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Transconductance gm of BJT
Short answer
gm is ratio of change in collector current to change in VBE: gm = ∂iC/∂vBE. From iC = IS exp(vBE/VT), gm = IC/VT. At room temperature (VT = 25 mV), gm = 40 IC (e.g., 40 mA/V for IC = 1 mA).
Define transconductance (gm). Derive gm for BJT.
Transconductance (gm) is the ratio of the change in output (collector) current to the small change in input voltage (VBE) that causes it, at a fixed VCE. It measures how effectively the input voltage controls the output current: gm = ∂iC/∂vBE (units: A/V or S, mA/V).
Derivation for BJT
In the active region the collector current is
iC = IS exp(vBE / VT)
Let vBE = VBE + vbe, where VBE is the dc bias and vbe the small signal:
iC = IS exp(VBE/VT) · exp(vbe/VT) = IC exp(vbe/VT)
For vbe ≪ VT (about < 10 mV), exp(x) ≈ 1 + x:
iC = IC (1 + vbe/VT) = IC + (IC/VT) vbe
The signal part is ic = (IC/VT) vbe, so
gm = ic/vbe = ∂iC/∂vBE |at Q = IC / VT
Or by differentiation: gm = d/dvBE [IS exp(vBE/VT)] = iC/VT = IC/VT.
Remarks
- At room temperature VT = 25 mV, so gm = 40 IC; for IC = 1 mA, gm = 40 mA/V.
- gm is proportional to the dc collector current only (independent of the transistor's size).
- Related quantities: rπ = β/gm and re = α/gm ≈ 1/gm.
- Voltage gain of CE stage: Av = -gm RC.
✓ Answer checked by Claude Opus 5.5 (high effort)
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