Chapter 5, Problem 5.58P

(a) Determine the Q-point values for the circuit in Figure P5.58. Assume $\beta =50$ . (b) Repeat part (a) if all resistor values are reduced by a factor of 3. (c) Sketch the load lines and plot the Q-point values for parts (a) and (b).

Figure P5.58

To determine

The Q -point of the circuit.

Answer to Problem 5.58P

  $\left(I_{CQ},V_{ECQ}\right)=\left(0.0888\text{mA},3.55\text{V}\right)$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\beta =50$

Calculation:

Converting the biasing circuit into its Thevenin equivalent:

  

The value of $V_{TH},R_{TH}$ is determined as follows:

  $\begin{array}{l}{V}_{TH}=\frac{R_1{V}^{-}+R_2{V}^{+}}{R_1+R_2}\\ V_{TH}=\frac{36\times \left(-5\right)+68\times 5}{68+36}\\ V_{TH}=1.54\text{V}\\ R_{TH}=\frac{R_1{R}_2}{R_1+R_2}\\ R_{TH}=\frac{36\times 68}{36+68}\\ R_{TH}=23.5\text{kΩ}\end{array}$

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

  $\begin{array}{l}{V}^{+}=\left(1+\beta \right)I_{BQ}{R}_E+V_{EB}\left(\text{on}\right)+I_{BQ}{R}_{TH}+V_{TH}\\ 5=51\times 30\times I_{BQ}+0.7+I_{BQ}\left(23.5\right)+1.54\\ I_{BQ}=\frac{2.76}{1553.5}\\ I_{BQ}=1.78\text{μA}\end{array}$

The value of collector and emitter current is:

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ I_{CQ}=50\times 0.00178\\ I_{CQ}=0.0888\text{mA}\\ I_{EQ}=\left(1+\beta \right)I_{BQ}\\ I_{EQ}=51\times 0.00178\\ I_{EQ}=0.0906\text{mA}\end{array}$

Applying Kirchhoff’s voltage law in collector-emitter loop:

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E+V_{ECQ}+I_{CQ}{R}_C+V^{-}\\ 5=\left(0.0906\right)\left(30\right)+V_{ECQ}+\left(0.0888\right)\left(42\right)-5\\ V_{ECQ}=10-2.718-3.7296\\ V_{ECQ}=3.55\text{V}\end{array}$

Hence, the Q -point is:

  $\left(I_{CQ},V_{ECQ}\right)=\left(0.0888\text{mA},3.55\text{V}\right)$

To determine

The Q -point of the circuit if resistor values are reduced by a given factor.

Answer to Problem 5.58P

  $\left(I_{CQ},V_{ECQ}\right)=\left(0.266\text{mA},3.56\text{V}\right)$

Explanation of Solution

Given Information:

  

  $\beta =50$

Calculation:

The new amplifier circuit is:

  $\begin{array}{l}{R}_1=\frac{68}{3}\\ R_1=22.7\text{kΩ}\\ R_2=\frac{36}{3}\\ R_2=12\text{kΩ}\\ R_C=\frac{42}{3}\\ R_C=14\text{kΩ}\\ R_E=\frac{30}{3}\\ R_E=10\text{kΩ}\end{array}$

  

Converting the biasing circuit into its Thevenin equivalent:

  

The value of $V_{TH},R_{TH}$ is determined as follows:

  $\begin{array}{l}{V}_{TH}=\frac{R_1{V}^{-}+R_2{V}^{+}}{R_1+R_2}\\ V_{TH}=\frac{12\times \left(-5\right)+22.7\times 5}{22.7+12}\\ V_{TH}=1.54\text{V}\\ R_{TH}=R_1\left|\right|R_2\\ R_{TH}=\frac{12\times 22.7}{12+22.7}\\ R_{TH}=7.85\text{kΩ}\end{array}$

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

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

  $\begin{array}{l}{V}^{+}=\left(1+\beta \right)I_{BQ}{R}_E+V_{EB}\left(\text{on}\right)+I_{BQ}{R}_{TH}+V_{TH}\\ 5=51\times 10\times I_{BQ}+0.7+I_{BQ}\left(7.85\right)+1.54\\ I_{BQ}=\frac{2.76}{517.85}\\ I_{BQ}=5.33\text{μA}\end{array}$

