Infinite Analog Calculator

An operational amplifier is an integrated circuit with wide applications in analog electronics. (Though it can also be used in digital circuits)

Ideally, its properties are:

  1. Infinite Input Impedance

  2. Zero Output Impedance

  3. Infinite Voltage Gain

One interesting application of the op amp is it can perform addition and subtraction.

Given the circuit, calculate V o V i \frac{-V_{o}}{V_{i}}

NOTE: The inputs of the opamp are the + and - sign, the output is connected to the vertex of the triangle.

infinite 1 cannot be determined 1/2 2

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Daniel Cabrales
Dec 28, 2015

The three properties of the ideal op amp will lead to these implications:

  1. Infinite Input Impedance - No current will flow into the input terminals

  2. Zero Output Impedance - The output will be unaffected by the load

  3. Infinite Voltage Gain - The voltages in the input terminals are equal

From the third implication, the current through each of the resistors in the input side are:

I i = V i V a R i I_{i} = \frac{V_{i} - V_{a}}{R_{i}}

V a = V b = 0 V V_{a} = V_{b} = 0 V

I i = V i 0 2 i R I_{i} = \frac{V_{i} - 0}{2^{i}R}

I i = V i 2 i R I_{i} = \frac{V_{i}}{2^{i}R} eq.(1)

By Kirchhoff's Current Law on b and by the first implication:

i = 0 I i I = 0 \sum_{i = 0}^\infty {I_{i}} - I' = 0

From eq.(1):

i = 0 V i 2 i R = I \sum_{i = 0}^\infty { \frac{V_{i}}{2^{i}R}} = I'

By infinite geometric series:

2 V i R = 0 V o R \frac{2V_{i}}{R} = \frac{0 - V_{o}}{R}

Hence,

V o V i = 2 \frac{-V_{o}}{V{i}} = 2

Just know a bit of electronics , you'll have an edge over others!!

Vignesh S - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...