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

Marks: 10M

Year: May 2015

Given:

$IDSS = 10mA$

$V_P = - 4V$

Solution:

We know that,

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

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

Put different values of VGS and obtain IDS.

Apply KVL from $R_2$ to ground through gate and source,

$VR_2 - VGS - IDS \times R_S = 0$...........(1)

Put $IDS = 0$ in equation (1) we get,

$VR_2 = VGS = \frac{VDD \times R_2}{R_1 + R_2}$

$VR_2 = VGS = \frac{16 \times 110K\Omega}{910K\Omega + 110K\Omega}$

$VR_2 = VGS = 1.72V$.......will point on X-axis.

Put $VGS = 0$ in equation (1) we get,

$IDS = \frac{VR_2}{R_S}$

$IDS = \frac{1.72V}{1.1K\Omega}$

$IDS = 1.56mA$.......will point on Y-axis.

From graph $IDSq = 3.2mA, VGSq = -1.7V$

Apply KVL from VDD to ground through drain and source,

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

$16 - 3.2m(2.2K\Omega + 1.1K\Omega) = VDSq$

$VDSq = 5.44V$

$VDSq \gt |V_P| \gt 4 V$, FET is operating in pinch off region.

Answer:

• $IDSq = 3.2mA$
• $VGSq = -1.7V$
• Operating in Pinch off region