Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Year: Dec 2016

Given:

$IDSS = 10mA$

$V_P = -4V$

Solution:

We know that $IDS = IDSS(1- \frac{VGS}{V_P})^2$

$IDS = 10m(1+ \frac{VGS}{4})^2$..............(1)

For self bias, $VGS = -IDS \times R_S$

Substitue VGS value in equation (1)

$IDS = 10m(1 - \frac{IDS \times R_S}{4})^2$

$16 IDS = 10m (4 - IDS \times 1.2K \Omega) …