For the R–C-coupled amplifier of Fig. 4.141, determinea. the voltages VB, VC, and VE for e...

Determine voltage at base \(\left(V_{B 1}\right)\) of transistor \(\left(Q_{1}\right)\).

$$ V_{B 1}=\frac{R_{2} V_{C C}}{R_{1}+R_{2}} $$

Replace \(20 \mathrm{~V}\) for \(V_{C C}, 18 \mathrm{k} \Omega\) for \(R_{1}, 4.7 \mathrm{k} \Omega\) for \(R_{2}\) in equation.

$$ \begin{aligned} V_{B 1} &=\frac{(4.7 \mathrm{k} \Omega)(20 \mathrm{~V})}{(4.7 \mathrm{k} \Omega)+(18 \mathrm{k} \Omega)} \\ &=\frac{\left(4.7 \times 10^{3} \Omega\right)(20 \mathrm{~V})}{\left(4.7 \times 10^{3} \Omega\right)+\left(18 \times 10^{3} \Omega\right)} \\ &=4.14 \mathrm{~V} \end{aligned} $$

Therefore, the voltage at the base terminal of the transistor \(\left(Q_{1}\right)\) is \(4.14 \mathrm{~V}\).

Determine voltage at emitter \(\left(V_{E 1}\right)\) of transistor \(\left(Q_{1}\right)\).

Consider \(V_{B E}=0.7 \mathrm{~V}\).

$$ \begin{aligned} &V_{B E}=V_{B}-V_{E} \\ &0.7=4.14-V_{E 1} \\ &V_{E 1}=4.14-0.7 \\ &V_{E 1}=3.44 \mathrm{~V} \end{aligned} $$

Therefore, the voltage at the emitter terminal of the transistor \(\left(Q_{1}\right)\) is \(3.44 \mathrm{~V}\).

Determine the collector current \(I_{C 1}\).

From Figure 1 , the collector current is same as the emitter current.

$$ I_{C_{1}} \cong I_{E_{1}} $$

Determine the emitter current.

$$ \begin{aligned} &V_{E 1}=I_{E 1} R_{E 1} \\ &I_{E 1}=\frac{V_{E 1}}{R_{E 1}} \end{aligned} $$

Replace \(3.44 \mathrm{~V}\) for \(V_{E 1}, 1 \mathrm{k} \Omega\) for \(R_{E 1}\) in equation.

$$ \begin{aligned} I_{E 1} &=\frac{3.44}{1 \mathrm{k} \Omega} \\ &=3.44 \mathrm{~mA} \end{aligned} $$

Apply Kirchhoff's voltage law at the collector terminal of \(\left(Q_{1}\right)\).

$$ \begin{aligned} &V_{C C}=I_{C l} R_{C 1}+V_{C 1} \\ &V_{C 1}=V_{C C}-I_{C 1} R_{C 1} \end{aligned} $$

Replace \(20 \mathrm{~V}\) for \(V_{C C^{+}} 3.44 \mathrm{~mA}\) for \(I_{C 1}\), and \(2.2 \mathrm{k} \Omega\) for \(R_{C 1}\) in equation.

$$ \begin{aligned} V_{C 1} &=20 \mathrm{~V}-(3.44 \mathrm{~mA})(2.2 \mathrm{k} \Omega) \\ &=20 \mathrm{~V}-\left(3.44 \times 10^{-3} \mathrm{~A}\right)\left(2.2 \times 10^{3} \Omega\right) \\ &=12.43 \mathrm{~V} \end{aligned} $$

Therefore, the collector voltage is \(12.43 \mathrm{~V}\).

Determine voltage at base \(\left(V_{B 2}\right)\) of transistor \(\left(Q_{2}\right)\).

$$ V_{B 2}=\frac{R_{2} V_{c C}}{R_{1}+R_{2}} $$

Replace \(20 \mathrm{~V}\) for \(V_{C C}, 22 \mathrm{k} \Omega\) for \(R_{1}, 3.3 \mathrm{k} \Omega\) for \(R_{2}\) in equation.

$$ \begin{aligned} V_{B 2} &=\frac{\left(3.3 \times 10^{3} \Omega\right)(20 \mathrm{~V})}{\left(3.3 \times 10^{3} \Omega\right)+\left(22 \times 10^{3} \Omega\right)} \\ &=2.61 \mathrm{~V} \end{aligned} $$

Therefore, the voltage at the base terminal \(V_{B 2}\) of the transistor \(\left(Q_{2}\right)\) is \(2.61 \mathrm{~V}\).

Determine voltage at emitter \(\left(V_{E 2}\right)\) of transistor \(\left(Q_{2}\right)\).

Consider \(V_{B E}=0.7 \mathrm{~V}\).

$$ \begin{aligned} &V_{B E}=V_{B}-V_{E} \\ &0.7=2.61-V_{E 2} \\ &V_{E 2}=2.61-0.7 \\ &V_{E 2}=1.91 \mathrm{~V} \end{aligned} $$

Therefore, the voltage at the emitter terminal \(V_{E 2}\) of the transistor \(\left(Q_{2}\right)\) is \(1.91 \mathrm{~V}\).

