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Output of three log amps with summer

Short answer

Each stage is a log amp giving V0i = -VT ln(Vi/(R Is)). The summing amplifier with feedback 2R and input resistances R has gain -2, so V0 = 2VT ln(V1 V2 V3/(R Is)^3). The output is proportional to the log of the product cubed.

Find the relationship of input and output voltage in the following figure. [Figure: three log amplifiers: V1, V2 and V3 each pass through a resistor R to the inverting input of their own op-amp (non-inverting inputs grounded), with transistors Q1, Q2 and Q3 (bases grounded) as the respective feedback elements. Each log-amp output passes through a resistor R to the inverting input of a summing op-amp with feedback resistor 2R; its non-inverting input is grounded; its output is V0.]

Each input stage is a log amplifier (transistor in the feedback path), and the last stage is an inverting summer with gain −2R/R = −2.

Log amplifier outputs

For a log amp with input resistor R and grounded-base transistor in feedback:

V01 = −V_T ln( V1 / (R·Is) )
V02 = −V_T ln( V2 / (R·Is) )
V03 = −V_T ln( V3 / (R·Is) )

where V_T = kT/q ≈ 26 mV at room temperature and Is is the reverse saturation current (transistors assumed matched).

Summing amplifier

Inputs through R, feedback 2R:

V0 = −(2R/R)(V01 + V02 + V03)
   = −2 (V01 + V02 + V03)

Substituting:

V0 = 2V_T [ ln(V1/RIs) + ln(V2/RIs) + ln(V3/RIs) ]
   = 2V_T ln( V1·V2·V3 / (R·Is)³ )

Answer: V0 = 2V_T · ln[ V1·V2·V3 / (R·Is)³ ]

This can also be written as V0 = V_T ln[ (V1·V2·V3)² / (R·Is)⁶ ], i.e. the output is proportional to the log of the square of the product of the three inputs. The circuit works only for positive V1, V2, V3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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