Understanding how the simple precision rectifier works

Looking for the idea

To see the idea in this circuit, it should be drawn more clearly in the following order: op-amp -> disturbance (diode) -> load. Then we can see that, in this "disturbed op-amp follower", the negative feedback is closed after the disturbance.

Here is my (and my students') Wikibooks story about this phenomenon observed in negative feedback circuits. To understand it, try a simple experiment - apply some negative input voltage (e.g., Vin = -1 V) and put yourself in the place of the op-amp. What will you do to make the voltage of the inverting input equal to the voltage of the non-inverting input (the "golden rule")? What will be the op-amp output voltage?

The classic "golden rule"

The so-called "golden rule" is introduced in the famous book of "The Art of Electronics". It says that an op-amp with negative feedback does its best to maintain equality between its two input voltages. In your circuit, it will make Vout = Vin. The diode is not perfect (it's not just a "short circuit")... and a voltage drop of about 0.7 V is lost across it. The op-amp "senses" it and additionally increases the magnitude of its output voltage (before the diode) with 0.7 V. In this way, the op-amp compensates the undesired VF = 0.7 V.

The great idea

So this is the great idea here - the op-amp compensates the voltage loss VF across the diode by an excessive output voltage VA = Vout + VF. You have to distingiish between VA and Vout (VA is the output voltage before the disturbance VF while Vout is the output voltage after the disturbance).

Implementation

Let's return to my example - Vin = -1 V. The op-amp will begin decreasing its output voltage VA towards the negative supply rail "observing" the difference Vin - Vout... and it will stop when Vout becomes equal to Vin (-1 V). Then the op-amp voltage VA will be -1.7 V and the circuit output voltage Vout will be -1 V.

When the input voltage is positive, the op-amp begins changing its output voltage towards the positive rail so the diode is backward biased and it does not conduct. The op-amp does not manage to make the voltage of the inverting input equal to the voltage of the non-inverting input... and it reaches the positive rail. However, Vout = 0 V...

More "golden rules"

Finally, let's summarize these particular observations into more "golden rules" for closing negative feedback in op-amp circuits:

• When a disturbance is undesired, close the negative feedback after it to eliminate the disturbance.

• To understand what op-amp does in negative feedback circuits, think of it not as a high-speed electronic device but as some kind of slow-thinking "being".

... and add this wisdom to our "collection of circuit principles".

See also

Circuit Principles (a collection of basic circuit ideas derived from specific circuit solutions)

When the diode becomes forward biased it effectively becomes a short and can be ignored.

Because of "OpAmp" action the OpAmp tries to drive the two terminals to the same level. When Vin < 0 the diode forward biases. The diode will have some finite amount of voltage drop but the OpAmp will null it out because the inverting terminal is sensing Vout dirrectly.

R1 is just there to drag Vout back to GND as Vin comes back to zero from the negative halve cycle.