For the transistor circuit shown in figure , evaluate `V_E , R_B and R_E`. Given `I_C = 1 mA`,
` V_(CE) = 3V , V_(BE) = 0.5 V , V_(CC )= 12 V and beta = 1000`

Explanation

Let the current through the various arms be as shown in Fig.

Here, `I_C=1mA=10^(-3)A, V_(CE)=3V, V_(BE)=0.5V, beta=100`
`I_B=I_C/beta=10^(-3)/100=10^(-5)A=.01mA`
`I_(E)=I_B+I_C=10^(-5)+10^(-3)~~10^(-3)A=I_C`
In closed circuit, `AB_(1)CEDA`
I_C R_C+V_(CE)+I_(E)R_(E)=V_(C C)`
or `I_C R_C+I_C R_(E)=V_(C C)-V_(CE)=12-3=9V [ :' I_(E)=I_C]`
or `I_C(R_C+R_(E))=9 or R_C+R_(E)=9/10^(-3)=9kOmega`
`R_(E)=9kOmega-R_=9kOmega-7.8kOmega=1.2kOmega=1.2xx10^(3)Omega`
`V_(E)=I_(E)R_(E)=I_C R_(E)=10^(-3)xx(1.2xx10^(3))=1.2V`
Let V= potential difference between H and J, then
`V=V_(E)+V_(BE)=1.2+0.5=1.7V`
`I=V/R=1.7/(20xx10^(3))A=0.085mA`
Now `V_(C C)=(I_B+I)R_B+V`
or `R_B=(V_(C C)-V)/(I_B+I)=((12-1.7)V)/((.01+0.085)mA)=(10.3V)/(0.095mA)=108.4kOmega`