Given:

VTN=1V Kn=0.5mA/V2 $\lambda=0.01V^{-1}$

To Find:

1. VGSQ & VDSQ

2. AV, Ri, R0

3. Equation ckt of above can be redrawn for DC analysis

Solution: In Mosfet, ID=IS=4mA

IG=0(reverse biased)

Applying KVL

-VGSQ+5V=0

VGSQ=+5V

VDSQ=VDD-IDRD

=5-(0.4)(10)

=1V

1. For Mosfet:

gm=Kn(VGS-VT)

=0.5(5-1)=2

Equation ckt for ac analysis: deactive all dc sources & replace all cap with short ckt.

$rds=\dfrac{V_{DS}}{I_D}=\dfrac1{0.4}=2.5\\ A_V=-gm(R_L||rd)\\ =-2(5||2.5)=2\left(\dfrac{5\times2.5}{37.5}\right)=-3.33V$

1. Ri = rgs||1M

$as \ rgs = \infty\\ \therefore R_i=1M\Omega$

1. $R_0=(rds||10||10)\\ =(2.5||5)=\left(\dfrac{2.5\times5}{37.5}\right)\Omega=\left(\dfrac53\right)\Omega$