Problem

For the collector-feedback configuration of Fig. 4.129, determine:a. IB.b. IC.c. VC.FIG. 4...

For the collector-feedback configuration of Fig. 4.129, determine:

a. I_B.
b. I_C.
c. V_C.

FIG. 4.129

Step-by-Step Solution

Solution 1

Step #1 of 3

Refer Figure $4.129$ from the textbook.

(a)

Base current is determined as follows:

$$ I_{B}=\frac{V_{C C}-V_{B E}}{R_{B}+\beta\left(R_{C}+R_{E}\right)} $$

Substitute the corresponding values to obtain $I_{B}$.

$$ \begin{aligned} I_{B} &=\frac{16 \mathrm{~V}-0.7 \mathrm{~V}}{270 \mathrm{k} \Omega+(120)(3.6 \mathrm{k} \Omega+1.2 \mathrm{k} \Omega)} \\ &=18.09 \mu \mathrm{A} \end{aligned} $$

Hence, the base current is $18.09 \mu \mathrm{A}$.

Step #2 of 3

(b)

Collector current is determined as follows:

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

Substitute the corresponding values in the equation.

$$ \begin{aligned} I_{C} &=(120)(18.09) \\ &=2.17 \mathrm{~mA} \end{aligned} $$

Hence, the corresponding value of current $I_{C}$ is $2.17 \mathrm{~mA}$.

Step #3 of 3

(c)

Voltage $V_{C}$ is determined as follows:

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

Substitute the corresponding values in the equation.

$$ \begin{aligned} V_{C} &=16 \mathrm{~V}-(2.17 \mathrm{~mA})(3.6 \mathrm{k} \Omega) \\ &=8.19 \mathrm{~V} \end{aligned} $$

Hence, the corresponding value of voltage $V_{C}$ is $8.19 \mathrm{~V}$.

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