Chapter 5, Problem D5.91DP

The multitransistor circuit in Figure 5.61 is to be redesigned. The bias voltages are to be ±3.3 V and the nominal transistor current gains are $\beta =120$ . Design a bias-stable circuit such that $I_{CQ1}=100\mu \text{A}$ , $I_{CQ2}=200\mu \text{A}$ , and $V_{CEQ1}\cong V_{CEQ2}\cong 3\text{V}$ .

To determine

The design parameters of the bias stable circuit.

Answer to Problem D5.91DP

The value of the resistance required to design the circuit are $R_2$ is $471.15\text{k}\Omega$ and $R_1$ is $553.27\text{k}\Omega$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

  

Figure 1

Calculation:

Mark the current and other parameters than redraw the circuit.

The required diagrams is shown in Figure 2

  

Figure 2

The value of the base current of the first transistor is calculated as,

  $I_{BQ1}=\frac{I_{CQ1}}{\beta }$

Substitute $100\text{μA}$ for $I_{CQ1}$ and $120$ for $\beta$ in the above equation.

  $\begin{array}{c}{I}_{BQ1}=\frac{100\text{μA}}{120}\\ =0.833\text{μA}\end{array}$

The value of the base current of the second transistor is calculated as,

  $I_{BQ2}=\frac{I_{CQ2}}{\beta }$

Substitute $200\text{μA}$ for $I_{CQ2}$ and $120$ for $\beta$ in the above equation.

  $\begin{array}{c}{I}_{BQ2}=\frac{200\text{μA}}{120}\\ =1.67\text{μA}\end{array}$

Apply KCL to the above circuit.

  $\begin{array}{c}{I}_{RC1}=I_{CQ1}-I_{BQ2}\\ =100\text{μA}-\text{1}\text{.67}\text{μA}\\ =98.33\text{μA}\end{array}$

The expression for the current $I_{RC1}$ is given by,

  $I_{RC1}=\frac{V^{+}-V_{CQ1}}{R_{C1}}$

Substitute $98.33\text{μA}$ for $I_{RC1}$ , $15\text{k}\Omega$ for $R_{C1}$ and $3.3\text{V}$ for $V_{CQ1}$ in the above equation in the above equation.

  $\begin{array}{l}98.33\text{μA}=\frac{3.3\text{V}-V_{CQ1}}{15\text{k}\Omega }\\ V_{CQ1}=1.825\text{V}\end{array}$

From KCL the expression for the emitter current of the second transistor is given by,

  $\begin{array}{c}{I}_{EQ2}=I_{BQ2}+I_{CQ2}\\ =1.67\text{μA}+200\text{μA}\\ =201.67\text{μA}\end{array}$

The expression to determine the voltage $V_{EQ2}$ is given by,

  $\begin{array}{c}{V}_{EQ2}=V_{EB}\left(\text{on}\right)+V_{CQ1}\\ =0.7\text{V}+1.825\text{V}\\ =2.525\text{V}\end{array}$

The expression to determine the value of the resistance $R_{E2}$ is given by,

  $\begin{array}{c}{R}_{E2}=\frac{V^{+}-V_{EQ2}}{I_{EQ2}}\\ =\frac{3.3\text{V}-2.525\text{V}}{201.67\text{μA}}\\ =3.483\text{k}\Omega \end{array}$

The expression to determine the value of the collector voltage is $V_{CQ2}$ is given by

  $\begin{array}{c}{V}_{CQ2}=V_{EQ2}-V_{ECQ2}\\ =2.525\text{V}-3\text{V}\\ =-0.475\text{V}\end{array}$

The expression to determine the value of the resistance $R_{C2}$ is given by,

  $R_{C2}=\frac{V_{CQ2}-V^{-}}{I_{CQ2}}$

Substitute $-0.475\text{V}$ for $V_{CQ2}$ , $-3.3\text{V}$ for $V^{-}$ and $200\text{μA}$ for in the above equation.

  $\begin{array}{c}{R}_{C2}=\frac{-0.475-\left(-3.3\text{V}\right)}{200\text{μA}}\\ =14.125\text{k}\Omega \end{array}$

By KCL in Figure 2 the expression for the current $I_{EQ1}$ is given by,

  $I_{EQ1}=I_{BQ1}+I_{CQ1}$

Substitute $0.833\text{μA}$ for $I_{BQ1}$ and $100\text{μA}$ for $I_{CQ1}$ in the above equation.

