---
title: "MOSFET ID and VDS (VDD 10 V, 6 kΩ, Vt=1 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:mosfet-id-vds-10m-6k"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-id-vds-10m-6k
site: BCT Engineering
---

# MOSFET ID and VDS (VDD 10 V, 6 kΩ, Vt=1 V)

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

- Asked in: 2075 Chaitra, 7 marks; 2074 Ashwin, 6 marks; 2068 Chaitra, 4 marks (asked 3 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-10m-6k
- 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 VDD = 10 V; gate divider R1 = 10 MΩ from VDD to gate and R2 = 10 MΩ from gate to ground; drain resistor RD = 6 kΩ; source resistor RS = 6 kΩ to ground]

## Short answer

VG=5 V, VGS=5−6ID. Using ID=0.5(VGS−1)² yields ID=0.5 mA (rejecting larger root). Then VDS=10−0.5×(6+6)=4 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 = 10 V, R1 = 10 MΩ, R2 = 10 MΩ, RD = 6 kΩ, RS = 6 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) = 10 × 10000/(10000 + 10000) = 5.000 V
VGS = VG - ID RS = 5.000 - 6 ID      (ID in mA, RS in kΩ)
```

#### Step 2: solve with the square law
```text
ID = k (VGS - Vt)² = 0.5 (4.000 - 6 ID)²
```
Expanding and collecting terms gives 18 ID² - 25 ID + 8 = 0, so
```text
ID = 0.889 mA   or   0.500 mA
```
For ID = 0.889 mA the gate-source voltage would be VGS = -0.33 V, which is below Vt (the MOSFET would be off), so this root is rejected.

#### Step 3: results
```text
ID  = 0.500 mA
VGS = 5.000 - 6 × 0.500 = 2.000 V
VDS = VDD - ID (RD + RS) = 10 - 0.500 × (6 + 6) = 4.000 V
```

#### Saturation check
Saturation requires VDS > VGS - Vt:
```text
VGS - Vt = 1.000 V ,   VDS = 4.000 V  >  1.000 V
```
So the MOSFET is in the saturation region and the assumption is valid.

**Answer: ID ≈ 0.50 mA, VGS ≈ 2.00 V, VDS ≈ 4.00 V.**

## Check yourself

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

1. What is VG with equal 10 MΩ divider resistors and VDD = 10 V?
   A. 10 V
   B. 5 V
   C. 2.5 V
   D. 0 V
   Answer: B. No gate current, so VG = VDD/2 = 5 V. (Key point: Gate voltage)

2. Why is the ID = 0.889 mA root rejected?
   A. It gives VGS = -0.33 V, below Vt, so the MOSFET would be off
   B. It exceeds VDD/RD
   C. It is too small
   D. It is greater than k
   Answer: A. A conducting enhancement MOSFET needs VGS > Vt = 1 V. (Key point: Rejecting root)

3. What are ID and VDS?
   A. 0.5 mA and 4 V
   B. 0.889 mA and 4.67 V
   C. 1 mA and 0 V
   D. 0.5 mA and 7 V
   Answer: A. VGS = 5 - 6(0.5) = 2 V; VDS = 10 - 0.5 × 12 = 4 V. (Key point: Results)

4. How is saturation confirmed?
   A. VDS > VGS - Vt (4 V > 1 V)
   B. VDS < VGS - Vt
   C. VGS < Vt
   D. ID > IDSS
   Answer: A. Saturation needs VDS above the overdrive voltage. (Key point: Saturation check)

## 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 (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×

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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×