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Transconductance and voltage gain of BJT

Short answer

Transconductance gm = ΔIC/ΔVBE = IC/VT, typically 80 mS for IC=2 mA, describes voltage-to-current conversion. Voltage gain Av = vout/vin; for common-emitter, Av = −gm (RC ∥ RL), phase shift 180°. Example: gm=80 mS, load 2 kΩ gives Av = −160.

Define transconductance and voltage gain with reference to BJT (explain with block diagram).

Block diagram

A BJT amplifier is treated as a two-port network:

        +--------------------------+
 vin o--|  Input   BJT amplifier   |--o vout
        |  port    (gm, ro, rpi)   |
 GND o--|                          |--o GND
        +--------------------------+
        iin                        iout

Transconductance (gm)

The ratio of the change in output (collector) current to the change in input (base-emitter) voltage at constant VCE:

gm = ΔIC/ΔVBE = IC/VT

with VT ≈ 25 mV at 300 K. Unit: siemens (S or A/V). Example: IC = 2 mA gives gm = 2 mA/25 mV = 80 mS. It describes how well the input voltage controls the output current. In the small-signal model, ic = gm vbe.

Voltage gain (Av)

The ratio of output signal voltage to input signal voltage:

Av = vout/vin

For a common-emitter amplifier with collector load RC and load RL (ro neglected):

vout = −ic (RC ∥ RL) = −gm vbe (RC ∥ RL)

Av = −gm (RC ∥ RL)

The minus sign indicates a phase reversal of 180°. Example: gm = 80 mS, RC ∥ RL = 2 kΩ gives Av = −160. In decibels, Av(dB) = 20 log |Av|.

✓ Answer checked by Claude Opus 5.5 (high effort)

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