Chapter 5, Problem D5.89DP

The emitter−follower circuit shown in Figure P5.89 is biased at $V^{+}=2.5\text{V}$ and $V^{-}=-2.5\text{V}$ . Design a bias−stable circuit such that the nominal Q−point values are $I_{CQ}\cong \text{5mA}$ and $V_{CEQ}\cong 2.5\text{V}$ . The transistor current gain values are in the range $100\le \beta \le 160$ . Select standard 5 percent tolerance resistance values in the final design. What is the range in Q−point values?

Figure P5.89

To determine

The design parameters of the circuit.

To select: Five percent tolerance values of resistances.

To find: The range of the Q- point values.

Answer to Problem D5.89DP

The design parameters of the circuit are:

  $R_1=8.65\text{k}\Omega$ , $R_2=19.61\text{k}\Omega$ it is closest to $19.61\text{k}\Omega$ and $R_E=0.5\text{k}\Omega$ .

Five percent tolerance resistance $R_1=8.2\text{k}\Omega$ and is closest to $8.65\text{k}\Omega$ , $R_2=20\text{k}\Omega$ it is closest to $19.61\text{k}\Omega$ and $R_E=0.51\text{k}\Omega$ it is closest to $0.5\text{k}\Omega$ .

The range of the current is $4.963\text{mA}\le I_{CQ}\le 3.236\text{mA}$ , range of the voltage is $\text{2}\text{.47}\text{V}\le V_{CEQ}\le 3.35\text{V}$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Figure 1

Calculation:

Mark the current as well as other parameters and then redraw the circuit.

The required diagram is shown in Figure 2

  

Figure 2

The Thevenin equivalent bias circuit is shown in Figure 3

  

Figure 3

The value of the base current is calculated as,

  $I_{BQ}=\frac{I_{CQ}}{\beta }$

Substitute $5\text{mA}$ for $I_{CQ}$ and $120$ for $\beta$ in the above equation.

  $\begin{array}{c}{I}_{BQ}=\frac{5\text{mA}}{120}\\ =41.67\text{μA}\end{array}$

Apply KVL in Figure 2

  $\begin{array}{l}-V^{+}+V_{CEQ}+I_{EQ}{R}_E+V^{-}=0\\ -{2.5}^{+}+{2.5}^{-}+5\text{mA}\left(R_E\right)+\left(-2.5\right)=0\\ R_E=\frac{2.5}{5\text{mA}}\\ R_E=0.5\text{k}\Omega \end{array}$

The expression for the value of the Thevenin resistance is given by,

  $R_{TH}=0.1\left(1+\beta \right)R_E$

Substitute $120$ for $\beta$ and $0.5\text{k}\Omega$ for $R_E$ in the above equation.

  $\begin{array}{c}{R}_{TH}=0.1\left(1+120\right)0.5\text{k}\Omega \\ =6.05\text{k}\Omega \end{array}$

Apply KVL to figure 3.

  $\begin{array}{l}-V_{TH}+R_{TH}+V_{BE}+I_{EQ}{R}_E+V^{-}=0\\ -V_{TH}+R_{TH}+V_{BE}+\left(1+\beta \right)I_{BQ}{R}_E+V^{-}=0\\ I_{BQ}=\frac{V_{TH}-V_{BE}-V^{-}}{R_{TH}+\left(1+\beta \right)R_E}\end{array}$ ……. (1)

Substitute $0.5\text{k}\Omega$ for $R_E$ , $6.05\text{k}\Omega$ for $R_{TH}$ , $120$ for $\beta$ , $0.7\text{V}$ for $V_{BE}$ and $-2.5\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{l}41.67\text{μA}=\frac{V_{TH}-0.7\text{V}-\left(-2.5\text{V}\right)}{6.05\text{k}\Omega +\left(1+120\right)0.5\text{k}\Omega }\\ V_{TH}=0.97\text{V}\end{array}$

The expression for the Thevenin voltage is evaluated as,

  $\begin{array}{l}{V}_{R2}=V_{TH}-V^{-}\\ V_{TH}=V_{R2}+V^{-}\\ V_{TH}=\left[\left(\frac{R_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)\right]+V^{-}\\ V_{TH}=\left[\left(\frac{R_{TH}}{R_1}\right)\left(V^{+}-V^{-}\right)\right]+V^{-}\end{array}$ …… (2)

Substitute $6.05\text{k}\Omega$ for $R_{TH}$ , $2.5\text{V}$ for $V^{+}$ , $0.97\text{V}$ for $V_{TH}$ and $-2.5\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{l}0.97\text{V}=\left[\left(\frac{6.05\text{k}\Omega }{R_1}\right)\left(2.5\text{V}-\left(-2.5\text{V}\right)\right)\right]+\left(-2.5\text{V}\right)\\ R_1=8.65\text{k}\Omega \end{array}$

The expression for the Thevenin resistance is given by,

  $\begin{array}{l}{R}_{TH}=\frac{R_1{R}_2}{R_1+R_2}\\ 6\text{k}\Omega =\frac{8.65\text{k}\Omega R_2}{8.65\text{k}\Omega +R_2}\\ R_2=19.6\text{k}\Omega \end{array}$

From the five percent tolerance resistance $R_1=8.2\text{k}\Omega$ and is closest to $8.65\text{k}\Omega$ , $R_2=20\text{k}\Omega$ it is closest to $19.61\text{k}\Omega$ and $R_E=0.51\text{k}\Omega$ it is closest to $0.5\text{k}\Omega$ .

