Skip to main content

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)

All Electronics Devices and Circuit chapter 2 questions →