Chapter 8, Problem 8.48P

Consider the class−AB output stage in Figure P8.48. The parameters are: $V^{+}=12V$ , $V^{-}=-12\text{V}$ , $R_L=100\Omega$ , and $I_{Biass}=5\text{mA}$ . The transistor and diode parameters are $I_S={10}^{-13}\text{A}$ . The transistor current gains are ${\beta }_n=100$ and ${\beta }_p=20$ 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) Repeat part (a) for ${\upsilon }_O=10\text{V}$ . What is the power delivered to the load and what is the power dissipated in each transistor?

Figure P8.48

To determine

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

Answer to Problem 8.48P

Thevalue of the voltage $V_{BB}$ is $1.28\text{V}$ , the value of the current $I_{C_2}$ is $49.95\times {10}^{-3}\text{A}$ , $I_{C_1}$ is $49.97\times {10}^{-3}\text{A}$ and the value of the current $I_{C_3}$ is $0.4995\times {10}^{-3}\text{A}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

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

The conversion from $\text{5}\text{mA}$ into $\text{A}$ is given by,

  $\text{5}\text{mA}=5\times {10}^{-3}\text{A}$

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

  $V_{BB}=2V_D=2V_T\ln \frac{I_{Bias}}{I_S}$

Substitute $0.026\text{V}$ for $V_T$ , $5\times {10}^{-3}\text{A}$ for $I_{Bias}$ and ${10}^{-13}\text{A}$ for $I_S$ in the above equation,.

  $\begin{array}{c}{V}_{BB}=2\left(0.026\text{V}\right)\ln \frac{5\times {10}^{-3}\text{A}}{{10}^{-13}\text{A}}\\ =1.281\text{V}\end{array}$

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

  $I_{C_3}=\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$

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

  $I_{C_2}={\beta }_n\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$ ...... (1)

The expression for the value of the current $I_{B2}$ is given by,

  $I_{B_2}=I_{C_3}$

Substitute $\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$ for $I_{C_3}$ in the above equation.

  $I_{B_2}=\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$

The expression for the value of the current $I_{E_1}$ is given by,

  $I_{E_1}=I_C\left(\frac{1+{\beta }_n}{{\beta }_n}\right)$

The expression for the value of the current $I_{E_3}$ is given by,

  $I_{E_3}=\left(\frac{1+{\beta }_p}{{\beta }_p}\right)I_{C_3}$ ...... (2)

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

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

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

  $V_{B{E}_3}=V_T\ln \left(\frac{I_{C_1}}{I_S}\right)$

The expression for the value of $I_{E1}$ is given by,

  $I_{E_1}=I_{E_3}+I_{C_2}$

Substitute $\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$ for $I_{C_2}$ in the above equation.

  $I_{E_1}=I_{E_3}+{\beta }_n\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_3}$

Substitute $I_C\left(\frac{1+{\beta }_n}{{\beta }_n}\right)$ for $I_{E_1}$ and $\left(\frac{1+{\beta }_p}{{\beta }_p}\right)I_{C_3}$ for $I_{E_3}$ in the above equation.

  $I_{C_1}\left(\frac{1+{\beta }_n}{{\beta }_n}\right)=\left(1+{\beta }_n\left(\frac{{\beta }_p}{1+{\beta }_p}\right)\right)\left(\frac{1+{\beta }_p}{{\beta }_p}\right)I_{C_3}$

Substitute $100$ for ${\beta }_n$ and $20$ for ${\beta }_p$ in the above equation.

  $\begin{array}{l}{I}_{C_1}\left(\frac{1+100}{100}\right)=\left(1+100\left(\frac{20}{1+20}\right)\right)\left(\frac{1+20}{20}\right)I_{C_3}\\ I_{C_1}=100.05I_{C_3}\end{array}$

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

  $V_{B{E}_1}=V_T\ln \left(\frac{I_{C_1}}{I_S}\right)$ ...... (3)

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

  $V_{B{E}_3}=V_T\ln \left(\frac{I_{C_3}}{I_S}\right)$ ...... (4)

The expression for the value of the base to emitter voltage $V_{BE1}$ is given by,

  $V_{B{E}_1}+V_{E{B}_3}=V_{BB}$

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

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

Substitute $100.05I_{C_3}$ for $I_{C_1}$ in the above equation.

