Chapter 5, Problem 15P

(a)    Determine the ratio v_o/i_s in the op amp circuit of Fig. 5.54.

(b)    Evaluate the ratio for R₁ = 20 kΩ, R₂ = 25 kΩ, R₃ = 40 kΩ.

Figure 5.54

For Prob. 5.15.

To determine

Calculate the ratio $\frac{v_o}{i_s}$ in the op amp circuit of Figure 5.52.

Answer to Problem 15P

The ratio $\frac{v_o}{i_s}$ is $\underset{\_}{-\left(R_1+R_3+\frac{R_1{R}_3}{R_2}\right)}$.

Explanation of Solution

Given data:

Refer Figure 5.54 in the textbook for the op amp circuit.

Calculation:

Modify the Figure 5.54 by indicating the node voltage $v_1$. The modified figure as shown in Figure 1.

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

$\begin{array}{l}\frac{v_1}{R_2}+\frac{v_1-v_o}{R_3}-i_s=0\\ i_s=\frac{v_1}{R_2}+\frac{v_1-v_o}{R_3}\\ i_s=\frac{v_1}{R_2}+\frac{v_1}{R_3}-\frac{v_o}{R_3}\end{array}$

$i_s=v_1\left(\frac{1}{R_2}+\frac{1}{R_3}\right)-\frac{v_o}{R_3}$ (1)

Consider the expression for the current $i_s$ since no current flows through the op amp.

$\begin{array}{l}{i}_s=\frac{0-v_1}{R_1}\\ i_s=\frac{-v_1}{R_1}\\ -v_1=i_s{R}_1\end{array}$

$v_1=-i_s{R}_1$ (2)

Substitute equation (2) in (1).

$\begin{array}{l}{i}_s=-i_s{R}_1\left(\frac{1}{R_2}+\frac{1}{R_3}\right)-\frac{v_o}{R_3}\\ i_s+i_s{R}_1\left(\frac{1}{R_2}+\frac{1}{R_3}\right)=-\frac{v_o}{R_3}\\ i_s+i_s\left(\frac{R_1}{R_2}+\frac{R_1}{R_3}\right)=-\frac{v_o}{R_3}\\ i_s\left(1+\frac{R_1}{R_2}+\frac{R_1}{R_3}\right)=-\frac{v_o}{R_3}\end{array}$

Simplify the equation as follows.

$\begin{array}{c}\frac{v_o}{i_s}=-R_3\left(1+\frac{R_1}{R_2}+\frac{R_1}{R_3}\right)\\ =-\left(R_3+\frac{R_1{R}_3}{R_2}+\frac{R_1{R}_3}{R_3}\right)\end{array}$

$\frac{v_o}{i_s}=-\left(R_1+R_3+\frac{R_1{R}_3}{R_2}\right)$ (3)

Conclusion:

Thus, the ratio $\frac{v_o}{i_s}$ is $\underset{\_}{-\left(R_1+R_3+\frac{R_1{R}_3}{R_2}\right)}$.

To determine

Find the value for the ratio $\frac{v_o}{i_s}$.

Answer to Problem 15P

The value for $\frac{v_o}{i_s}$ is $\underset{\_}{-92\text{kΩ}}$.

Explanation of Solution

Given data:

Consider that the resistors $R_1,R_2,\text{and}{R}_3$ are $20\text{kΩ, 25}\text{kΩ},\text{and}40\text{kΩ}$ respectively.

Calculation:

Substitute $20\text{kΩ}$ for $R_1$, $\text{25}\text{kΩ}$ for $R_2$, and $40\text{kΩ}$ for $R_3$ in equation (3).

$\begin{array}{c}\frac{v_o}{i_s}=-\left(20\text{kΩ}+40\text{kΩ}+\frac{\text{20}\text{kΩ}\times 40\text{kΩ}}{25\text{kΩ}}\right)\\ =-\left(60\text{kΩ}+\frac{\text{800}{\left(\text{kΩ}\right)}^2}{25\text{kΩ}}\right)\\ =-\left(60\text{kΩ}+32\text{kΩ}\right)\\ =-92\text{kΩ}\end{array}$

Conclusion:

Thus, the value for $\frac{v_o}{i_s}$ is $\underset{\_}{-92\text{kΩ}}$.