Sample Problem: Op Amps 8

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

125 calificaciones

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

125 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 circuits shown below.

It's an op amp circuit with a current source and

a voltage source in it, as well as three resistors.

Vout is the voltage measured across the load resistance which is

the 6 kilo Ohm resistor.

And Iout is the current which is flowing back into the op Amp

from the output side of the circuit.

So what we want to do is we want to use the ideal properties of op Amps along with

our tools such as nodal-analysis and mesh-analysis to solve the problem.

So the first thing we want to know is the properties or

are the properties of ideal op-amps.

First of all we have the symbol for the ideal op-amp.

It has an inverting and a non-inverting input.

Each one of those has a current associated with it.

And each one of them has a voltage associated with it as well.

So there's a voltage at the inverting input and

a voltage at the non-inverting input.

And for an ideal op-amp, we know that the currents,

both at the inverting input and the non-inverting input are equal to zero.

We also know that the voltages at the two inputs

are equal to each other, so V- = V+.

We use these two properties which are not the only two properties by op amps but

the two of the most important to this class for

solving problems and that are associate with linear circuits.

We'll use these two properties to solve our original circuit.

So let's do that.

The first thing we want to know is the voltage at this node at the bottom of our

circuit, which is the ground node, the voltage is equal to zero.

Which also tells us the voltage at the non inverting input is equal to zero.

Using the properties of the ideal op amp we know the voltage at

the inverting input is also zero which makes it it a zero volt

at this node at the inverting input of the op amp.

So, now we've used that first, or second property of the ideal op-amp, V- = V+.

Now we're going to use nodal analysis, and we're going to sum the currents

into this node which we've identified at zero volt potential.

And if we do that we'll get something that involves Vout is the unknown

variable in the equation along with other known voltages in the circuit.

So we're going to do Kirchhoff's current law at node one and with the sum of

voltage into, or sorry, we're going to sum the currents into, node 1.

So if we start with the four kilo Ohm resistor and

we want to sum that current it's going to be 12 volts- 0

volts / 4 kilo Ohms for the current

through the 4 kilo Ohm resistor flowing left to right into node 1.

We also have this 2 milliamps source which is flowing into node 1 as well.

So we're going to add the 2 milliamp source.

In addition to that, we have the current which is flowing

through the 3 kilo Ohm resistor back into node 1.

It's V out minus 0 over 3K.

And we have the current, which is flowing out of the inverting input,

which we know is zero, our first property of the op-amp, ideal op-amp.

want to add it in there for completeness, and there's our sum, so

this is Kirchhoff's Current Law at node one.

So we can see from this that we have an equation which has one unknown V out,

and we can solve for it.

If we solve for Vout, we end up with Vout = -15 volts.

Now, we want to find Iout.

And Iout is our current through our current back in to the Op Amp.

So we can use the node at the output, maybe we call this node 2.

And we can use Kirchhoff's Current Law at node 2 to find that current Iout.

So this is KCL at node 2 and again we're going to sum the currents into that node.

We could choose to sum them out of the nodes but for

consistency we're going to sum the currents into the nodes.

So starting with the current out through the 6 kilo Ohm resistor, we start with

the voltage on one side minus the voltage on the other, which is 0 volts- Vout.

We know Vout is -15 volts divided by 6 kilo Ohms.

So it's 0- a minus 15 for

Vout / 6 kilo Ohms.

That's the current flowing up through the 6 kilo Ohm resistor.

We also have the current flowing left to right through the 3k resistor,

which is 0 volts- Vout, Vout- 15 volts / 3K and

the other one we have is is we have Iout, and

in fact it should be a -Iout since we're assuming the currents into it.

We should have a -Io instead of a +1o for that current.

And those are all the currents into node two and the sum is equal to zero.

So as you can see we end up with equation which we only has one unknown,

which is Io and Io comes out to 7.5 milliamps.

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