- Asked 2/28
- 2068 Chaitra · 4 marks
- 2071 Shrawan · 3 marks
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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