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Short note: transconductance of JFET
Short answer
Transconductance gm=∂ID/∂VGS, given by gm=(2IDSS/|VP|)(1−VGS/VP)=gm0√(ID/IDSS) with gm0=2IDSS/|VP|. It's maximum at VGS=0 and zero at pinch-off. Voltage gain Av=−gm(RD∥rd∥RL).
Write a short note on transconductance of JFET.
Transconductance (gm) of a JFET is the ratio of the change in drain current to the change in gate-source voltage, at constant VDS. It measures how effectively the gate voltage controls the drain current: gm = ∂ID/∂VGS (siemens, mA/V). It is the FET's counterpart of the BJT's gm.
Expression
From Shockley's equation, ID = IDSS (1 - VGS/VP)²:
gm = dID/dVGS = 2 IDSS (1 - VGS/VP) × (-1/VP)
= (2 IDSS/|VP|) (1 - VGS/VP)
gm = gm0 (1 - VGS/VP) with gm0 = 2 IDSS/|VP| (the value at VGS = 0)
Eliminating VGS: since (1 - VGS/VP) = √(ID/IDSS),
gm = gm0 √(ID/IDSS)
Properties
- gm is maximum (gm0) at VGS = 0 and falls to zero at VGS = VP (cut-off).
- It depends on the dc bias point: it is higher at larger ID.
- Typical values are 1 to 10 mA/V (lower than for BJTs at the same current).
- Example: IDSS = 10 mA, VP = -4 V: gm0 = 5 mA/V; at VGS = -1 V, gm = 5 (1 - 0.25) = 3.75 mA/V.
- The voltage gain of a common-source amplifier is Av = -gm (RD || rd || RL).
✓ Answer checked by Claude Opus 5.5 (high effort)
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