Sample Problem: DC Op Amp 1

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En provenance du cours de Georgia Institute of Technology

Linear Circuits 1: DC Analysis

90 notes

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

90 notes

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.

À partir de la leçon

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

Rencontrer les enseignants

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 find v out in the circuit shown below,

it's a circuit with multiple resistors in it, one voltage source,

an independent 12 volt source Two op-amps.

And the output is measured across a load resistance which is 40 kilohms at

the output of the second op amp.

In order to solve this problem, we have to use the properties of an ideal op amp.

So we start with the symbol for an ideal op-amp.

It has an inverting input with a current associated with it,

i minus, and a voltage associated with it, v minus and

also has a non-inverting input with a current i plus and

a voltage v plus associated with it.

It has an output and a ground.

We know that the properties of the ideal op-amp

are that the currents at both of the inputs.

Are 0.

We also know that the voltages at the inverting and

the non-inverting input are equal.

Those are the two of properties of an ideal op-amp that we used

solve linear circuits.

So we're going to use those along with our knowledge of nodal analysis and mesh

analysis In order to solve this two op-amp circuit that we have in this problem.

So let's start with finding voltage at the input,

so if we can find the voltage at this non inverting input of the first op amp

then we know the volts is at the inverting input in the open,

and that will allow us to work further back into our circuit,

ultimately getting back to our voltage drop across to our 40 kilohm resistor.

So, a some are currents into this node,

into the V plus node, into our non-inverting input node.

Let's call this node 1.

So if we sum currents into node 1 then we have first of all

a current through the 40k ohm resistor which is minus 12 minus V plus

Divided by 40 kilohms.

We also have the current which is flowing up through the 20 kilohm resistor and

that current is going to be zero volts because

this is a ground point in our circuit the bottom of the circuit.

Zero volts minus v plus Over 20k.

And we have the current which is flowing out of the non-inverting input.

We know that through ideal op-amp put that current equals to 0.

And so the sum those using Kirchhoff's current law

That node one is equal to zero.

And so it's an equation which just has the one variable,

that is the voltage at the other non-inverting input of our first op-amp.

If we solve for that voltage the plus We get minus 4 volts.

So it's minus 4 volts at node 1 and it also

tells us that it's minus 4 volts at the inverting input as well.

So we need to work further back into our circuit.

So we rely on the first property of the op amp because a little bit further into

the circuit.

That is the current into each one of those inputs of the op amp

is always equals to zero.

If that is the case, then there is no current flowing through this ten kiloOhms

resistor because it is the current flowing into the.

Inverting input of our first op amp.

There's no current flowing through the ten column resistor

therefore there's no voltage drop across a ten column resistor.

So what that tells us is this voltage between the 10 and

20 kilom resistor at the top of our circuit is also a -4 volts at that point.

Now we can use that to work a little bit further into our circuit.

Let's look at our circuit a little bit closer.

If we look at V out, V out is measured across the 40 kilohm resistor.

It's a voltage at this output node of our second op amp.

We also, so there's no voltage drops between these two points along

the feedback path of our second map.

So, that also tells us that V out is the voltage at this point

in our circuit at the top of our circuit is equal to the voltage at

the inverting input of our second amp map.

We know that that voltage is the voltage at the noninverting

input of our second op amp as well using our second properties of an op amp.

So we have a V out which it can be measured,

at this point, in front of our second op amp.

So that helps us a little bit.

because what we can do is we can start summing voltages or

summing currents in this case into nodes and

come up with equations which allow us to solve for V out.

Let's see if we can do that.

Let's look at the top node which is at minus 4 volts first.

And let's call that node one, or sorry, so we've already added node one,

let's call this one node two.

So lets look at node two then we're going to use Kirchhoff's Current Law

at node two and we're going to sum the current into node two.

So it's a current, first of all,

through 10 kilohm resistor which we know is zero, it's the current through

the 20 kilohm resistor Flowing up through the 20k ohm resistor.

Let's call this node 3 down below.

So in this case the current up through the 20k ohm resistor is

going to be V sub 3- (-4v) divided

by 20 kiloms that's a current up through our 20 kilom resister.

We also have the current flowing through the 20 kilom resister at the top of

the circuit it's going to be v out minus a -4

divided by 20 kiloms for the current flowing into node 2 so

v out Minus a -4

volts divided by 20 kiloms.

V sub zero minus a -4 volts divided by 20 k.

And that's all the currents are flowing into node 2 because we know that

the current flowing through the 10 kilo amp is equal to zero when I add it

in there just for completeness and that is equal to zero.

So that's an equation which has two unknowns as V sub 3 is an unknown and

also has V out a unknown.

Let's look at this node where we have V out measured.

Let's call this node 4.

So let's sum the currents into node 4.

So all of these are summing currents into the nodes.

So if we send the current into node 4, we first of all have the current through

the 10 kilo of the 40 kilo resistor is going to be 0 volts minus v out over 40k.

We have the current flowing left or

right from node 3.

To node 4 through the 20k ohm resistor.

So that's going to be V3- V0

divided by 20k.

And we have the current which is flowing out of the noninverting input of

the second op amp and

we know that current is equal to zero through the properties of an ideal op amp.

Those are all the currents flowing into node four and the sums is equal to zero.

So now we have a second equation which is independent of our other equation that,

if we used these two equations simultaneously to solve for

v sub 3 and v out, we can do that.

And so, v out could be determined by finding these voltage levels.

So if we use these two equations to solve for V out simultaneously

then we end up with a V out which is equal to two thirds V sub F.

V sub 3 And that's equal to minus 16-fifths of a volt.

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