Chapter 3: Q. 3.14 (page 67)

Design a circuit to perform the following calculation:
$- V_{o} = \frac{1}{10} (5 V_{1} + 3 V_{2})$

Short Answer

Expert verified

The design of the circuit can perform the calculation $- V_{0} = \frac{1}{10} (5 V_{1} + 3 V_{2})$

Step by step solution

Step 1. Introduction

The operational amplifiers are also used as constant-current and constant-voltage sources. Operational amplifiers are used to perform mathematical operations in the electrical circuits. These operations include integration, multiplication, differentiation, and summation for precise measurement of voltage, current, and resistance.

The expression for the desired output is:

$\begin{array}{l} - V_{0} = \frac{1}{10} (5 V_{1} + 3 V_{2}) \\ V_{0} = - \frac{1}{10} (5 V_{1} + 3 V_{2}) \end{array}$ …… (I)

Here, the voltage at point 1 is $V_{1}$ and the voltage at point 2 is $V_{2}$ .

As the expression contains some mathematical operations, the circuit is designed with the use of an operational amplifier.

Step 2. Explanation

The figure represents the design of the circuit for the desired output.

The expression for the output voltage is;

$V_{0} = - R_{f} (\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}})$ …… (II)

Here, the resistance at branch is , the resistance at branch is , and is a sufficient resistance.

Substitute $6000 Ω$ for $R_{1}$ , $10000 Ω$ for $R_{2}$ , and $3000 Ω$ for $R_{f}$ in Equation (II).

$\begin{matrix} V_{0} = - 3000 Ω (\frac{V_{1}}{6000 Ω} + \frac{V_{2}}{10000 Ω}) \\ = - 3000 Ω (\frac{V_{1}}{6000 Ω} + \frac{V_{2}}{10000 Ω}) \times \frac{10}{10} \\ = - \frac{1}{10} (0.5 Ω \times 10 V_{1} + 0.3 Ω \times 10 V_{2}) \\ = - \frac{1}{10} (5 Ω V_{1} + 3 Ω V_{2}) \end{matrix}$