Chapter 8, Problem 8.26P

(a)

To determine

Thevalue of the voltage $V_{BB}$ and the power dissipated in the transistor.

Answer to Problem 8.26P

The value of the voltage $V_{BB}$ is $1.4\text{V}$ and the value of power dissipated in the transistor is $5\text{mW}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

  $1\text{mA}={10}^{-3}\text{A}$

The expression for the collector current is given by,

  $\begin{array}{l}{i}_{Cn}=I_S{e}^{\frac{V_{BE}}{V_T}}\\ V_{BE}=V_T\ln \left(\frac{i_{Cn}}{I_S}\right)\end{array}$

Substitute $0.026\text{V}$ for $V_T$ , $2\times {10}^{-15}\text{A}$ for $I_S$ and $1\text{mA}$ for $i_{Cn}$ in the above equation.

  $\begin{array}{c}{V}_{BE}=\left(0.026\text{V}\right)\ln \left(\frac{{10}^{-3}\text{A}}{2\times {10}^{-15}\text{A}}\right)\\ =0.7\text{V}\end{array}$

The expression for the base emitter voltage of the NPN transistor for no input voltage is given by,

  $V_{BEn}=\frac{V_{BB}}{2}-v_O$ …… (1)

The expression for the base emitter voltage for the PNP transistor for zero input voltage is given by,

  $V_{EBp}=v_O-\frac{V_{BB}}{2}$

From the above equation and equation (1).

  $\begin{array}{l}{V}_{BEn}-V_{EBp}=\frac{V_{BB}}{2}+\frac{V_{BB}}{2}\\ V_{BB}=2V_{BE}\end{array}$

Substitute $0.7\text{V}$ for $V_{BB}$ in the above equation.

  $\begin{array}{c}{V}_{BB}=2\left(0.7\text{V}\right)\\ =1.4\text{V}\end{array}$

The expression for the power dissipated in the transistor is given by,

  $P=V_{CC}{I}_C$

Substitute $5\text{V}$ for $V_{CC}$ and $1\text{mA}$ for $I_C$ in the above equation.

  $\begin{array}{c}P=\left(5\text{V}\right)\left(1\text{mA}\right)\\ =5\text{mW}\end{array}$

Conclusion:

Therefore, the value of the voltage $V_{BB}$ is $1.4\text{V}$ and the value of power dissipated in the transistor is $5\text{mW}$ .

(b)

To determine

The value of the power dissipated in $Q_n$ , $Q_p$ and $R_L$ .

Answer to Problem 8.26P

The value of $i_L$ is $-3.5\times {10}^{-3}\text{A}$ , $i_{Cn}$ is $3.71\times {10}^{-3}\text{A}$ , $v_I$ is $-3.54\text{V}$ , the power dissipated in the load is $12.25\text{W}$ , the power dissipated in the NPN transistor is $1.785\times {10}^{-3}\text{A}$ and in PNP transistor is $5.565\times {10}^{-3}\text{W}$ .

Explanation of Solution

Calculation:

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

  $i_L=\frac{V_O}{R_L}$

Substitute $-3.5\text{V}$ for $V_O$ and $1\text{k}\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{-3.5\text{V}}{1\text{k}\Omega }\\ =-3.5\text{mA}\end{array}$

The expression for the current through the PNP transistor is given by,

  $i_{Cp}=-i_L$

Substitute $3.5\text{mA}$ for $i_L$ in the above equation.

  $i_{Cp}=-3.5\text{mA}$

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

  $1\text{mA}={10}^{-3}\text{A}$

The conversion from $3.5\text{mA}$ into $\text{A}$ is given by,

  $3.5\text{mA}=3.5\times {10}^{-3}\text{A}$

The expression for the value of the base to emitter voltage of PNP transistor is given by,

  $V_{EBp}=V_T\ln \left(\frac{i_{Cn}}{I_S}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $2\times {10}^{-15}\text{A}$ for $I_S$ and $3.5\times {10}^{-3}\text{A}$ for $i_{Cn}$ in the above equation.

