Output of temperature-compensated log circuit

This is the temperature-compensated log amplifier. Q1 carries the signal current, while Q2 with A2 carries a fixed reference current. The divider R2–RTC sets the scale and cancels the effect of VT.

Assumptions

Q1 and Q2 are matched (same Is). The op-amps are ideal. The reference resistor (marked R2 at A2's input) is called R2' here to keep it separate from the divider R2. VB is small compared with VRef.

Currents

Q1: its collector is at A1's virtual ground (0 V), its base is grounded and its emitter is at VA.

IC1 = vi / R1
VBE1 = 0 − VA   →   VA = −VT ln( vi / (R1·Is) )

Q2: A2's inverting input is held at VB (virtual short with its + input). Q2's base is also at VB, and its emitter is at VA (driven by A2's output).

Iref = IC2 = (VRef − VB)/R2' ≈ VRef/R2'
VBE2 = VB − VA = VT ln( Iref / Is )

Solving for VB

VB = VA + VT ln(Iref/Is)
   = −VT ln( vi/(R1·Is) ) + VT ln( Iref/Is )
   = −VT ln( vi / (R1·Iref) )

Is has cancelled.

Output voltage

A1's output V0 feeds the divider, so VB = V0·RTC/(R2 + RTC). Therefore:

V0 = (1 + R2/RTC) · VB

V0 = −(1 + R2/RTC) · VT · ln( vi / (R1·Iref) )
   = −(1 + R2/RTC) · VT · ln( vi·R2' / (R1·VRef) )

In log10 form:

V0 = −2.303 (1 + R2/RTC) VT · log10( vi·R2' / (R1·VRef) )

Why it is temperature compensated

• Is cancels because Q1 and Q2 are matched.
• VT = kT/q rises with temperature (+0.33 %/°C). RTC has a positive temperature coefficient, so as T rises RTC rises and (1 + R2/RTC) falls. With R2 ≫ RTC the factor is about R2/RTC, so the product (R2/RTC)·VT stays nearly constant.

For example, with (1 + R2/RTC)·2.303VT = 1 V, the output falls by 1 V per decade of vi.