---
title: "Transconductance of E-MOSFET"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:emosfet-transconductance"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/emosfet-transconductance
site: BCT Engineering
---

# Transconductance of E-MOSFET

Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor

- Asked in: 2068 Chaitra, 4 marks; 2071 Shrawan, 3 marks (asked 2 times)
- Video: [MOSFET Transconductance and MOSFET Small Signal Model Explained](https://www.youtube.com/watch?v=utBRWK4GmmE) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/emosfet-transconductance
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Derive the expression to obtain the 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).

## Answer

**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
```text
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
```text
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: 
```text
iD = ID + id ,    id = 2k (VGSQ - Vt) vgs
```
Therefore
```text
gm = id/vgs = ∂ID/∂VGS = 2k (VGS - Vt)
```
Substituting (VGS - Vt) = √(ID/k):
```text
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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is gm defined?
   A. ∂ID/∂VGS at constant VDS
   B. ∂VGS/∂ID
   C. ∂ID/∂VDS
   D. ID/VDS
   Answer: A. gm is the change in drain current per unit gate-source voltage. (Key point: Definition)

2. Which term is dropped in the expansion of iD?
   A. k vgs^2
   B. 2k(VGSQ - Vt) vgs
   C. k(VGSQ - Vt)^2
   D. ID
   Answer: A. For small vgs the squared term is negligible, leaving id = 2k(VGSQ - Vt) vgs. (Key point: Small-signal approximation)

3. What is gm?
   A. 2k (VGS - Vt)
   B. k (VGS - Vt)
   C. k (VGS - Vt)^2
   D. 2k VGS
   Answer: A. id/vgs = 2k(VGS - Vt), which also equals 2√(k ID). (Key point: Result)

4. For k = 0.5 mA/V^2 and VGS - Vt = 2 V, what is gm?
   A. 1 mA/V
   B. 2 mA/V
   C. 4 mA/V
   D. 0.5 mA/V
   Answer: B. gm = 2 × 0.5 × 2 = 2 mA/V. (Key point: Numerical)

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