---
title: "JFET ID and VDS (VP -3.5 V, IDSS 10 mA), pinch-off check"
subject: "Electronics Devices and Circuit"
chapter: 3
key: "semester_3/electronics_devices_and_circuit:jfet-self-bias-vp-3p5"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-self-bias-vp-3p5
site: BCT Engineering
---

# JFET ID and VDS (VP -3.5 V, IDSS 10 mA), pinch-off check

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

- Asked in: 2078 Kartik, 7 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-self-bias-vp-3p5
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3

## Question

Find ID and VDS for the given circuit. Given data are VP = -3.5V, IDSS = 10 mA and assume all the capacitors are ideal and check whether transistor is operating in pinch off region or not?
[Figure: n-channel JFET amplifier with VDD = 20 V; gate divider of 910 kΩ from VDD to gate and 110 kΩ from gate to ground; drain resistor 2.2 kΩ; source resistor 1.1 kΩ with bypass capacitor C3; input coupling capacitor C1 and output coupling capacitor C2]

## Short answer

VG = 2.157 V, bias line VGS = 2.157 – 1.1ID. Solving ID = 10(1–VGS/–3.5)² yields ID = 3.31 mA, VGS = –1.49 V. VDS = 20 – 3.31(2.2+1.1) = 9.07 V. VDS > VGS–VP (2.01 V), so device is in pinch-off.

## Answer

The capacitors are open for dc, and C3 only bypasses RS for ac, so the dc analysis is that of a voltage-divider biased JFET.

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 = 10 mA, VP = -3.5 V, RD = 2.2 kΩ, RS = 1.1 kΩ, R1 = 910 kΩ, R2 = 110 kΩ (in the formulas below, resistances are in kΩ)

#### Step 1: gate voltage and bias line
```text
VG = VDD R2/(R1 + R2) = 20 × 110/(910 + 110) = 2.157 V
VGS = VG - ID RS = 2.157 - 1.1 ID      (ID in mA)
```

#### Step 2: Shockley equation
```text
ID = IDSS (1 - VGS/VP)² = 10 [1 - (2.157 - 1.1 ID)/(-3.5)]²
   = 10 (1.6162 - 0.3143 ID)²
```
Expanding gives the quadratic:
```text
0.9878 ID² - 11.1593 ID + 26.1225 = 0
ID = 7.986 mA   or   3.312 mA
```
#### Step 3: choose the valid root
The root 7.986 mA gives VGS = -6.63 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

```text
ID  = 3.312 mA
VGS = 2.157 - 1.1 × 3.312 = -1.486 V
VDS = VDD - ID (RD + RS) = 20 - 3.312 × (2.2 + 1.1) = 9.072 V
```

#### Pinch-off (saturation) check
The JFET is in the pinch-off region if VDS > VGS - VP:
```text
VGS - VP = -1.486 - (-3.5) = 2.014 V
VDS = 9.072 V  >  2.014 V
```
Yes, VDS is greater than VGS - VP, so the transistor operates in the pinch-off (saturation) region and the Shockley equation used is valid.

**Answer: ID ≈ 3.31 mA, VGS ≈ -1.49 V, VDS ≈ 9.07 V.**

## Check yourself

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

1. VG for R1 = 910 kΩ, R2 = 110 kΩ, VDD = 20 V is:
   A. 2.16 V
   B. 9.1 V
   C. 11 V
   D. 17.8 V
   Answer: A. VG = 20*110/1020 = 2.157 V. (Key point: Gate voltage)

2. Which bias line applies?
   A. VGS = 2.157 - 1.1 ID (ID in mA)
   B. VGS = 2.157 + 1.1 ID
   C. VGS = 20 - 2.2 ID
   D. VGS = -3.5 + ID
   Answer: A. VGS = VG - ID RS with RS = 1.1 kΩ. (Key point: Bias line)

3. Of ID = 7.986 mA and 3.312 mA, which is taken and why?
   A. 3.312 mA, since 7.986 mA gives VGS = -6.63 V, beyond VP
   B. 7.986 mA, as the larger is always right
   C. 7.986 mA since it exceeds IDSS
   D. 3.312 mA because it is below 5 mA
   Answer: A. The root must keep VGS between VP and 0; only 3.312 mA does (VGS = -1.49 V). (Key point: Valid root)

4. With VGS = -1.486 V and VP = -3.5 V, what is the pinch-off boundary?
   A. VGS - VP = 2.014 V; VDS = 9.07 V exceeds it
   B. VGS + VP = -4.99 V
   C. VDD = 20 V
   D. VP = -3.5 V; VDS is below it
   Answer: A. VDS > VGS - VP = 2.014 V, so the JFET is in pinch-off. (Key point: Pinch-off check)

## Close variants set in other exams

- [JFET ID and VDS (VP -5.5 V, IDSS 10 mA), pinch-off check](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-self-bias-vp-5p5) — Electronics Devices and Circuit, chapter 3, asked 3×
- [JFET ID and VDS (VP -5.5 V, IDSS 12 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-self-bias-vp-5p5-idss12) — Electronics Devices and Circuit, chapter 3, asked 3×

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