Let us select the transistor BC147B having

$P_{D(max)}$=250mW$ \ \ \ \ $h_{fc(typ)}$=330 $h_{fe}$=4.5k$ \ \ \ \ \ $h_{oe}=30\mu-mho$

i) Calculation of $R_{C}$

$A_{v}=\frac{-h_{fe}R_{C}}{h_{ic}}$ $\ \ \ \ \ $150=$\frac{330\times R_{C} }{4.5}k\Omega$

$R_{C}=2.045k\Omega$

Select $R_{C}=2.2 k\Omega$ $ \ \ \ \ \ \ .......... (i)$

ii)Selection of Q-point

$V_{CEQ}\geq V_{p}+V_{CE}$

$\geq 2.5v\sqrt 2 +03v$

$geq$3.83V

Select $V_{CEQ}$=4.5V

For normal operation let $V_{RE}$=15% of $V_{CC}$=1.35V

$V_{RC}=V_{CC}-V_{CEQ}-V_{RE}$

=9v-4.5V-1.35V

=3.15V

$I_{CQ}=\frac{V_{R_{c}}}{R_{C}}=\frac{3.15V}{2.2k\Omega}$=1.43mA

Selcet $I_{CQ}$=2mA .........(ii)

Q point will be (4.5 V.2 mA)

iii) Calculation of $R_{E}$

$R_{E}=\frac{V_{R_{E}}}{I_{CQ}}=\frac{1.35v}{2mA}$=0.675k

Select $R_{E}=680\Omega \ \ \ \ \ \ \ ...........(iii)$

iv) Selection of $R_{1} and R_{2}$

If S=10 then

$R_{B}=9\times R_{E}=9\times 680\Omega=6120\Omega$

and $V_{B}=V_{E}+V_{BE}$=1.35V+0.7V=2.05V

$R_{1}=\frac{R_{B}V_{CC}}{V_{B}}=\frac{6.12k\times 9V}{2.05V}=26.9k\Omega$

Select $R_{1}=22k\Omega$

As $R_{B}=R_{1}||R_{2}$ substituting values of $R_{B} and R_{1}$ in this equation we get $R_{2}=8.2k\Omega$

Select R1=27k$\Omega$ and $R_{2}=8.2k\Omega \ \ \ \ \ \ \ \ .......(iv)$

v) Selection of $C_{E}$

$X_{CE}=\frac{R_{E}}{10}=68\Omega$

$C_{E}=\frac{1}{2\pi \times 20Hz \times 68\Omega} =117\mu F \ \ \ \ \ \ ........(v)$

Select $C_{E}=220 \mu F$

vi) Selection of $C_{C}$

$R_{in1}=(R_{1})||R_{2}||h_{ie}$

=$(27k||8.2k)||4.5k$

=$2.62 k\Omega$

Output impedance of first stage $R_{01}$

$R_{01}=R_{C1}||R_{in2}$

The $R_{in2}$= input impedance of second stage = input impedance of first stage

$R_{01}=2.2k|| 2.62k=1.2k$

$C_{C}=\frac{1}{2\pi f(R_{01}+R_{in2})}$

$\frac{1}{2\pi \times 20 Hz \times (1.2k+2.62k)}=2.08\mu F$

Select $C_{c}=3.3\mu F \ \ \ \ \ ...........(vi)$