# Is there a current in $R_1$ in this precision rectifier if so, where does it go? So Wikipedia on Precision Rectifier says that :

... when the input is greater than zero, $$\D1\$$ is off, and $$\D2\$$ is on, so the output is zero because the other end of $$\R_{2}}\$$ is connected to the virtual ground and there is no current through $$\R_{2}}\$$

Now when we see $$\R_1\$$ it has a potential drop of $$\V_{in}\$$ across its ends , so a current must flow through it. But then where does it go? I cannot see a closed loop in which current can flow. It can only go through $$\D1\$$ but then it cannot flow back into op amp.

So my question is : In which path does the current flow? or If there is no current flowing through $$\R_1\$$ how is there a potential difference across it?

• No current can go into inputs of an ideal op amp but op amp can sink current. – Hedgehog Jan 5 at 14:07

## 1 Answer

But then where does it go? I cannot see a closed loop in which current can flow. It can only go through D1 but then it cannot flow back into op amp.

Sure it can (but not through D1). The op-amp output will be about 0.7 volts negative (in order to preserve the virtual ground at -Vin) and positive current will flow through R1, then through D2 and to the op-amp output: - 