---
title: "E-MOSFET ID, VDS and region (k 0.12, Vt 5 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:emosfet-id-vds-region-22m-18m"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/emosfet-id-vds-region-22m-18m
site: BCT Engineering
---

# E-MOSFET ID, VDS and region (k 0.12, Vt 5 V)

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

- Asked in: 2073 Chaitra, 6+2 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/emosfet-id-vds-region-22m-18m
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Determine ID and VDS for the given circuit and find the region of its operation. Given: k = 0.12 mA/V² and Vt = 5 V.
[Figure: enhancement MOSFET with VDD = 40 V; gate divider of 22 MΩ from VDD to gate and 18 MΩ from gate to ground; 3 kΩ drain resistor; 0.82 kΩ source resistor to ground]

## Short answer

VG=18V, VGS=18−0.82ID. Solve ID=0.12(VGS−5)², reject larger root. ID=6.72mA, VDS=40−6.72(3+0.82)=14.3V. VGS−Vt=7.49V, VDS>7.49, operates in saturation.

## Answer

**Given:** VDD = 40 V, R1 = 22 MΩ, R2 = 18 MΩ, RD = 3 kΩ, RS = 0.82 kΩ, k = 0.12 mA/V², Vt = 5 V (ID = k (VGS − Vt)²).

#### Step 1: Gate voltage
VG = 40 × 18/(22 + 18) = **18 V** (IG = 0)

#### Step 2: Solve for ID
VGS = 18 − 0.82 ID

ID = 0.12 (18 − 0.82 ID − 5)² = 0.12 (13 − 0.82 ID)²

0.0807 ID² − 3.5584 ID + 20.28 = 0

| Root | ID (mA) | VGS = 18 − 0.82 ID | Valid? |
|---|---|---|---|
| 1 | 37.4 | −12.6 V | No (VGS < Vt) |
| 2 | 6.72 | 12.49 V | Yes |

**ID = 6.72 mA**, VGS = 12.49 V

#### Step 3: VDS
VDS = VDD − ID (RD + RS) = 40 − 6.72 × (3 + 0.82) = 40 − 25.7 = **14.3 V**

#### Step 4: Region of operation
VGS − Vt = 12.49 − 5 = 7.49 V. Since VGS > Vt (a channel exists) and VDS = 14.3 V > 7.49 V, the MOSFET is in the **saturation (pinch-off) region**, so the square-law equation used is valid.

**Answer: ID = 6.72 mA, VDS = 14.3 V, saturation region.**

## Check yourself

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

1. Which equation do we solve for ID?
   A. ID = 0.12(13 - 0.82 ID)^2
   B. ID = 0.12(18 - 5)
   C. ID = 0.12(40 - 5)^2
   D. ID = 18/0.82
   Answer: A. VGS = 18 - 0.82 ID, so VGS - Vt = 13 - 0.82 ID inside the square law. (Key point: Bias equation)

2. Why is ID = 37.4 mA rejected?
   A. It gives VGS = -12.6 V, below Vt
   B. It is not a whole number
   C. It is too small
   D. It gives VDS = 0
   Answer: A. A negative VGS would mean the device is off, contradicting the assumed conduction. (Key point: Rejecting root)

3. VGS - Vt = 7.49 V and VDS = 14.3 V. The region is:
   A. Saturation
   B. Triode
   C. Cut-off
   D. Breakdown
   Answer: A. VDS > VGS - Vt with VGS > Vt means saturation. (Key point: Region)

## Close variants set in other exams

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