---
title: "Multiplier and divider using log-antilog"
subject: "Advanced Electronics"
chapter: 6
key: "semester_4/advanced_electronics:multiplier-divider-log-antilog"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/multiplier-divider-log-antilog
site: BCT Engineering
---

# Multiplier and divider using log-antilog

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2072 Kartik, 7 marks
- Video: [Analog Multiplier using Op-amp | Multiplier circuit using log antilog amplifier](https://www.youtube.com/watch?v=-2PWkOb1F1Y) — Electronics for You
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/multiplier-divider-log-antilog
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

Construct and explain the multiplier circuit using log and antilog amplifier and hence use the multiplier to realize the divider circuit.

## Short answer

For a multiplier, take log V1 and log V2, add them with a summer, then antilog to get Vo = V1 V2/(R Is). For a divider, replace the summer with a subtractor so antilog gives Vo = R Is V1/V2. Inputs must be positive.

## Answer

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

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

1. Log amp outputs:
```text
Va = −V_T ln(V1/(R·Is))
Vb = −V_T ln(V2/(R·Is))
```
2. Inverting summer (Rf = R):
```text
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):
```text
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:

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

```text
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.

## Check yourself

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

1. What is the multiplier output in the answer?
   A. Vo = V1 + V2
   B. Vo = V1/V2
   C. Vo = R Is V1 V2
   D. Vo = V1 V2/(R Is) = K V1 V2
   Answer: D. Antilog of VT ln(V1 V2/(R Is)^2) times R Is gives V1 V2/(R Is). (Key point: Product constant)

2. How is the divider made from the multiplier?
   A. Replace the log amps with antilog amps
   B. Add a second summer
   C. Increase Rf
   D. Replace the summer with a subtractor
   Answer: D. ln V1 - ln V2 = ln(V1/V2). (Key point: Divider change)

3. What is the divider output?
   A. Vo = (V1 - V2)
   B. Vo = V1/V2 / R Is
   C. Vo = K' V1/V2 with K' = R Is
   D. Vo = V1 V2
   Answer: C. Vo = R Is exp(Vc/VT) with Vc = VT ln(V1/V2). (Key point: Divider output)

4. In the feedback-multiplier method, what is Vo?
   A. Vo = -K Vz/Vx
   B. Vo = -Vz/(K Vx)
   C. Vo = -K Vz Vx
   D. Vo = Vz + K Vx
   Answer: B. The virtual-ground condition gives Vz = -K Vo Vx. (Key point: Feedback divider)

## Related in Advanced Electronics

- [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×
- [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×
- [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×
- [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×
- [Multifunction converter using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amultifunction-converter) — Advanced Electronics, chapter 6
- [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

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