---
title: "Transconductance of BJT, JFET and MOSFET"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:transconductance-bjt-jfet-mosfet"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/transconductance-bjt-jfet-mosfet
site: BCT Engineering
---

# Transconductance of BJT, JFET and MOSFET

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

- Asked in: 2073 Shrawan, 8 marks
- 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/transconductance-bjt-jfet-mosfet
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Derive expressions to obtain transconductance for BJT, JFET and MOSFET. Also prove that rπ = (β+1)re.

## 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.

## Answer

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.

## Check yourself

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

1. Differentiating IC = IS exp(VBE/VT) gives gm equal to:
   A. IC/VT
   B. VT/IC
   C. IC VT
   D. IC/2VT
   Answer: A. dIC/dVBE = IS exp(VBE/VT)/VT = IC/VT, about 38.5 mS at 1 mA. (Key point: BJT gm)

2. What is gm0 for a JFET?
   A. IDSS/VP
   B. 2 IDSS/|VP|, the gm at VGS = 0
   C. IDSS VP
   D. 2 VP/IDSS
   Answer: B. gm0 = 2IDSS/|VP|, giving gm = gm0(1 - VGS/VP). (Key point: JFET gm0)

3. For ID = k(VGS - Vt)^2, gm can also be written as:
   A. 2 sqrt(k ID)
   B. k ID
   C. sqrt(k ID)
   D. ID/k
   Answer: A. Since (VGS-Vt) = sqrt(ID/k), gm = 2k(VGS-Vt) = 2 sqrt(k ID). (Key point: MOSFET gm)

4. Which step proves rπ = (β+1) re?
   A. IE = (β+1) IB, so rπ = VT/IB = (β+1) VT/IE
   B. IC = IB/β
   C. IE = IC - IB
   D. rπ = VT/IC
   Answer: A. Substituting IB = IE/(β+1) into rπ = VT/IB gives (β+1)(VT/IE) = (β+1) re. (Key point: rπ = (β+1) re)

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