Multiplier and divider using log-antilog

A multiplier can be built by taking the log of each input, adding the logs and then taking the antilog, because ln V1 + ln V2 = ln(V1·V2). A divider uses subtraction instead: ln V1 − ln V2 = ln(V1/V2).

Building blocks

• Log amp (transistor in feedback): Vo = −V_T ln(Vi / (R·Is))
• Antilog amp (transistor at input): Vo = −R·Is·e^(Vi/V_T)

Multiplier circuit

V1 -->[LOG AMP 1]--Va--R--+
                          |
                     [INVERTING   ]
                     [SUMMER, Rf=R]--Vc-->[ANTILOG]--> Vo
                          |
V2 -->[LOG AMP 2]--Vb--R--+

1. Log amp outputs:

Va = −V_T ln(V1/(R·Is))
Vb = −V_T ln(V2/(R·Is))

2. Inverting summer (Rf = R):

Vc = −(Va + Vb) = V_T [ln(V1/RIs) + ln(V2/RIs)]
   = V_T ln( V1·V2 / (R·Is)² )

3. Antilog amp (transistor polarity chosen to accept this sign; output written in magnitude):

Vo = R·Is·e^(Vc/V_T) = R·Is · V1·V2/(R·Is)²
Vo = V1·V2 / (R·Is) = K·V1·V2,   K = 1/(R·Is)

So the output is proportional to the product of the two inputs.

Divider from the multiplier

Replace the summer by a difference amplifier (subtractor) so that the log of the divisor is subtracted. Because the log amps invert, Vb goes to the (+) input and Va to the (−) input:

V1 -->[LOG AMP 1]--Va--->(−)
                         [SUBTRACTOR]--Vc-->[ANTILOG]--> Vo
V2 -->[LOG AMP 2]--Vb--->(+)

Vc = Vb − Va = V_T [ln(V1/RIs) − ln(V2/RIs)]
   = V_T ln(V1/V2)
Vo = R·Is · e^(Vc/V_T) = R·Is · (V1/V2)
Vo = K'·V1/V2,   K' = R·Is

Another simple way: put the multiplier in the feedback path of an op-amp. With Vz at the input resistor and the multiplier output K·Vo·Vx fed back, the virtual-ground condition gives Vz = −K·Vo·Vx, so Vo = −Vz/(K·Vx), a divider.

Notes:

• Inputs must be positive (one-quadrant operation).
• Matched transistors on the same chip cancel Is; temperature compensation is needed for V_T.