The value of collector and emitter current is:

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ I_{CQ}=50\times 0.00533\\ I_{CQ}=0.266\text{mA}\\ I_{EQ}=\left(1+\beta \right)I_{BQ}\\ I_{EQ}=51\times 0.00533\\ I_{EQ}=0.272\text{mA}\end{array}$

Applying Kirchhoff’s voltage law in collector-emitter loop:

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E+V_{ECQ}+I_{CQ}{R}_C+V^{-}\\ 5=I_{EQ}{R}_E+V_{ECQ}+I_{CQ}{R}_C-5\\ 10=I_{CQ}{R}_C+V_{ECQ}+I_{EQ}{R}_E\\ 10=\left(0.266\right)\left(14\right)+V_{CEQ}+\left(0.272\right)\left(10\right)\\ V_{ECQ}=3.56\text{V}\end{array}$

Hence, the Q-point is:

  $\left(I_{CQ},V_{ECQ}\right)=\left(0.266\text{mA},3.56\text{V}\right)$

To determine

To plot: A DC load line and labelQ -point.

Answer to Problem 5.58P

A dc load line for part (a) along with Q -point is shown in Figure 1.

A dc load line for part (b) along with Q -point is shown in Figure 2.

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\beta =50$

Calculation:

a).

Converting the biasing circuit into its Thevenin equivalent:

  

Applying Kirchhoff’s voltage law in collector-emitter loop:

The load line equation is:

  $\begin{array}{l}{V}^{+}=I_E{R}_E+V_{EC}+I_C{R}_C+V^{-}\\ 5=I_C{R}_C+V_{EC}+I_E{R}_E-5\left(I_C\approx I_E\right)\\ V_{EC}=10-\left(R_C+R_E\right)I_C\end{array}$

It is equation of straight line with negative slope.

  $\begin{array}{l}{I}_C=-\left(\frac{1}{R_C+R_E}\right)V_{EC}+\frac{10}{R_C+R_E}\\ y=-mx+C\end{array}$

For $I_C=0$ ,

  $V_{EC}=10\text{V}$

For $V_{EC}=0$ ,

  $\begin{array}{l}{I}_C=\frac{10}{R_C+R_E}\\ I_C=\frac{10}{42+30}\\ I_C=138.8\text{μA}\end{array}$

The Q-point value is:

  $\left(I_{CQ},V_{ECQ}\right)=\left(88.8\text{μA},3.6\text{V}\right)$

The figure of DC load line is:

  

Figure 1

1unit along the voltage axis = 1 V

1unit along the current axis = 20 µA

(b)

Converting the biasing circuit into its Thevenin equivalent:

  

Applying Kirchhoff’s voltage law in collector-emitter loop:

The load line equation is:

  $\begin{array}{l}{V}^{+}=I_E{R}_E+V_{EC}+I_C{R}_C+V^{-}\\ 5=I_C{R}_C+V_{EC}+I_E{R}_E-5\left(I_C\approx I_E\right)\\ V_{EC}=10-\left(R_C+R_E\right)I_C\end{array}$

It is equation of straight line with negative slope.

  $\begin{array}{l}{I}_C=-\left(\frac{1}{R_C+R_E}\right)V_{EC}+\frac{10}{R_C+R_E}\\ y=-mx+C\end{array}$

For $I_C=0$ ,

  $V_{EC}=10\text{V}$

For $V_{EC}=0$ ,

  $\begin{array}{l}{I}_C=\frac{10}{R_C+R_E}\\ I_C=\frac{10}{14+10}\\ I_C=416.7\text{μA}\end{array}$

The Q-point value is:

  $\left(I_{CQ},V_{ECQ}\right)=\left(266\text{μA},3.6\text{V}\right)$

The figure of DC load line is:

  

1unit along the voltage axis = 1 V

1unit along the current axis = 50 µA