---
title: "MOSFET drain current and VDS (Vt=1 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:mosfet-bias-8m-5k-vt1"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-bias-8m-5k-vt1
site: BCT Engineering
---

# MOSFET drain current and VDS (Vt=1 V)

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

- Asked in: 2083 Baisakh, 7 marks; 2071 Chaitra, 6 marks (asked 2 times)
- 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-bias-8m-5k-vt1
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Find the drain current (ID) and drain to source voltage (VDS) for the following circuit. Given parameters are: Vt=1 V and k=0.5 mA/V².
[Figure: enhancement MOSFET with +12 V supply; gate divider of 8 MΩ from +12 V to gate and 8 MΩ from gate to ground; 5 kΩ drain resistor from +12 V to drain; 5 kΩ source resistor from source to ground]

## Short answer

VG=6 V, VGS=6−5ID. ID=0.5(VGS−1)² gives ID=0.754 mA (rejecting larger root). Then VDS=12−0.754×(5+5)=4.46 V.

## 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 = 12 V, R1 = 8 MΩ, R2 = 8 MΩ, RD = 5 kΩ, RS = 5 kΩ (resistances in kΩ in the formulas), Vt = 1 V, k = 0.5 mA/V².

#### Step 1: gate voltage
```text
VG = VDD R2/(R1 + R2) = 12 × 8000/(8000 + 8000) = 6.000 V
VGS = VG - ID RS = 6.000 - 5 ID      (ID in mA, RS in kΩ)
```

#### Step 2: solve with the square law
```text
ID = k (VGS - Vt)² = 0.5 (5.000 - 5 ID)²
```
Expanding and collecting terms gives 12.5 ID² - 26 ID + 12.5 = 0, so
```text
ID = 1.326 mA   or   0.754 mA
```
For ID = 1.326 mA the gate-source voltage would be VGS = -0.63 V, which is below Vt (the MOSFET would be off), so this root is rejected.

#### Step 3: results
```text
ID  = 0.754 mA
VGS = 6.000 - 5 × 0.754 = 2.228 V
VDS = VDD - ID (RD + RS) = 12 - 0.754 × (5 + 5) = 4.457 V
```

#### Saturation check
Saturation requires VDS > VGS - Vt:
```text
VGS - Vt = 1.228 V ,   VDS = 4.457 V  >  1.228 V
```
So the MOSFET is in the saturation region and the assumption is valid.

**Answer: ID ≈ 0.75 mA, VGS ≈ 2.23 V, VDS ≈ 4.46 V.**

## Check yourself

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

1. What is VG for VDD = 12 V and equal 8 MΩ resistors?
   A. 12 V
   B. 4 V
   C. 8 V
   D. 6 V
   Answer: D. VG = 12/2 = 6 V. (Key point: Gate voltage)

2. Which equation is solved for ID?
   A. ID = 0.5 (5 - 5 ID)^2
   B. ID = 0.5 (6 - 5 ID)
   C. ID = 0.5 (6 + 5 ID)^2
   D. ID = 5 ID
   Answer: A. VGS - Vt = 6 - 5 ID - 1 = 5 - 5 ID, then ID = k(VGS - Vt)^2. (Key point: Equation setup)

3. Which root is valid?
   A. 0.754 mA, giving VGS = 2.23 V
   B. 1.326 mA, giving VGS = -0.63 V
   C. Both
   D. Neither
   Answer: A. The 1.326 mA root gives VGS below Vt, so it is rejected. (Key point: Root choice)

4. What is VDS?
   A. 4.46 V
   B. 12 V
   C. 6 V
   D. 8.2 V
   Answer: A. VDS = 12 - 0.754 × 10 ≈ 4.46 V. (Key point: VDS)

## Close variants set in other exams

- [MOSFET ID and VDS (188 kΩ/47 kΩ, Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-id-vds-188k-vt1) — Electronics Devices and Circuit, chapter 3, asked 7×
- [MOSFET ID and VDS (VDD 10 V, 6 kΩ, Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-id-vds-10m-6k) — Electronics Devices and Circuit, chapter 3, asked 7×

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## Related in other subjects

- [Thevenin: 4 kΩ resistor, 12 V, 10 mA](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-12v-10ma) — Basic Electronics Engineering, chapter 1, asked 2×