Log and Antilog Amplifier

Log and antilog amplifiers exploit the exponential I-V relationship of a diode or transistor junction to compute logarithms and exponentials in hardware. Analog multipliers, RMS detectors, and audio companders in the NE570 compander IC all rely on this principle.

Core Concept

The log amplifier places a diode or BJT in the feedback loop of an inverting op-amp. The virtual ground forces the input current (Vin/R1) through the diode. The diode's forward voltage is proportional to the natural log of this current, so the output voltage is proportional to ln(Vin).

The transdiode configuration replaces the feedback diode with a BJT (BC547) wired as a diode (collector tied to base). This gives better logarithmic accuracy because the transistor has a more ideal exponential characteristic than a simple diode. Temperature affects VT = kT/q = 26 mV at 300K, so practical log amps include temperature compensation.

The antilog (exponential) amplifier is the mirror circuit: the diode or transistor sits at the input. The input voltage drives the diode forward voltage, setting the diode current exponentially. This current flows through the feedback resistor, producing an output proportional to e^(Vin/VT). Cascading a log amp followed by an antilog amp implements analog multiplication, because ln(A) + ln(B) = ln(AB).

Key Equations

Diode current (Shockley equation):

ID = Is * (e^(VD/VT) - 1) ≈ Is * e^(VD/VT) for VD >> VT

Log amplifier output:

Vout = -VT * ln(Vin / (Is * R1))

Where VT = kT/q = 26 mV at 300K, Is is diode saturation current (typically 10 nA for 1N4148), and R1 is the input resistor.

Antilog amplifier output:

Vout = -Is * Rf * e^(Vin/VT)

For the BC547 transdiode in feedback, Is is the BJT reverse saturation current Ico, typically 10 pA, giving better dynamic range than a diode.

Given:
  Log amplifier circuit
  R1 = 10 kΩ = 10,000 Ω
  Vin = 1V
  Is = 10 nA = 10 × 10^-9 A (1N4148 at 25°C)
  VT = 26 mV = 0.026V (at 300K)

Why this formula:
  Vout = -VT * ln(Vin / (Is * R1))
  Input current = Vin/R1 flows through diode

Formula:
  Vout = -VT * ln(Vin / (Is * R1))

Substitution:
  Input current = 1 / 10,000 = 100 µA = 10^-4 A
  Vout = -0.026 * ln(10^-4 / 10^-8)

Calculation:
  Ratio = 10^-4 / 10^-8 = 10^4 = 10000
  ln(10000) = ln(10^4) = 4 * ln(10) = 4 * 2.3026 = 9.21
  Vout = -0.026 * 9.21 = -0.239 V

Final Answer:
  Vout = -0.239 V ≈ -240 mV

Exam Tip: GATE tests whether the diode is in the feedback (log amp) or at the input (antilog amp). A common trap is temperature dependence: VT = kT/q changes with temperature, so the output of a log amp drifts unless compensated. Also, the input to a log amp must be a positive voltage only (the diode must be forward biased). Negative inputs are not valid and will cause the diode to cut off.

Key Properties

• Log amplifier: diode or BJT in feedback. Vout = -VT * ln(Vin/(IsR1)).
• Antilog amplifier: diode or BJT at input. Vout = -IsRf * e^(Vin/VT).
• VT = 26 mV at 300K. Output is temperature-dependent and drifts without compensation.
• The transdiode (BC547 with collector tied to base) gives wider dynamic range than a 1N4148 diode.
• Input voltage must be positive for the log amp (diode must be forward biased). Range is typically 1 mV to 10V.
• Two log amps plus a summing amp plus an antilog amp form an analog multiplier, used in the AD633 IC.
• Audio companders (NE570, NE571) use log/antilog stages to compress and expand dynamic range.

Quick Revision

• Log amp: diode in feedback. Output = -VT * ln(Vin/IsR1).
• Antilog amp: diode at input. Output = -Is*Rf * e^(Vin/VT).
• VT = kT/q = 26 mV at 300K.
• Input must be positive (diode forward biased).
• BC547 transdiode gives better accuracy than 1N4148.
• Cascading log + summing + antilog = analog multiplier.
• Temperature compensation is needed for accurate results.
• Exam trap: Swapping log and antilog circuits in a question. The position of the diode (feedback vs input) tells you which is which. Do not guess from the equation alone.