For the cascode amplifier of Fig. 4.143 determinea. the base and collector currents of eac...

Calculate emitter voltage \(\left(V_{E_{1}}\right)\) of transistor \(\left(Q_{1}\right)\).

$$ V_{E_{1}}=V_{B_{1}}-V_{B E_{1}} $$

Here,

Base emitter voltage \(\left(V_{B E_{1}}\right)\) of transistor \(\left(Q_{1}\right)\) is \(0.7 \mathrm{~V}\).

Substitute \(0.7 \mathrm{~V}\) for \(\left(V_{B E_{1}}\right)\) and \(4.48\) for \(\left(V_{B_{1}}\right)\) in the expression of emitter voltage \(\left(V_{E_{1}}\right)\).

$$ \begin{aligned} V_{E_{1}} &=4.48 \mathrm{~V}-0.7 \mathrm{~V} \\ &=3.78 \mathrm{~V} \end{aligned} $$

From the circuit collector current \(\left(I_{C_{1}}\right)\) is nearly same as emitter current \(\left(I_{E_{1}}\right)\).

$$ I_{E_{1}} \cong I_{E_{2}} \cong I_{C_{2}} \cong I_{C_{1}} $$

Calculate collector currents \(\left(I_{C_{1}}\right)\) and \(\left(I_{C_{2}}\right)\), emitter currents \(\left(I_{E_{1}}\right)\) and \(\left(I_{E_{2}}\right) .\)

$$ I_{E_{1}} \cong I_{E_{2}} \cong I_{c_{2}} \cong I_{C_{1}}=\frac{V_{E_{1}}}{R_{E_{1}}} $$

Here, emitter resistance \(\left(R_{E_{1}}\right)\) of transistor \(\left(Q_{1}\right)\) is \(1.1 \mathrm{k} \Omega\)

Substitute \(1.1 \mathrm{k} \Omega\) for \(\left(R_{E_{1}}\right)\) and \(3.78 \mathrm{~V}\) for \(\left(V_{E_{1}}\right)\) in the expression of collector current \(\left(I_{C_{\mathrm{i}}}\right)\)

$$ \begin{aligned} I_{E_{1}} \cong I_{E_{2}} \cong I_{C_{2}} \cong I_{C_{1}} &=\frac{3.78 \mathrm{~V}}{1.1 \mathrm{k} \Omega} \\ &=\frac{3.78 \mathrm{~V}}{1.1 \times 10^{3} \Omega} \\ &=3.44 \mathrm{~mA} \end{aligned} $$

Hence, collector current \(\left(I_{C_{1}}\right)\) is \(3.44 \mathrm{~mA}\) and collector current \(\left(I_{C_{2}}\right)\) is \(3.44 \mathrm{~mA}\)

Calculate base current \(\left(I_{B_{1}}\right)\) of transistor \(\left(Q_{1}\right)\).

$$ I_{B_{1}}=\frac{I_{C_{1}}}{\beta_{1}} $$

Here,

Forward common emitter gain \(\left(\beta_{1}\right)\) is 60 .

Substitute 60 for \(\left(\beta_{1}\right)\) and \(3.44 \mathrm{~mA}\) for \(\left(I_{C_{1}}\right)\) in the expression of base current \(\left(I_{B_{1}}\right)\).

$$ \begin{aligned} I_{B_{1}} &=\frac{3.44 \mathrm{~mA}}{60} \\ &=\frac{3.44 \times 10^{-3} \mathrm{~A}}{60} \\ &=57.33 \mu \mathrm{A} \end{aligned} $$

Hence, the base current \(\left(I_{B_{1}}\right)\) is \(57.33 \mu \mathrm{A}\).

Calculate base current \(\left(I_{B_{2}}\right)\) of transistor \(\left(Q_{2}\right)\).

$$ I_{B_{2}}=\frac{I_{C_{2}}}{\beta_{2}} $$

Here,

Forward common emitter gain \(\left(\beta_{2}\right)\) is 120 .

Substitute 120 for \(\left(\beta_{2}\right)\) and \(3.44 \mathrm{~mA}\) for \(\left(I_{C_{2}}\right)\) in the expression of base current \(\left(I_{B_{2}}\right)\).

$$ \begin{aligned} I_{B_{2}} &=\frac{3.44 \mathrm{~mA}}{120} \\ &=\frac{3.44 \times 10^{-3} \mathrm{~A}}{120} \\ &=28.67 \mu \mathrm{A} \end{aligned} $$

(b)

Calculate base voltage \(\left(V_{B_{i}}\right)\) of transistor \(\left(Q_{1}\right)\)

$$ V_{B_{1}}=\frac{R_{B_{3}} V_{C C}}{R_{B_{1}}+R_{B_{2}}+R_{B_{3}}} $$

Here,

Supply voltage \(\left(V_{C C}\right)\) is \(22 \mathrm{~V}\)

Base resistance \(\left(R_{B_{3}}\right)\) is \(3.3 \mathrm{k} \Omega\),

Base resistance \(\left(R_{B_{2}}\right)\) is \(4.7 \mathrm{k} \Omega\)

Base resistance \(\left(R_{B_{i}}\right)\) is \(8.2 \mathrm{k} \Omega\).

