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Precision half wave rectifier

Short answer

An op‑amp with diodes in feedback eliminates diode drop. For negative input, D2 conducts and Vout = –(Rf/R) Vin; for positive input, D1 clamps and output is zero. It rectifies millivolt‑level signals accurately.

Describe the operation of precision half wave rectifier with circuit diagram.

An ordinary diode rectifier cannot rectify signals smaller than the diode's cut-in voltage (about 0.6 V), because the diode does not conduct below it. A precision half-wave rectifier (super diode) places the diode inside the feedback loop of an op-amp. The effective cut-in voltage is divided by the open-loop gain, so even millivolt signals are rectified accurately.

Circuit diagram

                +---------[ Rf ]----------+
                |                         |
                |     D1                  |
                +-----|>|------+          |
                |              |          |
 Vin o---[ R ]--+--(-)\        |    D2    |
                       >-------+---|>|----+-----o Vo
           GND ----(+)/                   |
                                        [RL]
                                          |
                                         GND

D1 has its anode at the inverting node and its cathode at the op-amp output. D2 has its anode at the op-amp output and its cathode at the output node Vo. Rf runs from Vo back to the inverting node.

Operation (inverting version)

Positive half-cycle of Vin: The op-amp output goes negative. D1 turns on (D2 is off), closing the feedback loop and keeping the inverting input at virtual ground. No current flows to the load: Vout = 0.

Negative half-cycle of Vin: The op-amp output goes positive. D2 turns on (D1 is off). Now the op-amp works as an inverting amplifier with the feedback resistor Rf:

Vout = −(Rf/R) Vin

So the output is a positive half-wave, scaled by Rf/R (unity when Rf = R).

 Vin     _         _
       /   \     /   \
 -----/-----\---/-----\---/---> t
             \_/       \_/

 Vout         _         _
             / \       / \
 -----------/---\-----/---\---> t

The output is zero during the positive half-cycles of Vin; the negative half-cycles appear inverted (and scaled by Rf/R).

The diode drop is divided by the open-loop gain A (≈ 10⁵), so the circuit works for inputs of a few millivolts. Applications: signal detection, AC voltmeters, AM demodulation and instrumentation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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