Chapter 8, Problem 8.20P

(a)

To determine

The maximum and the minimum value of the output voltage and the minimum corresponding input voltages for the circuit to operate in the linear region for given value of $R_L$ .

Answer to Problem 8.20P

The maximum value of the output voltage is $4.8\text{V}$ and the minimum value of the output voltage is $-3.2\text{V}$ . The maximum value of the input voltage is $5.5\text{V}$ and the minimum value of the input voltage is $-2.5\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

  $v_O\left(\max \right)=V^{+}-V_{CE}\left(\text{sat}\right)$

Substitute $5\text{V}$ for $V_{GS}$ and $0.2\text{V}$ for $V_{CE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{v}_O\left(\max \right)=5\text{V}-0.2\text{V}\\ =\text{4}\text{.8}\text{V}\end{array}$

The expression for the drain to source saturation voltage is given by,

  $V_{DS\left(\text{sat}\right)}=V_{GS}-V_{TN}$

Substitute $0$ for $V_{GS}$ and $-1.8\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{DS\left(\text{sat}\right)}=0-\left(-1.8\text{V}\right)\\ =1.8\text{V}\end{array}$

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

  $v_{O\left(\min \right)}=V_{DS\left(\text{sat}\right)}+V^{-}$

Substitute $1.8\text{V}$ for $V_{DS\left(\text{sat}\right)}$ and $-5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{v}_{O\left(\min \right)}=1.8\text{V}+\left(-5\text{V}\right)\\ =-3.2\text{V}\end{array}$

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

  $v_{I\left(\min \right)}=V_{BE\left(ON\right)}+v_{O\left(\min \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(ON\right)}$ and $-3.2\text{V}$ for $v_{O\left(\min \right)}$ in the above equation.

  $\begin{array}{c}{v}_{I\left(\min \right)}=0.7\text{V}-3.2\text{V}\\ =-2.5\text{V}\end{array}$

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

  $v_{I\left(\max \right)}=V_{BE\left(ON\right)}+v_{O\left(\max \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(ON\right)}$ and $4.8\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}{v}_{I\left(\min \right)}=0.7\text{V}+4.8\text{V}\\ =5.5\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the output voltage is $4.8\text{V}$ and the minimum value of the output voltage is $-3.2\text{V}$ . The maximum value of the input voltage is $5.5\text{V}$ and the minimum value of the input voltage is $-2.5\text{V}$ .

(b)

To determine

The maximum and the minimum value of the output voltage and the minimum corresponding input voltages for the circuit to operate in the linear region for given value of $R_L$ .

Answer to Problem 8.20P

The maximum value of the output voltage is $4.8\text{V}$ and the minimum value of the output voltage is $-3.2\text{V}$ . The maximum value of the input voltage is $5.5\text{V}$ and the minimum value of the input voltage is $-2.5\text{V}$ .

Explanation of Solution

Calculation:

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

  $v_O\left(\max \right)=V^{+}-V_{CE}\left(\text{sat}\right)$

Substitute $5\text{V}$ for $V_{GS}$ and $0.2\text{V}$ for $V_{CE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{v}_O\left(\max \right)=5\text{V}-0.2\text{V}\\ =\text{4}\text{.8}\text{V}\end{array}$

The expression for the drain to source saturation voltage is given by,

  $V_{DS\left(\text{sat}\right)}=V_{GS}-V_{TN}$

Substitute $0$ for $V_{GS}$ and $-1.8\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{DS\left(\text{sat}\right)}=0-\left(-1.8\text{V}\right)\\ =1.8\text{V}\end{array}$

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

  $v_{O\left(\min \right)}=V_{DS\left(\text{sat}\right)}+V^{-}$

Substitute $1.8\text{V}$ for $V_{DS\left(\text{sat}\right)}$ and $-5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{v}_{O\left(\min \right)}=1.8\text{V}+\left(-5\text{V}\right)\\ =-3.2\text{V}\end{array}$

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

  $v_{I\left(\min \right)}=V_{BE\left(ON\right)}+v_{O\left(\min \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(ON\right)}$ and $-3.2\text{V}$ for $v_{O\left(\min \right)}$ in the above equation.

  $\begin{array}{c}{v}_{I\left(\min \right)}=0.7\text{V}-3.2\text{V}\\ =-2.5\text{V}\end{array}$

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

  $v_{I\left(\max \right)}=V_{BE\left(ON\right)}+v_{O\left(\max \right)}$

Substitute $0.7\text{V}$ for $V_{BE\left(ON\right)}$ and $4.8\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}{v}_{I\left(\min \right)}=0.7\text{V}+4.8\text{V}\\ =5.5\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the output voltage is $4.8\text{V}$ and the minimum value of the output voltage is $-3.2\text{V}$ . The maximum value of the input voltage is $5.5\text{V}$ and the minimum value of the input voltage is $-2.5\text{V}$ .

(c)

To determine

The smallest value of the $R_L$ possible and the conversion efficiency.

Answer to Problem 8.20P

The minimum value of the load resistance is $5.144\text{k}\Omega$ and the conversion efficiency is $10\%$ .

Explanation of Solution

Calculation:

The expression for the current through the transistor $Q_1$ when $V_O=-2\text{V}$ is given by,

  $I_D=-K_n{\left(V_{GS}-v_{TN}\right)}^2$

Substitute $12\text{mA}/{\text{V}}^2$ for $K_n$ , $0\text{V}$ for $V_{GS}$ and $-1.8\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{I}_D=-\left(12\text{mA}/{\text{V}}^2\right){\left(0-\left(-1.8\text{V}\right)\right)}^2\\ =-38.89\text{mA}\end{array}$

The expression for the minimum value of the load resistance is $R_{L\left(\min \right)}$ is given by,

  $R_{L\left(\min \right)}=\frac{V_O}{I_D}$

Substitute $-2\text{V}$ for $V_O$ and $38.89\text{mA}$ for $I_D$ in the above equation.

  $\begin{array}{c}{R}_{L\left(\min \right)}=\frac{-2\text{V}}{38.89\text{mA}}\\ =5.144\text{k}\Omega \end{array}$

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

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

Substitute $2\text{V}$ for $V_O$ and $5.144\text{k}\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{\overline{P}}_L=\frac{{\left(2\text{V}\right)}^2}{2\left(5.144\text{k}\Omega \right)}\\ =38.89\text{mW}\end{array}$

The expression for the average power supplied by the source is given by,

  ${\overline{P}}_S=I_D\left(V^{+}-V^{-}\right)$

Substitute $38.88\text{mA}$ for $I_D$ , $5\text{V}$ for $V^{+}$ and $-5\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{c}{\overline{P}}_S=\left(38.88\text{mA}\right)\left(5\text{V}-\left(-5\text{V}\right)\right)\\ =388.89\text{mW}\end{array}$

The expression for the value of power conversion efficiency is given by,

  $\%\eta =\left(\frac{{\overline{P}}_L}{{\overline{P}}_S}\right)100$

Substitute $388.89\text{mW}$ for $P_S$ and $38.89\text{mW}$ for ${\overline{P}}_L$ in the above equation.

  $\begin{array}{c}\%\eta =\left(\frac{38.89\text{mW}}{388.89\text{mW}}\right)100\\ =10\%\end{array}$

Conclusion:

Therefore, the minimum value of the load resistance is $5.144\text{k}\Omega$ and the conversion efficiency is $10\%$ .