For the fixed-bias configuration of Fig. 4.118 determine:a. b. c. d. VC.e. VB.f. VE.FIG. 4...

(b)

Write the expression of the collector current \(I_{C Q}\).

$$ I_{C Q}=\beta I_{B Q} $$

Substitute 120 for \(\beta\) and \(30 \mu \mathrm{A}\) for \(I_{B Q}\).

$$ \begin{aligned} I_{C Q} &=(120)\left(30 \times 10^{-6}\right) \\ &=3.6 \mathrm{~mA} \end{aligned} $$

Hence, the value of \(I_{C Q}\) is \(3.6 \mathrm{~mA}\).

(c)

Write the expression of the voltage \(V_{C E Q}\).

$$ V_{C E_{e}}=V_{C C}-I_{C Q} R_{C} $$

Substitute \(16 \mathrm{~V}\) for \(V_{C C}, 3.6 \mathrm{~mA}\) for \(I_{C Q}\) and \(1.8 \mathrm{k} \Omega\) for \(R_{C}\).

$$ \begin{aligned} V_{C E Q} &=16-\left(3.6 \times 10^{-3}\right)\left(1.8 \times 10^{3}\right) \\ &=16-6.48 \\ &=9.52 \mathrm{~V} \end{aligned} $$

Hence the value of \(V_{C E_{Q}}\) is \(9.52 \mathrm{~V}\).

(d)

Write the equation for the Voltage \(V_{C}\).

$$ V_{C}=V_{C E_{Q}} $$

Substitute \(9.52 \mathrm{~V}\) for \(V_{C E_{Q}}\).

$$ V_{C}=9.52 \mathrm{~V} $$

Hence the value of \(V_{C}\) is \(9.52 \mathrm{~V}\).

(e)

Write the equation for the Voltage \(V_{B}\).

$$ V_{B}=V_{B E} $$

Substitute \(0.7 \mathrm{~V}\) for \(V_{B E}\)

$$ V_{B}=0.7 \mathrm{~V} $$

Hence the value of \(V_{B}\) is \(0.7 \mathrm{~V}\).

(f) Since the emitter terminal of the transistor is connected to ground, the emitter voltage \(V_{E}\) is, \(V_{E}=0 \mathrm{~V}\)