Antilog amplifier with matched transistors

An antilog (exponential) amplifier gives an output proportional to the exponential of its input. With two matched transistors and two op-amps, the saturation current Is cancels out and a reference current sets the scale.

Circuit

 Vref-[R1]--+--(-)\
            |C     A1>------+
           Q1  GND-(+)/     |
       B=GND \E             |
              +-----node E--+
              |E
           Q2  \C                     Rf
 Vin-[R2]-+-B   +---------+--------/\/\/----+
          |               |                 |
        [R3]            (-)\                |
          |                 A2>-------------+--> Vo
         GND         GND-(+)/

Q1 and Q2 are a matched pair with their emitters tied at node E. A1 forces Q1 to carry Iref = Vref/R1. The base of Q2 receives Vb = Vin·R3/(R2 + R3) (a divider used for scaling and temperature compensation).

Derivation

The transistor law is IC = Is·exp(VBE/VT).

Q1 (base at 0 V, collector at virtual ground):

IC1 = Vref/R1,   VBE1 = 0 − VE
VE = −VT ln( Vref / (R1·Is) )

Q2 (base at Vb, emitter at VE):

VBE2 = Vb − VE = Vb + VT ln( Vref/(R1·Is) )
IC2  = Is · exp(VBE2/VT)
     = Is · exp(Vb/VT) · Vref/(R1·Is)
     = (Vref/R1) · exp(Vb/VT)

A2 works as a current-to-voltage converter. IC2 flows from the output through Rf:

Vo = Rf · IC2 = (Rf·Vref/R1) · exp(Vb/VT)

With Vb = K·Vin, where K = R3/(R2 + R3):

Vo = (Rf·Vref/R1) · exp( K·Vin / VT )
   = (Rf·Vref/R1) · 10^( K·Vin / (2.303 VT) )

Points to note

• Is (strongly temperature dependent) cancels because Q1 and Q2 are matched.
• VT = kT/q remains. Making R3 (or R2) a temperature-sensitive resistor keeps K/VT constant.
• At 300 K, 2.303VT ≈ 60 mV. With K = 1, the output rises ten times for every 60 mV increase of Vin.
• The output is always of one polarity, set by Vref.