Sample Problem: Op Amps 9

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

94 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

94 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 Circuits.

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, a 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

a current for the inverting input.

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

for the inverting and non-inverting inputs.

If we want 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 0.

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 are the two most important properties of the ideal op amp.

And those 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 voltages 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 of 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 kiloohm and

the 2 kiloohm, 4 kiloohm and the 8 kiloohm 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 -2 volts.

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

which gives us V out as the unknown in the equation.

So let's do that, we're going to do Kirchhoff's current law, and

we're going to call this node 1.

So we're going to do Kirchhoff's current law at node 1, and

we're going to sum the currents into node 1.

So we're going to sum the current through the 4 kiloohm resistor into that node,

through the 8 kiloohm resistor into that node, and also this current,

which we know is 0, flowing out of the op amp into that node.

So let's do that.

Starting with the current through the 4 kiloohm resistor,

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

/4 kiloohms, that gives us the current through the 4 kiloohm resistor.

We're going to add to that the current through the 8 killoohm resistor.

It's going to be the voltage at this point, which is V out,- (-2) volts

/8 kiloohms, and of course, we have the third current,

which we know is 0 coming out of the inverting input of the op amp.

I'm going to put it in there for completeness, and

the sum of those currents is equal to 0.

So we now we have an equation that has V out as 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 is I out, we want to find the current through the 8 kiloohm resistor.

And we know what that's going to be because we've 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 1.

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 1,

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

So it's -2- V out / 8 kiloohms.

And that will give us I out.

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

And so we can determine I out.

So I out, Is

equal to 28/8k,

which gives us 3.5 mA.

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