A standard diode rectifier loses 0.7V across the diode junction, which is unacceptable when rectifying signals below 1V. An active rectifier uses an op-amp to drive a diode, reducing the effective forward voltage to microvolts. Audio peak detectors and precision power supplies both rely on this technique.

Figure 1: Precision half-wave active rectifier. The op-amp feedback forces Vout to equal Vin during positive half cycles.

Core Concept

In a passive rectifier, the diode only conducts when the input exceeds 0.7V. The op-amp in an active rectifier amplifies the input before the diode sees it. When the input goes even slightly positive, the op-amp output swings high enough to forward-bias D1. The virtual short principle then forces the inverting input to match the non-inverting input, so Vout equals Vin exactly, minus a tiny error of Vd/Aol.

For a TL071 with open-loop gain Aol of 200,000, a diode drop of 0.6V becomes an effective error of only 3 µV at the output. This allows rectification of millivolt-level signals with high accuracy, which is impossible using a 1N4007.

During the negative half cycle, D1 is reverse biased. The op-amp output saturates at its negative rail. No current flows through RL, so Vout holds at 0V. One limitation is slew rate distortion: when the input crosses zero and D1 switches from reverse to forward bias, the op-amp must recover from saturation. The TL071 has a slew rate of 13 V/µs, which limits accurate rectification above about 50 kHz.

Key Equations

Effective forward voltage of the precision rectifier:

Veff = Vd / Aol where Vd is the diode forward voltage (~0.6V for 1N4148) and Aol is the op-amp open-loop gain (dimensionless).

Output voltage (positive half cycle):

Vout = Vin (for Vin > 0) with error ε = Vd/Aol. For TL071, ε ≈ 3 µV.

Maximum rectifiable frequency (limited by slew rate):

fmax = SR / (2π × Vpeak) where SR is slew rate in V/µs and Vpeak is the peak output voltage in volts.

Example

Given:
  Op-amp: TL071, Aol = 200,000, Slew Rate SR = 13 V/µs
  Input signal: 50 mV peak sine wave at 1 kHz
  Diode: 1N4148, Vd = 0.6 V
  Load: RL = 10 kΩ

Why this formula:
  Active rectifier effective error = Vd / Aol (op-amp compensates for diode drop).

Formula:
  Veff_error = Vd / Aol

Substitution:
  Veff_error = 0.6 / 200,000

Calculation:
  Veff_error = 3 × 10^-6 V = 3 µV

Also find max frequency before slew limiting:
  fmax = SR / (2π × Vpeak)
  fmax = (13 × 10^6) / (2 × 3.1416 × 0.050)
  fmax = 13,000,000 / 0.3142 = 41.4 MHz (well above 1 kHz, so no slew limiting here)

Final Answer:
  Effective diode error at output = 3 µV
  Circuit operates without slew distortion at 1 kHz for 50 mV input.

Exam Tip: GATE questions on active rectifiers test whether students know WHY the diode drop vanishes. The answer is always that the open-loop gain Aol is in the feedback loop and divides the diode voltage by Aol. A second trap: the output is 0V (not -Vin) during the negative half cycle in a half-wave active rectifier. For a full-wave version, a second op-amp stage with gain -1 is added. Also, slew rate limits high-frequency accuracy regardless of Aol.

Key Properties

The effective forward drop is Vd/Aol. For TL071 (Aol = 200,000), this is approximately 3 µV, compared to 600 mV for a bare 1N4148.
1N4148 is preferred over 1N4007 in active rectifiers because its fast switching time (4 ns) suits AC signals above a few kilohertz.
Slew rate of the op-amp limits the maximum input frequency. TL071 at 13 V/µs handles signals up to about 40 MHz at 50 mV peak.
During the negative half cycle, the op-amp output saturates to its negative rail. Recovery from saturation takes time and is the main source of crossover distortion at high frequencies.
A full-wave precision rectifier uses two op-amps: one half-wave stage and one summing inverter stage to reconstruct both half cycles.
The load resistor RL must not be too small. A 10 kΩ load is typical. Very low RL values demand high output current from the op-amp and cause additional errors.

Quick Revision

Active rectifier uses op-amp feedback to eliminate the 0.7V diode drop effectively.
Effective error = Vd / Aol. For Aol = 200,000 and Vd = 0.6V, error = 3 µV.
Positive half: D1 conducts, op-amp feedback holds Vout = Vin.
Negative half: D1 reverse biased, op-amp saturates, Vout = 0V.
Use 1N4148 (fast switching) not 1N4007 (slow recovery) for active rectifier diodes.
Slew rate limits high-frequency performance. fmax = SR / (2π × Vpeak).
Full-wave version needs two op-amp stages. Half-wave uses only one.
Exam trap: Students write Vout = Vin - Vd for active rectifiers. The correct answer for positive half cycles is Vout = Vin, because the feedback compensates for Vd. Vd/Aol is the error, not Vd itself.