Precision Rectifier

A precision rectifier uses an op-amp to overcome the 0.6V forward drop of a diode, allowing it to rectify signals as small as a few millivolts. It is used in true-RMS meters, signal demodulators, and peak detectors where standard diode rectifiers lose accuracy on small signals.

Core Concept

In a regular diode rectifier, the output is zero until the input exceeds 0.6V. This error is unacceptable for small signals. The precision rectifier places the diode inside the op-amp's feedback loop. When the input goes positive, the op-amp drives its output high enough to forward-bias D1, regardless of the diode's actual threshold. The output follows the input with near-zero error.

During the negative half-cycle, the super diode cuts off. The op-amp output swings to its negative rail, but D1 is reverse-biased, so no current flows. The output stays at 0V. The closed-loop forward drop is effectively Vdiode/Aol, which is microvolts for a gain of 100,000.

The TL071 is preferred because its FET input stage minimises offset and its 13 V/µs slew rate handles audio-frequency signals. For a precision full-wave rectifier, a second op-amp stage (summing amplifier) is added to invert and add both half-cycles together, producing a full rectified output with gain -2R/R = -2.

Key Equations

Effective forward voltage of the super diode:

Veff = Vdiode / Aol ≈ 0.6 / 100,000 = 6 µV

For positive input (ideal half-wave rectifier):

Vout = Vin (for Vin > 0)

Vout = 0 (for Vin < 0)

For the full-wave precision rectifier using two op-amps (standard two-op-amp topology):

Vout = |Vin|

Average output of a half-wave precision rectifier for sinusoidal Vin of amplitude Vm:

Vavg = Vm / pi = 0.318 * Vm

Given:
  Precision full-wave rectifier
  Vin = 50 mV peak sinusoidal
  R = 10 kΩ (all resistors equal)
  TL071 op-amp, Aol = 200,000

Why this formula:
  Full-wave rectifier output is |Vin|.
  Average value of |sin| waveform = 2Vm/pi

Formula:
  Vout_avg = 2 * Vm / pi

Substitution:
  Vout_avg = 2 * 50 mV / pi

Calculation:
  = 100 mV / 3.1416
  = 31.83 mV

  Compare with standard diode rectifier:
  For Vm = 50 mV < 0.6V, standard diode output = 0V
  Precision rectifier output = 31.83 mV

Final Answer:
  Average output = 31.83 mV
  A standard diode rectifier would give 0V for this signal.

Exam Tip: GATE asks why the op-amp output slews to negative rail during the negative half-cycle, and whether this causes a problem. The issue is called op-amp recovery time. When the input swings positive again, the op-amp must recover from negative saturation, causing a distortion at the zero-crossing. This is why fast op-amps (TL071, LF356) are used. Also remember: the effective forward drop is Vdiode/Aol, not zero. Students sometimes write zero, which is incorrect.

Key Properties

• Overcomes the 0.6V diode threshold by placing the diode inside the op-amp feedback loop.
• Effective threshold = Vdiode/Aol ≈ 6 µV for a gain of 100,000.
• Half-wave version passes only positive half-cycles to the output.
• Full-wave version requires two op-amps and gives Vout = |Vin|.
• TL071 is preferred (high slew rate 13 V/µs, low offset) over LM741 for small-signal rectification.
• Op-amp recovery from negative saturation limits maximum frequency of operation.
• Used in true-RMS AC measurement, envelope detectors, and audio signal processing.

Quick Revision

• Diode placed inside op-amp feedback loop.
• Rectifies signals below 0.6V accurately.
• Effective diode drop = Vdiode/Aol ≈ microvolts.
• Half-wave: one op-amp, one diode. Full-wave: two op-amps, two diodes.
• Average output of full-wave = 2Vm/pi = 0.636*Vm.
• Op-amp recovery from saturation limits high-frequency use.
• TL071 preferred for speed; use LM741 only for low-frequency signals.
• Exam trap: Students apply the ideal rectifier formula and write Vout = Vin - 0.6V for a precision rectifier. This is wrong. The whole point of the precision rectifier is that the op-amp eliminates the diode drop. Write Vout = Vin for positive half-cycles.