Chapter 9, Problem 9.51P

(a) Consider the ideal op-amp circuit shown in Figure P9.51. Determine thevoltage gains $A{v}_1=v_{O1}/v_I$ , and $A{v}_2=v_{O2}/v_I$ . What is the relationshipbetween $v_{O1}$ and $v_{O2}$ ? (b)For $R_2=60k\Omega ,R_1=20k\Omega$ , and $R=50k\Omega$ ,determine $v_{O1}$ and $v_{O2}$ for $v_1=-0.50V$ . (c) Determine $\left(v_{O1}-v_{O2}\right)$ for $v_I=+0.8V$ .

To determine

The expression for the voltage gain $A_{v1}$ and $A_{v2}$ , also determine the relationship between the $v_{O1}$ and $v_{O2}$ .

Answer to Problem 9.51P

The relation between the voltage $v_{O1}$ and $v_{O2}$ is $v_{O2}=-v_{O1}$ , the value of the gain $A_{v1}$ is $\left[1+\frac{R_2}{R_1}\right]$ and the value of the gain $A_{v2}$ is $-\left[1+\frac{R_2}{R_1}\right]$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Mark the voltages in the above circuit.

The required diagram is shown in Figure 2

  

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

  $A_{v1}=\frac{v_{O1}}{v_I}$

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

  $A_{v2}=\frac{v_{O2}}{v_I}$

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

  $v_2=v_I$

The expression for the value of the voltage $v_3$ is given by,

  $v_3=v_{O1}$

The expression for the voltage $v_6$ is given by,

  $v_6=0\text{V}$

The expression for the voltage $v_1$ is given by,

  $v_1=v_2$

The expression for the voltage $v_4$ is given by,

  $v_4=v_3$

Substitute $v_{O1}$ for $v_3$ in the above equation.

  $v_3=v_{O1}$

The expression for the value of the voltage $v_5$ is given by,

  $v_5=v_6$

Substitute $0\text{V}$ for $v_6$ in the above equation.

  $v_5=0\text{V}$

Apply KCL at node $v_1$ .

  $\frac{v_1}{R_1}+\frac{v_1-v_4}{R_2}=0$

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

  $\begin{array}{l}\frac{v_I}{R_1}+\frac{v_I-v_4}{R_2}=0\\ v_I\left[\frac{1}{R_1}+\frac{1}{R_2}\right]=\frac{v_4}{R_2}\\ v_4=v_I\left[1+\frac{R_2}{R_1}\right]\end{array}$

Substitute $v_{O1}$ for $v_4$ in the above equation.

  $\begin{array}{l}{v}_{O1}=v_I\left[1+\frac{R_2}{R_1}\right]\\ \frac{v_{O1}}{v_I}=\left[1+\frac{R_2}{R_1}\right]\end{array}$

Substitute $A_{v1}$ for $\frac{v_{O1}}{v_I}$ in the above equation.

  $A_{v1}=\left[1+\frac{R_2}{R_1}\right]$ ........... (1)

Apply KVL at node $v_5$ .

  $\begin{array}{l}\frac{v_5-v_4}{R}+\frac{v_5-v_{O2}}{R}=0\\ 2v_5-v_4-v_{O2}=0\\ v_{O2}=2v_5-v_4\end{array}$

Substitute $0$ for $v_5$ in the above equation.

  $\begin{array}{c}{v}_{O2}=2\left(0\right)-v_4\\ =-v_4\end{array}$

Substitute $v_I\left[1+\frac{R_2}{R_1}\right]$ for $v_4$ in the above equation.

  $\begin{array}{l}{v}_{O2}=-v_I\left[1+\frac{R_2}{R_1}\right]\\ \frac{v_{O2}}{v_I}=-\left[1+\frac{R_2}{R_1}\right]\end{array}$

Substitute $A_{v2}$ for $\frac{v_{O2}}{v_I}$ in the above equation.

  $A_{v2}=-\left[1+\frac{R_2}{R_1}\right]$

From above and from equation (1) the relation between the voltage $v_{O1}$ and $v_{O2}$ is given by,

  $v_{O2}=-v_{O1}$

Conclusion:

Therefore, the relation between the voltage $v_{O1}$ and $v_{O2}$ is $v_{O2}=-v_{O1}$ , the value of the gain $A_{v1}$ is $\left[1+\frac{R_2}{R_1}\right]$ and the value of the gain $A_{v2}$ is $-\left[1+\frac{R_2}{R_1}\right]$ .

To determine

The value of the voltage $v_{O1}$ and $v_{O2}$ .

Answer to Problem 9.51P

The value of the output voltage $v_{O1}$ is $-2\text{V}$ and $v_{O2}$ is $2\text{V}$ .

Explanation of Solution

Calculation:

The expression for the value of voltage $v_{O1}$ is given by,

  $v_{O1}=v_I\left(1+\frac{R_2}{R_1}\right)$

Substitute $-0.5\text{V}$ for $v_I$ , $60\text{k}\Omega$ for $R_2$ and $20\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{v}_{O1}=\left(-0.5\text{V}\right)\left(1+\frac{60\text{k}\Omega }{20\text{k}\Omega }\right)\\ =-2\text{V}\end{array}$

The expression for the value of the voltage $v_{O2}$ is given by,

  $v_{O2}=-v_{O1}$

Substitute $-2\text{V}$ for $v_{O1}$ in the above equation.

  $v_{O2}=2\text{V}$

Conclusion:

Therefore, the value of the output voltage $v_{O1}$ is $-2\text{V}$ and $v_{O2}$ is $2\text{V}$ .

To determine

The value of the difference $v_{O1}-v_{O2}$ .

Answer to Problem 9.51P

The value of the difference of voltage $v_{O1}-v_{O2}$ is $6.4\text{V}$ .

Explanation of Solution

Calculation:

The expression for the difference $v_{O1}-v_{O2}$ is given by,

  $\begin{array}{l}{v}_{O1}-v_{O2}=v_I\left[1+\frac{R_2}{R_1}\right]-\left(-v_I\left[\frac{R_2}{R_1}\right]\right)\\ v_{O1}-v_{O2}=2v_I\left[1+\frac{R_2}{R_1}\right]\end{array}$

Substitute $0.8\text{V}$ for $v_I$ , $60\text{k}\Omega$ for $R_2$ and $20\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{v}_{O1}-v_{O2}=2\left(0.8\text{V}\right)\left[1+\frac{60\text{k}\Omega }{20\text{k}\Omega }\right]\\ =6.4\text{V}\end{array}$

Conclusion:

Therefore, the value of the difference of voltage $v_{O1}-v_{O2}$ is $6.4\text{V}$ .