Chapter 9, Problem 9.6P

Assume the op-amps in Figure P9.6 are ideal. Find the voltage gain $A_v=v_o/v_I$ , and the input resistance $R_i$ of each circuit.

To determine

The voltage gain $A_v$ and the input resistance $R_i$ of each of the circuit.

Answer to Problem 9.6P

The value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Mark the values and redraw the circuit of part (a).

The required diagram is shown in Figure 2

  

The from the above circuit the expression for the value of the voltage $v_2$ is given by,

  $v_1=v_2$

Substitute $0$ for $v_2$ in the above equation.

  $v_1=0$

The value of the current $i_n$ is given by,

  $i_n=0$

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

  $A_v=\frac{v_O}{v_I}$

Apply KCL at the negative terminal.

  $i_1=\frac{v_I-v_1}{20\text{k}\Omega }$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{v_I-0}{20\text{k}\Omega }\\ =\frac{v_I}{20\text{k}\Omega }\end{array}$

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

  $i_2=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $0$ for $v_1$ , $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{20\text{k}\Omega }=\frac{0-v_O}{200\text{k}\Omega }\\ \frac{v_I}{20\text{k}\Omega }=\frac{-v_O}{200\text{k}\Omega }\\ \frac{v_O}{v_I}=-10\end{array}$

Substitute $\frac{v_O}{v_I}$ for $A_v$ in the above equation.

  $A_v=-10$

The expression for the value of the input resistance is given by,

  $R_i=\frac{v_I}{i_1}$

Substitute $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{v_I}{\frac{v_I}{20\text{k}\Omega }}\\ =20\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

To determine

The voltage gain $A_v$ and the input resistance $R_i$ of each of the circuit.

Answer to Problem 9.6P

The value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

Explanation of Solution

Calculation:

Mark the values and redraw the circuit of part (b).

The required diagram is shown in Figure 3

  

The from the above circuit the expression for the value of the voltage $v_2$ is given by,

  $v_1=v_2$

Substitute $0$ for $v_2$ in the above equation.

  $v_1=0$

The value of the current $i_n$ is given by,

  $i_n=0$

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

  $A_v=\frac{v_O}{v_I}$

Apply KCL at the negative terminal.

  $i_1=\frac{v_I-v_1}{20\text{k}\Omega }$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{v_I-0}{20\text{k}\Omega }\\ =\frac{v_I}{20\text{k}\Omega }\end{array}$

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

  $i_2=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $0$ for $v_1$ , $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{20\text{k}\Omega }=\frac{0-v_O}{200\text{k}\Omega }\\ \frac{v_I}{20\text{k}\Omega }=\frac{-v_O}{200\text{k}\Omega }\\ \frac{v_O}{v_I}=-10\end{array}$

Substitute $\frac{v_O}{v_I}$ for $A_v$ in the above equation.

  $A_v=-10$

The expression for the value of the input resistance is given by,

  $R_i=\frac{v_I}{i_1}$

Substitute $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{v_I}{\frac{v_I}{20\text{k}\Omega }}\\ =20\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

To determine

The voltage gain $A_v$ and the input resistance $R_i$ of each of the circuit.

Answer to Problem 9.6P

The value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

Explanation of Solution

Calculation:

Mark the values and redraw the circuit of part (c).

The required diagram is shown in Figure 4

  

The from the above circuit the expression for the value of the voltage $v_2$ is given by,

  $v_1=v_2$

Substitute $0$ for $v_2$ in the above equation.

  $v_1=0$

The value of the current $i_n$ is given by,

  $i_n=0$

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

  $A_v=\frac{v_O}{v_I}$

Apply KCL at the negative terminal.

  $i_1=\frac{v_I-v_1}{20\text{k}\Omega }$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{v_I-0}{20\text{k}\Omega }\\ =\frac{v_I}{20\text{k}\Omega }\end{array}$

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

  $i_2=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $0$ for $v_1$ , $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{20\text{k}\Omega }=\frac{0-v_O}{200\text{k}\Omega }\\ \frac{v_I}{20\text{k}\Omega }=\frac{-v_O}{200\text{k}\Omega }\\ \frac{v_O}{v_I}=-10\end{array}$

Substitute $\frac{v_O}{v_I}$ for $A_v$ in the above equation.

  $A_v=-10$

The expression for the value of the input resistance is given by,

  $R_i=\frac{v_I}{i_1}$

Substitute $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{v_I}{\frac{v_I}{20\text{k}\Omega }}\\ =20\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

To determine

The voltage gain $A_v$ and the input resistance $R_i$ of each of the circuit.

Answer to Problem 9.6P

The value of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .

Explanation of Solution

Calculation:

Mark the values and redraw the circuit of part (d).

The required diagram is shown in Figure 5

  

The from the above circuit the expression for the value of the voltage $v_2$ is given by,

  $v_1=v_2$

Substitute $0$ for $v_2$ in the above equation.

  $v_1=0$

The value of the current $i_n$ is given by,

  $i_n=0$

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

  $A_v=\frac{v_O}{v_I}$

Apply KCL at the negative terminal.

  $i_1=\frac{v_I-v_1}{20\text{k}\Omega }$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{v_I-0}{20\text{k}\Omega }\\ =\frac{v_I}{20\text{k}\Omega }\end{array}$

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

  $i_1=i_2+i_3$

Apply KVL at the inverting terminal.

  $i_2+i_3=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $i_1$ for $i_2+i_3$ in the above equation.

  $i_1=\frac{v_1-v_O}{200\text{k}\Omega }$

Substitute $0$ for $v_1$ , $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{20\text{k}\Omega }=\frac{0-v_O}{200\text{k}\Omega }\\ \frac{v_I}{20\text{k}\Omega }=\frac{-v_O}{200\text{k}\Omega }\\ \frac{v_O}{v_I}=-10\end{array}$

Substitute $\frac{v_O}{v_I}$ for $A_v$ in the above equation.

  $A_v=-10$

The expression for the value of the input resistance is given by,

  $R_i=\frac{v_I}{i_1}$

Substitute $\frac{v_I}{20\text{k}\Omega }$ for $i_1$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{v_I}{\frac{v_I}{20\text{k}\Omega }}\\ =20\text{k}\Omega \end{array}$

Conclusion:

Therefore, thevalue of the voltage gain is $-10$ and the value of the resistance is $20\text{k}\Omega$ .