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Four-quadrant multiplier circuit and derivation

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.

Draw the detailed circuit diagram for four quadrant multiplier and derive its input and output relationship.

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

           +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

ΔI = I · tanh(V / 2V_T)

Lower pair (Q1, Q2), input Vy:

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:

(IC3 − IC4) =  I1 · tanh(Vx / 2V_T)
(IC6 − IC5) =  I2 · tanh(Vx / 2V_T)

Output current difference:

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:

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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