0
827views
Design single stage CE voltage amplifier as to give the output voltage of $10V_{pp}$ across the load resistor 2K$\Omega$. The frequency range operation 30 Hz -19KHz. The source resistance 600$\Omega$
1 Answer
0
4views

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 $C_{2}=\frac{10}{2\pi R_{L}f_{0}}=\frac{10}{2\pi\times2K\times 30}$

$C_{2}=26.5\mu F=22 \mu F \ \ \ \ \ ....(xii)$

Please log in to add an answer.