Chapter 5, Problem 5.55P

a.

To determine

The R_TH and the V_TH for the given circuit.

Answer to Problem 5.55P

Explanation of Solution

Given:

The circuit is given as:

  

Assume $\beta =100$ .

Redrawing the given circuit by replacing voltage sources by the short circuit:

  

Hence, the equivalent Thevenin resistance is:

  $\begin{array}{l}{R}_{TH}=R_1\parallel R_2\\ R_{TH}=\frac{R_1\times R_2}{R_1+R_2}\mathrm{........}\left(1\right)\\ R_{TH}=\frac{12\times {10}^3\times 2\times {10}^3}{12\times {10}^3+2\times {10}^3}\\ R_{TH}=\frac{24\times {10}^3}{14}\\ \therefore R_{TH}=1.714K\Omega \end{array}$

The current through the resistors $R_1$ and $R_2$:

  $\begin{array}{l}I=\frac{5+5}{R_1+R_2}\mathrm{...........}\left(2\right)\\ I=\frac{5+5}{\left(12+2\right)\times {10}^3}\\ I=\frac{10\times {10}^{-3}}{14}\\ I=0.714\text{mA}\end{array}$

Hence, the Thevenin voltage is

  $\begin{array}{l}{V}_{TH}=I{R}_2-V_{EE}\mathrm{..........}\left(3\right)\\ V_{TH}=\left(0.714\times {10}^{-3}\right)\left(2\times {10}^3\right)-5\\ V_{TH}=1.428-5\\ \therefore V_{TH}=-3.572V\end{array}$

b.

To determine

The range of the I_CQ and V_CEQ.

Answer to Problem 5.55P

The range of the given values are:

  $\begin{array}{l}{I}_{CQ}=1.4\text{mA}\\ V_{CEQ}=2.3\text{V}\end{array}$

Explanation of Solution

Given:

The circuit is given as:

  

Assume, $\beta =100$ .

Redrawing the Thevenin equivalent circuit for the given circuit as:

  

Apply Kirchhoff s voltage law to the above circuit we get

  $\begin{array}{l}{V}_{TH}+I_{BQ}{R}_{TH}+V_{BE\left(on\right)}+I_{EQ}{R}_E-5=0\\ I_{BQ}{R}_{TH}+\left(1+\beta \right)I_{BQ}{R}_E=5-V_{TH}-V_{BE\left(on\right)}\\ I_{BQ}=\frac{5-V_{TH}-V_{BE(on)}}{R_{TH}+\left(1+\beta \right)R_E}\mathrm{............}\left(4\right)\end{array}$

Substitute the values in the above expression , we get

  $\begin{array}{l}{I}_{BQ}=\frac{5-3.572-0.7}{1.714\times {10}^3+\left(1+100\right)0.5\times {10}^3}\\ I_{BQ}=\frac{0.728}{52.214\times {10}^3}\\ I_{BQ}=13.94\text{μA}\end{array}$

The common emitter mode current gain is:

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\mathrm{..........}\left(5\right)\\ I_{CQ}=\left(100\right)\left(13.94\times {10}^{-6}\right)\\ I_{CQ}=1.4\text{mA}\end{array}$

The quiescent collector emitter voltage is

  $\begin{array}{l}{V}_{CEQ}=5+5-\left(R_C+R_E\right)I_{CQ}\mathrm{..........}\left(6\right)\\ V_{CEQ}=10-\left(5+0.5\right)\times {10}^3\times 1.4\times {10}^{-3}\\ V_{CEQ}=10-7.7\\ \therefore V_{CEQ}=2.3V\end{array}$

c.

To determine

To draw: The load line and the plot Q-point.

Explanation of Solution

Given:

The circuit is given as:

  

Assume, $\beta =100$ .

The equation for the load line is :

  $\begin{array}{l}{V}_{CE}=5+5-\left(R_C+R_E\right)I_C\mathrm{..........}\left(7\right)\\ V_{CE}=10-\left(5+0.5\right)\times {10}^3{I}_C\\ V_{CE}=10-5.5\times {10}^3{I}_C\end{array}$

Hence, the coordinates of the end points of the load line are :

Let $V_{CE}=0V$ In the above expression is

  $\begin{array}{l}0=10-5.5\times {10}^3{I}_C\\ I_C=\frac{10}{5.5\times {10}^3}\\ I_C=1.818\text{mA}\end{array}$

Let $I_C=0\text{A}$ In the Above expression is

  $\begin{array}{l}{V}_{CE}=10-5.5\times {10}^3\left(0\right)\\ V_{CE}=10-0\\ V_{CE}=10V\end{array}$

Hence, the co-ordinates of the two extremities of the load line is

  $\left(0V,1.818mA\right)\text{and}\left(10V,0mA\right)$ .

Therefore, the plot for load line is drawn is:

  

d.

To determine

The range of I_CQ and V_CEQ and the load line corresponding to the minimum and the maximum resistor value and the plot Q-points.

