Chapter 3, Problem 92P

To determine

Design a problem to provide better understanding regarding transistors.

Explanation of Solution

Given data:

Refer Figure 3.128 in the textbook for the transistor circuit.

Formula used:

Write the expression for collector current in transistor.

$I_C=\beta I_B$        (1)

Here,

$\beta$ is the common-emitter current gain, and

$I_B$ is the base current.

Calculation:

Let us assume that the value of resistance $R_1$, $R_2$, $R_3$, and $R_4$ are $20\Omega$, $30\Omega$, $40\Omega$, and $10\Omega$ respectively, the value common-emitter current gain $\left(\beta \right)$ is $50$, and the value of voltages $V_{\text{CC}}$, $v_s$, and $V_{\text{BE}}$ are $40\text{V}$, $30\text{V}$, and $0.7\text{V}$ respectively.

The given circuit with the assumed value is shown in Figure 1.

Apply Kirchhoff’s voltage law to loop 1 with current $i_1$.

$-v_s+20i_1+V_{\text{BE}}+10\left(i_1-i_3\right)=0$

Substitute $0.7\text{V}$ for $V_{\text{BE}}$ and $30\text{V}$ for $v_s$.

$\begin{array}{l}-30+20i_1+0.7+10\left(i_1-i_3\right)=0\\ 20i_1+10i_1-10i_3=30-0.7\\ 30i_1-10i_3=29.3\end{array}$

$3i_1-i_3=2.93$        (2)

Apply Kirchhoff’s voltage law to loop 2 with current $i_2$.

$30i_2+V_{\text{CB}}=0$        (3)

Apply Kirchhoff’s voltage law to output loop in Figure 1.

$\begin{array}{l}10\left(i_3-i_1\right)-V_{\text{CE}}+40i_3+40=0\\ 10i_3-10i_1+40i_3-V_{\text{CE}}=-40\end{array}$

$-10i_1+50i_3-V_{\text{CE}}=-40$        (4)

Write the constraint equations.

$I_B=i_1-i_2$        (5)

$I_C=i_2-i_3$        (6)

$V_{\text{CE}}=V_{\text{BE}}+V_{\text{CB}}$        (7)

Substitute equations (5) and (6) in (1).

$\beta \left(i_1-i_2\right)=i_2-i_3$

Substitute $50$ for $\beta$.

$\begin{array}{l}50\left(i_1-i_2\right)=i_2-i_3\\ i_3=i_2-50i_1+50i_2\end{array}$

$i_3=-50i_1+51i_2$        (8)

Rearrange equation (7).

$V_{\text{CB}}=V_{\text{CE}}-V_{\text{BE}}$        (9)

Substitute equation (9) in (3).

$30i_2+V_{\text{CE}}-V_{\text{BE}}=0$

Substitute $0.7\text{V}$ for $V_{\text{BE}}$.

$\begin{array}{l}30i_2+V_{\text{CE}}-0.7=0\\ 30i_2=0.7-V_{\text{CE}}\\ i_2=\frac{0.7}{30}-\frac{V_{\text{CE}}}{30}\end{array}$

$i_2=0.023-0.033V_{\text{CE}}$        (10)

Substitute equation (8) in (2).

$\begin{array}{l}3i_1-\left(-50i_1+51i_2\right)=2.93\\ 3i_1+50i_1-51i_2=2.93\\ 53i_1-51i_2=2.93\\ 53i_1=2.93+51i_2\end{array}$

$i_1=0.055+0.962i_2$        (11)

Substitute equation (10) in (11).

$\begin{array}{c}{i}_1=0.055+0.962\left(0.023-0.033V_{\text{CE}}\right)\\ =0.055+0.022-0.0317V_{\text{CE}}\end{array}$

$i_1=0.077-0.0317V_{\text{CE}}$        (12)

Substitute equation (10) and (12) in (8).

$\begin{array}{c}{i}_3=-50\left(0.077-0.0317V_{\text{CE}}\right)+51\left(0.023-0.033V_{\text{CE}}\right)\\ =-3.85+1.585V_{\text{CE}}+1.173-1.683V_{\text{CE}}\end{array}$

$i_3=-2.677-0.098V_{\text{CE}}$        (13)

Substitute equation (12) and (13) in (4).

$\begin{array}{l}-10\left(0.077-0.0317V_{\text{CE}}\right)+50\left(-2.677-0.098V_{\text{CE}}\right)-V_{\text{CE}}=-40\\ -0.77+0.317V_{\text{CE}}-133.85-4.9V_{\text{CE}}-V_{\text{CE}}=-40\\ -5.583V_{\text{CE}}=-40+133.85+0.77\\ -5.583V_{\text{CE}}=94.62\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{V}_{\text{CE}}=-\frac{94.62}{5.583}\text{V}\\ =-16.95\text{V}\end{array}$

Conclusion:

Therefore, the problem has been designed to provide better understanding regarding transistors.