Chapter 9, Problem 9.14P

(a) The input to the circuit shown in Figure P9.14 is $v_I=-0.2V$ . (i) Whatis $v_o$ ? (ii) Determine $i_2,i_0$ , and $i_L$ (b) Repeat part (a) for $v_I=+0.05V$ .(c) Repeat part (a) for $v_I=8\sin \omega tmV$ .

To determine

The value of the output voltage $v_O$ , current $i_2$ , $i_O$ and $i_L$ .

Answer to Problem 9.14P

The value of the voltage $v_O$ is $3\text{V}$ , current $i_L$ is $-0.75\text{mA}$ , current $i_2$ is $-0.2\text{mA}$ and current $i_o$ is $0.95\text{mA}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Apply KCL at the inverting terminal.

  $\begin{array}{l}\frac{v_I-0}{R_1}=\frac{0-v_O}{R_2}\\ \frac{v_I}{R_1}=\frac{-v_O}{R_2}\\ v_O=-\left(\frac{R_2}{R_1}\right)v_I\end{array}$

Substitute $-0.20\text{V}$ for $v_I$ , $15\text{k}\Omega$ for $R_2$ and $1\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{v}_O=-\left(\frac{15\text{k}\Omega }{1\text{k}\Omega }\right)\left(-0.20\text{V}\right)\\ =3\text{V}\end{array}$

The expression for the value of the current $i_2$ is given by,

  $i_2=\frac{-v_O}{15\text{k}\Omega }$

Substitute $3\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_2=\frac{-3\text{V}}{15\text{k}\Omega }\\ =-0.2\text{mA}\end{array}$

The expression for the value of the current $i_L$ is given by,

  $i_L=\frac{v_O}{4\text{k}\Omega }$

Substitute $3\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{-3\text{V}}{4\text{k}\Omega }\\ =-0.75\text{mA}\end{array}$

The expression for the value of the current $i_O$ is given by,

  $i_O=i_L-i_2$

Substitute $-0.75\text{mA}$ for $i_L$ and $-0.2\text{mA}$ fir $i_2$ in the above equation.

  $\begin{array}{c}{i}_O=-0.75\text{mA}-0.2\text{mA}\\ =0.95\text{mA}\end{array}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $3\text{V}$ , current $i_L$ is $-0.75\text{mA}$ , current $i_2$ is $-0.2\text{mA}$ and current $i_o$ is $0.95\text{mA}$ .

To determine

The value of the output voltage $v_O$ , current $i_2$ , $i_O$ and $i_L$ for $v_I$

  $+0.05\text{V}$ .

Answer to Problem 9.14P

The value of the voltage $v_O$ is $-0.75\text{V}$ , current $i_L$ is $-187.5\text{μA}$ , current $i_2$ is $50\text{μA}$ and current $i_o$ is $-\text{237}\text{.5}\text{μA}$ .

Explanation of Solution

Calculation:

The expression for the output voltage is given by,

  $v_O=-\left(\frac{R_2}{R_1}\right)v_I$

Substitute $0.05\text{V}$ for $v_I$ , $15\text{k}\Omega$ for $R_2$ and $1\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{v}_O=-\left(\frac{15\text{k}\Omega }{1\text{k}\Omega }\right)\left(0.05\text{V}\right)\\ =-0.75\text{V}\end{array}$

The expression for the value of the current $i_2$ is given by,

  $i_2=\frac{-v_O}{15\text{k}\Omega }$

Substitute $0.75\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_2=\frac{-0.75\text{V}}{15\text{k}\Omega }\\ =-0.05\text{mA}\end{array}$

The expression for the value of the current $i_L$ is given by,

  $i_L=\frac{v_O}{4\text{k}\Omega }$

Substitute $-0.75\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{-0.75\text{V}}{4\text{k}\Omega }\\ =-0.1875\text{mA}\end{array}$

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

  $1\text{mA}={10}^3\text{μA}$

The conversion from $0.1875\text{mA}$ into $\text{μA}$ is given by,

  $0.1875\text{mA}=0.1875\times {10}^3\text{μA}$

The conversion from $0.05\text{mA}$ into $\text{μA}$ is given by,

  $0.05\text{mA}=0.05\times {10}^3\text{μA}$

The expression for the value of the current $i_O$ is given by,

  $i_O=i_L-i_2$

Substitute $-187.5\text{μA}$ for $i_L$ and $50\text{μA}$ fir $i_2$ in the above equation.

  $\begin{array}{c}{i}_O=\left(-187.5\text{μA}\right)-50\text{μA}\\ =-\text{237}\text{.5}\text{μA}\end{array}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $-0.75\text{V}$ , current $i_L$ is $-187.5\text{μA}$ , current $i_2$ is $50\text{μA}$ and current $i_o$ is $-\text{237}\text{.5}\text{μA}$ .

To determine

The value of the output voltage $v_O$ , current $i_2$ , $i_O$ and $i_L$ for $v_I$

  $8\sin \omega t\text{mV}$ .

Answer to Problem 9.14P

The value of the voltage $v_O$ is $-0.12\sin \omega t\text{V}$ , current $i_L$ is $-0.03\sin \omega t\text{mA}$ , current $i_2$ is $-0.008\sin \omega t\text{mA}$ and current $i_o$ is $-0.038\sin \omega t\text{mA}$ .

Explanation of Solution

Calculation:

The conversion from $1\text{mV}$ into $\text{V}$ is given by,

  $1\text{mV}={10}^{-3}\text{V}$

The conversion from $8\sin \omega t\text{mV}$ into $\text{V}$ is given by,

  $8\sin \omega t\text{mV}=\left(8\sin \omega t\right){10}^{-3}\text{V}$

The expression for the output voltage is given by,

  $v_O=-\left(\frac{R_2}{R_1}\right)v_I$

Substitute $\left(8\sin \omega t\right){10}^{-3}\text{V}$ for $v_I$ , $15\text{k}\Omega$ for $R_2$ and $1\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{v}_O=-\left(\frac{15\text{k}\Omega }{1\text{k}\Omega }\right)\left(\left(8\sin \omega t\right){10}^{-3}\text{V}\right)\\ =-0.12\sin \omega t\text{V}\end{array}$

The expression for the value of the current $i_2$ is given by,

  $i_2=\frac{-v_O}{15\text{k}\Omega }$

Substitute $-0.12\sin \omega t\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_2=\frac{--0.12\sin \omega t\text{V}}{15\text{k}\Omega }\\ =-0.008\sin \omega t\text{mA}\end{array}$

The expression for the value of the current $i_L$ is given by,

  $i_L=\frac{v_O}{4\text{k}\Omega }$

Substitute $-0.12\sin \omega t\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{-0.12\sin \omega t\text{V}}{4\text{k}\Omega }\\ =-0.03\sin \omega t\text{mA}\end{array}$

The expression for the value of the current $i_O$ is given by,

  $i_O=i_L-i_2$

Substitute $-0.03\sin \omega t\text{mA}$ for $i_L$ and $-0.008\sin \omega t\text{mA}$ fir $i_2$ in the above equation.

  $\begin{array}{l}{i}_O=-0.03\sin \omega t\text{mA}-\left(-0.008\sin \omega t\text{mA}\right)\\ =-0.038\sin \omega t\text{mA}\end{array}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $-0.12\sin \omega t\text{V}$ , current $i_L$ is $-0.03\sin \omega t\text{mA}$ , current $i_2$ is $-0.008\sin \omega t\text{mA}$ and current $i_o$ is $-0.038\sin \omega t\text{mA}$ .