Given the information provided in Fig. 4.124, determine:a. β.b. VCC.c. RB.FIG. 4.124

Calculate the value of forward common emitter current gain, \(\beta\).

\(\beta=\frac{I_{C}}{I_{B}}\)

Substitute \(20 \mu \mathrm{A}\) for \(I_{B}\) and \(3.09 \mathrm{~mA}\) for \(I_{C}\).

\(\beta=\frac{3.09 \times 10^{-3}}{20 \times 10^{-6}}\)

\(=154.5\)

Therefore, the value of forward common emitter current gain, \(\beta\) is \(154.5\).

(b)

Apply Kirchhoff's voltage law across the output terminal.

\(-V_{c C}+I_{C} R_{C}+V_{C E}+V_{E}=0\)

\(V_{C C}=I_{C} R_{C}+V_{C E}+V_{E}\)

Substitute \(7.3 \mathrm{~V}\) for \(V_{C E}, 3.09 \mathrm{~mA}\) for \(I_{C}, 2.7 \mathrm{k} \Omega\) for \(R_{C}, 2.1 \mathrm{~V}\) for \(V_{E} .\)

$$ \begin{aligned} V_{C C} &=(3.09 \mathrm{~mA})(2.7 \mathrm{k} \Omega)+7.3 \mathrm{~V}+2.1 \mathrm{~V} \\ &=8.34+7.3+2.1 \\ &=17.74 \mathrm{~V} \end{aligned} $$

Therefore, the value of supply voltage, \(V_{C C}\) is \(17.74 \mathrm{~V}\).

(c)

Calculate the value of base resistance \(\left(R_{B}\right)\).

$$ \begin{aligned} R_{B} &=\frac{V_{R_{8}}}{I_{B}} \\ &=\frac{V_{C C}-V_{B E}-V_{E}}{I_{B}} \end{aligned} $$

Substitute \(17.74 \mathrm{~V}\) for \(V_{C C}, 0.7 \mathrm{~V}\) for \(V_{B E}, 2.1 \mathrm{~V}\) for \(V_{E}\) and \(154.5\) for \(\beta\).

\(R_{E}=\frac{17.74-0.7-2.1}{20 \times 10^{-6}}\)

$$ \begin{aligned} &=\frac{14.94}{20 \times 10^{-6}} \\ &=747 \times 10^{3} \\ &=747 \mathrm{k} \Omega \end{aligned} $$

Therefore, the value of base resistance, \(R_{B}\) is \(747 \mathrm{k} \Omega\).