---
title: "MOSFET ID and VDS (188 kΩ/47 kΩ, Vt=1 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:mosfet-id-vds-188k-vt1"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-id-vds-188k-vt1
site: BCT Engineering
---

# MOSFET ID and VDS (188 kΩ/47 kΩ, Vt=1 V)

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

- Asked in: 2080 Baisakh, 8 marks; 2078 Bhadra, 8 marks (asked 2 times)
- 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-id-vds-188k-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 = 0.5mA/V².
[Figure: MOSFET with +20 V supply; gate divider of 188 kΩ from +20 V to gate and 47 kΩ from gate to ground; 1.85 kΩ drain resistor from +20 V to drain; 1.65 kΩ source resistor from source to ground]

## Short answer

VG=4 V from divider; VGS=4−1.65ID. Solving ID=0.5(VGS−1)² yields ID=0.97 mA (rejecting larger root), VDS=20−0.97×(1.85+1.65)=16.6 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 = 20 V, R1 = 188 kΩ, R2 = 47 kΩ, RD = 1.85 kΩ, RS = 1.65 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) = 20 × 47/(188 + 47) = 4.000 V
VGS = VG - ID RS = 4.000 - 1.65 ID      (ID in mA, RS in kΩ)
```

#### Step 2: solve with the square law
```text
ID = k (VGS - Vt)² = 0.5 (3.000 - 1.65 ID)²
```
Expanding and collecting terms gives 1.3612 ID² - 5.95 ID + 4.5 = 0, so
```text
ID = 3.398 mA   or   0.973 mA
```
For ID = 3.398 mA the gate-source voltage would be VGS = -1.61 V, which is below Vt (the MOSFET would be off), so this root is rejected.

#### Step 3: results
```text
ID  = 0.973 mA
VGS = 4.000 - 1.65 × 0.973 = 2.395 V
VDS = VDD - ID (RD + RS) = 20 - 0.973 × (1.85 + 1.65) = 16.595 V
```

#### Saturation check
Saturation requires VDS > VGS - Vt:
```text
VGS - Vt = 1.395 V ,   VDS = 16.595 V  >  1.395 V
```
So the MOSFET is in the saturation region and the assumption is valid.

**Answer: ID ≈ 0.97 mA, VGS ≈ 2.39 V, VDS ≈ 16.60 V.**

## Check yourself

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

1. What is VG?
   A. 4 V
   B. 3 V
   C. 5 V
   D. 20 V
   Answer: A. VG = 20 × 47/235 = 4 V. (Key point: Gate voltage)

2. Which equation is solved?
   A. ID = 0.5 (3 - 1.65 ID)^2
   B. ID = 0.5 (4 - 1.65 ID)
   C. ID = 0.5 (4 + 1.65 ID)^2
   D. ID = 0.5 (1 - 1.65 ID)^2
   Answer: A. VGS - Vt = 4 - 1.65 ID - 1. (Key point: Square-law equation)

3. Why is 3.398 mA rejected?
   A. VGS would be -1.61 V, below Vt
   B. It is greater than 1 mA
   C. It makes VDS negative
   D. It is below IDSS
   Answer: A. VGS = 4 - 1.65 × 3.398 = -1.61 V, below Vt, so the device would be off. (Key point: Valid root)

4. What is VDS for ID = 0.973 mA?
   A. 16.6 V
   B. 3.4 V
   C. 20 V
   D. 10 V
   Answer: A. VDS = 20 - 0.973 × 3.5 ≈ 16.6 V. (Key point: VDS)

## Close variants set in other exams

- [MOSFET drain current and VDS (Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-bias-8m-5k-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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