Chapter 3, Problem 91P

To determine

Calculate the value of $I_B$, $V_{CE}$, and $v_o$ for the transistor circuit in the Figure 3.127.

Answer to Problem 91P

The value of $I_B$, $V_{CE}$, and $v_o$ for the transistor circuit are $0.8078\mu \text{A}$, $8.345\text{V}$ and $48.79\text{mV}$ respectively.

Explanation of Solution

Given data:

Refer Figure 3.127 in the textbook for the transistor circuit.

The common-emitter current gain $\beta$ is $150$.

The base-emitter voltage $V_{BE}$ is $0.7\text{V}$.

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.

Write the expression for emitter current of transistor.

$I_E=\left(1+\beta \right)I_B$ (2)

Calculation:

The Thevenin resistance of the input circuit is the parallel combination of $6\text{kΩ}$ and $2\text{kΩ}$ resistors.

$\begin{array}{c}{R}_{\text{Th}}=\left(6\text{kΩ}\right)\parallel \left(2\text{kΩ}\right)\\ =\frac{\left(6\text{kΩ}\right)\left(2\text{kΩ}\right)}{\left(6\text{kΩ}\right)+\left(2\text{kΩ}\right)}\\ =\frac{12{\left(\text{kΩ}\right)}^2}{8\text{kΩ}}\\ =1.5\text{kΩ}\end{array}$

The Thevenin voltage of the input circuit is the voltage across $2\text{kΩ}$ resistor.

$\begin{array}{c}{V}_{\text{Th}}=\left(3\text{V}\right)\left(\frac{2\text{kΩ}}{2\text{kΩ}+6\text{kΩ}}\right)\\ =\left(3\text{V}\right)\left(\frac{2\text{kΩ}}{8\text{kΩ}}\right)\\ =\left(\frac{3\times 2}{8}\right)\text{V}\\ =\text{0}\text{.75}\text{V}\end{array}$

Modify the Figure 3.127 with the incorporation of Thevenin equivalent circuit as shown in Figure 1.

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

$\left(-0.75\text{V}\right)+\left(1.5\text{kΩ}\right)I_B+V_{BE}+\left(400\Omega \right)I_E=0$ (3)

Substitute equation (2) in (3).

$\left(-0.75\text{V}\right)+\left(1.5\text{kΩ}\right)I_B+V_{BE}+\left(400\Omega \right)\left(1+\beta \right)I_B=0$

Substitute $150$ for $\beta$ and $0.7\text{V}$ for $V_{BE}$.

$\begin{array}{l}\left(-0.75\text{V}\right)+\left(1.5\text{kΩ}\right)I_B+\left(0.7\text{V}\right)+\left(400\Omega \right)\left(1+150\right)I_B=0\left\{\because 1\text{kΩ}=1\times {10}^3\Omega \right\}\\ \left(-0.75\text{V}\right)+\left(1.5\times {10}^3\text{Ω}\right)I_B+\left(0.7\text{V}\right)+\left(60,400\Omega \right)I_B=0\\ \left(61,900\Omega \right)I_B=\left(0.75\text{V}\right)-\left(0.7\text{V}\right)\\ \left(61,900\Omega \right)I_B=0.05\text{V}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}{I}_B=\frac{0.05\text{V}}{61,900\Omega }\\ =0.8078\times {10}^{-6}\text{A}\left\{\because 1\mu \text{A=1}\times {\text{10}}^{-6}\text{A}\right\}\\ =0.8078\mu \text{A}\end{array}$

From Figure 1, write the expression for voltage $v_o$.

$v_o=\left(400\Omega \right)I_E$ (4)

Substitute equation (2) in (4).

$v_o=\left(400\Omega \right)\left(1+\beta \right)I_B$ (5)

Substitute $150$ for $\beta$ and $0.8078\times {10}^{-6}\text{A}$ for $I_B$ in equation (5) to find $v_o$.

$\begin{array}{c}{v}_o=\left(400\Omega \right)\left(1+150\right)\left(0.8078\times {10}^{-6}\text{A}\right)\\ =\left(60,400\Omega \right)\left(0.8078\times {10}^{-6}\text{A}\right)\\ =0.04879\text{V}\left\{\because 1\text{mV=1}\times {\text{10}}^{-3}\text{V}\right\}\\ =48.79\text{mV}\end{array}$

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

$\left(9\text{V}\right)-v_o-V_{CE}-\left(5\text{k}\Omega \right)I_C=0$ (6)

Substitute equation (1) in (6).

$\begin{array}{l}\left(9\text{V}\right)-v_o-V_{CE}-\left(5\text{k}\Omega \right)\beta I_B=0\\ \left(9\text{V}\right)-v_o-V_{CE}-\left(5\times {10}^3\Omega \right)\beta I_B=0\end{array}$

Substitute $0.04879\text{V}$ for $v_o$, $150$ for $\beta$ and $0.8078\times {10}^{-6}\text{A}$ for $I_B$.

$\begin{array}{l}\left(9\text{V}\right)-\left(0.04879\text{V}\right)-V_{CE}-\left(5\times {10}^3\Omega \right)\left(150\right)\left(0.8078\times {10}^{-6}\text{A}\right)=0\\ \left(9\text{V}\right)-\left(0.04879\text{V}\right)-V_{CE}-\left(0.60585\text{V}\right)=0\\ V_{CE}=\left(9\text{V}\right)-\left(0.04879\text{V}\right)-\left(0.60585\text{V}\right)\\ V_{CE}=8.345\text{V}\end{array}$

Conclusion:

Therefore, the value of $I_B$, $V_{CE}$, and $v_o$ for the given transistor circuit are $0.8078\mu \text{A}$, $8.345\text{V}$ and $48.79\text{mV}$ respectively.