Chapter 3, Problem 89P

For the transistor circuit shown in Fig. 3.125, find I_B and V_CE. Let β = 100, and V_BE = 0.7 V.

Figure 3.125

To determine

Find $I_B$ and $V_{CE}$ for the transistor circuit in the Figure 3.125.

Answer to Problem 89P

The value of $I_B$ and $V_{CE}$ for the given circuit are $22.5\mu \text{A}$ and $12.75\text{V}$ respectively.

Explanation of Solution

Given data:

Refer Figure 3.125 in the textbook for the transistor circuit.

The common-emitter current gain $\beta$ is $100$.

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.

Calculation:

Modify the given figure with the representation of currents and voltages as shown in Figure 1.

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

$\begin{array}{l}-\left(2.25\text{V}\right)-\left(0.7\text{V}\right)+\left(100\text{kΩ}\right)I_B+V_{BE}=0\\ -\left(2.25\text{V}\right)-\left(0.7\text{V}\right)+\left(100\text{kΩ}\right)I_B+V_{BE}=0\left\{\because 1\text{kΩ}=1000\Omega \right\}\\ -\left(2.95\text{V}\right)+\left(100\times {10}^3\Omega \right)I_B+V_{BE}=0\\ \left(100\times {10}^3\Omega \right)I_B+V_{BE}=2.95\text{V}\end{array}$

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

$\begin{array}{l}\left(100\times {10}^3\Omega \right)I_B+0.7\text{V}=2.95\text{V}\\ \left(100\times {10}^3\Omega \right)I_B=\left(2.95\text{V}\right)-\left(0.7\text{V}\right)\\ \left(100\times {10}^3\Omega \right)I_B=2.25\text{V}\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{I}_B=\frac{2.25\text{V}}{100\times {10}^3\Omega }\\ =2.25\times {10}^{-5}\text{A}\\ I_B=22.5\mu \text{A}\left\{\because 1\mu \text{A}=1\times {10}^{-6}\text{A}\right\}\end{array}$

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

$\left(15\text{V}\right)-\left(1\text{kΩ}\right)I_C-V_{CE}=0\left\{\because 1\text{kΩ}=1000\Omega \right\}$

$\left(15\text{V}\right)-\left(1\times {10}^3\text{Ω}\right)I_C-V_{CE}=0$ (2)

Substitute equation (1) in (2).

$\left(15\text{V}\right)-\left(1\times {10}^3\text{Ω}\right)\left(\beta I_B\right)-V_{CE}=0$

Substitute $100$ for $\beta$ and $22.5\mu \text{A}$ for $I_B$.

$\begin{array}{l}\left(15\text{V}\right)-\left(1\times {10}^3\text{Ω}\right)\left(100\times 22.5\mu \text{A}\right)-V_{CE}=0\left\{\because 1\mu \text{A}=1\times {10}^{-6}\text{A}\right\}\\ \left(15\text{V}\right)-\left(1\times {10}^3\text{Ω}\right)\left(100\times 22.5\times {10}^{-6}\text{A}\right)-V_{CE}=0\\ \left(15\text{V}\right)-\left(2.25\text{V}\right)-V_{CE}=0\\ V_{CE}=12.75\text{V}\end{array}$

Conclusion:

Therefore, the value of $I_B$ and $V_{CE}$ for the given circuit are $22.5\mu \text{A}$ and $12.75\text{V}$ respectively.