---
title: "Four- vs single-quadrant multiplier"
subject: "Advanced Electronics"
chapter: 6
key: "semester_4/advanced_electronics:four-vs-single-quadrant-multiplier"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/four-vs-single-quadrant-multiplier
site: BCT Engineering
---

# Four- vs single-quadrant multiplier

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2073 Shrawan, 1+7 marks
- Video: [#223: Basics of the Gilbert Cell | Analog Multiplier | Mixer | Modulator](https://www.youtube.com/watch?v=7nmmb0pqTU0) — w2aew
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/four-vs-single-quadrant-multiplier
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

How four quadrant multiplier differs from single quadrant multiplier? Derive output voltage of four quadrant multiplier with necessary diagrams.

## Short answer

A single-quadrant multiplier, usually log-antilog, accepts only positive inputs, while a four-quadrant Gilbert cell accepts any signs and gives Vo = K Vx Vy. Derivation uses IEE tanh(Vx/2VT) tanh(Vy/2VT), approximating to K Vx Vy with K = IEE RL/(4VT^2).

## Answer

#### Difference between four-quadrant and single-quadrant multiplier

| Point | Single-quadrant | Four-quadrant |
|---|---|---|
| Allowed inputs | Vx > 0 and Vy > 0 only | Vx and Vy of any sign |
| Output sign | Always one polarity | Follows sign of Vx·Vy |
| Typical circuit | Log–antilog multiplier | Gilbert cell (transconductance) |
| Use | DC/positive signals | AC signals, modulators, phase detectors |

A log amplifier cannot accept negative input (ln of a negative number is undefined), so a log-antilog multiplier is only single-quadrant. A four-quadrant multiplier works in all four quadrants of the Vx–Vy plane.

#### Output voltage of four-quadrant multiplier (Gilbert cell)

```text
         +Vcc            +Vcc
          RL              RL
          |  Io1     Io2  |
          +--+--  ...  +--+----> Vo
         Q3  Q4       Q5  Q6    <- Vx (cross-coupled)
           \/           \/
           I1           I2
           Q1 ---------- Q2     <- Vy
                  |
                 IEE
                  |
                -VEE
```

Q3, Q5 collectors join to give Io1; Q4, Q6 collectors join to give Io2.

For a differential pair: ΔI = I·tanh(V/2V_T).

1. Lower pair:
```text
I1 − I2 = IEE·tanh(Vy/2V_T)
```
2. Upper pairs (both driven by Vx, outputs cross-coupled):
```text
IC3 − IC4 = I1·tanh(Vx/2V_T)
IC6 − IC5 = I2·tanh(Vx/2V_T)
```
3. Output:
```text
ΔIo = (IC3 + IC5) − (IC4 + IC6)
    = (I1 − I2)·tanh(Vx/2V_T)
    = IEE·tanh(Vx/2V_T)·tanh(Vy/2V_T)
```
4. For small inputs, tanh(u) ≈ u:
```text
ΔIo ≈ IEE·Vx·Vy / (4V_T²)
Vo  = RL·ΔIo = K·Vx·Vy,  K = IEE·RL/(4V_T²)
```

Because tanh is odd, a change of sign of Vx or Vy changes the sign of Vo, so all four quadrants are covered. Practical ICs (AD633) linearise the cell and give Vo = Vx·Vy/10 V.

## Check yourself

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

1. What input condition does a single-quadrant multiplier need?
   A. Inputs of any sign
   B. Only AC inputs
   C. Both inputs positive
   D. Both inputs negative
   Answer: C. A log-antilog multiplier takes only positive inputs. (Key point: Input restriction)

2. Why is a log-antilog multiplier only single-quadrant?
   A. Transistors are too slow
   B. It needs a negative supply
   C. The log of a negative number is undefined
   D. The antilog can't give positive output
   Answer: C. A log amp cannot accept a negative input. (Key point: Why log is one-quadrant)

3. What is the output of the Gilbert cell for small inputs?
   A. Vo = K Vx/Vy
   B. Vo = K Vx Vy, with K = IEE RL/(4 VT^2)
   C. Vo = K(Vx + Vy)
   D. Vo = K ln(Vx Vy)
   Answer: B. tanh(u) is about u for small u. (Key point: Gilbert cell output)

4. Where is a four-quadrant multiplier needed?
   A. AC signals, modulators and phase detectors
   B. Only DC positive signals
   C. Only for logs
   D. Voltage regulators
   Answer: A. These signals have both polarities. (Key point: Applications)

## Related in Advanced Electronics

- [Four-quadrant multiplier: concept and output](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afour-quadrant-multiplier-concept) — Advanced Electronics, chapter 6
- [Four-quadrant multiplier circuit and derivation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afour-quadrant-multiplier-derive) — Advanced Electronics, chapter 6
- [DIAC drawback; TRIAC four-quadrant operation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adiac-drawback-triac-quadrants) — Advanced Electronics, chapter 7

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