Log amplifier transfer function, matched transistors

A log amplifier gives an output proportional to the logarithm of its input. In the matched-transistor version, a second identical transistor carrying a reference current cancels the saturation current Is, which depends strongly on temperature.

Circuit

 Vin-[R1]-+-(-)\   Q1: C at (-), E at Vo1
          |  A1 >--Vo1--[R3]--+--[R4]--+
 GND-----(+)/                 |        |
                             (-)\      |
                                 A3>---+--> Vo
                             (+)/
 Vref-[R2]-+-(-)\   Q2        |
           |  A2 >--Vo2--[R3]-+--[R4]--GND
 GND------(+)/
 (bases of Q1, Q2 grounded; Q1, Q2 matched)

A1 and A2 are log stages. A3 is a difference amplifier of gain R4/R3.

Derivation

The transistor law gives VBE = VT ln(IC/Is).

Stage A1:

IC1 = Vin/R1
Vo1 = −VBE1 = −VT ln( Vin / (R1·Is) )

Stage A2:

IC2 = Vref/R2
Vo2 = −VBE2 = −VT ln( Vref / (R2·Is) )

Difference amplifier A3:

Vo = (R4/R3)(Vo2 − Vo1)
   = (R4/R3)·VT [ ln(Vin/(R1·Is)) − ln(Vref/(R2·Is)) ]
   = (R4/R3) · VT · ln( (Vin·R2) / (Vref·R1) )

Transfer function

With R1 = R2:

Vo = (R4/R3) · VT · ln( Vin / Vref )
   = 2.303 (R4/R3) VT · log10( Vin / Vref )

• Is has cancelled, because Q1 and Q2 are matched and at the same temperature.
• At 300 K, VT ≈ 25.9 mV. With R4/R3 = 1, the output changes by about 60 mV per decade of Vin.
• When Vin = Vref, the output is 0.
• Vin and Vref must be positive (for npn transistors).
• VT = kT/q still varies with temperature. It is compensated by making R3 a temperature-sensitive resistor.