Precision rectifier

[likephysics]

I am bit confused about precision rectifier/superdiode circuit using opamp.

The first ckt in the wikipedia link - http://en.wikipedia.org/wiki/Precision_rectifier

When the input is positive, say 1v, the opamp tries to make the -ve terminal equal to its +ve terminal. So to do that, the output has to be 1.6v (1V+diode drop). This part I understand.

Lets say the input is -1v. Now the opamp tries to make the -ve terminal equal to its +ve terminal. To do that the output has to be -1.6V.
But then the output would be -1v. Where am I going wrong?

[es1]

the feedback loop doesn't close for voltages below zero because the cap is grounded and the diode doesn't reverse conduct

so the output of the opamp will just stay stuck at its low rail.

but this is what you want since the output is not suppose to follow the input during the negative part of the cycle in a rectifier

[likephysics]

the feedback loop doesn't close for voltages below zero because the cap is grounded and the diode doesn't reverse conduct

so the output of the opamp will just stay stuck at its low rail.

but this is what you want since the output is not suppose to follow the input during the negative part of the cycle in a rectifier

So the op amp doesn't try to make the 2 inputs equal?
If RL was connected to a -ve voltage source, you would see -1v at the output (when input is -1v)?

[AlephZero]

So the op amp doesn't try to make the 2 inputs equal?

When the diode is reverse biased, the op amp can "try" to make the inputs equal as much as it likes, but it can't succeed. If Vout is not at zero volts, there would be a current flowing through RL whcih has nowhere else to go. It can't flow into the op-amp output becase of the diode, and the op-amp input won't source or sink any current either.

What actually happems is that the op-amp saturates, and the two inputs are then at different voltages. That is why the "simple circuit" is a poor design compared with the "improved circuit". Wnen the input goes positive again, It takes a finite amount of time for the amp to recover from being saturated.


If RL was connected to a -ve voltage source, you would see -1v at the output (when input is -1v)?

Yes, you would see -1V on the output when the input was below -1V.