Matched-transistor antilog amp and temperature

An antilog (exponential) amplifier produces an output proportional to the exponential of its input. Using two matched transistors with a reference current cancels the saturation current Is. A temperature-sensitive resistor then compensates VT.

Circuit

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

A1 forces Q1's collector current to Iref = Vref/R1 by driving the common emitter node E. The base of Q2 gets Vb = Vin·RTC/(R2 + RTC).

Input–output relation

Using VBE = VT ln(IC/Is):

Q1:

IC1 = Vref/R1,  base at 0 V
VE  = −VT ln( Vref/(R1·Is) )

Q2:

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

A2 (current-to-voltage):

Vo = Rf · IC2

Result:

Vo = (Rf·Vref/R1) · exp[ (RTC/(R2 + RTC)) · Vin / VT ]

For example, with K = RTC/(R2 + RTC) and 2.303VT/K = 1 V:

Vo = (Rf·Vref/R1) · 10^(Vin)

Minimizing the effect of temperature

1. Matched pair. Q1 and Q2 are made on the same chip and kept at the same temperature. Their Is values are equal and cancel, as shown above. Is is the most temperature-sensitive term, roughly doubling every 10 °C.
2. Compensating VT = kT/q. The exponent is K·Vin/VT, and VT rises about 0.33 %/°C. RTC is a resistor with a positive temperature coefficient (about +0.3 %/°C, with R2 ≫ RTC). Then K ≈ RTC/R2 rises with T at the same rate as VT, so K/VT, and therefore the exponent, stays constant.
3. A stable reference Vref and low-bias-current op-amps keep the scale factor Rf·Vref/R1 constant.

With these measures the antilog output stays accurate over a wide temperature range.