- 2070 Chaitra · 2+5 marks
Four-quadrant multiplier: concept and output
Short answer
A four-quadrant multiplier correctly gives Vo = K·Vx·Vy for any sign of Vx and Vy, enabling AC signal processing. The Gilbert cell uses two cross-coupled differential pairs to obtain ΔIo = IEE·tanh(Vx/2VT)·tanh(Vy/2VT); for small signals Vo ≈ (IEE·RL/4VT²)·Vx·Vy, working in all four quadrants.
What do you understand by four-quadrant multiplier? Draw the circuit diagram and derive expression for its output voltage.
Four-quadrant multiplier
A multiplier gives Vo = K·Vx·Vy. If both inputs can be positive or negative and the output takes the correct sign, the multiplier works in all four quadrants of the Vx–Vy plane and is called a four-quadrant multiplier.
Vy
II | I
Vx<0,Vy>0 | Vx>0,Vy>0
Vo < 0 | Vo > 0
------------+------------ Vx
Vx<0,Vy<0 | Vx>0,Vy<0
Vo > 0 | Vo < 0
III | IV
A log-antilog multiplier works only in quadrant I (one-quadrant). A four-quadrant multiplier is needed for AC signals, e.g. in modulators, phase detectors and power measurement.
Circuit (Gilbert cell)
+Vcc +Vcc
RL RL
| Io1 Io2 |
+--+-- ... +--+----> Vo
Q3 Q4 Q5 Q6 <- Vx (cross-coupled)
\/ \/
I1 I2
Q1 ---------- Q2 <- Vy
|
IEE
|
-VEE
Collectors of Q3, Q5 give Io1; collectors of Q4, Q6 give Io2.
Derivation
Differential pair with tail current I and input V: ΔI = I·tanh(V/2V_T).
Lower pair: I1 − I2 = IEE·tanh(Vy/2V_T)
Upper pairs: IC3 − IC4 = I1·tanh(Vx/2V_T)
IC6 − IC5 = I2·tanh(Vx/2V_T)
ΔIo = (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 signals, tanh(u) ≈ u:
ΔIo ≈ IEE·Vx·Vy/(4V_T²)
Vo = RL·ΔIo = (IEE·RL/4V_T²)·Vx·Vy = K·Vx·Vy
The product of two odd functions keeps the correct sign for every combination of input polarities, so the circuit is four-quadrant. IC versions (AD633, MC1495) give Vo = Vx·Vy/10.
✓ Answer checked by Claude Opus 5.5 (high effort)
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