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)

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

$16 IDS = 10m (4 - IDS \times 1.2K \Omega) ^2$

$16 IDS = 10m(16 - 9.6K \times IDS + 1.44M \times IDS^2 )$

$16 IDS = 160m - 96 \times IDS + 14.4K \times IDS^2$

$0 = 160m - 112 \times IDS + 14.4K \times IDS^2$

$IDS = 5.89mA$ or $IDS = 1.88mA$

$VGS = - IDS \times R_S = - IDS \times 1.2k\Omega$

Select $IDSq$ with $VGS$ is less negative than $V_P$.

Hence $IDSq$ should be 1.88mA with this $VGS = - 2.256V$ which is less negative than $V_P = - 4V.$

$IDSq = 1.88mA$

Apply KVL from VDD to ground through drain and source.

$VDD - IDSq (R_S + R_D) -VDSq = 0$

$15 - 1.88m (1.5K + 1.2K) = VDSq$

$VDSq = 9.924V$

Answer:

- VDSq = 9.924V

- IDSq = 1.88mA