---
title: "JFET ID and VDS (IDSS 10 mA, VP -4 V, 25 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:jfet-divider-bias-idss10-vp4"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-divider-bias-idss10-vp4
site: BCT Engineering
---

# JFET ID and VDS (IDSS 10 mA, VP -4 V, 25 V)

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

- Asked in: 2068 Baisakh, 5 marks; 2070 Chaitra, 5 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/jfet-divider-bias-idss10-vp4
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Find ID and VDS for the given circuit.
[Figure: n-channel JFET with +25 V supply; gate divider of 1.5 MΩ from +25 V to gate and 300 kΩ from gate to ground; 2.2 kΩ drain resistor from +25 V to drain; 1 kΩ source resistor to ground; VP = -4 V, IDSS = 10 mA]

## Short answer

VG = 4.167 V from divider; VGS = 4.167 – ID·1kΩ. Solving ID = 10(1 – VGS/(–4))² yields ID = 5.26 mA. VDS = 25 – 5.26×(2.2k + 1k) = 8.15 V.

## Answer

Method: find the gate voltage, write VGS in terms of ID, combine this with Shockley's equation, solve the quadratic and reject the root that is not physically possible. The gate current is zero.

#### Given
VDD = 25 V, IDSS = 10 mA, VP = -4 V, RD = 2.2 kΩ, RS = 1 kΩ, R1 = 1.5 MΩ, R2 = 300 kΩ (in the formulas below, resistances are in kΩ)

#### Step 1: gate voltage and bias line
```text
VG = VDD R2/(R1 + R2) = 25 × 300/(1500 + 300) = 4.167 V
VGS = VG - ID RS = 4.167 - 1 ID      (ID in mA)
```

#### Step 2: Shockley equation
```text
ID = IDSS (1 - VGS/VP)² = 10 [1 - (4.167 - 1 ID)/(-4)]²
   = 10 (2.0417 - 0.25 ID)²
```
Expanding gives the quadratic:
```text
0.625 ID² - 11.2083 ID + 41.684 = 0
ID = 12.669 mA   or   5.264 mA
```
#### Step 3: choose the valid root
The root 12.669 mA gives VGS = -8.50 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

```text
ID  = 5.264 mA
VGS = 4.167 - 1 × 5.264 = -1.098 V
VDS = VDD - ID (RD + RS) = 25 - 5.264 × (2.2 + 1) = 8.154 V
```

**Answer: ID ≈ 5.26 mA, VGS ≈ -1.10 V, VDS ≈ 8.15 V.**

## Check yourself

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

1. What is VG?
   A. 4.167 V
   B. 25 V
   C. 5 V
   D. 3 V
   Answer: A. VG = 25 × 300/1800 = 4.167 V. (Key point: VG)

2. What is VGS in terms of ID (mA)?
   A. 4.167 - ID
   B. 4.167 + ID
   C. -ID
   D. 4.167 - 2.2 ID
   Answer: A. RS = 1 kΩ, so VGS = VG - ID RS. (Key point: Bias line)

3. Why is 12.669 mA rejected?
   A. VGS = -8.50 V, below VP = -4 V
   B. It exceeds 25 mA
   C. It is non-physical because negative
   D. It is less than IDSS
   Answer: A. Valid VGS lies between VP and 0. (Key point: Root rejection)

4. What is VDS?
   A. 8.15 V
   B. 14 V
   C. 25 V
   D. 5.26 V
   Answer: A. VDS = 25 - 5.264 × 3.2 ≈ 8.15 V. (Key point: VDS)

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