- 2073 Shrawan · 8 marks
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)
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