---
title: "Short note: transconductance of JFET"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:jfet-transconductance-note"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-transconductance-note
site: BCT Engineering
---

# Short note: transconductance of JFET

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

- Asked in: 2078 Kartik, 3 marks; 2075 Ashwin, 4 marks; 2068 Baisakh, 2 marks; 2070 Ashad, 4 marks; 2070 Chaitra, 3 marks (asked 5 times)
- Video: [JFET Transconductance](https://www.youtube.com/watch?v=-IkZbHTFdzk) — BillKleitz
- 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/jfet-transconductance-note
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Write a short note on 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).

## Answer

**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:
```text
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),
```text
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).

## Check yourself

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

1. What is JFET transconductance gm?
   A. dID/dVGS at constant VDS
   B. dVGS/dID
   C. dID/dVDS
   D. VDS/ID
   Answer: A. gm measures how effectively VGS controls ID. (Key point: Definition)

2. Which expression is correct for gm?
   A. gm0 (1 - VGS/VP), with gm0 = 2 IDSS/|VP|
   B. IDSS/VP
   C. gm0 (VGS/VP)
   D. 2 IDSS VP
   Answer: A. Differentiating Shockley's equation gives gm = (2 IDSS/|VP|)(1 - VGS/VP). (Key point: Expression)

3. Where is gm maximum?
   A. At VGS = VP
   B. At VGS = 0
   C. At cut-off
   D. At ID = 0
   Answer: B. gm = gm0 at VGS = 0 and falls to zero at VGS = VP. (Key point: Variation)

4. For IDSS = 10 mA and VP = -4 V, what is gm at VGS = -1 V?
   A. 5 mA/V
   B. 3.75 mA/V
   C. 2.5 mA/V
   D. 6.25 mA/V
   Answer: B. gm0 = 5 mA/V, gm = 5(1 - 0.25) = 3.75 mA/V. (Key point: Numerical)

## Related in Electronics Devices and Circuit

- [n-channel DMOSFET and JFET transconductance](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Admosfet-working-jfet-transconductance) — Electronics Devices and Circuit, chapter 3
- [Transconductance of BJT, JFET and MOSFET](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Atransconductance-bjt-jfet-mosfet) — Electronics Devices and Circuit, chapter 3