written 7.8 years ago by | modified 2.9 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M
Year: Dec 2016
written 7.8 years ago by | modified 2.9 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M
Year: Dec 2016
written 7.8 years ago by |
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 $IDS = 1.88mA$
$VGS = - IDS \times R_S = - IDS \times 1.2k\Omega$
IDS = 5.89mA | IDS = 1.88mA |
---|---|
VGS = - $5.89mA \times 1.2K\Omega$ | VGS = - $1.88mA \times 1.2K\Omega$ |
VGS = - 7.068V | VGS = -2.256V |
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