Chapter 5, Problem 34P

Given the op amp circuit shown in Fig. 5.72, express v_o in terms of v₁ and v₂.

Figure 5.72

For Prob. 5.34.

To determine

State the expression for the voltage $v_o$ in terms of $v_1$ and $v_2$.

Answer to Problem 34P

The expression for the voltage $v_o$ is $\underset{\_}{\frac{R_3+R_4}{R_3\left(R_1+R_2\right)}\left(v_1{R}_2+v_2{R}_1\right)}$.

Explanation of Solution

Given data:

Refer Figure 5.72 in the textbook for the op amp circuit.

Calculation:

Consider that the given circuit is a non-inverting amplifier.

Modify to the Figure 5.72 by indicating the node voltage $v_a$. The modified figure as shown in Figure 1.

In Figure 1, write the expression by the use of Kirchhoff's current law.

$\frac{v_{in}-v_1}{R_1}+\frac{v_{in}-v_2}{R_2}=0$

Re-arrange the equation.

$\begin{array}{l}-\frac{v_1-v_{in}}{R_1}-\frac{v_2-v_{in}}{R_2}=0\\ -\left(\frac{v_1-v_{in}}{R_1}+\frac{v_2-v_{in}}{R_2}\right)=0\end{array}$

$\frac{v_1-v_{in}}{R_1}+\frac{v_2-v_{in}}{R_2}=0$ (1)

Consider the expression for the voltage $v_a$ by using virtual ground concept of op amp.

$v_a=v_{in}$

Substitute $v_a$ for $v_{in}$ in equation (1).

$\frac{v_1-v_a}{R_1}+\frac{v_2-v_a}{R_2}=0$ (2)

Consider the expression for voltage $v_a$ by using voltage divider rule.

$v_a=\frac{R_3}{R_3+R_4}{v}_o$ (3)

Re-arrange equation (2).

$\frac{v_1-v_a}{R_1}+\frac{v_2-v_a}{R_2}=0$

Multiply $R_1$ on both sides of the equation.

$\begin{array}{l}{R}_1\left(\frac{v_1-v_a}{R_1}+\frac{v_2-v_a}{R_2}\right)=R_1\times 0\\ v_1-v_a+\frac{R_1}{R_2}\left(v_2-v_a\right)=0\\ v_1-v_a+\frac{R_1}{R_2}{v}_2-\frac{R_1}{R_2}{v}_a=0\end{array}$

$v_1+\frac{R_1}{R_2}{v}_2=v_a\left(1+\frac{R_1}{R_2}\right)$ (4)

Substitute equation (3) in (4).

$\begin{array}{l}{v}_1+\frac{R_1}{R_2}{v}_2=\left(\frac{R_3}{R_3+R_4}{v}_o\right)\left(1+\frac{R_1}{R_2}\right)\\ v_o=\frac{R_3+R_4}{R_3\left(1+\frac{R_1}{R_2}\right)}\left(v_1+\frac{R_1}{R_2}{v}_2\right)\\ v_o=\frac{R_3+R_4}{R_3\left(\frac{R_1+R_2}{R_2}\right)}\left(\frac{v_1{R}_2+v_2{R}_1}{R_2}\right)\\ v_o=\frac{R_3+R_4}{R_3\left(R_1+R_2\right)}\left(v_1{R}_2+v_2{R}_1\right)\end{array}$

Conclusion:

Thus, the expression for the voltage $v_o$ is $\underset{\_}{\frac{R_3+R_4}{R_3\left(R_1+R_2\right)}\left(v_1{R}_2+v_2{R}_1\right)}$.