Chapter 5, Problem 5.52P

For the circuit shown in Figure P5.52, let $\beta =125$ . (a) Find $I_{CQ}$ and $V_{CEQ}$ . Sketch the load line and plot the Q-point. (b) If the resistors $R_1$ and $R_2$ vary by ±5 percent, determine the range in $I_{CQ}$ and $V_{CEQ}$ . Plot the various Q-points on the load line.

Figure P5.52

To determine

The I_CQ and V_CEQ and then sketching the load line and plotting the Q-point.

Answer to Problem 5.52P

The I_CQ and V_{CEQ :}

I_CQ =0.9128mA.

V_CEQ =14.87V.

Explanation of Solution

Given:

The value of the attenuation factor, $\beta =125$ .

The circuit diagram is shown below:

  

The Thevenin resistance is evaluated as:

  $\begin{array}{c}{R}_{th}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(58\times {10}^3\right)\times \left(42\times {10}^3\right)}{\left(58\times {10}^3\right)+\left(42\times {10}^3\right)}\\ =\frac{2436\times {10}^3}{100}\\ =24.36\text{k}\Omega \end{array}$

Applying the voltage division rule to evaluate the Thevenin voltage:

  $\begin{array}{c}{V}_{th}=\frac{V_{CC}{R}_2}{R_1+R_2}\mathrm{......}\left(2\right)\\ =\frac{24\left(42\times {10}^3\right)}{58\times {10}^3+\left(42\times {10}^3\right)}\\ =\frac{1008}{100}\\ =10.08\text{V}\end{array}$

Redrawing the given circuit:

  

Applying the Kirchhoff’s voltage law in the base-emitter loop:

  $\begin{array}{c}-V_{th}+I_{BQ}{R}_{th}+V_{BE}+I_{EQ}{R}_E=0\\ I_{BQ}{R}_{th}+\left(1+\beta \right)I_{BQ}{R}_E=V_{th}-V_{BE}\\ I_{BQ}=\frac{V_{th}-V_{BE}}{R_{th}+\left(1+\beta \right)R_E}\mathrm{.......}\left(3\right)\\ =\frac{10.08-0.7}{24.36\times {10}^3+\left(1+125\right)10\times {10}^3}\\ =\frac{9.38}{1284.36\times {10}^3}\\ =7.303\text{μA}\end{array}$

By the expression of the common emitter current gain:

  $\begin{array}{c}{I}_{CQ}=\beta I_{BQ}\mathrm{......}\left(4\right)\\ =125\left(7.30\times {10}^{-6}\right)\\ =0.9128\text{mA}\end{array}$

The quiescent collector emitter voltage is given as:

  $\begin{array}{c}{V}_{CEQ}=V_{CC}-I_{CQ}{R}_E\mathrm{.....}\left(5\right)\\ V_{CEQ}=24-\left(0.913\times {10}^{-3}\right)\left(10\times {10}^3\right)\\ =24-9.13\\ =14.87\text{V}\end{array}$

Hence, the load line equation is given as:

  $\begin{array}{l}{V}_E=V_{CC}-I_C{R}_E\mathrm{.......}\left(6\right)\\ V_{CE}=24-10\times {10}^3{I}_C\end{array}$

The coordinates of the end points of the load line.

Since, V_CE =0-V.

Hence,

  $\begin{array}{l}0=24-10\times {10}^3{I}_C\\ I_C=2.4\text{mA}\end{array}$

Now, substituting I_C =0A, then:

  $\begin{array}{c}{V}_{CE}=24-0\\ =24\text{V}\end{array}$

Hence, the coordinates of the two extremities of the load line is:

  $\left(0\text{V},2.4\text{mA}\right)\text{ and }\left(24\text{V, 0mA}\right)$

Sketching the load line graph:

  

To determine

The range of the I_CQ and V_CEQ and plotting the various Q-point on the load line.

Explanation of Solution

Given:

The value of the attenuation factor, $\beta =125$ .

The circuit diagram is shown below:

  

The resistor R₁ and R₂ vary by the $\pm 5\%$ .

Now, evaluating the range of the resistor:

  $\begin{array}{c}{R}_1=\left(58+5\%\text{of 58}\right)\text{k}\Omega \\ \text{=}\left(58+2.9\right)\text{k}\Omega \\ =60.9\text{k}\Omega \\ R_1=\left(58-5\%\text{of 58}\right)\text{k}\Omega \\ \text{=}\left(58-2.9\right)\text{k}\Omega \\ =55.1\text{k}\Omega \end{array}$

Now, evaluating the range of the R₂:

  $\begin{array}{c}{R}_2=\left(42+5\%\text{of 42}\right)\text{k}\Omega \\ \text{=}\left(42+2.1\right)\text{k}\Omega \\ =44.1\text{k}\Omega \\ R_1=\left(42-5\%\text{of 42}\right)\text{k}\Omega \\ \text{=}\left(42-2.1\right)\text{k}\Omega \\ =39.9\text{k}\Omega \end{array}$

