Chapter 5, Problem 5.57P

(a) Determine the Q-point values for the circuit in Figure P5.57. 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 P5.57

To determine

The Q -point of the circuit.

Answer to Problem 5.57P

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

Explanation of Solution

Given Information:

  

  $\beta =50$

Calculation:

Converting the biasing circuit into its Thevenin equivalent:

  

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

  $\begin{array}{l}{V}_{TH}=\frac{R_2}{R_1+R_2}\times V_{CC}\\ V_{TH}=\frac{36\times 10}{68+36}\\ V_{TH}=3.46\text{V}\\ R_{TH}=R_1\left|\right|R_2\\ R_{TH}=\frac{36\times 68}{36+68}\\ R_{TH}=23.5\text{kΩ}\end{array}$

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

  $\begin{array}{l}{V}_{TH}=R_{TH}{I}_{BQ}+V_{BE}\left(\text{on}\right)+\left(1+\beta \right)I_{BQ}{R}_E\\ 3.46=23.5I_{BQ}+0.7+\left(51\times 30\right)I_{BQ}\\ I_{BQ}=\frac{3.46-0.7}{23.5+1530}\\ I_{BQ}=0.00178\text{mA}\end{array}$

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:

  $\begin{array}{l}{V}_{CC}=I_{CQ}{R}_C+V_{CEQ}+I_{EQ}{R}_E\\ 10=\left(0.0888\right)\left(42\right)+V_{CEQ}+\left(0.0906\right)\left(30\right)\\ V_{CEQ}=3.55\text{V}\end{array}$

Hence, the Q-point is:

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

To determine

The Q -point of the circuit.

Answer to Problem 5.57P

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

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:

  $\begin{array}{l}{V}_{TH}=\frac{R_2}{R_1+R_2}\times V_{CC}\\ V_{TH}=\frac{12\times 10}{22.7+12}\\ V_{TH}=3.46\text{V}\\ R_{TH}=R_1\left|\right|R_2\\ R_{TH}=\frac{12\times 22.7}{12+22.7}\\ R_{TH}=7.85\text{kΩ}\end{array}$

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

  $\begin{array}{l}{V}_{TH}=R_{TH}{I}_{BQ}+V_{BE}\left(\text{on}\right)+\left(1+\beta \right)I_{BQ}{R}_E\\ 3.46=7.85I_{BQ}+0.7+\left(51\times 10\right)I_{BQ}\\ I_{BQ}=\frac{3.46-0.7}{7.85+510}\\ I_{BQ}=0.00533\text{mA}\end{array}$

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:

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

Hence, the Q-point is:

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

To determine

To sketch: A DC load line and label Q -point.

Explanation of Solution

Given Information:

  

  $\beta =50$

Calculation:

a).

Converting the biasing circuit into its Thevenin equivalent:

  

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

The load line equation is:

  $\begin{array}{l}{V}_{CC}=I_C{R}_C+V_{CE}+I_E{R}_E\\ V_{CE}=V_{CC}-\left(R_C+R_E\right)I_C\left(I_C\approx I_E\right)\end{array}$

It is equation of straight line with negative slope.

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

For $I_C=0$ ,

  $\begin{array}{l}{V}_{CE}=V_{CC}\\ V_{CE}=10\text{V}\end{array}$

For $V_{CE}=0$ ,

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

The Q-point value is:

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

The figure of DC load line is:

  

1unit along the voltage axis = 1 V

1unit along the current axis = 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:

  $\begin{array}{l}{V}_{CC}=I_C{R}_C+V_{CE}+I_E{R}_E\\ V_{CE}=V_{CC}-\left(R_C+R_E\right)I_C\left(I_C\approx I_E\right)\end{array}$

It is equation of straight line with negative slope.

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

For $I_C=0$ ,

  $\begin{array}{l}{V}_{CE}=V_{CC}\\ V_{CE}=10\text{V}\end{array}$

For $V_{CE}=0$ ,

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

The Q -point value is:

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

The figure of DC load line is:

  

1unit along the voltage axis = 1 V

1unit along the current axis = 50 µA