Substitute \(20 \mathrm{~V}\) for \(V_{C C}, 3.3 \mathrm{k} \Omega\) for \(\left(R_{B_{3}}\right), 4.7 \mathrm{k} \Omega\) for \(\left(R_{B_{2}}\right)\) and \(8.2 \mathrm{k} \Omega\) for \(\left(R_{B_{4}}\right)\) in the expression of base voltage \(\left(V_{B_{1}}\right)\)

$$ \begin{aligned} V_{B_{1}} &=\frac{(3.3 \mathrm{k} \Omega)(20 \mathrm{~V})}{(3.3 \mathrm{k} \Omega)+(4.7 \mathrm{k} \Omega)+(8.2 \mathrm{k} \Omega)} \\ &=\frac{\left(3.3 \times 10^{3} \Omega\right)(20 \mathrm{~V})}{\left(3.3 \times 10^{3} \Omega\right)+\left(4.7 \times 10^{3} \Omega\right)+\left(8.2 \times 10^{3} \Omega\right)} \\ &=4.48 \mathrm{~V} \end{aligned} $$

Calculate base voltage \(\left(V_{B_{2}}\right)\) of transistor \(\left(Q_{1}\right)\)

$$ V_{B_{2}}=\frac{\left(R_{B_{2}}+R_{B_{y}}\right) V_{C C}}{R_{B_{1}}+R_{B_{2}}+R_{B_{3}}} $$

Substitute \(20 \mathrm{~V}\) for \(V_{C C^{+}} 3.3 \mathrm{k} \Omega\) for \(\left(R_{B_{3}}\right), 4.7 \mathrm{k} \Omega\) for \(\left(R_{B_{2}}\right)\) and \(8.2 \mathrm{k} \Omega\) for \(\left(R_{B_{1}}\right)\) in the expression of base voltage \(\left(V_{B_{2}}\right) .\)

$$ \begin{aligned} V_{B_{2}} &=\frac{\{(4.7 \mathrm{k} \Omega)+(8.2 \mathrm{k} \Omega)\}(20 \mathrm{~V})}{\left(3.3 \mathrm{k} \Omega^{3}\right)+(4.7 \mathrm{k} \Omega)+(8.2 \mathrm{k} \Omega)} \\ &=\frac{\left\{\left(4.7 \times 10^{3} \Omega\right)+\left(8.2 \times 10^{3} \Omega\right)\right\}(20 \mathrm{~V})}{\left(3.3 \times 10^{3} \Omega\right)+\left(4.7 \times 10^{3} \Omega\right)+\left(8.2 \times 10^{3} \Omega\right)} \\ &=10.86 \mathrm{~V} \end{aligned} $$

Calculate emitter voltage \(\left(V_{E_{1}}\right)\) of transistor \(\left(Q_{1}\right)\).

$$ V_{E_{1}}=V_{B_{1}}-V_{B E_{1}} $$

Here,

Base emitter voltage \(\left(V_{B E_{1}}\right)\) of transistor \(\left(Q_{1}\right)\) is \(0.7 \mathrm{~V}\).

Substitute \(0.7 \mathrm{~V}\) for \(\left(V_{B E_{1}}\right)\) and \(4.48\) for \(\left(V_{B_{1}}\right)\) in the expression of emitter voltage \(\left(V_{E_{1}}\right)\).

$$ \begin{aligned} V_{E_{1}} &=4.48 \mathrm{~V}-0.7 \mathrm{~V} \\ &=3.78 \mathrm{~V} \end{aligned} $$

Calculate emitter voltage \(\left(V_{E_{2}}\right)\) of transistor \(\left(Q_{2}\right)\)

$$ V_{E_{2}}=V_{B_{2}}-V_{B E_{2}} $$

Substitute \(0.7 \mathrm{~V}\) for \(\left(V_{B E_{2}}\right)\) and \(10.86\) for \(\left(V_{B_{2}}\right)\) in the expression of emitter voltage \(\left(V_{E_{2}}\right)\)

$$ \begin{aligned} V_{E_{1}} &=10.86 \mathrm{~V}-0.7 \mathrm{~V} \\ &=10.16 \mathrm{~V} \end{aligned} $$

Calculate collector voltage \(\left(V_{c_{1}}\right)\) of transistor \(Q_{1}\)

$$ V_{C_{1}}=V_{B_{2}}-V_{B E_{2}} $$

Substitute \(0.7 \mathrm{~V}\) for \(\left(V_{B E_{2}}\right)\) and \(10.86 \mathrm{~V}\) for \(\left(V_{B_{2}}\right)\) in the expression of collector voltage \(\left(V_{C_{1}}\right)\)

$$ \begin{aligned} V_{E_{1}} &=10.86 \mathrm{~V}-0.7 \mathrm{~V} \\ &=10.16 \mathrm{~V} \end{aligned} $$

Calculate collector voltage \(\left(V_{C_{2}}\right)\) of transistor \(Q_{2}\)

$$ V_{C_{2}}=V_{C C}-I_{C_{2}} R_{C_{2}} $$

Here,

Collector resistance \(\left(R_{C_{2}}\right)\) of transistor \(Q_{2}\) is \(2.2 \mathrm{k} \Omega\)

Substitute \(22 \mathrm{~V}\) for \(\left(V_{C C}\right), 2.2 \mathrm{k} \Omega\) for \(\left(R_{C_{2}}\right)\) and \(3.44 \mathrm{~mA}\) for \(I_{C_{2}}\) in the expression of collector voltage \(\left(V_{C_{2}}\right)\)

$$ \begin{aligned} V_{C_{2}} &=22 \mathrm{~V}-(3.44 \mathrm{~mA})(2.2 \mathrm{k} \Omega) \\ &=22 \mathrm{~V}-\left(3.44 \times 10^{-3} \mathrm{~A}\right)\left(2.2 \times 10^{3} \Omega\right) \\ &=14.43 \mathrm{~V} \end{aligned} $$