Now taking $60.9\text{k}\Omega \text{and }44.1\text{k}\Omega$

as R₁ and R₂ respectively:

Evaluating the Thevenin resistance using the equation 1:

  $\begin{array}{c}{R}_{th}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(60.9\times {10}^3\right)\times \left(44.1\times {10}^3\right)}{\left(60.9\times {10}^3\right)+\left(44.1\times {10}^3\right)}\\ =\frac{2685.69\times {10}^3}{105}\\ =25.578\text{k}\Omega \end{array}$

Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:

  $\begin{array}{c}{V}_{th}=\frac{V_{CC}{R}_2}{R_1+R_2}\mathrm{......}\left(2\right)\\ =\frac{24\left(44.1\times {10}^3\right)}{60.9\times {10}^3+\left(44.1\times {10}^3\right)}\\ =\frac{1058.4}{105}\\ =10.08\text{V}\end{array}$

By the use of the equation 3:

  $\begin{array}{c}{I}_{BQ}=\frac{10.08-0.7}{25.58\times {10}^3+\left(1+125\right)\times 10\times {10}^3}\\ =\frac{9.38}{1285.58\times {10}^3}\\ =7.296\text{μA}\end{array}$

Now, from the equation 4:

  $\begin{array}{l}{I}_{CQ}=125\left(7.286\times {10}^{-6}\right)\\ I_{CQ}=0.912\text{mA}\\ \text{From equation 5:}\\ V_{CEQ}=24-\left(0.912\times {10}^{-3}\right)\left(10\times {10}^3\right)\\ V_{CEQ}=24-9.12\\ V_{CEQ}=14.88\text{V}\end{array}$

Now taking $60.9\text{k}\Omega \text{and 39}\text{.9k}\Omega$

as R₁ and R₂ respectively:

Evaluating the Thevenin resistance using the equation 1:

  $\begin{array}{c}{R}_{th}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(60.9\times {10}^3\right)\times \left(39.9\times {10}^3\right)}{\left(60.9\times {10}^3\right)+\left(39.9\times {10}^3\right)}\\ =\frac{2429.91\times {10}^3}{100.8}\\ =24.11\text{k}\Omega \end{array}$

Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:

  $\begin{array}{c}{V}_{th}=\frac{V_{CC}{R}_2}{R_1+R_2}\mathrm{......}\left(2\right)\\ =\frac{24\left(44.1\times {10}^3\right)}{60.9\times {10}^3+\left(44.1\times {10}^3\right)}\\ =\frac{957.6}{100.8}\\ =9.5\text{V}\end{array}$

By the use of the equation 3:

  $\begin{array}{c}{I}_{BQ}=\frac{9.5-0.7}{24.11\times {10}^3+\left(1+125\right)\times 10\times {10}^3}\\ =\frac{8.8}{1284.58\times {10}^3}\\ =6.85\text{μA}\end{array}$

Now, from the equation 4:

  $\begin{array}{l}{I}_{CQ}=125\left(6.85\times {10}^{-6}\right)\\ I_{CQ}=0.856\text{mA}\\ \text{From equation 5:}\\ V_{CEQ}=24-\left(0.856\times {10}^{-3}\right)\left(10\times {10}^3\right)\\ V_{CEQ}=24-8.56\\ V_{CEQ}=15.43\text{V}\end{array}$

Now taking $55.1\text{k}\Omega \text{and 39}\text{.9k}\Omega$

as R₁ and R₂ respectively:

Evaluating the Thevenin resistance using the equation 1:

  $\begin{array}{c}{R}_{th}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(55.1\times {10}^3\right)\times \left(39.9\times {10}^3\right)}{\left(55.1\times {10}^3\right)+\left(39.9\times {10}^3\right)}\\ =\frac{2198.49\times {10}^3}{95}\\ =23.14\text{k}\Omega \end{array}$

Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:

  $\begin{array}{c}{V}_{th}=\frac{V_{CC}{R}_2}{R_1+R_2}\mathrm{......}\left(2\right)\\ =\frac{24\left(39.9\times {10}^3\right)}{55.1\times {10}^3+\left(39.9\times {10}^3\right)}\\ =\frac{957.6}{95}\\ =10.08\text{V}\end{array}$

By the use of the equation 3:

  $\begin{array}{c}{I}_{BQ}=\frac{10.08-0.7}{23.14\times {10}^3+\left(1+125\right)\times 10\times {10}^3}\\ =\frac{9.38}{1283.14\times {10}^3}\\ =7.31\text{μA}\end{array}$

Now, from the equation 4:

  $\begin{array}{l}{I}_{CQ}=125\left(7.31\times {10}^{-6}\right)\\ I_{CQ}=0.914\text{mA}\\ \text{From equation 5:}\\ V_{CEQ}=24-\left(0.914\times {10}^{-3}\right)\left(10\times {10}^3\right)\\ V_{CEQ}=24-9.14\\ V_{CEQ}=14.86\text{V}\end{array}$

Hence, the plot for the various Q points on the load line is: