written 5.5 years ago by |
i)Selection of operating point
The maximum voltage developed across the transistor collector emitter is
$V_{CEQ}1+V_{0}$=1+10=11 V
Also the quiescent voltage
$V_{CEQ}\geq V_{p}+V_{CE}$
$\geq 5+0.3$
$V_{CEQ}\geq5.3 V=6V \ \ \ \ \ .......(i)$
$I_{Lp}=\frac{V_{P}}{R_{L}}=\frac{5}{2K}$=2.5 mA
Select $I_{CQ}\ge 1_{L(P)}=5 mA \ \ \ \ \ ......(ii) $
The operating point is located at (6 V.5mA)
ii)Selection of $V_{CC}$
With given operating point (6 V.5 mA)
$V_{CC}\geq 2 V_{CEQ} \geq 2\times 6 =12 V=12 V \ \ \ \ \ \ ......(iii)$
iii)Selection of transistor
Power of dissipation =$6\times 5$ mA=30mW
Select the transistor 2N2976 with following specification ......(iv)
$P_{D(max)}=250mW$ $V_{CE(max)}$=45v $I_{C(max)}$=30mA
$V_{C(max}$=45V. $h_{FE}$=60/240
iv) Selection of $R_{E}$
For normal operation of the transistor with maximum stability Select $V_{RE}$=15% of $V_{CC}$=0.15 \times 12=1.8v
$R_{E}=\frac{V_{RE}}{I_{E}}=\frac{1.8}{5mA}=360\Omega \ \ \ \ ....(v)$
v)Selection of $R_{C}$
The maximum voltage across collector resistor $R_{C}$ is
$V_{RC}=V_{CC}-V_{CEQ}-V_{RE}$
=12v-6v-1.8v
$V_{RC}$=4.2V
$R_{C}=\frac{V_{RC}}{I_{C}}=\frac{4.2}{5mA}=840\Omega \ \ \ \ ....(vi)$
vi) Selection of $R_{1}and R_{2}$
Assume stability factor of 10
$R_{B}=9 R_{E}=9\times 360=3.24k\Omega$
and $V_{B}=V_{E}+V_{BE}$=1.8+0.7=2.5V
Applying voltage divider rule at base of transistor
$R_{1}=\frac{R_{B}V_{CC}}{V_{B}}=\frac{3.24K\times 12V}{2.5}$
$R_{1}=15.55K\Omega=15K\Omega \ \ \ \ .....(viii)$
But $R_{B}=R_{1}||R_{2}$
3.24K=$\frac{15K\times R_{2}}{15K+R_{2}}$
$48.6\times10^{6}+3.24K\times R_{2}=15K\times R_{2}$
$R_{2}=\frac{48.6\times10^{6}}{11.76\times10^{3}}=4.13K\Omega=4.2K\Omega \ \ \ \ \ ....(ix)$
vii)Selection of $C_{E}$
$X_{CE}=\frac{R_{E}}{10}$=$\frac{360}{10}=36\Omega; X_{CE}=\frac{1}{2\pi f_{0}C_{E}}$
$C_{E}=\frac{1}{2\pi\times 30Hz\times 36\Omega}$
=147.36$\mu F=147\mu F \ \ \ \ \ ........(x)$
viii)Calculation of coupling capacitors ($C_{1}and C_{2}$)
The value of coupling capacitor
$C_{1}=\frac{1}{2\pi X_{CI}f_{0}}$
But $X_{CI}=\frac{R_{S}+R_{i}}{10}=\frac{R_{s}+(R_{1}||R_{2}||h_{ie})}{10}$
$\frac{600+(15K||4.2K||2.1K)}{10}$
$\frac{1.88K}{10}=188\Omega$
$C_{1}=\frac{1}{2\pi \times 188\times30}=28.2\mu F=33\mu F \ \ \ \ \ \ ......(ix)$
and $C_{2}=\frac{10}{2\pi R_{L}f_{0}}=\frac{10}{2\pi\times2K\times 30}$
$C_{2}=26.5\mu F=22 \mu F \ \ \ \ \ ....(xii)$