Chapter 5, Problem 50P

Design a circuit to amplify the difference between two inputs by 2.5.

1. (a) Use only one op amp.
2. (b) Use two op amps.

To determine

Design an amplifier circuit that contains only one op amp to amplify the required difference between two inputs.

Explanation of Solution

Calculation:

The difference amplifier is drawn and it is shown in Figure 1.

Write the expression for the output voltage $v_o$ of the difference amplifier in Figure 1.

$v_o=\frac{R_2}{R_1}\left(v_2-v_1\right)$ (1)

Consider that the required value of difference is 2.5.

Consider the expression for the output voltage of the difference amplifier with the difference of 2.5 between two inputs.

$v_o=2.5\left(v_2-v_1\right)$ (2)

Compare Equations (1) and (2).

$\frac{R_2}{R_1}=2.5$

$R_2=2.5R_1$ (3)

Choose the value of resistor $R_1$ as $100\text{k}\Omega$.

Substitute $100\text{k}\Omega$ for $R_1$ in Equation (3).

$\begin{array}{c}{R}_2=2.5\left(100\text{k}\Omega \right)\\ =250\text{k}\Omega \end{array}$

Conclusion:

Thus, the required op amp circuit is designed and obtained values are $100\text{k}\Omega$ and $250\text{k}\Omega$ for $R_1$ and $R_2$ respectively.

To determine

Design an amplifier circuit that contains two op amps to amplify the required difference between two inputs.

Explanation of Solution

Calculation:

Consider that the required value of difference is 2.5. To design the required circuit by cascading the inverting amplifier with summer amplifier.

The required difference amplifier with two op amps is drawn and it is shown in Figure 2.

The output voltage of an inverting amplifier is,

$\begin{array}{c}{v}_{o1}=-\frac{R}{R}\left(v_1\right)\\ =-v_1\end{array}$

Write the expression for the output voltage $v_o$ of the summer amplifier in Figure 1.

$v_o=-\frac{R}{\frac{R}{2.5}}\left(-v_1\right)-\frac{R}{\frac{R}{2.5}}\left(v_2\right)$ (4)

Modify Equation (4).

$\begin{array}{c}{v}_o=-\frac{2.5R}{R}\left(-v_1\right)-\frac{2.5R}{R}\left(v_2\right)\\ =\frac{2.5R}{R}\left(v_1\right)-2.5v_2\\ =2.5\left(v_1-v_2\right)\end{array}$

Hence, Equation (4) satisfies the Equation (2).

Re-arrange Equation (4).

$v_o=-\left(\frac{R}{\frac{R}{2.5}}\left(-v_1\right)+\frac{R}{\frac{R}{2.5}}\left(v_2\right)\right)$

$v_o=-\frac{R_f}{R_1}\left(-v_1\right)-\frac{R_f}{R_2}\left(v_2\right)$

Here,

$R_f=R$ (5)

$R_1=\frac{R}{2.5}$ (6)

And

$R_2=\frac{R}{2.5}$.                                                                                      (7)

Choose the value of resistor R.

$R=100\text{k}\Omega$.

Substitute $100\text{k}\Omega$ for R in Equation (5).

$R_f=100\text{k}\Omega$

Substitute $100\text{k}\Omega$ for R in Equation (6).

$\begin{array}{c}{R}_1=\frac{100\text{k}\Omega }{2.5}\\ =40\text{k}\Omega \end{array}$

Substitute $100\text{k}\Omega$ for R in Equation (7).

$\begin{array}{c}{R}_2=\frac{100\text{k}\Omega }{2.5}\\ =40\text{k}\Omega \end{array}$

Therefore, the value of resistors $R_f=100\text{k}\Omega$, $R_1=40\text{k}\Omega$, and $R_2=40\text{k}\Omega$.

Conclusion:

Thus, the required op amp circuit is designed and obtained values are $100\text{k}\Omega$, $40\text{k}\Omega$ and $40\text{k}\Omega$ for R, $R_1$ and $R_2$ respectively.