The circuit diagram for cs - cs Amplifier is as shown below.

Let $A_2$ is gain of BJT Ampli

" $A_1$ " " " FET "

Taking $A_2 = 50 A_1$ = 10

$\therefore$ A = $A_1 x A_2$ = 10 x 50 = 500.

Designing of 2nd stage.

1) Selection of biasing ckt - we will use voltage divider biasing.

2) Selection of Rc

Rc = $\frac{|Av| . hie} {hfe}$

= $\frac{50 x 4.5k}{290}$

Rc = 0.775k$\Omega$

$\therefore$ taking Rc = 810 $\Omega$ / 1/4 w

3) Selection of Q point

Given Vcc = 6V

VCEQ = $\frac{Vcc}{2}$ = 3V

VRE = $\frac{vcc}{10}$ = $\frac{6}{10}$ = 0.6 v.

$V_RC = V_CC - V_CE - V_RE$

= 6 - 3 - 0.6

VRC = 2.4 v

$\therefore$ ICQ Rc = 2.4

ICQ = $\frac{2.4}{RC}$ = $\frac{2.4}{0.81}$ = 3MA

4) Selection of RE

VRE = ICQ RE

RE = $\frac{VRE}{ICQ}$ = $\frac{0.6v}{3mA}$ = 200 $\Omega$

$\therefore$ Selecting RE = 200 $\Omega$ / 1/4 w.

5) Selection of biasing Resi.

S = $\frac{(1+\beta)}{1+\beta\frac{RE}{rth}}$

8 = $\frac{1+290}{1+\frac{290 x 0.2}{RTH}}$

8 + $\frac{464}{RTH}$ = 291

RTH = $\frac{464}{283}$ = 1.63k

$\frac{R_1 R_2}{R_1 + R_2}$ = 1.63k ------ (A)

$V_B = V_BE + V_RE$

= 0.7 + 0.6

VB = 1.3v

$\frac{vcc R2}{R_1 + R_2}$ = 1.3

$\frac{R_2}{R_1 + R_2}$ = $\frac{1.3}{6}$

$\frac{R_2}{R_1 + R_2}$ = 0.21k ------ (B)

Putting the value of B in (A)

$R_1$ x 0.21 = 1.63

$R_1$ = 7.76 k $\Omega$

$\therefore$ Selecting $R_1$ = 8.1k / 1/4w

Putting selected value of $R_1$ in (B)

$\frac{R_2}{8.1 + R_2}$ = 0.21

0.79 R_2 = 1.7

$R_2$ = 2.15 k

$\therefore$ Selecting $R_2$ = 2.2 k / 1/4 w

Designing of 1st stage

6) We will design JFET Ampli for ZTD using self bias method.

| Vasq | = | vp | - 0.63

| VasQ | = 3.37 v

$\therefore$ VasQ = -3.37 v

7) Calculation of ID and gm

ID = IDss ( 1- $\frac{Vasq}{vp})^2$

= 7 ( 1 - $\frac{-3.37}{-4})^2$

ID = 0.18 MA

gm = gmo ( 1- $\frac{vasq}{vp})$

= 5000 ( 1 - $\frac{-3.37}{vp})$

gm = 800 $\mu$ $\mho$

$\therefore$ gm = 0.8 m $\mho$

8) Selection of R4

$\because$ we want Ri > 1 M $\Omega$ $\therefore$ we will select

R4 = 1M $\Omega$ / 1/4 w

9) Selection of Rs.

for self bias ckt

Rs = | $\frac{vasq}{IO}$ |

= $\frac{3.37}{0.18}$

Rs = 18.72 k

$\therefore$ Rs = 20k / 1/4 w.

10) Selecting of RD.

RD = $\frac{lAil}{gm}$

= $\frac{10}{0.8}$

RD = 12.5 k

$\therefore$ Selecting RD = 15k / 1/4 w.

11) Selection of capacitors.

Cc1 = $\frac{1}{2\pi fc Rg}$

$\because$ fl is not given assuming 20 Hz

Cc1 = $\frac{1}{2 IT 20x 1m}$ = 0.793 nf

$\therefore$ Selecting Cc1 = 0.79nf/25v

Cs = $\frac{10}{2\pi flRS}$ = $\frac{10}{2x\pi x 20 x 20 x 10^3}$

Cs = 3.97 x $10^-6$ f

Selecting Cs = 3.9 $\mu$ f / 25 v

CE = $\frac{10}{2\pi flRE}$

= $\frac{10}{2 x \pi x 20 x 200}$

CE = 397 $\mu$ f

$\therefore$ Selecting CE = 390 = $\mu$ f / 25v

Cc2 = $\frac{1}{2 \pi fl Req}$

Req = RO1 + Ri2

Ro1 = RD = 15k

Ei2 = R1 || R2 || hie

= 8.1 || 2.2 || 4.5

= 1.25k

Cc2 = $\frac{1}{2 x \pi x 20 x 16.25 x 10^3}$

Cc2 = 48.9 $\mu$ f

$\therefore$ Selecting Cc2 = 50 $\mu$ f / 25v.