? -1V(Vin) + 1V(Vout) = 0V. Okay sick.

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Now here we have added two diodes to create an active half-wave inverting rectifier. D1 is the top diode.

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When Vin is positive, we get the expected inverted output at Vout. However, when Vin is negative, the voltage gets locked at one forward diode drop. So I understand that Vout raises to + 0.7V (the diodes Vf), D1 is now forward biased and our inverting terminal is at GND. Now what happens when we lower Vin below -0.7V to say -5V? Why does the Vout not equally & inversely raise to keep the inverting terminal at ground? Vout right now is at +0.7V (or D1's Vf), but Vin is at -5V. Wouldn't the voltage at our inverting terminal be -5V? What am I missing?


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    \$\begingroup\$ Wait.....so 5mA needs to flow through R1 (left) for there to be a voltage drop of 5V across it. When D1 is NOT present, the only way for this to happen is if 5mA flows through R2 (middle) from the output of the opamp, which of course corresponds to 5V (because R1 = 1K). But, when the diode is there, the voltage only needs to raise to .0.7V to bring the inverting terminal to GND, and the 5mA can just flow through the diode through R1 satisfying the 5V drop across R1!? Did I get it!? \$\endgroup\$ Commented May 5, 2023 at 16:43

1 Answer 1


Assume op-amp and diode are ideal.

The output does everything to keep inverting input at 0V.

As the ideal diode has 0.7V drop for any amount of current such as 1mA or 10mA, all that matters is that output is at 0.7V for the inverting input to be 0V.

So, no matter how negative the input voltage is, and how much current flows in the input resistance, the op-amp output will go positive and will provide all the current it needs to keep inverting input at 0V.

But because there is a diode (instead of a simple resistor), the inverting input will stay at 0V, if output voltage stays at 0.7V.

That's because an ideal diode has always 0.7V in forward bias with any current, the voltage does not depend on current.

Compare that to the standard amplifier, where instead of diode, there is a resistor, and resistors have this linear relationship between voltage and current.

  • \$\begingroup\$ Can you please explain further? \$\endgroup\$ Commented May 5, 2023 at 18:13

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