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Transconductance of E-MOSFET

Short answer

In saturation, ID = k (VGS – Vt)². Differentiating: gm = ∂ID/∂VGS = 2k (VGS – Vt). Substituting (VGS – Vt) = √(ID/k) gives gm = 2√(k ID) = 2 ID/(VGS – Vt).

Derive the expression to obtain the transconductance of E-MOSFET.

Transconductance gm is the rate of change of the drain current with the gate-source voltage at constant VDS, gm = ∂ID/∂VGS, for a small-signal input.

Derivation

In the saturation region of an n-channel enhancement MOSFET the drain current is

ID = k (VGS - Vt)²       (k in A/V²; equivalently ID = ½ kn'(W/L)(VGS - Vt)²)

Applying a small signal vgs: VGS = VGSQ + vgs, so

iD = k (VGSQ + vgs - Vt)²
   = k (VGSQ - Vt)² + 2k (VGSQ - Vt) vgs + k vgs²

For a small signal, the last term (k vgs²) is negligible:

iD = ID + id ,    id = 2k (VGSQ - Vt) vgs

Therefore

gm = id/vgs = ∂ID/∂VGS = 2k (VGS - Vt)

Substituting (VGS - Vt) = √(ID/k):

gm = 2√(k ID) = 2 ID/(VGS - Vt)

(With the other convention ID = ½ kn'(W/L)(VGS - Vt)², the result is gm = kn'(W/L)(VGS - Vt) = √(2 kn'(W/L) ID).)

Remarks

  • gm increases with the overdrive voltage (VGS - Vt) and with √ID.
  • Example: k = 0.5 mA/V², VGS - Vt = 2 V gives gm = 2 mA/V.
  • The common-source gain is Av = -gm (RD || ro || RL).

✓ Answer checked by Claude Opus 5.5 (high effort)

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