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DC design
i)Selection of operating point Q
The operating point will be selected approximately at the center of dc load line to get output signal without distortion
$V_{CEQ}\frac{V_{CC}}{2}$=\frac{12}{2}=6 V .......(i)
and $I_{CEQ}\ge I_L{P}=\frac{V_{OP}}{R_{L}}=\frac{3}{5K}$=0.6mA
Hence select $I_{CQ}$=1mA .......(ii)
Therefore operating point Q(6V,1mA)
ii)Selection of transistor
Power dissipation =$6\times1$mA=6mW
Select the transistor BC146 with following specifications ......(iii)
$P_{D(max)}$=50mW, $V_{CE(max)}$=20V, I_{C(max)}=50mA
$V_{CB(max)}$=20V, $h_{FE}=80$
iii)Selection of $R_{E}$
For better stabilization use the voltage across emitter resistor approximately equal to $(1/10)^{th}$ of $V_{cc}$ or $5V_{BE}$
$V_{BE}=(1/10)^{th}V_{CC}=\frac{12}{10}$=1.2V
$R_{E}=\frac{V_{RE}}{I_{E}}=\frac{1.2}{1 mA}=1.2k\Omega \ \ \ \ .. . .. (iv)$
iv)Selection of $R_{C}$
The voltage across the collector resistor $R_{C}$
$V_{RC}=V_{CC}-V_{CEQ}-V_{RE}$
=12-6-1.2=4.8V
$R_{C}=\frac{V_{RC}}{I_{C}}=\frac{4.8V}{1mA}=4.8k\Omega \ \ \ \ ....(v)$
v)Selection of R_{1} and R_{2}
For better stability, the current flowing through resistor $R_{2}$ is approximately greater than 10 times base current or $(1/10)^{th}$ of $I_{C}$
$I_{2}=\frac{I_{C}}{10}=\frac{1mA}{10}-0.1mA$
and the voltage at base of transistor is
$V_{B}=V_{BE}+V_{RE}$ ($I_{B}$ is very small)
=0.7+1.2
$V_{B}$=1.9V
$R_{2}=\frac{V_{B}}{I_{2}}=\frac{1.9V}{0.1mA}=19k\Omega \ \ \ \ .......(vi)$
The value of current through $R_{1}$ is
$I_{1}=I_{B}+I_{2}=I_{2}$
(since $I_{B}=I_{c}/h_{FE}=1 mA/80=12.5\mu$ A very less hence neglected)
$R_{1}=\frac{V_{CC}-V_{B}}{I_{2}}$
$\frac{12-1.9}{0.1mA}=101 k\Omega=100k\Omega \ \ \ ....(vii)$
and parallet combination $R_{1}$ and $R_{2}$ is
$\frac{R_{1}R_{2}}{R_{1}+R_{2}}$
=$\frac{100K\times 16K}{100K+19K}=15.9k\Omega \ \ \ \ .....(viii)$
AC design
All the capacitive effects are considered at the lower 3 dB frequency.Since with increase in frequency the reactance of capacitor decreases. At high frequency it is almost short circuited.
vi) The effective ac load
$R_{Lte}$=$R_{C}||R_{L}=4.8K||5K$
$\frac{4.8k\times 5k}{4.8K+5K}=2.4k\Omega \ \ \ ....(ix)$
vii) Selection of emitter bypass capacitor $(C_{E})$
The reacatnce value of bypass capacitor can be calculated as
$X_{CE}=\frac{R_{E}}{10}=\frac{1.2K}{10}=120\Omega$
$C_{E}=\frac{1}{2\pi X_{CE}f_{0}}=\frac{1}{2\pi\times 120\times 20}$
$C_{E}=66.6\mu F=66\mu F \ \ \ \ ...... (x)$
viii) Selection fo coupling capacitor $(C_{1} and C_{2})$
The value of coupling capacitor $C_{1}$ is
$C_{1}=\frac{1}{2\pi X_{ci}f_{0}}$
But $X_{Ci}=\frac{R_{s}+R_{1}||R_{2}||h_{fe}}{10}$
=$\frac{600+(100K||19K||2.1K)}{10}$
$\frac{2.45K}{10}=245\Omega$
$C_{1}=\frac{1}{2\pi 245\times 20}=32\mu F \ \ \ ......(xi)$
and $C_{2}=\frac{10}{2\pi (R_{C}+R_{L})f_{0}}=\frac{10}{2\pi(4.8K+5K)\times 20}$
$\frac{10}{2\pi\times 9.8K\times20}$
$C_{2}=8.12\mu F=10\mu F \ \ \ ....(xii)$