**1 Answer**

written 2.8 years ago by |

- We know the gain of a BJT transistor given y the formula

$A_v=\dfrac{V_o}{V_i}$

Hence the Output Voltage is the product of Gain and Input Voltage

$V_{o}=300V$

Wr know the Stability factor for a BJT transistor is given as

$S=\dfrac{(\beta+1){(1+\dfrac{R_1}{R_L})}}{({\beta+1)}+\dfrac{R_1}{R_L}}$

- From data sheet of BJT147B transistor we have Gain value of 150 to 460

we can take gain value as 150 here

$10=\dfrac{(151){(1+\dfrac{R_1}{10000})}}{({151)}+\dfrac{R_1}{10000}}=\dfrac{(151){(1+{0.0001R_1}{})}}{({151)}+{0.0001R_1}{}}$

$=1510+0.0001R_1=151+0.0151R_1$

$=0.015R_1=1359$

$R_1=90.6K\Omega$

Let $V_{cc}=10V;V_{CE}=4V;I_C=10mA$

- The Collector current is product of Current gain and Base current

$I_b=\dfrac{I_{c}}{{\beta}}$

$I_b=\dfrac{{2\times 10^{-3}}}{{150}}=0.01mA$

- We know that by applying in Collector Emitter junction we have

$V_{CE}=V_{CC}-I_C(R_C+R_L)$

$4=10-2(R_c+1000)=-2R_c+2010$

$-2R_c=-2006$

$R_C=2.006K\Omega$

- By ohms law the Base voltage is given by

$V_B=I_C\times R_L=2V$

- By voltage division we have the base voltage as

$V_B=\dfrac{R_{2}V_{cc}}{R_1+R_2}$

$2=\dfrac{10R_2}{90600+R_2}$

$8R_2=181200$

$R_2=22.650K\Omega$