  $\begin{array}{l}{V}_T\ln \left(\frac{100.05I_{C_3}}{I_S}\right)+V_T\ln \left(\frac{I_{C_3}}{I_S}\right)=V_{BB}\\ I_{C_3}-\frac{I_S}{\sqrt{100.05}}\sqrt{\exp \left(\frac{V_{BB}}{V_T}\right)}\end{array}$

Substitute $1.281\text{V}$ for $V_{BB}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{I}_{C_3}=\frac{{10}^{-13}\text{A}}{\sqrt{100.05}}\sqrt{\exp \left(\frac{1.281\text{V}}{0.026\text{V}}\right)}\\ =0.4995\times {10}^{-3}\text{A}\end{array}$

Substitute $0.4995\times {10}^{-3}\text{A}$ for $I_{C_3}$ and $20$ for ${\beta }_p$ in equation (2).

  $\begin{array}{c}{I}_{E_3}=\left(\frac{1+20}{20}\right)\left(0.4995\times {10}^{-3}\text{A}\right)\\ =0.5245\times {10}^{-3}\text{A}\end{array}$

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

  $I_{C_1}=100.05I_{C_3}$

Substitute $0.4995\times {10}^{-3}\text{A}$ for $I_{C_3}$ in the above equation.

  $\begin{array}{c}{I}_{C_1}=100.05\left(0.4995\times {10}^{-3}\text{A}\right)\\ =49.97\times {10}^{-3}\text{A}\end{array}$

Substitute $0.4995\times {10}^{-3}\text{A}$ for $I_{C_3}$ and $20$ for ${\beta }_p$ in equation (1).

  $I_{C_2}=20\left(\frac{20}{1+20}\right)\left(0.4995\times {10}^{-3}\text{A}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $49.97\times {10}^{-3}\text{A}$ for $I_{C_1}$ and ${10}^{-13}\text{A}$ for $I_S$ in equation (3).

  $\begin{array}{c}{V}_{B{E}_1}=0.026\text{V}\left[\ln \left(\frac{49.97\times {10}^{-3}\text{A}}{{10}^{-13}\text{A}}\right)\right]\\ =0.70037\text{V}\end{array}$

Substitute $0.026\text{V}$ for $V_T$ , $0.4995\times {10}^{-3}\text{A}$ for $I_{C_3}$ and ${10}^{-13}\text{A}$ for $I_S$ in equation (4).

  $\begin{array}{c}{V}_{B{E}_3}=0.026\text{V}\left[\ln \left(\frac{0.4995\times {10}^{-3}\text{A}}{{10}^{-13}\text{A}}\right)\right]\\ =0.58062\text{V}\end{array}$

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

  $V_{BB}=V_{B{E}_1}+V_{B{E}_3}$

Substitute $0.70037\text{V}$ for $V_{B{E}_1}$ and $0.58062\text{V}$ for $V_{B{E}_3}$ in the above equation.

  $\begin{array}{c}{V}_{BB}=0.70037\text{V}+0.58062\text{V}\\ =1.28\text{V}\end{array}$

Conclusion:

Therefore, the value of the voltage $V_{BB}$ is $1.28\text{V}$ , the value of the current $I_{C_2}$ is $49.95\times {10}^{-3}\text{A}$ , $I_{C_1}$ is $49.97\times {10}^{-3}\text{A}$ and the value of the current $I_{C_3}$ is $0.4995\times {10}^{-3}\text{A}$ .

To determine

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

Answer to Problem 8.48P

The value of power delivered $Q_1$ is $0.34\text{W}$ , the power delivered to transistor $Q_2$ is $0.352\text{W}$ and $Q_2$ is $3.4\text{W}$ . The power delivered to the load is $1\text{W}$ , the value of collector current for transistor $Q_1$ is $0.1\text{A}$ , $i_{C_2}$ is $15.98\times {10}^{-3}\text{A}$ , $i_{C_3}$ is $0.1589\times {10}^{-3}\text{A}$ , the voltage $V_{BB}$ is $1.2694\text{V}$ , the value of the voltage $V_{B{E}_1}$ is $0.7184\text{V}$ and the voltage $V_{E{B}_3}$ is $0.55099\text{V}$ .

Explanation of Solution

Calculation:

The expression for the value of the current $i_{E_1}$ is given by,

  $i_{E_1}=\frac{v_O}{R_L}$

Substitute $10\text{V}$ for $v_O$ and $100\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{i}_{E_1}=\frac{10\text{V}}{100\Omega }\\ =0.1\text{A}\end{array}$

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

  $V_{BB}=2\left(V_T\right)\ln \left(\frac{I_{Bias}}{I_S}\right)$

Substitute ${10}^{-13}\text{A}$ for $I_S$ , $0.026\text{V}$ for $V_T$ and $5\times {10}^{-3}\text{A}$ for $I_{Bias}$ in the above equation.

