Chapter 5, Problem D5.59P

(a) For the circuit shown in Figure P5.59, design a bias−stable circuit such that $I_{CQ}=0.8\text{mA}$ and $V_{CEQ}=5\text{V}$ . Le $\beta =100$ . (b) Using the results of part (a), determine the percentage change in $I_{CQ}$ if $\beta$ is in the range $75\le \beta \le 150$ . (c) Repeat parts (a) and (b) if $R_E=1\text{kΩ}$ .

Figure P5.59

To determine

The design parameters of a bias stable circuit.

Answer to Problem D5.59P

  $R_C=5.75\text{ k}\Omega$ , $R_1\text{=44}\text{.14 k}\Omega$ and $R_2=5.7\text{ k}\Omega$ .

Explanation of Solution

Given Information:

  $\beta =100,V_{CEQ}=5\text{ V, }{I}_{CQ}=0.8\text{ mA}$

The given circuit is shown below.

  

Calculation:

Find the equivalent Thevenin voltage and equivalent Thevenin resistance of the bias circuit. Then, find the required resistor values using the equations for Thevenin voltage and Thevenin resistance.

Using KVL, write the equation,

  $\begin{array}{l}{V}_{CC}=V_{CEQ}+I_{EQ}{R}_E+I_{CQ}{R}_C\\ V_{CC}=V_{CEQ}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+R_C{I}_{CQ}\end{array}$

  $\begin{array}{l}{R}_C=\frac{V_{CC}-V_{CEQ}-\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E}{I_{CQ}}\\ R_C=\frac{10-5-\left(\frac{101}{100}\right)0.8\times 0.5}{0.8}\text{ k}\Omega \end{array}$

Then the collector resistance is

  $R_C=5.75\text{ k}\Omega$

Thevenin equivalent resistance is

  $R_{TH}=0.1\left(1+\beta \right)R_E=0.1\times 101\times 0.5\text{ k}\Omega =5.\text{05 k}\Omega$

Applying Kirchhoff’s voltage law around the B-E loop,

  $\begin{array}{l}{V}_{TH}=V_{BE(on)}+I_{EQ}{R}_E+I_{BQ}{R}_{TH}\\ V_{TH}=V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{CQ}{R}_{TH}\\ V_{TH}=V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{CQ}{R}_{TH}\\ V_{TH}=0.7+\left(\frac{1+100}{100}\right)\times 0.8\times 0.5+\left(\frac{1}{100}\right)\times 0.8\times 5.05\text{ V}\\ V_{TH}=1.144\text{ V}\end{array}$

Thevenin resistance is,

  $\begin{array}{l}{R}_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\\ \left(\frac{R_1{R}_2}{R_1+R_2}\right)=5.05\to (1)\end{array}$

Thevenin voltage is,

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{CC}$

Using equation (1), rewrite the above equation as,

  $\begin{array}{l}{V}_{TH}=\frac{1}{R_1}\left(\frac{R_1{R}_2}{R_1+R_2}\right)V_{CC}\\ 1.144=\frac{1}{R_1}\times 5.05\times 10\\ R_1=\frac{50.5}{1.144}\text{ k}\Omega \end{array}$

  $R_1\text{=44}\text{.14 k}\Omega$

Substituting in equation (1),

  $\begin{array}{l}\left(\frac{44.14R_2}{44.14+R_2}\right)=5.05\\ R_2=\frac{44.14\times 5.05}{44.14-5.05}\text{ k}\Omega \end{array}$

  $R_2=5.7\text{ k}\Omega$

To determine

The percent change in Q -point collector current values for a known range of current gain.

Answer to Problem D5.59P

  $\Delta I_{CQ}=6.6625\%$

Explanation of Solution

Given Information:

  $75\le \beta \le 150,V_{CEQ}=5\text{ V, }{I}_{CQ}=0.8\text{ mA}$

The given circuit is shown below.

  

Calculation:

Then find the equivalent Thevenin voltage and equivalent Thevenin resistance. Calculate the collector current. The values from part (a) are $R_C=5.75\text{ k}\Omega$ , $R_1\text{=44}\text{.14 k}\Omega$ and $R_2=5.7\text{ k}\Omega$ .

