Design Vo = (Vx/Vy)^α·Vz with log-antilog

The idea is to work in the logarithmic domain:

ln V0 = α (ln Vx − ln Vy) + ln Vz
V0 = exp[ α ln(Vx/Vy) + ln Vz ] = (Vx/Vy)^α · Vz

So the circuit must find logs, subtract, multiply by α, add, and take the antilog.

Building blocks used (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):

• Log amp (L): output = −K ln(Vin), for Vin > 0
• Antilog amp (E): output = exp(Vin/K)
• Inverting amp of gain −m: output = −m × input

Block diagram

 Vx->[LOG A1]--Vx'-->(-)[A4: diff amp]
 Vy->[LOG A2]--Vy'-->(+)[ gain 1     ]--Vd
                                          |
             [A5: non-inverting, gain α]<-+
                          |
                          Ve
                          v
                    (+)[A6: diff amp]--Vs
 Vz->[LOG A3]--Vz'->(-)[ gain 1     ]   |
                                       v
                          [ANTILOG A7] --> V0

Design steps

1. Log amplifiers (transistor in feedback, matched, all with the same K):

Vx' = −K ln Vx,  Vy' = −K ln Vy,  Vz' = −K ln Vz

2. Difference amplifier A4 (all four resistors R = 10 kΩ), with Vy' on the (+) input and Vx' on the (−) input:

Vd = Vy' − Vx' = K (ln Vx − ln Vy) = K ln(Vx/Vy)

3. Non-inverting amplifier A5 with gain α = 1 + Rf/R1:

Ve = α K ln(Vx/Vy) = K ln( (Vx/Vy)^α )

Since α > 1, a non-inverting amplifier can give it directly:

Rf = (α − 1) R1

For example, α = 2.5 with R1 = 10 kΩ gives Rf = 15 kΩ. Make Rf variable to set any α.

4. Difference amplifier A6 (all resistors R), with Ve on the (+) input and Vz' on the (−) input:

Vs = Ve − Vz' = K ln( (Vx/Vy)^α ) + K ln Vz
   = K ln( (Vx/Vy)^α · Vz )

5. Antilog amplifier A7:

V0 = exp(Vs/K) = (Vx/Vy)^α · Vz

Check

Take Vx = 4 V, Vy = 2 V, Vz = 1.5 V, α = 2. Then V0 = (2)² × 1.5 = 6 V.

Notes: Vx, Vy and Vz must be positive. The log amps use matched transistors, so Is cancels, and temperature-compensating resistors, so the drift of VT is compensated. The output is scaled by the reference voltage (1 V here).