# Op-Amp Precision Diode: Eliminating Voltage Drop

• Bassalisk
In summary, this circuit has a voltage follower that is corrected for the voltage drop between the input and output.f

#### Bassalisk

[PLAIN]http://pokit.org/get/3deef3b185bf6b18007a25ddc18ce7d5.jpg [Broken]

So I have another thread running on op-amps, and while waiting for mr. Jiggy to come back online, I want to ask another op-amp question.

This precision diode. I understand what is the point, and why it is so useful. Eliminating that pesky voltage drop on diode, when rectifying <0.6 voltages is very nice.

But all in all I do not understand how it works. I've searched the internet high and low, very little answers.

So here is my current viewing:(and questions)

From the picture I see that the input is connected to the non-inverting lead. This makes the output conduct when the sine wave is in the positive part.

Because of that feedback, we have a simple voltage follower, correct?

But here is the thing. Don't we have that voltage drop, we are trying to eliminate still at the output?

How is this negative feedback correcting?

Any help would be appreciated, these op-amps are so versatile I cannot believe it!

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The + and - op-amp inputs will be at the same voltage, provided that the + input is positive. The op-amp output terminal supplies a positive current, and so will be 0.6V above those inputs due to the diode. But the output from the circuit you have shown will be 0.6V below the op-amp output -- in other words, it will be the same as the voltages at the op-amp inputs. So, there is no (significant) drop between the input and output of the circuit.

The + and - op-amp inputs will be at the same voltage, provided that the + input is positive. The op-amp output terminal supplies a positive current, and so will be 0.6V above those inputs due to the diode. But the output from the circuit you have shown will be 0.6V below the op-amp output -- in other words, it will be the same as the voltages at the op-amp inputs. So, there is no (significant) drop between the input and output of the circuit.

But why do input pins have to be at the same potential(voltage). I just can't understand that. I have another post running to answer that question...

What causes this behaviour of op-amp?

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