Chapter 5, Problem D5.62P

For the circuit shown in Figure P5.61, the bias voltages are changed to $V^{+}=\text{3V}$ and $V^{-}=-\text{3V}$ . (a) Design a bias−stable circuit for $\beta =120$ such that $V_{CEQ}=2.8\text{V}$ . Determine $I_{CQ}$ , $R_1$ , and $R_2$ . (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 P5.61

To determine

The design parameters of the circuit and the collector current at Q -point.

Answer to Problem D5.62P

  $I_{CQ}=1.453\text{ mA}$ , $R_1\text{=14}\text{.21 k}\Omega$ and $R_2=2.92\text{ k}\Omega$ .

Explanation of Solution

Given Information:

  $\beta =120,V_{CEQ}=2.8\text{ V, }{V}^{+}=3\text{ V and }{V}^{-}=-3\text{ V}$

The given circuit is shown below.

  

Calculation:

First, redraw the circuit with a Thevenin equivalent circuit in the base. Then find the Thevenin equivalent voltage and resistance. Calculate the transistor currents and voltages. Then find the required resistor values using the equations for Thevenin voltage and Thevenin resistance. The below figure shows the circuit with the Thevenin equivalent circuit at the base of the transistor.

  

Calculation:

Applying Kirchhoff’s law around C-E loop

  $\begin{array}{l}{V}^{+}=V_{CEQ}+I_{EQ}{R}_E+I_{CQ}{R}_C+V^{-}\\ V^{+}=V_{CEQ}+\left(\left(\frac{1+\beta }{\beta }\right)R_E+R_C\right)I_{CQ}+V^{-}\\ I_{CQ}=\frac{V^{+}-V_{CEQ}-V^{-}}{\left(\frac{1+\beta }{\beta }\right)R_E+R_C}=\frac{3-2.8-\left(-3\right)}{\left(\frac{1+120}{120}\right)0.2+2}\text{ mA}\end{array}$

  $I_{CQ}=1.453\text{ mA}$

Thevenin equivalent resistance is

  $R_{TH}=0.1\left(1+\beta \right)R_E=0.1\times 121\times 0.2\text{ k}\Omega \text{= 2}\text{.42 k}\Omega$

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

  $\begin{array}{l}{V}_{TH}=V_{BE(on)}+I_{EQ}{R}_E+I_{BQ}{R}_{TH}+V^{-}\\ V_{TH}=V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{CQ}{R}_{TH}+V^{-}\\ V_{TH}=V^{-}+V_{BE(on)}+\left(\frac{1+\beta }{\beta }\right)I_{CQ}{R}_E+\left(\frac{1}{\beta }\right)I_{cQ}{R}_{TH}\\ V_{TH}=-3+0.7+\left(\frac{1+120}{120}\right)\times 1.453\times 0.2+\left(\frac{1}{120}\right)\times 1.453\times 2.42\\ V_{TH}=-1.978\text{ V}\end{array}$

Thevenin resistance is,

  $\begin{array}{l}{R}_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\\ \left(\frac{R_1{R}_2}{R_1+R_2}\right)=2.42\to (1)\end{array}$

Thevenin voltage is,

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)+V^{-}$

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

  $\begin{array}{l}{V}_{TH}=\frac{1}{R_1}\left(\frac{R_1{R}_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)+V^{-}\\ -1.978=\frac{1}{R_1}\times 2.42\times 6-3\\ R_1=\frac{14.52}{1.022}\text{ k}\Omega \end{array}$

  $R_1\text{=14}\text{.21 k}\Omega$

Substituting in equation (1),

  $\left(\frac{14.21R_2}{14.21+R_2}\right)=2.42\text{ k}\Omega$

  $R_2=2.92\text{ k}\Omega$

To determine

The range of Q -point values, $I_{CQ}$ and $V_{CEQ}$ for the percent change of bias resistors.

To plot: Various Q -pints on the load line.

Answer to Problem D5.62P

The range of values is

  $\text{1}\text{.0906 mA}\le I_{CQ}\le \text{1}\text{.8452 mA}$ and $\text{1}\text{.9375 V}\le V_{CEQ}\le 3.599\text{ V}$

Th plot is shown below.

