--- title: "Precision half wave rectifier" subject: "Electronics Devices and Circuit" chapter: 5 key: "semester_3/electronics_devices_and_circuit:precision-half-wave-rectifier" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/precision-half-wave-rectifier site: BCT Engineering --- # Precision half wave rectifier Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits - Asked in: 2082 Shrawan, 3 marks; 2078 Bhadra, 4 marks; 2076 Ashwin, 3 marks; 2074 Chaitra, 4 marks; 2073 Chaitra, 5 marks (asked 5 times) - Video: [What is Precision Rectifier? Precision Rectifier Explained](https://www.youtube.com/watch?v=5HweBajP-5g) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/precision-half-wave-rectifier - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5 ## Question Describe the operation of precision half wave rectifier with circuit diagram. ## 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. ## Answer 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 ```text +---------[ 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). ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can this circuit rectify millivolt signals when an ordinary diode cannot? A. The diode is in the feedback loop, so its cut-in voltage is divided by the open-loop gain B. The diode is replaced by a resistor C. The supply voltage is raised D. The input is amplified before the diode Answer: A. The effective cut-in voltage is about 0.6 V divided by the op-amp open-loop gain (about 10^5). (Key point: Why precision rectifier) 2. In the inverting version, what happens in the positive half-cycle of Vin? A. D2 conducts and Vout = Vin B. The output is amplified C. The op-amp output goes negative, D1 conducts (D2 off), keeping the inverting input at virtual ground and Vout = 0 D. Both diodes are off Answer: C. D1 closes the feedback loop but no current reaches the load, so the output is zero. (Key point: Positive half-cycle) 3. In the negative half-cycle, what is the output? A. Zero B. Vout = +Vin C. Vout = -Vin/2 D. Vout = -(Rf/R) Vin, a positive half-wave scaled by Rf/R Answer: D. D2 conducts, D1 is off, and the op-amp acts as an inverting amplifier giving a positive output of gain Rf/R (unity if Rf = R). 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