---
title: "Short note: ac equivalent of common source MOSFET amplifier"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:mosfet-cs-ac-equivalent-note"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-cs-ac-equivalent-note
site: BCT Engineering
---

# Short note: ac equivalent of common source MOSFET amplifier

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

- Asked in: 2072 Kartik, 4 marks
- Video: [MOSFET: Common Source (CS) configuration : DC & AC analysis](https://www.youtube.com/watch?v=ruOMOjSXqSY) — Shrenik Jain
- 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/mosfet-cs-ac-equivalent-note
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Write a short note on ac equivalent circuit of common source amplifier using MOSFET.

## Short answer

Replace MOSFET with its small-signal model: open gate, dependent source gm·vgs, drain resistance ro. Input resistance is gate bias RG. Output resistance is RD||ro. Voltage gain Av=−gm(RD||ro||RL); if source unbypassed, Av=−gmRD/(1+gmRS).

## Answer

In a common-source (CS) MOSFET amplifier the input is applied to the gate, the output is taken from the drain, and the source is common (usually bypassed to ground). The ac equivalent circuit is obtained by short-circuiting the dc supply and the large capacitors and replacing the MOSFET with its small-signal model.

#### Small-signal model of the MOSFET
- The gate current is zero, so the gate-source path is an open circuit.
- The drain current is a voltage-controlled current source gm·vgs.
- ro is the output (drain-source) resistance.

gm = 2k (VGSQ − Vt) = 2 √(k IDQ),  ro = 1/(λ IDQ)

#### AC equivalent circuit
```text
 vi o---+----o G          D o----+--------+--------+--------+----o vo
        |    +                   |        |        |        |
      [RG]  vgs        gm·vgs (↓)       [ro]     [RD]     [RL]
        |    -                   |        |        |        |
 GND ---+----o S          S o----+--------+--------+--------+---- GND
```

Here RG = R1 ∥ R2 is the gate bias resistance, and RD and RL are the drain and load resistors, which are in parallel with ro for ac. The gate (input) and drain (output) are separate nodes; the source is at ac ground.

#### Results
- Input resistance: Ri = RG (very large, in the MΩ range)
- Output resistance: Ro = RD ∥ ro ≈ RD
- Voltage gain: Av = vo/vi = −gm (ro ∥ RD ∥ RL) ≈ −gm (RD ∥ RL)

The negative sign shows a 180° phase shift between input and output. If the source resistor RS is not bypassed, the gain falls to Av = −gm (RD ∥ RL)/(1 + gm RS).

## Check yourself

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

1. How is the ac equivalent circuit obtained?
   A. Short dc supply and large capacitors, replace MOSFET by small-signal model
   B. Open the supply and shorts the load
   C. Remove all resistors
   D. Keep the dc supply
   Answer: A. The dc supply and big capacitors are ac shorts, and the MOSFET is replaced by gm vgs with ro. (Key point: Building the ac circuit)

2. The CS MOSFET voltage gain is approximately:
   A. +gm RD
   B. -gm (RD || RL)
   C. -1/gm
   D. gm/RD
   Answer: B. Av = -gm(ro||RD||RL) ≈ -gm(RD||RL); the minus sign is a 180° phase shift. (Key point: Voltage gain)

3. If RS is not bypassed, the gain becomes:
   A. -gm(RD||RL)/(1 + gm RS)
   B. -gm(RD||RL)(1 + gm RS)
   C. unchanged
   D. zero
   Answer: A. RS adds negative feedback, reducing the gain by 1 + gm RS. (Key point: Unbypassed RS)

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