Skip to main content

Differential amplifier using BJT

Short answer

A BJT differential amplifier uses two matched transistors with a common emitter resistor (or current source). A differential input produces amplified outputs at the collectors: one in phase, one out of phase. The output voltage is proportional to (v1 – v2) with high common-mode rejection.

Draw the circuit diagram of differential amplifier using BJT (and indicate its input and output voltage waveforms).

A differential amplifier amplifies the difference between two input signals and rejects the signal common to both.

Circuit (BJT, dual input, dual output)

                 +VCC
          +-------+-------+
          |               |
        [RC]            [RC]
          |               |
 vo1 o----+               +----o vo2
          |               |
         C|               |C
 v1 o----B| Q1        Q2  |B----o v2
         E|               |E
          |               |
          +-------+-------+
                  |
                [RE]  (or constant current source)
                  |
                -VEE

Two matched transistors Q1 and Q2 share a common emitter resistor RE (or a constant current source). The inputs v1 and v2 are applied to the bases; the outputs vo1 and vo2 are taken from the collectors.

Input and output waveforms

For a differential input, v1 = +vd/2 and v2 = −vd/2 (sine waves 180° out of phase):

  • vo1 is an amplified sine wave 180° out of phase with v1.
  • vo2 is an amplified sine wave 180° out of phase with v2, and hence in phase with v1.
  • The differential output taken between the collectors, vo = vo2 − vo1 = Ad vd, has twice the amplitude of either single collector output.
 v1:      /\    /\         input 1
         /  \  /  \
        /    \/    \

 v2:    \    /\    /       input 2 (180° out of phase with v1)
         \  /  \  /
          \/    \/

 vo1:   \    /\    /       amplified, 180° out of phase with v1
         \  /  \  /
          \/    \/

 vo2:     /\    /\         amplified, in phase with v1
         /  \  /  \
        /    \/    \

Output voltage: vo = Ad (v1 − v2) + Ac (v1 + v2)/2, and CMRR = Ad/Ac.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 4 questions →