Thevenin voltage is,

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{CC}=\left(\frac{5.7}{44.14+5.7}\right)10\Omega$

this yields,

  $V_{TH}=1.144\text{ V}$

Thevenin resistance is,

  $R_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)=\left(\frac{44.14\times 5.7}{44.14+5.7}\right)=5.05\Omega$

Applying Kirchhoff’s voltage law around the B-E loop,

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

  $\begin{array}{l}{I}_{CQ}=\beta \left(\frac{V_{TH}-V_{BE(on)}}{\left(1+\beta \right)R_E+R_{TH}}\right)\\ I_{CQ}=\beta \left(\frac{0.444}{\left(1+\beta \right)0.5+5.05}\right)\end{array}$

For $\beta =75,$

  $I_{CQ}=75\left(\frac{0.444}{\left(1+75\right)0.5+5.05}\right)$

  $I_{CQ}=0.7735\text{ mA}$

For $\beta =150,$

  $I_{CQ}=150\left(\frac{0.444}{\left(1+150\right)0.5+5.05}\right)$

  $I_{CQ}=0.8268\text{ mA}$

So, the percentage change in $I_{CQ}$ is

  $\Delta I_{CQ}=\frac{0.8268-0.7735}{0.8}\times 100\%$

  $\Delta I_{CQ}=6.6625\%$

To determine

The design parameters of the circuit and the percent change in Q -point collector current value for a known range of current gain.

Answer to Problem D5.59P

  $R_C=5.24\text{ k}\Omega$ , $R_1\text{=63}\text{.57 k}\Omega$ and $R_2=12.01\text{ k}\Omega$ .

  $\Delta I_{CQ}=6.675\%$

Explanation of Solution

Given Information:

  $\beta =100,V_{CEQ}=5\text{ V, }{I}_{CQ}=0.8\text{ mA, }{R}_E=1\text{ k}\Omega$

The given circuit is shown below.

  

Calculation:

Find the equivalent Thevenin voltage and equivalent Thevenin resistance. Then, find the required resistor values using the equations for Thevenin voltage and Thevenin resistance.

Using KVL, write the equation,

  $\begin{array}{l}{V}_{CC}=V_{CEQ}+I_{EQ}{R}_E+I_{CQ}{R}_C\\ V_{CC}=V_{CEQ}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+R_C{I}_{CQ}\\ R_C=\frac{V_{CC}-V_{CEQ}-\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E}{I_{CQ}}\\ R_C=\frac{10-5-\left(\frac{101}{100}\right)0.8\times 1}{0.8}\text{ k}\Omega \end{array}$

Then the collector resistance is

  $R_C=5.24\text{ k}\Omega$

Thevenin equivalent resistance is

  $R_{TH}=0.1\left(1+\beta \right)R_E=0.1\times 101\times 1\text{ k}\Omega \text{= 10}\text{.1 k}\Omega$

Applying Kirchhoff’s voltage law around the B-E loop,

  $\begin{array}{l}{V}_{TH}=V_{BE(on)}+I_{EQ}{R}_E+I_{BQ}{R}_{TH}\\ V_{TH}=V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{CQ}{R}_{TH}\\ V_{TH}=V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{CQ}{R}_{TH}\\ V_{TH}=0.7+\left(\frac{1+100}{100}\right)\times 0.8\times 1+\left(\frac{1}{100}\right)\times 0.8\times 10.1\text{ V}\\ V_{TH}=1.5888\text{ V}\end{array}$

Thevenin resistance is,

  $\begin{array}{l}{R}_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\\ \left(\frac{R_1{R}_2}{R_1+R_2}\right)=10.1\to (1)\end{array}$

Thevenin voltage is,

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{CC}$

Using equation (1), rewrite the above equation as,

  $\begin{array}{l}{V}_{TH}=\frac{1}{R_1}\left(\frac{R_1{R}_2}{R_1+R_2}\right)V_{CC}\\ 1.5888=\frac{1}{R_1}\times 10.1\times 10\\ R_1=\frac{101}{1.5888}\text{ k}\Omega \end{array}$

  $R_1\text{=63}\text{.57 k}\Omega$

Substituting in equation (1),

  $\begin{array}{l}\left(\frac{63.57R_2}{63.57+R_2}\right)=10.1\\ R_2=\frac{63.57\times 10.1}{63.57-10.1}\text{ k}\Omega \end{array}$

  $R_2=12.01\text{ k}\Omega$

Applying Kirchhoff’s voltage law around the B-E loop,

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

  $\begin{array}{l}{I}_{CQ}=\beta \left(\frac{V_{TH}-V_{BE(on)}}{\left(1+\beta \right)R_E+R_{TH}}\right)\\ I_{CQ}=\beta \left(\frac{0.8888}{\left(1+\beta \right)+10.1}\right)\end{array}$

For $\beta =75,$

  $I_{CQ}=75\left(\frac{0.8888}{\left(1+75\right)+10.1}\right)$

  $I_{CQ}=0.7742\text{ mA}$

For $\beta =150,$

  $I_{CQ}=150\left(\frac{0.8888}{\left(1+150\right)+10.1}\right)$

  $I_{CQ}=0.8276\text{ mA}$

So, the percentage change in $I_{CQ}$ is

  $\Delta I_{CQ}=\frac{0.8276-0.7742}{0.8}\times 100\%$

  $\Delta I_{CQ}=6.675\%$