1) Fixed Bias ckt

Applying kvc in $c_1 - s \ ckt$

$-V_{GG} - I_G \ R_G - V_{as} = 0$

But $I_G = 0$

$\therefore V_{as} = -V_{GG}$

• Drain current is given as.

$I_D = K_n (V_{as }- V_{Tn})^2$

Applying kyl in D - S ckt

VDD - ID RD - VDS = 0

VDS = VDD - ID RD

2) Self bias.

Applying KVL in g-s ckt

-IGRG - VAS - IDRS = 0

But IG = 0

$\therefore$ $V-{as} $- -ID RS.

Putting this in equation of ID.

$ID = Kn \ (V_{as} = VTn)^2$

Solving this to get value of iD.

Applying KVL in D.S ckt

VDS = VDD - ID ( RD + RS)

3) Voltage divider bias.

VGs = VG - VS.

VG = $\frac{VDD \ R2}{R_1 + R_2}$ VS = Id RS

$\therefore$ $V_{as} = \frac{VDDR2}{R_1 + R_2}$ -Id RS

Putting the expression of $V_{as}$ in the equation of ID

$I_D = K_n (V_{as} = VTn)^2$

Solving above equation to get value of ID.

Applying KVL in D-S ckt

VDD - ID RD - Vos - ID RD = 0

VDS = VDD - Id (RD + RJ)

Given

$\theta_{JA}$ = 1.2˚ c/w

$\theta_{JC}$ = 0.5˚ c/w

$\theta_{cs}$ = 0.5˚ c/w

Tamb = 40˚ c

Tj(max) = 200˚ c

PD(max) = $\frac{Tdev - Tamb}{\theta \ sk + \theta \ jc + \theta \ cs}$

$\frac{200 - 40}{0.5 + 0.5 + 1.2}$

PD(max) = 72.72 w

For Hartely OSC

f = $\frac{1}{2\pi \ \sqrt {Leq \ C}}$

Leq = $L_1 + L_2$ = 2mH

c = 0.2 $\mu$f

f = $\frac{1}{2 \ \pi $\sqrt {2mH x 0.2 \mu \ f}}$ $f_2 \ 8 KH_z$