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Matched-transistor log amp and temperature effects

Short answer

Matched transistors cancel Is, so output is Vo = K VT ln(Vin/Vref). To minimize temperature drift due to VT, the gain K is made temperature-dependent by using a resistor with +0.33%/°C temperature coefficient, keeping K VT constant. The reference current sets the zero-crossing point.

Explain the operation of log amplifier using matched transistor. Explain how the effect of temperature can be minimized in this circuit.

A log amplifier produces Vo ∝ ln(Vin). The basic one-transistor log amp gives Vo = −VT ln(Vin/(R1·Is)). This has two temperature-dependent terms: Is (which rises very steeply with temperature) and VT = kT/q. The matched-transistor log amp removes Is and a temperature-sensitive resistor removes the VT effect.

Circuit

 Vin-[R1]-+-(-)\  Q1 (C at -, E at out, B GND)
          |  A1 >---Vo1---[R3]--+--[R4]--+
 GND-----(+)/                   |        |
                               (-)\      |
                                   A3>---+--> Vo
                               (+)/
 Vref-[R2]-+-(-)\  Q2            |
           |  A2 >--Vo2--[R3]----+--[R4]--GND
 GND------(+)/
   Q1, Q2: matched pair on one chip (same temperature)
   R3: temperature-sensitive resistor (RTC)

Operation

  1. A1 with Q1 (log of input). The current Vin/R1 flows through Q1, and A1's output settles at −VBE1:
Vo1 = −VT ln( Vin / (R1·Is) )
  1. A2 with Q2 (log of reference). It is driven by a fixed current Vref/R2:
Vo2 = −VT ln( Vref / (R2·Is) )
  1. A3 subtracts the two outputs with gain K = R4/R3:
Vo = K (Vo2 − Vo1) = K·VT·ln( Vin·R2 / (Vref·R1) )

With R1 = R2: Vo = K·VT·ln(Vin/Vref).

The output follows the logarithm of the input ratio. It is about 60 mV per decade per unit K at room temperature.

How the temperature effect is minimized

  1. Is is removed by matching. Q1 and Q2 are made on the same chip (a matched pair), so they have the same Is and the same temperature. In the subtraction, ln(1/Is) − ln(1/Is) = 0, so the large drift of Is disappears.
  2. VT is compensated by a temperature-sensitive gain. The output is still proportional to VT = kT/q, which rises by about 0.33 %/°C at 25 °C. So the gain K is made to fall at the same rate. R3 (or RTC in a non-inverting gain stage, K = 1 + R2/RTC) is a resistor with a positive temperature coefficient of about +0.33 %/°C. As T rises, RTC rises and K falls, so K·VT stays constant.
  3. Using a reference current (Vref/R2) also fixes the zero point of the log curve (Vo = 0 at Vin = Vref).
  4. Low-bias-current, low-offset op-amps reduce errors at small input currents.

The result is a log amplifier that is accurate over several decades of input with little drift.

✓ Answer checked by Claude Opus 5.5 (high effort)

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