In short: Place the rectifying diodes inside the feedback loop of an op-amp, so the amplifier's huge open-loop gain divides the diode drop down to microvolts, then combine a half-wave stage with a summing amplifier (gain 1 and 2) to get an exact |Vin| even for millivolt signals.
Why an ordinary diode bridge fails on small signals
A silicon diode needs roughly 0.6 V before it conducts, a Schottky about 0.2 to 0.3 V. Feed a 50 mV peak sensor signal into a conventional full-wave rectifier and the output is a flat line: the signal never climbs past the knee. Even at 1 V amplitude a bridge loses two drops (about 1.2 V), so the error is 60 percent of the peak. Audio level meters, AC voltmeters, RF detectors and lock-in front ends all need the output to track the input down to the noise floor, which is the job of a precision rectifier, sometimes called a "super diode".
The super-diode idea: put the diode inside the loop
Connect a diode in series with the output of an op-amp and close the feedback from the diode's cathode back to the inverting input. When the op-amp output is below the diode's threshold the diode is off and the amplifier just ramps its output upward. As soon as the output exceeds the load voltage plus Vd, the diode conducts and the loop closes. Feedback forces the load to equal the input.
The key number is the effective threshold: Vd divided by the open-loop gain. With Vd = 0.6 V and Aol = 100,000 (100 dB) at DC, the input must exceed only 6 µV before conduction starts. That is the whole trick; the diode drop is still there, but the loop hides it.
Key fact: Effective turn-on threshold = Vd / Aol(f). It is tiny at DC but grows with frequency because Aol falls at 20 dB per decade beyond the op-amp's dominant pole.
The classic two op-amp full-wave circuit
The most-taught topology uses one op-amp as an inverting half-wave rectifier and a second as a summing amplifier. All resistors are R except one input resistor of the summer, which is R/2.
- A1 (half-wave stage): Inverting gain of -1 with two diodes. For Vin > 0, D1 conducts and the node Vh = -Vin. For Vin < 0, D2 conducts and keeps A1's output one diode drop below its inverting node, holding the virtual ground; D1 blocks and Vh = 0.
- A2 (summer): Vout = -(Vin + 2·Vh), with Vin through R and Vh through R/2, feedback R.
Check the algebra. For Vin = +1 V: Vh = -1 V, so Vout = -(1 - 2) = +1 V. For Vin = -1 V: Vh = 0, so Vout = -(-1) = +1 V. The output is |Vin| with unity gain. For a 2 V peak sine you get a clean 2 V peak, 2 V-high humps at twice the input frequency.
Where the textbook circuit breaks: dynamics
The DC analysis flatters the circuit. Real limits show up at the zero crossing, where A1 must swing its output from one diode conduction state to the other. When Vin changes sign, A1's output travels from about +0.6 V to -0.6 V, a 1.2 V excursion, entirely in open loop, at the amplifier's slew rate. A general-purpose LM358 slews at about 0.3 V/µs, so each crossing costs 4 µs of dead time. At 20 kHz the half-period is only 25 µs, so a sixth of every hump is distorted. A TL072 (13 V/µs) shrinks that to under 0.1 µs.
Second, gain-bandwidth. A 1 MHz GBW amplifier has Aol = 10 at 100 kHz, so the effective threshold becomes 0.6 V / 10 = 60 mV. This is why a precision rectifier that works perfectly at 50 Hz fails on a 100 kHz, 20 mV signal.
Watching it work
The animation below follows a small sinusoid through the rectifier. The negative half simply flips up, and the positive half passes unchanged, with no dead zone at the crossings.
Design checklist with real numbers
| Parameter | Typical choice | Reason |
|---|---|---|
| Resistors | 10 kΩ, 0.1 % (R/2 as two 10 k in parallel... use 5 kΩ matched) | Ratio mismatch of 1 % gives ~1 % rectification error and a visible "odd-even hump" difference |
| Op-amp | TL072, OPA2134, or a 10 V/µs+ part | Slew and GBW set the usable frequency |
| Diodes | 1N4148 or Schottky BAT54 | Low junction capacitance and fast recovery |
| Supply | ±12 V | Headroom for the 1.2 V excursion plus signal |
Rule of thumb: For less than 1 % error at frequency f and amplitude Vp, you want Aol(f) > Vd / (0.01 · Vp), so a 100 mV signal at 10 kHz needs Aol above 600, meaning GBW of at least 6 MHz.
Simulating and validating it
Model the circuit in SPICE before building it. In Cadence Spectre or Siemens Questa ADMS, run a transient analysis with a 50 mV, 1 kHz source and sweep frequency from 100 Hz to 200 kHz; measure the error of Vout against abs(Vin). Add Monte Carlo on the resistors with 1 % tolerance to see the ripple asymmetry. For integrated implementations, slew rate and settling drive the amplifier sizing, which is a task analog IC teams at design houses such as Avecas VLSI design services handle daily. Single-supply designs often replace the diodes with a current-mode or a transconductance-based rectifier, which also eliminates the dead time.
Common pitfalls: forgetting D2, which lets A1 saturate to the rail and then recover slowly; using electrolytic coupling that shifts the DC level; and ignoring input bias current, which creates a DC offset after rectification that an averager will read as signal.
If you want to practise this on real interview-style problems, see our analog design courses, or explore hands-on training at ChipXpert VLSI training.
Ready to design op-amp circuits with confidence? Enrol in the CourseTron Analog Circuit Design course for guided SPICE labs and interview practice.
Frequently Asked Questions
Why does a precision rectifier work below the diode drop?
The diode sits inside the op-amp's feedback loop, so the amplifier output rises by whatever extra voltage the diode needs. The input only has to supply that drop divided by the open-loop gain, typically microvolts.
What limits the frequency of a precision rectifier?
Slew rate during the zero-crossing transition and the falling open-loop gain, which raises the effective threshold as Vd / Aol(f).
Why is the summer input resistor R/2?
The half-wave signal must be weighted twice as heavily as Vin so that, for positive inputs, 2·(-Vin) + Vin leaves +Vin.
Can I use Schottky diodes?
Yes. They reduce the output swing A1 must make at each crossing and cut recovery time, improving high-frequency accuracy.