Chapter 8, Problem 8.49P

For the class−AB output stage in Figure 8.36, the parameters are: $V^{+}=24\text{V}$ , $V^{-}=-24\text{V}$ , $R_L=20\Omega$ , and $I_{Biass}=\text{10mA}$ . The diode and transistor parameters are $I_S=2\times {10}^{-12}\text{A}$ . The transistor current gains are ${\beta }_n=20$ and ${\beta }_p=5$ for the npn and pnp devices, respectively. (a) For ${\upsilon }_O=0$ , determine $V_{BB}$ , and the quiescent collector current and base−emitter voltage for each transistor. (b) An average power of 10 watts is to be delivered to the load. Determine the quiescent collector current in each transistor and the instantaneous power dissipated in ${\text{Q}}_2$ , ${\text{Q}}_5$ , and $R_L$ when the output voltage is at its peak negative amplitude.

To determine

The value of $V_{BB}$ , the quiescent collector current and the base emitter voltage for each transistor.

Answer to Problem 8.49P

The value of the $V_{BB}$ is $1.74191\text{V}$ , $V_{B{E}_1}$ is $0.58065\text{V}$ , $V_{B{E}_3}$ is $0.50276\text{V}$ , $V_{B{E}_2}$ is $0.6585\text{V}$ , $I_{C_2}$ is $0.2\text{A}$ , $I_{C_1}$ is $10\times {10}^{-3}\text{A}$ and $I_{C_3}$ is $0.5\times {10}^{-3}\text{A}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The conversion from $1\text{mA}$ into $\text{A}$ is given by

  $1\text{mA}={\text{10}}^{-3}\text{A}$

The conversion from $10\text{mA}$ into $\text{A}$ is given by

  $10\text{mA}=10\times {\text{10}}^{-3}\text{A}$

The expression for the value of voltage $V_{BB}$ is given by

  $V_{BB}=3V_T\ln \left(\frac{I_{Bias}}{I_S}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $10\times {10}^{-3}\text{A}$ for $I_{Bias}$ and $2\times {10}^{-12}\text{A}$ for $I_S$ in the above equation.

  $\begin{array}{c}{V}_{BB}=3\left(0.026\text{V}\right)\ln \left(\frac{10\times {10}^{-3}\text{A}}{2\times {10}^{-12}\text{A}}\right)\\ =1.742\text{V}\end{array}$

The expression for the value of the voltage $V_{B{E}_1}$ is given by,

  $V_{B{E}_1}=V_T\ln \left(\frac{I_{C_1}}{I_S}\right)$ …… (1)

The expression for the value of the voltage $V_{B{E}_2}$ is given by,

  $V_{B{E}_2}=V_T\ln \left(\frac{I_{C_2}}{I_S}\right)$ …… (2)

The expression for the value of the voltage $V_{B{E}_3}$ is given by,

  $V_{B{E}_3}=V_T\ln \left(\frac{I_{C_3}}{I_S}\right)$ …… (3)

The expression for the value of $I_{C_1}$ is given by,

  $I_{C_1}=\frac{I_{C_2}}{{\beta }_n}$ …… (4)

The expression for the value of $I_{C_3}$ is given by,

  $I_{C_3}=\frac{I_{C_2}}{{\beta }_n^2}$ …… (5)

The expression for the value of the voltage $V_{BB}$ is given by,

  $V_{BB}=V_{B{E}_1}+V_{B{E}_2}+V_{B{E}_3}$ …… (6)

Substitute $V_T\ln \left(\frac{I_{C_2}}{I_S}\right)$ for $V_{B{E}_2}$ , $V_T\ln \left(\frac{I_{C_3}}{I_S}\right)$ for $V_{B{E}_3}$ and $V_T\ln \left(\frac{I_{C_1}}{I_S}\right)$ for $V_{B{E}_1}$ in the above equation.

  $V_{BB}=V_T\ln \left(\frac{I_{C_1}}{I_S}\right)+V_T\ln \left(\frac{I_{C_2}}{I_S}\right)+V_T\ln \left(\frac{I_{C_3}}{I_S}\right)$

Substitute $\frac{I_{C_2}}{{\beta }_n}$ for $I_{C_1}$ , $\frac{I_{C_2}}{{\beta }_n^2}$ for $I_{C_3}$ in the above equation.

  $\begin{array}{l}{V}_{BB}=V_T\ln \left(\frac{\frac{I_{C_2}}{{\beta }_n}}{I_S}\right)+V_T\ln \left(\frac{I_{C_2}}{I_S}\right)+V_T\ln \left(\frac{\frac{I_{C_2}}{{\beta }_n^2}}{I_S}\right)\\ V_T\ln \left(\frac{I_{C_2}^3}{{\beta }_n^3{I}_s^3}\right)=V_{BB}\\ I_{C_2}={\beta }_n{I}_S\sqrt[3]{\exp \left(\frac{V_{BB}}{V_T}\right)}\end{array}$

Substitute $20$ for ${\beta }_n$ , $2\times {10}^{-12}\text{A}$ for $I_S$ , $1.742\text{V}$ for $V_{BB}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{I}_{C_2}=\left(20\right)\left(2\times {10}^{-12}\text{A}\right)\sqrt[3]{\exp \left(\frac{1.742\text{V}}{0.026\text{V}}\right)}\\ =0.2\text{A}\end{array}$

Substitute $0.2\text{A}$ for $I_{C_2}$ and $20$ for ${\beta }_n$ in equation (4).

