---
title: "JFET ID and VDS (IDSS 15 mA, VP -4.5 V)"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:jfet-divider-bias-idss15"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-divider-bias-idss15
site: BCT Engineering
---

# JFET ID and VDS (IDSS 15 mA, VP -4.5 V)

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

- Asked in: 2082 Shrawan, 8 marks
- 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-idss15
- 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= 15mA and VP = -4.5V. 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; VGS = 4 – 1.65 ID. From ID = 15(1 – VGS/(–4.5))², the valid root is ID = 3.78 mA. VDS = 20 – 3.78×(1.85 + 1.65) = 6.76 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 = 15 mA, VP = -4.5 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 = 15 [1 - (4.000 - 1.65 ID)/(-4.5)]²
   = 15 (1.8889 - 0.3667 ID)²
```
Expanding gives the quadratic:
```text
2.0167 ID² - 21.7778 ID + 53.5185 = 0
ID = 7.017 mA   or   3.782 mA
```
#### Step 3: choose the valid root
The root 7.017 mA gives VGS = -7.58 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

```text
ID  = 3.782 mA
VGS = 4.000 - 1.65 × 3.782 = -2.240 V
VDS = VDD - ID (RD + RS) = 20 - 3.782 × (1.85 + 1.65) = 6.763 V
```

**Answer: ID ≈ 3.78 mA, VGS ≈ -2.24 V, VDS ≈ 6.76 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. 20 V
   D. 5 V
   Answer: A. VG = 20 × 47/235 = 4 V. (Key point: VG)

2. Which equation combines the bias line with Shockley?
   A. ID = 15 (1 - (4 - 1.65 ID)/(-4.5))^2
   B. ID = 15 (4 - 1.65 ID)
   C. ID = 15 (1 + VG/VP)
   D. ID = 4.5 (1 - VGS)^2
   Answer: A. Substitute VGS = 4 - 1.65 ID into ID = IDSS(1 - VGS/VP)^2. (Key point: Equation)

3. Which root is valid?
   A. 3.782 mA (VGS = -2.24 V)
   B. 7.017 mA (VGS = -7.58 V)
   C. Both
   D. Neither
   Answer: A. 7.017 mA gives VGS below VP = -4.5 V. (Key point: Valid root)

4. What is VDS?
   A. 6.76 V
   B. 13.2 V
   C. 20 V
   D. 3.5 V
   Answer: A. VDS = 20 - 3.782 × 3.5 ≈ 6.76 V. (Key point: VDS)

## Close variants set in other exams

- [JFET ID and VDS (IDSS 18 mA, VP -5 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-divider-bias-idss18) — Electronics Devices and Circuit, chapter 3, asked 5×

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