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For the transistor circuit shown in Fig.14.19, evaluate VE, RB, and RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V, and VCC = 12 V, \beta = 100.

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Explanation:-

As we know the base current is very small So,

I_{C}=I_{E}\\ R_{C}=7.8 k\Omega\\ from \, th\, fig. I_{C}\left ( R_{c}+R_{E} \right )+V_{CE}=12\\ \left ( R_{E}+R_{C} \right )\times 1 \times 10^{-3}+3=12\\ \left ( R_{E}+R_{C} \right )=9 \times 10^{3}=9k\Omega\\ R_{E}=9-7.8=1.2k\Omega\\ V_{E}=I_{E}\times R_{E}\\ =1 \times 10^{-3} \times 1.2 \times 10^{3}=1.2V\\ Voltage V_{B}=V_{E}+V_{BE}=1.2+0.5=1.7V\\ Current \: I=\frac{V_{B}}{20 \times 10^{3}}=\frac{1.7}{20 \times 10^{3}}=0.085mA\\ Resistance \: R_{B}=\frac{12-1.7}{\frac{I_{C}}{\beta }+0.085}\times 10^{3}=\frac{10.3}{0.01+0.085} \: \: \: \: \: \: \left [ Given \beta =100 \right ]\\ =108 k\Omega\\

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