Sample Problem: Energy Storing Elements and Op Amps 2

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Do curso por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 classificações

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

97 classificações

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Na lição

Module 5

This module introduces capacitors and inductors, which are called "reactive" circuit elements.

Sample Problem: Equivalent Capacitance3:09

Sample Problem: Energy Storing Elements, Inductor3:11

Sample Problem: Energy Storing Elements, Capacitor1:05

Sample Problem: Energy Storing Elements 17:32

Sample Problem: Energy Storing Elements 23:47

Sample Problem: Energy Storing Elements 33:45

Sample Problem: Energy Storing Element and Op Amps 16:36

Sample Problem: Energy Storing Elements and Op Amps 24:53

Sample Problem: Energy Storing Elements and Op Amps 35:56

Sample Problem: Energy Storing Elements and Op Amps 45:02

Conheça os instrutores

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Energy Storage Elements & Operational Amplifiers.

The problem is to find V out in terms of V in for the circuit shown below.

We have a circuit that has an opamp in it, it has a resistor and a capacitor in it,

has a input voltage, and it has an output voltage at the output of the opamp.

What we want to do is we want to find this output voltage in terms

of this input voltage.

And we do that by knowing the properties of an ideal opamp and

also by using our prior knowledge of nodal analysis to solve the problem.

So we'll start with the circuit symbol for the ideal opamp.

The ideal opamp has two inputs,

it has an inverting input and it has a non-inverting input.

We know that these input currents for an ideal opamp are both equal to zero.

So that's one of the properties that we use of an ideal opamp.

The other one is that the voltage is at this inputs

are always equal to one another.

So V minus is always equal to V plus.

So we use these two properties with our circuit and our prior knowledge of nodel

analysis to find the output voltage in terms of the input voltage.

So how are we going to do that?

First we thing we recognize is that the voltage at the non-inverting input of

the opamp is zero faults, because it's time to ground.

We also know through our properties of ideal opamps that the voltage at

the non-inverting input end is equal to the voltage of the inverting input.

So this voltage at the upper node at the inverting input is also zero volts.

Now we can use Kirchhoff's current law and

do nodal analysis on this upper node, let’s call it node 1.

And we can sum the currents into that node or

we can choose to sum the currents out, it's our choice.

So if we do this, first of all looking at the current that

flows through the capacitor, we use our IV characteristic for the capacitor.

That is the current of the capacitor is equal to C dv dt.

And our voltage is our difference between v

sub n and the lowest voltage which is zero volts.

So it's V sub n- 0, dt.

And we also sum more came from the resistor and through the resistor.

It's going to be V out, which is the voltage at the upper node,

minus zero divided by R And

we have one other input to this node and

it's the current coming out of the inverting input of the opamp.

We know using our properties of our ideal opamp that that current is equal to zero.

We'll include it for completeness.

And the sum of those is equal to zero through Kirchhoff's Current Law.

So now if we simply this

a little bit more we have

C d Vn/dt + Vo/R =0.

So if we solve for V oout of this equation,

it's a v out of t, it's equal to what?

-RC Times derivative of V sub n/dt.

And so if you look at this relationship between our output voltage and

our input voltage, the output voltage is the time derivative of

the input voltage times this multiplication factor RC,

and it's inverted as well, so negative sign out front.

So in this case this circuit that we have,

that we've been analyzing is called a differentiator.

It takes an input voltage.

It takes a time derivative of it, multiplies it by

a amplification factor to get the output voltage V out of t.

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