For the transistor circuit shown in figure, evaluate $V_E, R_B, R_E$, given $I_C=1 \mathrm{~mA}, V_{C E}=3 \mathrm{~V}, V_{B E}=0.5 \mathrm{~V}$ and $V_{C C}=12 \mathrm{~V}, \beta=100$.
Consider the fig. (b) given here to solve this problem
$$\begin{aligned} I_C & \approx I_E \quad \text{[As base current is very small.]}\\ R_C & =7.8 \mathrm{k} \Omega \end{aligned}$$
From the figure, $\quad I_C\left(R_C+R_E\right)+V_{C E}=12$
$$\begin{aligned} \left(R_E+R_C\right) \times 1 \times 10^{-3}+3 & =12 \\ R_E+R_C & =9 \times 10^3=9 \mathrm{k} \Omega \\ R_E & =9-7.8=1.2 \mathrm{k} \Omega \\ V_E & =I_E \times R_E \\ & =1 \times 10^{-3} \times 1.2 \times 10^3=1.2 \mathrm{~V} \\ \text { Voltage } V_B & =V_E+V_{B E}=1.2+0.5=1.7 \mathrm{~V} \\ \text { Current } I & =\frac{V_B}{20 \times 10^3}=\frac{1.7}{20 \times 10^3}=0.085 \mathrm{~mA} \\ \text { Resistance } R_B & =\frac{12-1.7}{\frac{I_C}{\beta}+0.085}=\frac{10.3}{0.01+0.085} \quad [\text{Given, }\beta=100]\\ & =108 \mathrm{k} \Omega \end{aligned}$$
In the circuit shown in fig. (a), find the value of R$_C$.
Consider the fig. (b) to solve this question,
$$ \begin{aligned} I_E & =I_C+I_B \text { and } I_C=\beta I_B \quad\text{.... (i)}\\ I_C R_C+V_{C E}+I_E R_E & =V_{C C} \quad\text{.... (ii)}\\ R I_B+V_{B E}+I_E R_E & =V_{C C} \quad\text{.... (iii)}\\ \because\quad I_E & \approx I_C=\beta I_B \end{aligned}$$
From Eq. (iii),
$$\left(R+\beta R_E\right) I_B=V_{C C}-V_{B E}$$
$$\begin{aligned} \Rightarrow \quad I_B & =\frac{V_{\mathrm{CC}}-V_{\mathrm{BE}}}{R+\beta \cdot R_E} \\ & =\frac{12-0.5}{80+1.2 \times 100}=\frac{11.5}{200} \mathrm{~mA} \end{aligned}$$
$$\begin{aligned} &\text { From Eq. (ii), }\\ &\begin{aligned} \left(R_C+R_E\right) & =\frac{V_{\mathrm{CE}}-V_{\mathrm{BE}}}{I_C}=\frac{V_{\mathrm{CC}}-V_{\mathrm{CE}}}{\beta I_B} \quad\left(\because I_C=\beta I_B\right)\\ \left(R_C+R_E\right) & =\frac{2}{11.5}(12-3) \mathrm{k} \Omega=1.56 \mathrm{k} \Omega \\ R_C+R_E & =1.56 \\ R_C & =1.56-1=0.56 \mathrm{k} \Omega \end{aligned} \end{aligned}$$