  $\begin{array}{c}{I}_{EQ1}=0.833\text{μA}+100\text{μA}\\ =100.833\text{μA}\end{array}$

The expression to determine the value of the emitter voltage $V_{EQ1}$ is given by,

  $\begin{array}{c}{V}_{EQ1}=V_{CQ}-V_{CEQ1}\\ =1.825\text{V}-3\text{V}\\ =-1.175\text{V}\end{array}$

The expression to determine the value of the resistance $R_{E1}$ is given by,

  $R_{E1}=\frac{-V^{-}+V_{EQ1}}{I_{EQ2}}$

Substitute $-1.175\text{V}$ for $V_{EQ1}$ , $-3.3\text{V}$ for $V^{-}$ and $100.833\text{μA}$ for $I_{EQ1}$ in the above equation.

  $\begin{array}{c}{R}_{E1}=\frac{-\left(-3\text{V}\right)-1.175\text{V}}{100.833\text{μA}}\\ =21.07\text{k}\Omega \end{array}$

The expression for the Thevenin voltage is evaluated as,

  $\begin{array}{l}{V}_{R2}=V_{TH}-V^{-}\\ V_{TH}=V_{R2}+V^{-}\\ V_{TH}=\left[\left(\frac{R_2}{R_1+R_2}\right)\left(V^{+}-V^{-}\right)\right]+V^{-}\\ V_{TH}=\left[\left(\frac{R_{TH}}{R_1}\right)\left(V^{+}-V^{-}\right)\right]+V^{-}\end{array}$ …… (1)

The Thevenin equivalent base circuit is shown in Figure 3.

  

Figure 3

Apply KVL to the above circuit.

  $\begin{array}{l}-V_{TH}+R_{TH}{I}_{BQ1}+V_{BE}+I_{EQ1}{R}_{E1}+V^{-}=0\\ V_{TH}=R_{TH}{I}_{BQ1}+V_{BE}+\left(1+\beta \right)I_{BQ1}{R}_{E1}+V^{-}\\ V_{TH}=\left(R_{TH}+\left(1+\beta \right)R_{E1}\right)I_{BQ}+V_{BE}+V^{-}\\ V_{TH}=\left(R_{TH}+\left(1+\beta \right)R_{E1}\right)I_{BQ}+V_{BE}\end{array}$ …… (2)

For bias stable voltage the value of the Thevenin equivalent resistance is given by,

  $R_{TH}=0.1\left(1+\beta \right)R_{E1}$

Substitute $120$ for $\beta$ and $21\text{k}\Omega$ for $R_{E1}$ in the above equation.

  $\begin{array}{c}{R}_{TH}=0.1\left(1+120\right)21\text{k}\Omega \\ =254.1\text{k}\Omega \end{array}$

Substitute $0.833\text{μA}$ for $I_{BQ}$ , $0.7\text{V}$ for $V_{TH}$ , $254.1\text{k}\Omega$ for $R_{TH}$ , $120$ for $\beta$ and $21\text{k}\Omega$ for $R_{E1}$ in equation (2).

  $\begin{array}{c}{V}_{TH}=\left(254.1\text{k}\Omega +\left(1+120\right)21\text{k}\Omega \right)\left(0.833\text{μA}\right)+0.7\text{V}\\ =-0.27\text{V}\end{array}$

Substitute $3.3\text{V}$ for $V^{+}$ , $-0.27\text{V}$ for $V_{TH}$ , $254.1\text{k}\Omega$ for $R_{TH}$ , $120$ for $\beta$ and $3.3\text{V}$ for $V^{-}$ in equation (1).

  $\begin{array}{l}-0.27\text{V}=\left[\left(\frac{254\text{k}\Omega }{R_1}\right)\left(3.3\text{V}-\left(-3.3\text{V}\right)\right)\right]-3.3\text{V}\\ R_1=553.27\text{k}\Omega \end{array}$

The expression to determine the value of the Thevenin equivalent resistance is given by,

  $R_{TH}=\frac{R_1{R}_2}{R_1+R_2}$

Substitute $254\text{k}\Omega$ for $R_{TH}$ and $553.27\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{l}254\text{k}\Omega =\frac{\left(553.27\text{k}\Omega \right)R_2}{553.27\text{k}\Omega +R_2}\\ R_2=471.15\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistance required to design the circuit are $R_2$ is $471.15\text{k}\Omega$ and $R_1$ is $553.27\text{k}\Omega$ .