  $\begin{array}{c}{V}_{BB}=2\left(0.026\text{V}\right)\ln \left(\frac{5\times {10}^{-3}\text{A}}{{10}^{-13}\text{A}}\right)\\ =1.2694\text{V}\end{array}$

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

  $V_{B{E}_1}=V_T\ln \left(\frac{i_{E_1}}{I_S}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $0.1\text{A}$ for $i_{E_1}$ and ${10}^{-13}\text{A}$ for $I_S$ in equation (4).

  $\begin{array}{c}{V}_{B{E}_1}=0.026\text{V}\left[\ln \left(\frac{0.1\text{A}}{{10}^{-13}\text{A}}\right)\right]\\ =0.7184\text{V}\end{array}$

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

  $V_{B{E}_3}=V_{BB}-V_{B{E}_1}$

Substitute $0.7184\text{V}$ for $V_{B{E}_1}$ and $1.2694\text{V}$ for $V_{BB}$ in the above equation.

  $\begin{array}{c}{V}_{B{E}_3}=1.2694\text{V}-0.7184\text{V}\\ =\text{0}\text{.55099}\text{V}\end{array}$

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

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

Substitute ${10}^{-13}\text{A}$ for $I_S$ , $0.026\text{V}$ for $V_T$ and $\text{0}\text{.55099}\text{V}$ for $V_{B{E}_3}$ in the above equation.

  $\begin{array}{c}{I}_{C_3}={10}^{-13}\text{A}\exp \left(\frac{0.55099\text{V}}{0.026\text{V}}\right)\\ =0.1598\times {10}^{-3}\text{A}\end{array}$

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

  ${\overline{P}}_L=\frac{v_O^2}{R_L}$

Substitute $10\text{V}$ for $v_o$ and $100\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{\overline{P}}_L=\frac{\left(10\text{V}\right)}{100\Omega }\\ =1\text{W}\end{array}$

The expression for the power delivered in the transistor of $Q_1$ is given by,

  $P_{Q_1}=i_{E_1}\left(I_{C_3}\right)\left[v_O-\left(V_{BE}-V^{+}\right)\right]$

Substitute $0.1\text{A}$ for $i_{E_1}$ , $0.1598\times {10}^{-3}\text{A}$ for $I_{C_3}$ , $12\text{V}$ for $V^{+}$ , $0.7\text{V}$ for $V_{BE}$ and $10\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{P}_{Q_1}=\left(0.1598\times {10}^{-3}\text{A}\right)\left(0.1\text{A}\right)\left[10\text{V}-\left(0.7\text{V}-12\text{V}\right)\right]\\ =0.34\text{W}\end{array}$

The expression for the power delivered in the transistor of $Q_3$ is given by,

  ${\overline{P}}_{Q_3}=\left(I_{C_3}\right)\left[v_O-\left(V_{BE}-V^{+}\right)\right]$

Substitute $0.1598\times {10}^{-3}\text{A}$ for $I_{C_3}$ , $12\text{V}$ for $V^{+}$ , $0.7\text{V}$ for $V_{BE}$ and $10\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{\overline{P}}_{Q_3}=\left(0.1598\times {10}^{-3}\text{A}\right)\left[10\text{V}-\left(0.7\text{V}-12\text{V}\right)\right]\\ =3.4\text{W}\end{array}$

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

  $i_{C_2}={\beta }_n{i}_{C_3}$

Substitute $0.1598\times {10}^{-3}\text{A}$ for $I_{C_3}$ and $100$ for ${\beta }_n$ in the above equation.

  $\begin{array}{c}{i}_{C_2}=100\left(0.1598\times {10}^{-3}\text{A}\right)\\ =15.98\times {10}^{-3}\text{A}\end{array}$

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

  $P_{Q_2}=i_{C_2}\left[v_O-V^{-}\right]$

Substitute $15.98\times {10}^{-3}\text{A}$ for $i_{C_2}$ , $10\text{V}$ for $v_O$ and $-12\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{c}{P}_{Q_2}=\left(15.98\times {10}^{-3}\text{A}\right)\left[10\text{V}-\left(-12\text{V}\right)\right]\\ =0.352\text{W}\end{array}$

Conclusion:

Therefore, the value of power delivered $Q_1$ is $0.34\text{W}$ , the power delivered to transistor $Q_2$ is $0.352\text{W}$ and $Q_2$ is $3.4\text{W}$ . The power delivered to the load is $1\text{W}$ , the value of collector current for transistor $Q_1$ is $0.1\text{A}$ , $i_{C_2}$ is $15.98\times {10}^{-3}\text{A}$ , $i_{C_3}$ is $0.1589\times {10}^{-3}\text{A}$ , the voltage $V_{BB}$ is $1.2694\text{V}$ , the value of the voltage $V_{B{E}_1}$ is $0.7184\text{V}$ and the voltage $V_{E{B}_3}$ is $0.55099\text{V}$ .