---
title: "BJT bipolar vs FET unipolar; MOSFET transconductance"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:bjt-bipolar-fet-unipolar-gm"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-bipolar-fet-unipolar-gm
site: BCT Engineering
---

# BJT bipolar vs FET unipolar; MOSFET transconductance

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

- Asked in: 2078 Bhadra, 2+3 marks
- Video: [What is FET (Field Effect Transistor) BJT Vs FET (Basic Electronics)](https://www.youtube.com/watch?v=pd79hGW5ixI) — Gautam Varde
- 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/bjt-bipolar-fet-unipolar-gm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Why BJT is called bipolar and FET is called unipolar device? Derive mathematically the transconductance of MOSFET.

## Short answer

BJT is bipolar because current involves both electrons and holes; FET is unipolar because current uses only majority carriers. MOSFET transconductance in saturation: gm = 2k(VGS–Vt) = 2√(k·ID) = 2ID/(VGS–Vt).

## Answer

#### Bipolar vs unipolar
- A **BJT** is called **bipolar** because its current is carried by **both types of charge carriers**, electrons and holes: the emitter injects majority carriers into the base, where they are minority carriers, and the current in the device depends on minority carriers as well as majority carriers (e.g. electrons and holes in an npn).
- A **FET** is called **unipolar** because the current is carried by **only one type of carrier**, the majority carriers (electrons in an n-channel, holes in a p-channel) flowing through the channel; minority carriers do not take part.

#### Transconductance of MOSFET (n-channel enhancement, saturation)
```text
ID = k (VGS - Vt)²           (k in A/V²)
```
gm is defined as gm = ∂ID/∂VGS at constant VDS. Differentiating:
```text
gm = d/dVGS [k (VGS - Vt)²] = 2k (VGS - Vt)
```
Using (VGS - Vt) = √(ID/k):
```text
gm = 2√(k ID) = 2 ID/(VGS - Vt)
```
(If ID = ½ kn'(W/L)(VGS - Vt)², then gm = kn'(W/L)(VGS - Vt).)

For a depletion MOSFET or JFET, ID = IDSS(1 - VGS/VP)² gives gm = (2 IDSS/|VP|)(1 - VGS/VP). Example: k = 0.5 mA/V², VGS - Vt = 2 V gives gm = 2 mA/V.

## Check yourself

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

1. Why is a BJT called bipolar?
   A. It has two terminals
   B. Current involves both electrons and holes (majority and minority carriers)
   C. It uses two supplies
   D. It conducts in both directions
   Answer: B. BJT current depends on both types of carriers, including minority carriers in the base. (Key point: Why BJT is bipolar)

2. A FET is unipolar because:
   A. Current is carried by only one type of carrier, the majority carriers in the channel
   B. It has one gate
   C. It uses one supply
   D. It has only one p-n junction
   Answer: A. Only majority carriers (electrons in n-channel, holes in p-channel) flow in the channel. (Key point: Why FET is unipolar)

3. Differentiating ID = k(VGS - Vt)^2 w.r.t. VGS gives:
   A. gm = k (VGS - Vt)
   B. gm = 2k (VGS - Vt)
   C. gm = k (VGS - Vt)^2
   D. gm = 2k/(VGS - Vt)
   Answer: B. d/dVGS of k(VGS-Vt)^2 is 2k(VGS-Vt), which also equals 2 sqrt(k ID). (Key point: MOSFET gm derivation)

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