Refer to Fig. 5.15 (b). We note at the outset that since we are required to design for $V_c=+5 \mathrm{~V}$, the CBJ will be reverse biased and the BJT will be operating in the active mode. To obtain a voltage $V_C=+5 \mathrm{~V}$ the voltage drop across $R_C$ must be $15-5=10 \mathrm{~V}$. Now, since $I_C=2 \mathrm{~mA}$, the value of $R_C$ should be selected according to

$$
R_C=\frac{10 \mathrm{~V}}{2 \mathrm{~mA}}=5 \mathrm{k} \Omega
$$


Since $v_{R F}=0.7 \mathrm{~V}$ at $i_C=1 \mathrm{~mA}$, the value of $v_{B E}$ at $i_C=2 \mathrm{~mA}$ is

$$
V_{B E}=0.7+V_T \ln \left(\frac{2}{1}\right)=0.717 \mathrm{~V}
$$


Since the base is at 0 V , the emitter voltage should be

$$
V_E=-0.717 \mathrm{~V}
$$


For $\beta=100, \alpha=100 / 101=0.99$. Thus the emitter current should be

$$
I_E=\frac{I_C}{\alpha}=\frac{2}{0.99}=2.02 \mathrm{~mA}
$$


Now the value required for $R_E$ can be determined from

$$
\begin{aligned}
R_E & =\frac{V_E-(-15)}{I_E} \\
& =\frac{-0.717+15}{2.02}=7.07 \mathrm{k} \Omega
\end{aligned}
$$