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Design a single stage common emitter amplifier for the quiescent condition of (7.0 v, 2 mA) If bandwidth is 25Hz-20 KHz and $R_{i}\geq 5k\Omega$ Also calculate the stability factor
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i)Selection of $V_{CC}$

With given operating point (7.0 v. 2 mA)

Select $V_{cc}\geq2 V_{CEQ}\geq 2 \times 7 =14 v ==15 v$ ........(i)

ii) Selection of transistor

Minimum power dissipation =$7 \times 2 mA =14 mW$

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$P_{D(max)}$ =50mW. $V_{CE(max)}$=20V, $I_{C(max)}=50mA$

$V_{CB(max)}$=20v, $h_{FE}$=80

iii) Selection of $R_{C}$ and $R_{E}$

Writing loop equation at the output side of Fig. Ex. 4.16

$V_{cc}=V_{CE}=I_{C}(R_{C}+R_{g})$

15-7=2 mA$(R_{C}+R_{g})$

$R_{C}+R_{E}=4k\Omega$

$R_{E}=\frac{20}{100}\times 4k\Omega =800 \Omega \ \ \ \ ......(iii)$

$R_{C}=4K\Omega-800 \Omega=3.2 k\Omega \ \ \ \ \ ......(iv)$

iv) 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 current through $R_{2}$ is $(1/10)^{th} of I_{C}$

$I_{2}=\frac{I_{C}}{10}=\frac{mA}{10}$=0.2mA

The voltage at base of transistor is,

$V_{B}=V_{BE}+V_{RE}$

0.7+1.5

$V_{B}=2.2V$

$R_{2}$ is given by

$R_{2}=\frac{V_{B}}{I_{2}}=\frac{2.2}{0.2mA}=11k\Omega \ \ \ \ ....(v)$

$R_{1}$ is given by

$R_{1}=\frac{V_{CC}-V_{B}}{I_{2}}=\frac{15-22}{0.2mA}=64k\Omega \ \ \ \ (vi)$

(Since $I_{1}=I_{B}+I_{2}=I_{2},I_{B}=I_{C}/h_{FE}=2 mA/80=25uA very less$)

Parallel combination of $R_{1}$and $R_{2}$ is

$R_{B}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{64K\times 11K}{64K+11K}=9.3k\Omega=10k\Omega \ \ \ ....(vii)$

(v) Selection of emitter bypass capacitor $(C_{B})$

The value of $C_{E}$ is calculated at lower end of bandwidth i.e f=25HZ

$X_{CE}=\frac{h_{ie}}{1+h_{fe}}$

But $h_{ie}=h_{fe}\times f_{e}=80\times \frac{25mV}{1mA}=2 k\Omega$

$X_{CE}\frac{2\times 10^{3}}{1+80}$=24.69

$C_{E}=\frac{1}{2\pi X_{CE}f_{0}}$

=$\frac{1}{2\pi\times 24.69\times 25}=2578\mu F=260\mu F \ \ \ .....(viii)$

vi) Calculation of $C_{1} and C_{2}$

$C_{1}=\frac{1}{2\pi X_{cI}f_{0}}$

But $X_{C1}=\frac{R_{i}}{10}=\frac{5K}{10}=500\Omega$

$C_{1}=\frac{1}{2\pi \times 500 \times 25}=12.73\mu F=10\mu F \ \ \ ....(ix)$

Also, $C_{2}=\frac{1}{2\pi X_{C2}f_{0}}$

But $X_{c2}=(R_{C}+R_{L})/10$

=(3.2K+5K)/10=820\Omega

$C_{2}=\frac{1}{2\pi\times 820\times 25}$

$C_{2}=7.76\mu F=10\mu F \ \ \ .....(x)$

vii)Calculation of stability factor

stability factor is given by

S=$\frac{1+h_{FE}}{1+h_{FE}[R_{E}/(R_{E}+R_{B})]}$

=$\frac{1+80}{1+80[800/(800+9400)]}$

=11.13=11 which is better.. . . . . .(xi)

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