Sample Problem: Non-inverting Op Amp

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 calificaciones

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.

De la lección

Module 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

Conoce a los instructores

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 operational amplifiers, and

we're going to work with non-inverting amplifiers.

The problem is to find V out in terms of the input voltages for

the circuit shown below.

So we have a circuit that has an off amp in it, It has two resistors,

an R sub 1, which is tied to the inverting input of the op amp, and

also an R sub f, which is the feedback resistor that

feedback the voltage from the output to the input of the op amp.

We also have an input voltage.

In our case, our input voltage is V sub 1,

which is tied to the non-inverting input of our op-amp.

And so, what we're going to do is, we're going to find V out,

in terms of V sub one for this problem.

And so, we have to use a few properties of the op-amp,

in the words, solve this problem.

Whenever we're working op-amp problems, we always want to understand what

the properties of the op-amp are, and so, if we draw a general op-amp.

Symbol.

And we also, typically draw a ground into that op-amp as well.

We can draw our circuit, as well there's some current, i-,

which is the inverting input current, is the current i+,

which is the non-inverting input current.

There's also a voltage associated with the non-inverting input,

the plus, and with the inverting input V minus.

And what we know about op amps and ideal op amps, is that I minus,

so the current flowing into the inverting input of the op amp

is equal to the current that flows into the non-inverting

input of the op amp and that they're both equal to zero.

In other words, there's no current that flows into an ideal op amp.

That's not true about the output of the op amp,

it could have current flow in either direction,

depending on the application and what's inside of this operational amplifier.

The other thing that we always want to know and to work these problems with op

amp center, is want to understand that the voltage at the inverting

input is equal to the voltage at the non inverting input.

So we're going to use these two properties, and

not the only two properties of op amp center important but for, this class, and

for linear circuits, these are our two most important properties.

We're going to use those to solve our problem, and

we're also going to lean on some of the tools that we already have learned

from nodal analysis and mesh analysis to solve these op amp problems.

Let's take this one as an example.

So we have this op amp circuit that has a DC source tied to the non-inverting input.

And what this circuit's going to do ultimately is it's going to take this

input voltage and it's going to amplify it to some output voltage level.

The amplification can be greater than one or can be less than one or can be equal to

one depending on the op amp and also the elements that we've used in our circuit.

So let see how that is possible.

So, what we're going to do is, we're going to use our properties of the op amp.

And if we do that, then we know the voltage at the non

inverting input to the op amp it's equal to V sub 1.

And if we use our second property of op amps,

we also know that the voltage at the inverting input is

V sub one as well, so it's V sub one at both points.

So that helps us a lot, because now we can use some of the tools that we already have

in our toolbox to solve this problem.

And namely, we're going to use notable analysis at that node

of the inverting input that we just assigned the voltage of V sub 1,

so itsthis top node in our circuit.

So if we sum the currents into that node, so let's call it Node 1,

And we sum the voltages into that or sum the currents into that node.

So, we're going to use Kirchhoff's current law and nodal analysis.

Then our equation's going to be for, first of all,

the current flowing from ground through R1 into that node, is going to be zero,

for the voltage at the ground, minus V sub 1 divided by R1.

We also have the current flowing to the feedback path into that node as well.

It's going to be V0- V sub 1 divided by R sub f

And we also have this current, which is flowing out of the op amp directly

into that node, and we know that current is equal to zero.

I'm going to write down, because it is the third

branch into that node.

We're going to give it a value or 0 amps and

so that's equal to 0 through Kirchhoff's Current Law.

So what that gives, it gives us is an equation with V sub 1 in it,

which is our input voltage and V out at.

And in fact, once we re-write this equation, we can get V out in terms of V1.

So let's work on this.

First of all, we need a common denominator, and

then we can start adding things together.

So if we do that, We'd

end up with a V1 times 1 over R1 + 1 over R sub f.

V = to V out divided by R sub f.

So if we continue with this.

We find that V out is equal

to 1 + R sub f divided by R sub 1 times,

our input voltage, V sub 1.

So our output voltage is a function of the input voltage.

It's not negated so it's not inverting, it's not an inverted signal.

Otherwise, we'd have a negative sign out front.

And the value of the amplification factor can be controlled by

the values of our feedback resistor and the values of our R sub 1,

which is attached to the inverting terminal of our op amp.

So, this is the example of a non-inverting amplifier.

It's one of the basic op amp configurations.

Four op amp circuits.

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