  $\begin{array}{c}{I}_{C_1}=\frac{0.2\text{A}}{20}\\ =10\times {10}^{-3}\text{A}\end{array}$

Substitute $0.2\text{A}$ for $I_{C_2}$ and $20$ for ${\beta }_n$ in equation (5).

  $\begin{array}{c}{I}_{C_3}=\frac{0.2\text{A}}{\left(20\right)2}\\ =0.5\times {10}^{-3}\text{A}\end{array}$

Substitute $10\times {10}^{-3}\text{A}$ for $I_{C_1}$ and $2\times {10}^{-12}\text{A}$ for $I_S$ and $0.026\text{V}$ for $V_T$ in equation (1).

  $\begin{array}{c}{V}_{B{E}_1}=\left(0.026\text{V}\right)\ln \left(\frac{10\times {10}^{-3}\text{A}}{2\times {10}^{-12}\text{A}}\right)\\ =0.58065\text{V}\end{array}$

Substitute $0.2\text{A}$ for $I_{C_2}$ and $2\times {10}^{-12}\text{A}$ for $I_S$ and $0.026\text{V}$ for $V_T$ in equation (2).

  $\begin{array}{c}{V}_{B{E}_2}=\left(0.026\text{V}\right)\ln \left(\frac{0.2\text{A}}{2\times {10}^{-12}\text{A}}\right)\\ =0.6585\text{V}\end{array}$

Substitute $0.5\times {10}^{-3}\text{A}$ for $I_{C_3}$ and $2\times {10}^{-12}\text{A}$ for $I_S$ and $0.026\text{V}$ for $V_T$ in equation (3).

  $\begin{array}{c}{V}_{B{E}_3}=\left(0.026\text{V}\right)\ln \left(\frac{0.5\times {10}^{-3}\text{A}}{2\times {10}^{-12}\text{A}}\right)\\ =0.50276\text{V}\end{array}$

Substitute $0.58065\text{V}$ for $V_{B{E}_1}$ , $0.6585\text{V}$ for $V_{B{E}_2}$ and $0.50276\text{V}$ for $V_{B{E}_3}$ in equation (6).

  $\begin{array}{c}{V}_{BB}=0.58065\text{V}+0.6585\text{V}+0.50276\text{V}\\ =1.74191\text{V}\end{array}$

Conclusion:

Therefore, the value of the current $i_{C_2}$ is $9.39\times {10}^{-3}\text{A}$ , $i_{C_1}$ is $4.47\times {10}^{-3}\text{A}$ , $i_{C_3}$ is $2.267\times {10}^{-3}\text{A}$ , $i_{C_4}$ is $45.3\times {10}^{-3}\text{A}$ and $i_{C_5}$ is $0.952\text{A}$ . The value of the power dissipated in $Q_2$ is $4.132\text{W}$ , $Q_5$ is $13.3\text{W}$ and in $R_L$ is $20\text{W}$ .

To determine

The value of the quiescent collector current in each transistor and the value of the collector quiescent current and the instantaneous power dissipated in $Q_2$ , $Q_5$ and $R_L$ .

Answer to Problem 8.49P

The is $45.3\times {10}^{-3}\text{A}$ and $i_{C_5}$ is $0.952\text{A}$ . The value of the power dissipated in $Q_2$ is $4.132\text{W}$ , $Q_5$ is $13.3\text{W}$ and in $R_L$ is $20\text{W}$ .

Explanation of Solution

Calculation:

The expression for the value of maximum output voltage is given by,

  $v_{O\left(\max \right)}^2=\sqrt{2R_L{\overline{P}}_L}$

Substitute $20\Omega$ for $R_L$ and $10\text{W}$ for ${\overline{P}}_L$ in the above equation.

  $\begin{array}{c}{v}_{O\left(\max \right)}^2=\sqrt{2\left(20\Omega \right)\left(10\text{W}\right)}\\ =20\text{V}\end{array}$

The expression for the value of the current delivered to the load is given by,

  $i_{O\left(\max \right)}=\frac{-v_{O\left(\max \right)}}{R_L}$

Substitute $20\text{V}$ for $v_{O\left(\max \right)}$ and $20\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{i}_{O\left(\max \right)}=\frac{-20\text{V}}{20\Omega }\\ =-1\text{A}\end{array}$

Apply KCL at the output node.

  $i_{C_5}+\frac{i_{C_5}}{{\beta }_n}\left[\frac{{\beta }_n}{1+{\beta }_n}\right]+\frac{i_{C_4}}{{\beta }_n}\left[\frac{1+{\beta }_p}{{\beta }_P}\right]+i_{O\left(\max \right)}=0$

Substitute $1\text{A}$ for $i_{O\left(\max \right)}$ , $20$ for ${\beta }_n$ and $5$ for ${\beta }_P$ in the above equation.

