Chapter 5, Problem 24P

In the circuit shown in Fig. 5.62, find k in the voltage transfer function v_o = kv_s.

Figure 5.62

For Prob. 5.24.

To determine

Derive the expression for k in the voltage transfer function $v_o=k{v}_s$.

Answer to Problem 24P

The expression for k is $\underset{\_}{R_f\left[\left(\frac{R_3}{R_3+R_4}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{1}{R_2}\right]}$.

Explanation of Solution

Given data:

Refer to Figure 5.62 in the textbook for the given op amp circuit.

Calculation:

Modify the Figure 5.62 by indicating the node voltages $v_1$ and $v_2$. The modified figure is shown in Figure 1.

From the properties of ideal op amp, consider that the voltage across the two input terminals of op amp is equal to zero.

$v_1=v_2$ (1)

Apply Kirchhoff's current law to the node $v_1$ in Figure 1.

$\begin{array}{l}\frac{v_1-0}{R_1}+\frac{v_1-v_s}{R_2}+\frac{v_1-v_o}{R_f}=0\\ \frac{v_1-0}{R_1}+\frac{v_1-v_s}{R_2}+\frac{v_1-v_o}{R_f}=0\\ \frac{v_1}{R_1}+\frac{v_1}{R_2}-\frac{v_s}{R_2}+\frac{v_1}{R_f}-\frac{v_o}{R_f}=0\end{array}$

$v_1\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{v_s}{R_2}=\frac{v_o}{R_f}$ (2)

Apply Kirchhoff's current law to the node $v_2$ in Figure 1.

$\begin{array}{l}\frac{v_2-0}{R_3}+\frac{v_2-v_s}{R_4}=0\\ \frac{v_2}{R_3}+\frac{v_2}{R_4}-\frac{v_s}{R_4}=0\\ \frac{v_2{R}_4+v_2{R}_3}{R_3{R}_4}=\frac{v_s}{R_4}\\ v_2\left(\frac{R_3+R_4}{R_3{R}_4}\right)R_4=v_s\end{array}$

Simplify the equation as follows.

$v_2=\frac{R_3}{R_3+R_4}{v}_s$ (3)

Substitute equation (1) in (3).

$v_1=\frac{R_3}{R_3+R_4}{v}_s$ (4)

Substitute equation (4) in (2).

$\begin{array}{l}\left(\frac{R_3}{R_3+R_4}{v}_s\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{v_s}{R_2}=\frac{v_o}{R_f}\\ v_o=R_f\left(\left(\frac{R_3}{R_3+R_4}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{1}{R_2}\right)v_s\end{array}$

$v_o=R_f\left[\left(\frac{R_3}{R_3+R_4}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{1}{R_2}\right]v_s$ (5)

Consider the expression for the voltage transfer function.

$v_o=k{v}_s$ (6)

Compare equations (5) and (6).

$k=R_f\left[\left(\frac{R_3}{R_3+R_4}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{1}{R_2}\right]$

Conclusion:

Thus, the expression for k is $\underset{\_}{R_f\left[\left(\frac{R_3}{R_3+R_4}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_f}\right)-\frac{1}{R_2}\right]}$.