--- title: "Four-quadrant multiplier circuit and derivation" subject: "Advanced Electronics" chapter: 6 key: "semester_4/advanced_electronics:four-quadrant-multiplier-derive" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/four-quadrant-multiplier-derive site: BCT Engineering --- # Four-quadrant multiplier circuit and derivation Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications - Asked in: 2074 Asoj, 7 marks - Video: [Gilbert multiplier in English](https://www.youtube.com/watch?v=vTs0pt__qnA) — BE Technical - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/four-quadrant-multiplier-derive - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6 ## Question Draw the detailed circuit diagram for four quadrant multiplier and derive its input and output relationship. ## Short answer The Gilbert cell uses a lower differential pair and cross-coupled upper pairs, giving output current difference IEE tanh(Vx/2VT) tanh(Vy/2VT). For small inputs, Vo = K Vx Vy with K = IEE RL/(4VT^2). Because tanh is odd, it works in all four quadrants. ## Answer A **four-quadrant multiplier** gives Vo = K·Vx·Vy for both positive and negative values of Vx and Vy. The standard circuit is the **Gilbert cell** (variable-transconductance multiplier), used in ICs such as AD633 and MC1496. #### Circuit ```text +Vcc +Vcc | | RL RL | Io1 Io2 | +---+ +---+---> Vo (differential) | | | | Q3 Q4 Q5 Q6 upper pairs: input Vx \ / \ / (Q3,Q6 bases at +Vx/2, \/ \/ Q4,Q5 bases at -Vx/2) I1 I2 | | Q1 ------ Q2 lower pair: input Vy \ / \ / IEE (tail current source) | -VEE ``` Collectors of Q3 and Q5 are joined (current Io1), and collectors of Q4 and Q6 are joined (current Io2). #### Derivation For a BJT differential pair with tail current I and differential input V, the collector current difference is ```text ΔI = I · tanh(V / 2V_T) ``` **Lower pair (Q1, Q2), input Vy:** ```text I1 − I2 = IEE · tanh(Vy / 2V_T) ``` **Upper pairs:** Q3–Q4 is fed by I1 and Q5–Q6 by I2, both driven by Vx but cross-connected: ```text (IC3 − IC4) = I1 · tanh(Vx / 2V_T) (IC6 − IC5) = I2 · tanh(Vx / 2V_T) ``` **Output current difference:** ```text ΔIo = Io1 − Io2 = (IC3 + IC5) − (IC4 + IC6) = (IC3 − IC4) − (IC6 − IC5) = (I1 − I2) · tanh(Vx / 2V_T) = IEE · tanh(Vx/2V_T) · tanh(Vy/2V_T) ``` For small inputs (|V| << 2V_T), tanh(u) ≈ u: ```text ΔIo ≈ IEE · Vx · Vy / (4 V_T²) Vo = RL · ΔIo = (IEE · RL / 4V_T²) · Vx · Vy Vo = K · Vx · Vy, K = IEE·RL / (4 V_T²) ``` Since tanh is an odd function, the sign of ΔIo follows the signs of both Vx and Vy, so the circuit works in **all four quadrants**. In practical ICs, emitter-degeneration resistors and pre-distortion (log) circuits extend the linear range so that K becomes a fixed constant, usually 1/10 V⁻¹ (Vo = Vx·Vy/10). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the differential output current of a BJT differential pair with tail current I and input V? A. I ln(V/VT) B. I tanh(V/2VT) C. I exp(V/VT) D. I V/VT Answer: B. ΔI = I tanh(V/2VT). (Key point: Differential pair law) 2. What is the output current difference of the Gilbert cell? A. IEE ln(Vx Vy) B. IEE tanh(Vx/2VT) tanh(Vy/2VT) C. IEE tanh(Vx + Vy) D. IEE (Vx + Vy)/VT Answer: B. The lower pair gives IEE tanh(Vy/2VT), which multiplies tanh(Vx/2VT) in the upper pairs. (Key point: Cell result) 3. For small inputs, what is K in Vo = K Vx Vy? A. IEE RL / (4 VT^2) B. IEE RL / VT C. RL / IEE D. IEE VT / RL Answer: A. tanh(u) is about u, so ΔIo = IEE Vx Vy/(4 VT^2), then multiplied by RL. (Key point: Small signal constant) 4. Why does the Gilbert cell work for both signs of Vx and Vy? A. Because it uses logs B. Because Vcc is negative C. Because RL is negative D. tanh is an odd function, so the sign of the output follows the signs of both inputs Answer: D. Each tanh changes sign with its input. (Key point: Why four-quadrant) ## 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- vs single-quadrant multiplier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afour-vs-single-quadrant-multiplier) — 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 ## Related in other subjects - [Type-E four-quadrant chopper](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Atype-e-four-quadrant-chopper) — Power Electronics, chapter 4, asked 4× - [Four quadrant operation of a hoist](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afour-quadrant-hoisting-mechanism) — Utilization of Electrical Energy, chapter 2 - [Four quadrant operation in DC motor control](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afour-quadrant-dc-motor-control) — Utilization of Electrical Energy, chapter 3 - [Static Ward-Leonard four quadrant control](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Astatic-ward-leonard-four-quadrant) — Utilization of Electrical Energy, chapter 3