- 2070 Asar · 7 marks
Multifunction converter using log-antilog
Short answer
A multifunction converter uses log amps to take logs of Vz and Vx, subtract them, apply gain m, add log of Vy, then antilog. Its output is Vo = Vy·( Vz / Vx )^m, performing multiplication, division, squaring, square-root, etc. by selecting inputs and setting m via a resistor divider or amplifier.
How a multifunction converter can be constructed using log and antilog amplifiers? Explain with circuit realization.
A multifunction converter is a log-antilog circuit whose output is
Vo = Vy · (Vz / Vx)^m
By choosing the inputs and the exponent m, the same circuit performs multiplication, division, squaring, square-root and other powers (e.g. the Burr-Brown 4302 IC).
Circuit realization
Vz -->[LOG AMP]--V1-->(+)
[SUBTRACTOR]--V3-->[GAIN m]--V4
Vx -->[LOG AMP]--V2-->(−) (pot/R ratio)
|
Vy -->[LOG AMP]--V5---------------->[SUMMER]<-+
|
V6
|
[ANTILOG AMP]--> Vo
Working and derivation
Take each log amp as V = V_T ln(Vin/Vref), with Vref = R·Is (signs handled by inverting stages).
- Log of Vz and Vx:
V1 = V_T ln(Vz/Vref)
V2 = V_T ln(Vx/Vref)
- Subtractor:
V3 = V1 − V2 = V_T ln(Vz/Vx)
- Gain stage (voltage divider or amplifier, gain m):
V4 = m·V_T ln(Vz/Vx) = V_T ln[(Vz/Vx)^m]
- Add log of Vy:
V6 = V4 + V_T ln(Vy/Vref)
= V_T ln[ (Vy/Vref)·(Vz/Vx)^m ]
- Antilog amp:
Vo = Vref · e^(V6/V_T)
Vo = Vy · (Vz/Vx)^m
Functions obtained
| Setting | Output | Function |
|---|---|---|
| m = 1, Vx = 1 V | Vy·Vz | Multiplier |
| m = 1, Vy = 1 V | Vz/Vx | Divider |
| m = 2, Vx = Vy = 1 V | Vz² | Squarer |
| m = 0.5, Vx = Vy = 1 V | √Vz | Square root |
| 0.2 < m < 5 | Vy(Vz/Vx)^m | General power/root |
m < 1 is set with a resistor divider; m > 1 with a non-inverting amplifier. Matched transistors cancel Is, and all inputs must be positive.
✓ Answer checked by Claude Opus 5.5 (high effort)
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