---
title: "Multifunction converter using log-antilog"
subject: "Advanced Electronics"
chapter: 6
key: "semester_4/advanced_electronics:multifunction-converter"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/multifunction-converter
site: BCT Engineering
---

# Multifunction converter using log-antilog

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2070 Asar, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/multifunction-converter
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

How a multifunction converter can be constructed using log and antilog amplifiers? Explain with circuit realization.

## 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.

## Answer

A **multifunction converter** is a log-antilog circuit whose output is

```text
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

```text
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).

1. Log of Vz and Vx:
```text
V1 = V_T ln(Vz/Vref)
V2 = V_T ln(Vx/Vref)
```
2. Subtractor:
```text
V3 = V1 − V2 = V_T ln(Vz/Vx)
```
3. Gain stage (voltage divider or amplifier, gain m):
```text
V4 = m·V_T ln(Vz/Vx) = V_T ln[(Vz/Vx)^m]
```
4. Add log of Vy:
```text
V6 = V4 + V_T ln(Vy/Vref)
   = V_T ln[ (Vy/Vref)·(Vz/Vx)^m ]
```
5. Antilog amp:
```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the output of the multifunction converter?
   A. Vo = Vy + m Vz
   B. Vo = Vz^m / Vy
   C. Vo = Vy (Vz/Vx)^m
   D. Vo = Vy (Vx/Vz)^m
   Answer: C. Subtract log Vx from log Vz, multiply by m, add log Vy, then take the antilog. (Key point: Output form)

2. Which setting gives a squarer?
   A. m = 1, Vx = 1 V
   B. m = 2, Vx = Vy = 1 V
   C. m = 0.5, Vx = Vy = 1 V
   D. m = 1, Vy = 1 V
   Answer: B. Vo = Vz^2. (Key point: Squarer setting)

3. Which setting makes it a divider Vz/Vx?
   A. m = 0.5 and Vx = 1 V
   B. m = 1 and Vy = 1 V
   C. m = 1 and Vx = 1 V
   D. m = 2 and Vy = 1 V
   Answer: B. Vo = (1)(Vz/Vx). (Key point: Divider setting)

4. How is the exponent m set?
   A. Divider for m < 1, non-inverting amplifier for m > 1
   B. Changing the transistor
   C. Changing Vref
   D. Adding a diode
   Answer: A. The gain stage in the log domain sets m. (Key point: Setting m)

## Related in Advanced Electronics

- [RMS of sinusoid using log-antilog amplifiers](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Arms-sinusoid-log-antilog) — Advanced Electronics, chapter 6
- [Short note: Applications of log-antilog amplifiers](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ashort-note-log-antilog-applications) — Advanced Electronics, chapter 6, asked 2×
- [Design Vo = (Vx/Vy)^α·Vz with log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adesign-vx-vy-power-alpha) — Advanced Electronics, chapter 6, asked 2×
- [Log-antilog circuit: Vo in terms of Vi and VR](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-antilog-circuit-vo-vi) — Advanced Electronics, chapter 5, asked 3×
- [Need for log-antilog circuits; analog multiplier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Awhy-log-antilog-multiplier) — Advanced Electronics, chapter 6, asked 2×
- [Multiplier and divider using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amultiplier-divider-log-antilog) — Advanced Electronics, chapter 6

## Related in other subjects

- [Classification of amplifiers by operating region](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aamplifier-classification-operating-region) — Electronics Devices and Circuit, chapter 4