  $\begin{array}{l}{i}_{C_5}+\frac{i_{C_5}}{20}\left[\frac{20}{1+20}\right]+\frac{i_{C_4}}{20}\left[\frac{1+5}{5}\right]+\left(-1\text{A}\right)=0\\ i_{C_5}=0.952\text{A}\end{array}$

The expression for the value of the current $i_{C_4}$ is given by,

  $i_{C_4}=\frac{1}{{\beta }_n+1}\left(i_{C_5}\right)$

Substitute $0.952\text{A}$ for $i_{C_5}$ and $20$ for ${\beta }_n$ in the above equation.

  $\begin{array}{c}{i}_{C_4}=\frac{1}{20+1}\left(0.952\text{A}\right)\\ =45.3\times {10}^{-3}\text{A}\end{array}$

The expression for the value of the current $i_{C_3}$ is given by,

  $i_{C_3}=\frac{i_{C_5}}{{\beta }_n\left(1+{\beta }_n\right)}$

Substitute $0.952\text{A}$ for $i_{C_5}$ and $20$ for ${\beta }_n$ in the above equation.

  $\begin{array}{c}{i}_{C_3}=\frac{0.952\text{A}}{20\left(1+20\right)}\\ =2.267\times {10}^{-3}\text{A}\end{array}$

The expression for the value of the voltage $V_{BB}$ is given by,

  $V_{EBB}=V_T\ln \frac{i_{C_3}}{I_S}$

Substitute $2\times {10}^{-12}\text{A}$ for $I_S$ , $0.026\text{V}$ for $V_T$ and $2.267\times {10}^{-3}\text{A}$ for $i_{C_3}$ in the above equation.

  $\begin{array}{c}{V}_{EBB}=\left(0.026\text{V}\right)\ln \frac{2.267\times {10}^{-3}\text{A}}{2\times {10}^{-12}\text{A}}\\ =\mathrm{0.5420.6}\text{A}\end{array}$

The expression for the value of the current $i_{C_2}$ is given by,

  $i_{C_2}=\sqrt{{\beta }_n}{I}_S\sqrt{\exp \left(\frac{V_{BB}-V_{EBB}}{V_T}\right)}$

Substitute $20$ for ${\beta }_n$ , $2\times {10}^{-12}\text{A}$ for $I_S$ , $0.026\text{V}$ for $V_T$ , for $V_{EBB}$ and $1.742\text{V}$ for $V_{BB}$ in the above equation.

  $\begin{array}{c}{i}_{C_2}=\sqrt{20}\left(2\times {10}^{-12}\text{A}\right)\sqrt{\exp \left(\frac{1.742\text{V}-0.54206\text{V}}{0.026\text{V}}\right)}\\ =93.9\times {10}^{-3}\text{A}\end{array}$

The expression for the value of the current $i_{C_1}$ is given by,

  $i_{C_1}=\frac{i_{C_2}}{{\beta }_n}\left(\frac{{\beta }_n}{1+{\beta }_n}\right)$

Substitute $93.9\times {10}^{-3}\text{A}$ for $i_{C_2}$ and $20$ for ${\beta }_n$ in the above equation.

  $\begin{array}{c}{i}_{C_1}=\frac{\left(93.9\times {10}^{-3}\text{A}\right)}{1+20}\\ =4.47\times {10}^{-3}\text{A}\end{array}$

The expression for the value of power dissipated in the transistor $Q_2$ is given by,

  $P_{Q_2}=i_{C_2}\left(V^{+}-\left(-v_{O\left(\max \right)}\right)\right)$

Substitute $9.39\times {10}^{-3}\text{A}$ for $i_{C_2}$ , $24\text{V}$ for $V^{+}$ and $-20\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}{P}_{Q_2}=9.39\times {10}^{-3}\text{A}\left(24\text{V}-\left(-20\text{V}\right)\right)\\ =4.132\text{W}\end{array}$

The expression for the value of power dissipated in the transistor $Q_2$ is given by,

  $P_{Q_5}=i_{C_5}\left(\frac{v_{O\left(\max \right)}}{2}-\left(V^{-}\right)\right)$

Substitute $0.952\text{A}$ for $i_{C_2}$ , $-24\text{V}$ for $V^{-}$ and $-20\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}{P}_{Q_2}=0.952\text{A}\left(\frac{-20\text{V}}{2}-\left(-24\text{V}\right)\right)\\ =13.3\text{W}\end{array}$

Conclusion:

Therefore, the value of the current $i_{C_2}$ is $9.39\times {10}^{-3}\text{A}$ , $i_{C_1}$ is $4.47\times {10}^{-3}\text{A}$ , $i_{C_3}$ is $2.267\times {10}^{-3}\text{A}$ , $i_{C_4}$ is $45.3\times {10}^{-3}\text{A}$ and $i_{C_5}$ is $0.952\text{A}$ . The value of the power dissipated in $Q_2$ is $4.132\text{W}$ , $Q_5$ is $13.3\text{W}$ and in $R_L$ is $20\text{W}$ .