Answer to Problem 5.55P

The range of the data:

  $\begin{array}{l}1.33\text{mA}\le I_{CQ}\le 1.473\text{mA}\\ 1.915\text{V}\le V_{CEQ}\le 2.685\text{V}\end{array}$

Explanation of Solution

Given:

The circuit is given as:

  

Assume, $\beta =100$ .

The resistors $R_C$ and $R_E$ vary by $\pm 5$ percent .

The values of the resistor are given as:

  $\begin{array}{l}{R}_E=\left(0.5+5\%of0.5\right)\text{kΩ}\\ R_E=\left(0.5+0.025\right)\text{kΩ}\\ R_E=0.525K\Omega \\ R_E=\left(0.5-5\%of0.5\right)\text{kΩ}\\ R_E=\left(5-0.025\right)\text{kΩ}\\ R_E=0.475\text{kΩ}\end{array}$

Using the equation (7) when $R_C=5.25\text{kΩ},R_E=0.525\text{kΩ}and{V}_{CEQ}=2.3\text{V}$ :

  $\begin{array}{l}{V}_{CEQ}=10-\left(R_C+R_E\right)I_{CQ}\\ I_{CQ}=\frac{10-V_{CEQ}}{R_C+R_E}\\ I_{CQ}=\frac{10-2.3}{\left(4.75+0.475\right)\times {10}^3}\\ I_{CQ}=\frac{7.7}{5.225\times {10}^3}\\ I_{CQ}=1.473\text{mA}\end{array}$

Thus, the range of $I_{CQ}$ is

  $1.33\text{mA}\le I_{CQ}\le 1.473\text{mA}$

Load line equation for maximum resistor values using

  $\begin{array}{l}{V}_{CE}=10-\left(R_C+R_E\right)I_C\\ V_{CE}=10-\left(5.25+0.525\right)\times {10}^3{I}_C\\ V_{CE}=10-5.775\times {10}^3{I}_C\end{array}$

Put $V_{CE}=0V$ In the above expression:

  $\begin{array}{l}0=10-5.775\times {10}^3{I}_C\\ I_C=\frac{10}{5.775\times {10}^3}\\ I_C=1.731\text{mA}\end{array}$

Put $I_C=0\text{mA}$ in the above expression, we get

  $\begin{array}{l}{V}_{CE}=10-\left(5.25+0.525\right)\times {10}^3\left(0\right)\\ V_{CE}=10-0\\ V_{CE}=10\text{V}\end{array}$

Hence, the co-ordinates of the two extremities of the load line is

  $\left(0V,1.731\text{mA}\right)\text{and}\left(10\text{V},0\text{mA}\right)$ and the plot for land line for the ,maximum resistance and Q point is :

  

From equation (7) when $R_E=5.25\text{kΩ},R_E=0.525\text{kΩ}\text{and}{I}_{CQ}=1.4\text{mA}$

  $\begin{array}{l}{V}_{CEQ}=10-\left(R_C+R_E\right)I_{CQ}\\ V_{CEQ}=10-\left(5.25+0.525\right)\times {10}^3\times 1.4\times {10}^{-3}\\ V_{CEQ}=10-8.085\\ V_{CEQ}=1.915\text{V}\end{array}$

From equation (7) when $R_C=4.75\text{kΩ},R_E=0.475\text{kΩ}\text{and}{I}_{CQ}=1.4\text{mA}$ :

  $\begin{array}{l}{V}_{CEQ}=10-\left(R_C+R_E\right)I_{CQ}\\ V_{CEQ}=10-\left(4.75+0.475\right)\times {10}^3\times 1.4\times {10}^{-3}\\ V_{CEQ}=10-7.315\\ V_{CEQ}=2.685\text{V}\end{array}$

Hence , the range of $V_{CEQ}$ is

  $1.915\text{V}\le V_{CEQ}\le 2.685\text{V}$

Load line equation for maximum resistor values using

  $\begin{array}{l}{V}_{CE}=10-\left(R_C+R_E\right)I_C\\ V_{CE}=10-\left(4.75+0.475\right)\times {10}^3{I}_C\\ V_{CE}=10-5.225\times {10}^3{I}_C\end{array}$

Put $V_{CE}=0V$ in the above expression , we get

  $\begin{array}{l}0=10-5.225\times {10}^3{I}_C\\ I_C=\frac{10}{5.225\times {10}^3}\\ I_C=1.914\text{mA}\end{array}$

Put $I_C=0\text{mA}$ In the above expression :

  $\begin{array}{l}{V}_{CE}=10-\left(4.75+0.475\right)\times {10}^3\left(0\right)\\ V_{CE}=10-0\\ V_{CE}=10\text{V}\end{array}$

Thus the co-ordinates of the two extremities of the load line is :

  $\left(0\text{V},1.914\text{mA}\right)\text{and}\left(10\text{V},0\text{mA}\right)$ and the plot for load line for the minimum resistance and Q point is: