Skip to main content

Transconductance of BJT, JFET and MOSFET

Short answer

BJT: gm = IC/VT. JFET: gm = (2IDSS/|VP|)(1–VGS/VP) = gm0√(ID/IDSS). MOSFET (saturation): gm = 2k(VGS–Vt) = 2√(k·ID). Proof: rπ = VT/IB = VT(β+1)/IE = (β+1)re.

Derive expressions to obtain transconductance for BJT, JFET and MOSFET. Also prove that rπ = (β+1)re.

Transconductance is the ratio of the change in output (collector or drain) current to the change in input control voltage: gm = ∂I_out / ∂V_in at the Q point.

BJT

In the active region, IC = IS exp(VBE/VT), VT = kT/q ≈ 26 mV at 300 K.

gm = dIC/dVBE = (1/VT) · IS exp(VBE/VT) = IC / VT

At IC = 1 mA, gm = 1/26 = 38.5 mS.

JFET

Shockley equation: ID = IDSS (1 − VGS/VP)²

gm = dID/dVGS = 2 IDSS (1 − VGS/VP) · (−1/VP) = (2 IDSS/|VP|)(1 − VGS/VP)

Define gm0 = 2 IDSS/|VP| (value at VGS = 0). Then

gm = gm0 (1 − VGS/VP) = gm0 √(ID/IDSS)

MOSFET (enhancement, saturation)

ID = k (VGS − Vt)²

gm = dID/dVGS = 2k (VGS − Vt) = 2 √(k ID)

(If the equation is written ID = (K/2)(VGS − Vt)², then gm = K(VGS − Vt).)

Proof that rπ = (β + 1) re

Definitions: rπ = ∂VBE/∂IB (small-signal base-emitter resistance) and re = ∂VBE/∂IE = VT/IE.

Using the BJT relations IE = IB + IC = (β + 1) IB, so IB = IE/(β + 1).

rπ = VT / IB = VT (β + 1)/IE = (β + 1) · (VT/IE)

rπ = (β + 1) re

Also, rπ = β/gm since gm = IC/VT = β IB/VT, giving rπ = VT/IB = β/gm. This is consistent with the result above, since re ≈ 1/gm when α ≈ 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Electronics Devices and Circuit chapter 3 questions →