A JFET has the following parameters: VGS(off) = -8 V: IDSS = 30 mA, and VGS = -4 V. The drain current will be: 

This question was previously asked in
MPPGCL JE Electrical 19 March 2019 Shift 2 Official Paper
View all MPPGCL Junior Engineer Papers >
  1. 7.5 μA
  2. 7.5 mA
  3. 5.7 mA
  4. 5.7 μA

Answer (Detailed Solution Below)

Option 2 : 7.5 mA
Free
MPPGCL JE Electrical Full Test 1
100 Qs. 100 Marks 120 Mins

Detailed Solution

Download Solution PDF

Concept

The drain current in a JFET is given by:

 

where, ID = Drain current

IDSS = Saturation drain current

VGS = Gate -source voltage

VP = Pinch off voltage

Calculation

Given, Vp = - 8 V

IDSS = 30 mA

VGS = - 4 V

ID = 7.5 mA

Latest MPPGCL Junior Engineer Updates

Last updated on May 29, 2025

-> MPPGCL Junior Engineer result PDF has been released at the offiical website.

-> The MPPGCL Junior Engineer Exam Date has been announced.

-> The MPPGCL Junior Engineer Notification was released for 284 vacancies.

-> Candidates can apply online from 23rd December 2024 to 24th January 2025.

-> The selection process includes a Computer Based Test and Document Verification.

-> Candidates can check the MPPGCL JE Previous Year Papers which helps to understand the difficulty level of the exam.

Hot Links: teen patti casino download teen patti rules teen patti gold apk download online teen patti real money teen patti lucky