---
title: "JFET ID and VDS (IDSS 18 mA, VP -5 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:jfet-divider-bias-idss18"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-divider-bias-idss18
site: BCT Engineering
---

# JFET ID and VDS (IDSS 18 mA, VP -5 V)

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

- Asked in: 2081 Baisakh, 8 marks; 2073 Shrawan, 8 marks; 2069 Chaitra, 8 marks; 2072 Kartik, 8 marks (asked 4 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-idss18
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

The n-channel JFET in the figure below has IDSS = 18mA and Vp = -5V. Determine the values of ID and VDS.
[Figure: JFET 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=18(1−VGS/(−5))² gives quadratic, valid root ID=4.02 mA. Then VDS=20−4.02×(1.85+1.65)=5.92 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 = 20 V, IDSS = 18 mA, VP = -5 V, RD = 1.85 kΩ, RS = 1.65 kΩ, R1 = 188 kΩ, R2 = 47 kΩ (in the formulas below, resistances are in kΩ)

#### Step 1: gate voltage and bias line
```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)
```

#### Step 2: Shockley equation
```text
ID = IDSS (1 - VGS/VP)² = 18 [1 - (4.000 - 1.65 ID)/(-5)]²
   = 18 (1.8 - 0.33 ID)²
```
Expanding gives the quadratic:
```text
1.9602 ID² - 22.384 ID + 58.32 = 0
ID = 7.397 mA   or   4.022 mA
```
#### Step 3: choose the valid root
The root 7.397 mA gives VGS = -8.21 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

```text
ID  = 4.022 mA
VGS = 4.000 - 1.65 × 4.022 = -2.636 V
VDS = VDD - ID (RD + RS) = 20 - 4.022 × (1.85 + 1.65) = 5.923 V
```

**Answer: ID ≈ 4.02 mA, VGS ≈ -2.64 V, VDS ≈ 5.92 V.**

## Check yourself

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

1. What is VG for VDD = 20 V, R1 = 188 kΩ, R2 = 47 kΩ?
   A. 4 V
   B. 5 V
   C. 20 V
   D. 16 V
   Answer: A. VG = 20 × 47/235 = 4 V, since IG = 0. (Key point: Gate voltage)

2. What is VGS in terms of ID?
   A. VG - ID RS
   B. VG + ID RS
   C. VG - ID RD
   D. -ID RD
   Answer: A. The source is at ID RS above ground, so VGS = VG - ID RS = 4 - 1.65 ID. (Key point: Bias line)

3. Why is ID = 7.397 mA rejected?
   A. It gives VGS = -8.21 V, below VP = -5 V (channel cut off)
   B. It is greater than 5 mA
   C. It is negative
   D. It exceeds IDSS
   Answer: A. A valid solution requires VP ≤ VGS ≤ 0. (Key point: Rejecting root)

4. What is VDS for ID = 4.022 mA?
   A. 5.92 V
   B. 12.6 V
   C. 13.6 V
   D. 20 V
   Answer: A. VDS = 20 - 4.022 × (1.85 + 1.65) ≈ 5.92 V. (Key point: VDS)

## Close variants set in other exams

- [JFET ID and VDS (IDSS 15 mA, VP -4.5 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-divider-bias-idss15) — Electronics Devices and Circuit, chapter 3, asked 5×

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