Sample Problem: DC Op Amp 5

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

192 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

192 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 amplifier circuit.

And the problem is to determine V out and I out in the circuit shown below.

The circuit has an op amp, it has two voltage sources of 12 volt and

a 2 volt source, and three resistors associated with it.

What we want to do is we want to use the properties of an ideal op amp in order to

help us solve this circuit.

So if we look at an ideal op amp, first of all, the symbol for

the ideal op-amp has two inputs, it has an inverting input and a non-inverting input.

And each one of those inputs have a current associated with it.

So we have a current for the non-inverting input and the current for

the inverting input.

We also have voltage associated with each one of these inputs as well,

for the inverted and non-inverting inputs.

For an ideal operational amplifier, we know that the current

going into the inverting input is equal to the current

going into the non-inverting input and that both of them are equal to zero.

We also know that the voltage at the inverting input

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

So there are other properties of the ideal op amp that are of interest but for

this class, these two most important properties of the ideal op amp.

And these are the ones we need in order to solve our problem.

So if we go back to our problem here and we use these properties, it'll allow us to

work the problem knowing voltage is at certain points in the circuit.

For instance, we know what the voltage is at this point.

It's -2 volts at the non-inverting input ff the op amp which

means it's also -2 volts at the inverting input of the op amp,

which means that it's -2 volts at this node between the 4 kilo ohm and

the 2 kilo ohm, 4 kilo ohm and the 8 kilo ohm resistor.

So having known that, we can use that in order to solve the problem.

The way we do that is we use Kirchhoff's current laws and we'll sum the currents

into this top node that we've identified as a node at a -2 volts.

And if we do that, then we will come up with an equation which

gives us V out is the unknown in the equation.

So let's do that.

Well, the Kirchhoff's Current Law, I'm going to call this node one.

So we're going to do Kirchhoff's Current Law node one and

we're going to sum the currents into node one.

So we're going to sum the current through the four kilo ohm resistor into that node

through the 8 kilo ohm resistor into that node and

also this current which we know is zero flowing out of the op amp into that node.

So let's do that.

Starting with the current through the 4 kilo ohm resistor,

it's going to be 12 volts- (-2 volts).

Divided by 4 kilo ohms.

That gives us the current through the 4 kilo ohm resistor.

I'm going to add to that the current through the 8 kilo ohm resistor.

It's going to be the voltage at this point which is V out minus a -2 volts

Divided by 8 kilo ohms and of course we have the third current which you

know is zero coming out of the inverting input of the op amp.

I'm going to put it in there for

completeness and the some of those currents is equal to zero.

So now we have an equation that has V out is the only unknown in the equation.

If we solve for v out, We end

up with a v out which is equal to -30 volts.

So now that's one of our variables that we want to determine in this problem.

The other one is I out,

we want to find the current through the a kilo ohm resistor and

we know what that's going to be because we're already investigated that.

But as it's described and defined in this problem I out is in the opposite

direction of the current that we were summing into node one.

So I out, if we're going to find that current, we know I out,

it's going to be equal to the voltage at this point, node one,

which is -2 volts minus this voltage which is V out.

So it's -2- V out / 8 kilo ohms.

And that will give us I out.

We know what V out is, it's -30 volts.

And so we can determine I out.

So I out Is

equal to 28 divided by 8k,

which gives us 3.5 milliamps.

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