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

Marks: 10M

Year: May 2016

Given:

VGS(th) = $V_T$ = 3V

Id(ON) = 5mA

VGS(ON) = 6V

Solution:

We know that,

$IDS = k_n(VGS - V_T)^2$

$k_n = \frac{IDS(ON)}{[VGS(ON) - V_T]^2} = \frac{5mA}{[6-3]^2}$

$k_n = 0.55mA/V^2$

Hence, $IDS = 0.55mA/V^2 \times (VGS - 3)^2$ .........(1)

But voltage divider biasing equation is,

$VR_2 - VGS - IDS \times R_S = 0$

Hence, $VGS = VR_2 - IDS \times 0.75K\Omega$

$VR_2 = \frac{VDD \times R_2}{R_1 + R_2} = \frac{24 \times 6.8M\Omega}{10M\Omega + 6.8M\Omega}$

$VR_2 = 9.71V$

Hence, $VGS = 9.71 - IDS \times 0.75K\Omega$ ........(2)

Put equation (2) in (1),

$IDS = 0.55mA/V^2 \times (9.71 - IDS \times 0.75K\Omega - 3)^2$

$IDS = 0.55m \times (6.71 - IDS \times 0.75K\Omega)^2$

$IDS = 4.94mA or 16.15mA$

Select IDS with which VGS is greater than or equal to $VGS(T)$

Hence, $IDSq = 4.94mA$

with $IDSq = 4.94mA$ in equation (2),

$VGSq = 6.005V$

Apply KVL from VDD to Drain terminal,

$V_D = VDD - IDSq \times R_D$

$V_D = 24 - 4.94mA \times 2.2K\Omega = 13.132V$

Apply KVL from Source to ground terminal,

$V_S = IDSq \times R_S = 4.94mA \times 0.75K\Omega$

$V_S = 3.70V$

Answers:

1. $IDSq = 4.94mA$
2. $VGSq = 6.005V$
3. $V_D = 13.132V$
4. $V_S = 3.70V$