Now consider the value of $\beta$ is 100.

The value for the Thevenin resistance is calculated as,

  $\begin{array}{c}{R}_{TH}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(8.2\text{k}\Omega \right)\left(20\text{k}\Omega \right)}{\left(8.2\text{k}\Omega \right)+\left(20\text{k}\Omega \right)}\\ =5.815\text{k}\Omega \end{array}$

Substitute $5.815\text{k}\Omega$ for $R_{TH}$ , $2.5\text{V}$ for $V^{+}$ , $8.2\text{k}\Omega$ for $R_1$ and $-2.5\text{V}$ for $V^{-}$ in equation (2).

  $\begin{array}{c}{V}_{TH}=\left[\left(\frac{5.815\text{k}\Omega }{8.2\text{k}\Omega }\right)\left(2.5\text{V}-\left(-2.5\text{V}\right)\right)\right]+\left(-2.5\text{V}\right)\\ =1.045\text{V}\end{array}$

Substitute $5.815\text{k}\Omega$ for $R_{TH}$ , $1.045\text{V}$ for $V_{TH}$ , $0.7\text{V}$ for $V_{BE}$ and $-2.5\text{V}$ for $V^{-}$ in equation (1).

  $\begin{array}{c}{I}_{BQ}=\frac{\left(1.045\text{V}\right)-\left(0.7\text{V}\right)-\left(-2.5\text{V}\right)}{5.815\text{k}\Omega +\left(1+100\right)\left(0.51\text{k}\Omega \right)}\\ =49.63\text{μA}\end{array}$

The expression for the collector current is given by,

  $I_{CQ}=\beta I_{BQ}$

Substitute $100$ for $\beta$ and $49.63\text{μA}$ for $I_{BQ}$ in the above equation.

  $\begin{array}{c}{I}_{CQ}=100\left(49.63\text{μA}\right)\\ =4.693\text{mA}\end{array}$

Apply KVL in Figure 2

  $\begin{array}{l}-V^{+}+V_{CEQ}+I_{CQ}{R}_E+V^{-}=0\\ -{2.5}^{+}+V_{CEQ}+\left(4.963\text{μA}\right)\left(0.51\text{k}\Omega \right)+\left(-2.5\right)=0\\ V_{CEQ}=2.47\text{V}\end{array}$

Consider the value of $\beta$ is 160.

The value for the Thevenin resistance is calculated as,

  $\begin{array}{c}{R}_{TH}=\frac{R_1{R}_2}{R_1+R_2}\\ =\frac{\left(8.2\text{k}\Omega \right)\left(20\text{k}\Omega \right)}{\left(8.2\text{k}\Omega \right)+\left(20\text{k}\Omega \right)}\\ =5.815\text{k}\Omega \end{array}$

Substitute $5.815\text{k}\Omega$ for $R_{TH}$ , $2.5\text{V}$ for $V^{+}$ , $8.2\text{k}\Omega$ for $R_1$ and $-2.5\text{V}$ for $V^{-}$ in equation (2).

  $\begin{array}{c}{V}_{TH}=\left[\left(\frac{5.815\text{k}\Omega }{8.2\text{k}\Omega }\right)\left(2.5\text{V}-\left(-2.5\text{V}\right)\right)\right]+\left(-2.5\text{V}\right)\\ =1.045\text{V}\end{array}$

Substitute $160$ for $\beta$ , $5.815\text{k}\Omega$ for $R_{TH}$ , $1.045\text{V}$ for $V_{TH}$ , $0.7\text{V}$ for $V_{BE}$ and $-2.5\text{V}$ for $V^{-}$ in equation (1).

  $\begin{array}{c}{I}_{BQ}=\frac{\left(1.045\text{V}\right)-\left(0.7\text{V}\right)-\left(-2.5\text{V}\right)}{5.815\text{k}\Omega +\left(1+10\right)\left(0.51\text{k}\Omega \right)}\\ =32.36\text{μA}\end{array}$

The expression for the collector current is given by,

  $I_{CQ}=\beta I_{BQ}$

Substitute $160$ for $\beta$ and $32.36\text{μA}$ for $I_{BQ}$ in the above equation.

  $\begin{array}{c}{I}_{CQ}=160\left(32.36\text{μA}\right)\\ =3.236\text{mA}\end{array}$

Apply KVL in Figure 2

  $\begin{array}{l}-V^{+}+V_{CEQ}+I_{CQ}{R}_E+V^{-}=0\\ -{2.5}^{+}+V_{CEQ}+\left(3.236\text{mA}\right)\left(0.51\text{k}\Omega \right)+\left(-2.5\right)=0\\ V_{CEQ}=3.35\text{V}\end{array}$

The Q- point values of the current and voltage for the range of $\beta$ between $100$ to 160 is given by,

  $\begin{array}{l}4.963\text{mA}\le I_{CQ}\le 3.236\text{mA}\\ \text{2}\text{.47}\text{V}\le V_{CEQ}\le 3.35\text{V}\end{array}$