Problem

Using the characteristics of Fig. 4.121, determine the following for an emitter-bias confi...

Using the characteristics of Fig. 4.121, determine the following for an emitter-bias configuration if a Q-point is defined at and

a. R_C if V_CC = 24 V and R_E = 1.2 kΩ.
b. β at the operating point.
c. R_B.
d. Power dissipated by the transistor.
e. Power dissipated by the resistor R_C.

FIG. 4.121

Step-by-Step Solution

Solution 1

Step #1 of 7

(a)

Refer figure $4.121$ from the text book.

The value of saturation voltage for amplifier is,

$$ \begin{aligned} V_{C E_{\mathrm{stt}}} &=V_{C C} \\ &=24 \mathrm{~V} \end{aligned} $$

In the transfer characteristic graph, the line obtained from $V_{C E_{m}}$ point, and $\mathrm{Q}$ -point will intersect the $I_{C}$ axis at $I_{C_{\mathrm{m}}}$.

Step #2 of 7

Redraw the transfer characteristic for values.

$G:\Users\Abhinav Rai\diagrams\10774-4-13P_A.jpg$

Figure 1: transfer characteristic for amplifier

Step #3 of 7

From figure 1, the value of saturation current $I_{C_{\mathrm{san}}}$ is $6.8 \mathrm{~mA}$.

From figure 1 , the base current value for corresponding value of collector current at $4 \mathrm{~mA}$ is

$30 \mu \mathrm{A}$

Calculate the collector resistance is calculated as follows:

$I_{C_{\mathrm{at}}}=\frac{V_{C C}}{R_{C}+R_{E}}$

Substitute $6.8 \mathrm{~mA}$ for $I_{C_{\mathrm{m}}}, 24 \mathrm{~V}$ for $V_{C C}$ and $1.2 \mathrm{k} \Omega$ for $R_{E}$ in the equation.

$6.8 \mathrm{~mA}=\frac{24 \mathrm{~V}}{R_{C}+1.2 \mathrm{k} \Omega}$

$R_{C}+1.2 \mathrm{k} \Omega=\frac{24 \mathrm{~V}}{6.8 \mathrm{~mA}}$

$R_{C}=3.529 \mathrm{k} \Omega-1.2 \mathrm{k} \Omega$

$R_{C}=2.33 \mathrm{k} \Omega$

Therefore, the collector resistance, $R_{C}$ is $2.33 \mathrm{k} \Omega$.

Step #4 of 7

(b)

From figure 1 , the base current value for corresponding value of collector current at $4 \mathrm{~mA}$ is

$$ 30 \mu \mathrm{A} $$

Calculate the Gain $\beta$.

$$ \begin{aligned} \beta &=\frac{I_{C}}{I_{B}} \\ &=\frac{4 \mathrm{~mA}}{30 \mu \mathrm{A}} \\ &=133.33 \end{aligned} $$

Therefore, the value of gain is $133.33$.

Step #5 of 7

(c)

Calculate the base resistance.

$\begin{array}{rl}R_{B} & =\frac{V_{R_{n}}}{I_{B}} \\ & =\frac{V_{C C}-V_{B E}-V_{E}}{I_{B}} \\ & =\frac{24 \mathrm{~V}-0.7 \mathrm{~V}-(4 \mathrm{~mA})(1.2 \mathrm{k} \Omega)}{30 \mu \mathrm{A}} \\ & =\frac{18.5 \mathrm{~V}}{30 \mu \mathrm{A}} \\ =6 & 6.67 \mathrm{k} \Omega\end{array}$

Therefore, the base resistance, $R_{B}$ is $616.67 \mathrm{k} \Omega$.

Step #6 of 7

(d)

Calculate the dissipated power $P_{D}$.

$$ \begin{aligned} P_{D} &=V_{C E_{Q}} I_{C_{Q}} \\ &=(10 \mathrm{~V})(4 \mathrm{~mA}) \\ &=40 \mathrm{~mW} \end{aligned} $$

Therefore, the dissipated power, $P_{D} 40 \mathrm{~mW}$.

Step #7 of 7

(e)

Calculate the Power $P$.

$$ P=I_{C}^{2} R_{C} $$

Substitute $4 \mathrm{~mA}$ for $I_{C}$ and $2.33 \mathrm{k} \Omega$ for $R_{C}$ in the equation.

$$ \begin{aligned} P &=(4 \mathrm{~mA})^{2}(2.33 \mathrm{k} \Omega) \\ &=37.28 \mathrm{~mW} \end{aligned} $$

Therefore, the value of Power $P$ is $37.28 \mathrm{~mW}$.

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