---
title: "Need for log-antilog circuits; analog multiplier"
subject: "Advanced Electronics"
chapter: 6
key: "semester_4/advanced_electronics:why-log-antilog-multiplier"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/why-log-antilog-multiplier
site: BCT Engineering
---

# Need for log-antilog circuits; analog multiplier

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2073 Chaitra, 2+4 marks; 2071 Chaitra, 1+6 marks (asked 2 times)
- 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/why-log-antilog-multiplier
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

Why log antilog amplifier circuits are required? Draw the circuit diagram of antilog (analog) multiplier and explain it.

## Short answer

Logarithms convert multiplication into addition, enabling analog multiplication/division with linear circuits. The multiplier uses two log amplifiers for -K ln V1 and -K ln V2, an inverting summer to get K ln(V1V2), and an antilog to output V1·V2.

## Answer

#### Why log–antilog circuits are required
Logarithms turn multiplication into addition and powers into multiplication by a constant:

```text
ln(xy)  = ln x + ln y
ln(x/y) = ln x − ln y
ln(xⁿ)  = n ln x
```

Adders, subtractors and gain stages are easy to build with op-amps. So log and antilog amplifiers let simple linear circuits perform non-linear operations:
- Analog **multiplication and division** of signals
- **Powers and roots** (square, square root, xⁿ), RMS and vector magnitude
- **Compression** of signals with a very wide dynamic range (5–6 decades) into a small voltage range, for example in sonar, radar, audio and dB meters
- Linearizing sensors with exponential responses, and pH or light measurement

#### Antilog (analog) multiplier
**Building blocks used** (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):
- Log amp (L): output = −K ln(Vin), for Vin > 0
- Antilog amp (E): output = exp(Vin/K)
- Inverting amp of gain −m: output = −m × input

```text
 V1 -> [LOG A1] --Va--[R]--+
                           |       R
 V2 -> [LOG A2] --Vb--[R]--+--+--/\/\/--+
                              |         |
                            (-)\  A3    |
                       GND--(+)  >------+--> Vs
 Vs -> [ANTILOG A4] -> Vo
```

**Working:**
1. **Log stages:** A1 and A2 (each with a transistor in feedback) give:

```text
Va = −K ln V1
Vb = −K ln V2
```

2. **Inverting summer A3** with all resistors R:

```text
Vs = −(Va + Vb) = K (ln V1 + ln V2) = K ln(V1·V2)
```

3. **Antilog stage A4:**

```text
Vo = exp(Vs/K) = exp( ln(V1·V2) ) = V1·V2
```

With a general reference voltage the scaled result is Vo = V1·V2/Vref. For example, V1 = 2 V and V2 = 3 V give Vo = 6 V (Vref = 1 V).

**Division:** replace the summer with a difference amplifier. Then Vs = K ln(V1/V2) and Vo = V1/V2.

**Limitation:** the inputs must be positive (one-quadrant operation). Matched transistors and temperature-compensating resistors are needed for accuracy.

## Check yourself

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

1. Why are log-antilog circuits used for multiplication?
   A. Because they are cheaper than resistors
   B. Because ln(xy) = ln x + ln y, so multiplication becomes addition, which op-amps do easily
   C. Because logs amplify signals
   D. Because they remove noise
   Answer: B. Adders are easy to build, so addition replaces multiplication. (Key point: Why logs)

2. In the antilog multiplier, what does the inverting summer A3 give if Va = -K ln V1 and Vb = -K ln V2?
   A. Vs = K ln(V1 V2)
   B. Vs = -K ln(V1 V2)
   C. Vs = K ln(V1/V2)
   D. Vs = K(V1 + V2)
   Answer: A. Vs = -(Va + Vb) = K(ln V1 + ln V2). (Key point: Summer output)

3. How is the circuit changed to divide?
   A. Replace the antilog with a log
   B. Add a second antilog
   C. Swap the transistors
   D. Replace the summer with a difference amplifier
   Answer: D. ln V1 - ln V2 = ln(V1/V2). (Key point: Division)

4. What is a limitation of the log-antilog multiplier?
   A. It cannot divide
   B. It needs no reference
   C. Inputs must be positive (one-quadrant)
   D. It works only for AC
   Answer: C. The log of a non-positive number is undefined. (Key point: Limitation)

## Related in Advanced Electronics

- [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
- [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×
- [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×
- [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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