Determine the collector current \(I_{C 2}\).

From Figure 1, the collector current is same as the emitter current.

$$ I_{C 2} \cong I_{E 2} $$

Determine the emitter current.

$$ \begin{aligned} &V_{E 2}=I_{E 2} R_{E 2} \\ &I_{E 2}=\frac{V_{E 2}}{R_{E 2}} \end{aligned} $$

Replace \(1.91\) for \(V_{E 2}, 1.2 \mathrm{k} \Omega\) for \(R_{E 2}\) in equation.

$$ \begin{aligned} I_{E 2} &=\frac{1.91}{1.2 \mathrm{k} \Omega} \\ &=1.59 \mathrm{~mA} \end{aligned} $$

Apply Kirchhoff's voltage law at the collector terminal of \(\left(Q_{2}\right)\).

$$ \begin{aligned} &V_{C C}=I_{C 2} R_{C 2}+V_{C 2} \\ &V_{C 2}=V_{C c}-I_{c 2} R_{C 2} \end{aligned} $$

Replace \(20 \mathrm{~V}\) for \(V_{C C}, 1.59 \mathrm{~mA}\) for \(I_{C 2}\), and \(2.2 \mathrm{k} \Omega\) for \(R_{C 2}\) in equation.

$$ \begin{aligned} V_{C 2} &=V_{C c}-I_{C 2} R_{C 2} \\ V_{C 2} &=20 \mathrm{~V}-(1.59 \mathrm{~mA})(2.2 \mathrm{k} \Omega) \\ &=20 \mathrm{~V}-\left(1.59 \times 10^{-3} \mathrm{~A}\right)\left(2.2 \times 10^{3} \Omega\right) \\ &=16.502 \mathrm{~V} \end{aligned} $$

Therefore, the collector \(V_{C 2}\) voltage is \(16.502 \mathrm{~V}\).

(b)

Determine the collector current \(I_{C 1}\).

From Figure 1, the collector current is same as the emitter current.

$$ I_{C} \cong I_{E} $$

Determine the emitter current.

$$ \begin{gathered} V_{E}=I_{E} R_{E} \\ I_{E 1}=\frac{V_{E 1}}{R_{E 1}} \end{gathered} $$

Replace \(3.44 \mathrm{~V}\) for \(V_{E 2}, 1 \mathrm{k} \Omega\) for \(R_{E 2}\) in equation.

$$ \begin{aligned} I_{E 1} &=\frac{3.44}{1 \mathrm{k} \Omega} \\ &=3.44 \mathrm{~mA} \\ I_{C 1} &=3.44 \mathrm{~mA} \end{aligned} $$

Therefore, the emitter current \(I_{E 1}\) of transistor \(\left(Q_{1}\right)\) is \(3.44 \mathrm{~mA}\).

Therefore, the collector current \(I_{C 1}\) of transistor \(\left(Q_{1}\right)\) is \(3.44 \mathrm{~mA}\).

Determine the base current of transistor \(\left(Q_{1}\right)\).

$$ I_{B 1}=\frac{I_{C 1}}{\beta} $$

Consider \(\beta=160\).

$$ \begin{aligned} I_{B 1} &=\frac{3.44 \mathrm{~mA}}{160} \\ &=0.0215 \mathrm{~mA} \end{aligned} $$

Therefore, the base current \(I_{B 1}\) of transistor \(\left(Q_{1}\right)\) is \(0.0215 \mathrm{~mA}\).

Determine the collector current \(I_{C 2}\).

From Figure 1 , the collector current is same as the emitter current.

$$ I_{C 2} \cong I_{E 2} $$

Determine the emitter current.

$$ \begin{gathered} V_{E}=I_{E} R_{E} \\ I_{E 2}=\frac{V_{E 2}}{R_{E 2}} \end{gathered} $$

Replace \(1.91\) for \(V_{E 2}, 1.2 \mathrm{k} \Omega\) for \(R_{E 2}\) in equation.

$$ \begin{aligned} I_{E 2} &=\frac{1.91}{1.2 \mathrm{k} \Omega} \\ &=1.59 \mathrm{~mA} \\ I_{C 2} &=1.59 \mathrm{~mA} \end{aligned} $$

Therefore, the emitter current \(I_{E 2}\) of transistor \(\left(Q_{2}\right)\) is \(1.59 \mathrm{~mA}\).

Therefore, the collector current \(I_{C 2}\) of transistor \(\left(Q_{2}\right)\) is \(1.59 \mathrm{~mA}\).

Determine the base current of transistor \(\left(Q_{2}\right)\).

$$ I_{B 2}=\frac{I_{C 2}}{\beta} $$

Consider \(\beta=90\)

$$ \begin{aligned} I_{B 2} &=\frac{1.59 \mathrm{~mA}}{90} \\ &=0.0176 \mathrm{~mA} \end{aligned} $$