---
title: "MOSFET ID, VGS, VDS and saturation check"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:mosfet-saturation-check-10m-6m8"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-saturation-check-10m-6m8
site: BCT Engineering
---

# MOSFET ID, VGS, VDS and saturation check

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

- Asked in: 2081 Bhadra, 8 marks
- Video: [MOSFET Circuits in DC, Part 1](https://www.youtube.com/watch?v=mszImxmcbww) — unwired
- 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-saturation-check-10m-6m8
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Find the value of ID, VGS, VDS and check whether the MOSFET operated in the saturation region or not.
Given: Vt= 3V, k= 0.556 mA/V²
[Figure: enhancement MOSFET with 20 V supply; gate divider of 10 MΩ from 20 V to gate and 6.8 MΩ from gate to ground; 2.2 kΩ drain resistor from 20 V to drain; 750 Ω source resistor from source to ground]

## Short answer

VG = 8.095 V; VGS = 8.095 – 0.75 ID. Using ID = 0.556(VGS – 3)² gives ID = 3.47 mA, VGS = 5.50 V, VDS = 20 – 3.47(2.2+0.75) = 9.78 V. Since VDS > VGS – Vt, the MOSFET operates in saturation.

## Answer

The enhancement MOSFET is biased by a voltage divider (no gate current). In saturation ID = k (VGS - Vt)², with k in mA/V².

#### Given
VDD = 20 V, R1 = 10 MΩ, R2 = 6.8 MΩ, RD = 2.2 kΩ, RS = 0.75 kΩ (resistances in kΩ in the formulas), Vt = 3 V, k = 0.556 mA/V².

#### Step 1: gate voltage
```text
VG = VDD R2/(R1 + R2) = 20 × 6800/(10000 + 6800) = 8.095 V
VGS = VG - ID RS = 8.095 - 0.75 ID      (ID in mA, RS in kΩ)
```

#### Step 2: solve with the square law
```text
ID = k (VGS - Vt)² = 0.556 (5.095 - 0.75 ID)²
```
Expanding and collecting terms gives 0.3128 ID² - 5.2494 ID + 14.4346 = 0, so
```text
ID = 13.320 mA   or   3.465 mA
```
For ID = 13.320 mA the gate-source voltage would be VGS = -1.89 V, which is below Vt (the MOSFET would be off), so this root is rejected.

#### Step 3: results
```text
ID  = 3.465 mA
VGS = 8.095 - 0.75 × 3.465 = 5.496 V
VDS = VDD - ID (RD + RS) = 20 - 3.465 × (2.2 + 0.75) = 9.778 V
```

#### Saturation check
Saturation requires VDS > VGS - Vt:
```text
VGS - Vt = 2.496 V ,   VDS = 9.778 V  >  2.496 V
```
So the MOSFET is in the saturation region and the assumption is valid.

**Answer: ID ≈ 3.47 mA, VGS ≈ 5.50 V, VDS ≈ 9.78 V.**

## Check yourself

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

1. Why is VG found simply as VDD*R2/(R1+R2) for this MOSFET?
   A. Because the gate draws no current, so the divider is unloaded
   B. Because the source resistor is shorted
   C. Because the drain current is zero
   D. Because VGS equals VDS
   Answer: A. The insulated gate carries no current, so the divider R1, R2 behaves as an unloaded voltage divider giving VG = 8.095 V. (Key point: Gate voltage from divider)

2. Which relation links VGS to ID in the self-bias circuit?
   A. VGS = VG + ID RS
   B. VGS = VG - ID RS
   C. VGS = VDD - ID RD
   D. VGS = VG - ID (RD + RS)
   Answer: B. The source sits at ID*RS above ground, so VGS = VG - ID*RS. (Key point: Bias line VGS = VG - ID RS)

3. The quadratic gives ID = 13.32 mA or 3.465 mA. Why is 13.32 mA rejected?
   A. It exceeds VDD
   B. It gives VGS = -1.89 V, below Vt, so the MOSFET would be off
   C. It is not an integer
   D. It makes VDS negative only
   Answer: B. With ID = 13.32 mA, VGS = 8.095 - 0.75*13.32 is negative, below Vt, contradicting conduction. (Key point: Rejecting the invalid root)

4. What condition confirms the MOSFET is in saturation?
   A. VDS < VGS - Vt
   B. VDS > VGS - Vt
   C. VGS < Vt
   D. VDS = VDD
   Answer: B. Saturation needs VDS > VGS - Vt; here 9.778 V > 2.496 V. (Key point: Saturation check)

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