  $\begin{array}{c}{V}_{EBp}=\left(0.026\text{V}\right)\ln \left(\frac{2\times {10}^{-15}\text{A}}{3.5\times {10}^{-3}\text{A}}\right)\\ =0.74\text{V}\end{array}$

The expression to determine the value of the base emitter voltage of the NPN transistor is given by,

  $V_{BEn}+V_{BEp}=V_{BB}$

Substitute $0.74\text{V}$ for $V_{BEp}$ and $1.4\text{V}$ for $V_{BB}$ in the above equation.

  $\begin{array}{l}{V}_{BEn}+0.74\text{V}=1.4\text{V}\\ V_{BEn}=0.66\text{V}\end{array}$

The expression for the collector current of the NPN transistor is given by,

  $i_{Cn}=I_S{e}^{\frac{V_{BE}}{V_T}}$

Substitute $0.66\text{V}$ for $V_{BE}$ , $0.026\text{V}$ for $V_T$ and $2\times {10}^{-15}\text{A}$ for $I_S$ in the above equation.

  $\begin{array}{c}{i}_{Cn}=\left(2\times {10}^{-15}\text{A}\right)e^{\frac{0.66\text{V}}{0.026\text{V}}}\\ =0.21\times {10}^{-3}\text{A}\end{array}$

The expression for the value of collector current of PNP transistor is given by,

  $i_{Cp}=i_{Cn}+i_L$

Substitute $0.21\text{mA}$ for $i_{Cn}$ and $3.5\text{mA}$ for $i_L$ in the above equation.

  $\begin{array}{c}{i}_{Cp}=0.21\times {10}^{-3}\text{A}+3.5\times {10}^{-3}\text{A}\\ =3.71\times {10}^{-3}\text{A}\end{array}$

Apply KVL at the input terminals of the given figure.

  $\begin{array}{l}-v_I+\frac{V_{BB}}{2}-v_{EBp}+v_O=0\\ v_I=\frac{V_{BB}}{2}-v_{EBp}+v_O\end{array}$

Substitute $1.4\text{V}$ for $V_{BB}$ , $0.74\text{V}$ for $V_{EBp}$ and $3.5\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{v}_I=\frac{1.4\text{V}}{2}-\left(0.74\text{V}\right)+\left(3.5\text{V}\right)\\ =-3.54\text{V}\end{array}$

The expression for the power dissipated in the load is given by,

  $P_L=v_L{i}_L$

Substitute $3.5\times {10}^{-3}\text{A}$ for $i_L$ and $3.5\text{V}$ for $v_L$ in the above equation.

  $\begin{array}{c}{P}_L=\left(3.5\text{V}\right)\left(3.5\times {10}^{-3}\text{A}\right)\\ =12.25\text{W}\end{array}$

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

  $P_N=\left(V_{CC}-V_O\right)i_{Cn}$

Substitute $5\text{V}$ for $V_{CC}$ , $0.21\times {10}^{-3}\text{A}$ for $i_{Cn}$ and $-3.5\text{V}$ for $V_O$ in the above equation.

  $\begin{array}{c}{P}_N=\left(5\text{V}-\left(-3.5\text{V}\right)\right)\left(0.21\times {10}^{-3}\text{A}\right)\\ =1.785\times {10}^{-3}\text{A}\end{array}$

The expression for the power dissipated in the PNP transistor is given by,

  $P_P=\left(-V_O-\left(-V_{CC}\right)\right)i_{C_p}$

Substitute $3.5\text{V}$ for $V_O$ , $5\text{V}$ for $V_{CC}$ and $3.71\times {10}^{-3}\text{A}$ for $i_{Cp}$ in the above equation.

  $\begin{array}{c}{P}_P=\left(-3.5\text{V}-\left(-5\text{V}\right)\right)\left(3.71\times {10}^{-3}\text{A}\right)\\ =5.565\times {10}^{-3}\text{W}\end{array}$

Conclusion:

Therefore, the value of $i_L$ is $-3.5\times {10}^{-3}\text{A}$ , $i_{Cn}$ is $3.71\times {10}^{-3}\text{A}$ , $v_I$ is $-3.54\text{V}$ , the power dissipated in the load is $12.25\text{W}$ , the power dissipated in the NPN transistor is $1.785\times {10}^{-3}\text{A}$ and in PNP transistor is $5.565\times {10}^{-3}\text{W}$ .