  

Explanation of Solution

Given Information:

  $\begin{array}{l}\beta =120,V^{+}=3\text{ V and }{V}^{-}=-3\text{ V}\\ \text{Change of bias resistence is }\pm 5\%\end{array}$

The given transistor circuit is shown below.

  

Calculaion:

First, redraw the circuit with a Thevenin equivalent circuit in the base. Then find the Thevenin equivalent voltage and resistance. Calculate the transistor currents and voltages at Q -point.

  

Thevenin resistance is,

  $R_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\to (1)$

Thevenin voltage is,

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)+V^{-}\to (2)$

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

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

Substituting, equation (1) and equation (2) in equation (3) we obtain,

  $\begin{array}{l}{I}_{BQ}=\frac{\left(\frac{R_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)-V_{BE(on)}}{\left(1+\beta \right)R_E+\left(\frac{R_1{R}_2}{R_1+R_2}\right)}\\ I_{BQ}=\frac{\left(6R_2\right)-0.7\left(R_1+R_2\right)}{\left(R_1+R_2\right)\left(1+120\right)0.2+R_1{R}_2}\end{array}$

  $\begin{array}{l}{I}_{BQ}=\frac{5.3R_2-0.7R_1}{24.2\left(R_1+R_2\right)+R_1{R}_2}\\ I_{CQ}=\beta I_{BQ}=120\left(\frac{5.3R_2-0.7R_1}{24.2\left(R_1+R_2\right)+R_1{R}_2}\right)\to (3)\end{array}$

Applying Kirchhoff’s law around C-E loop,

  $\begin{array}{l}{V}^{+}=V_{CEQ}+I_{EQ}{R}_E+I_{CQ}{R}_C+V^{-}\\ V_{CEQ}=V^{+}-V^{-}-\left[\left(\left(1+\beta \right)R_E+\beta R_C\right)\right]\left(\frac{I_{CQ}}{\beta }\right)\\ V_{CEQ}=6-\left(24.2+240\right)\left(\frac{I_{CQ}}{120}\right)\\ V_{CEQ}=6-\left(\frac{264.2}{120}\right)I_{CQ}\to (4)\end{array}$

Then substitute each possible combination of resistance values in the above equations

Case 1:

  $R_1\text{=14}\text{.21}+5\%=14.92\text{ k}\Omega \text{ and }{R}_2=2.92+5\%=3.066\text{ k}\Omega$

Using equation (3) and equation (4),

  $I_{CQ}=\text{1}\text{.4510 mA}$ and $V_{CEQ}=\text{2}\text{.805 V}$

Case 2:

  $R_1\text{=14}\text{.21}+5\%=14.92\text{ k}\Omega \text{ and }{R}_2=2.92-5\%=2.774\text{ k}\Omega$

Using equation (3) and equation (4),

  $I_{CQ}=\text{1}\text{.0906 mA}$ and $V_{CEQ}=3.599\text{ V}$

Case 3:

  $R_1\text{=14}\text{.21}-5\%=13.499\text{ k}\Omega \text{ and }{R}_2=2.92+5\%=3.066\text{ k}\Omega$

Using equation (3) and equation (4),

  $I_{CQ}=\text{1}\text{.8452 mA}$ and $V_{CEQ}=\text{1}\text{.9375 V}$

Case 4:

  $R_1\text{=14}\text{.21}-5\%=13.499\text{ k}\Omega \text{ and }{R}_2=2.92-5\%=2.774\text{ k}\Omega$

Using equation (3) and equation (4),

  $I_{CQ}=\text{1}\text{.4617 mA}$ and $V_{CEQ}=\text{ 2}\text{.782 V}$

Then,

  $\text{1}\text{.0906 mA}\le I_{CQ}\le \text{1}\text{.8452 mA}$ and $\text{1}\text{.9375 V}\le V_{CEQ}\le 3.599\text{ V}$

Using equation (4), the load line is,

  $V_{CE}=6-\left(\frac{264.2}{120}\right)I_C$

For $V_{CE}=0,I_C=\text{ 2}\text{.7252 mA and }{I}_C=0,V_{CE}=\text{6 V}$

Now plot the load line and mark the Q -point values,