Sample Problem: Op Amps 6

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來自 Georgia Institute of Technology 的課程

Linear Circuits 1: DC Analysis

125 個評分

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

125 個評分

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.

從本節課中

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

與講師見面

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.

So we have a circuit that has one voltage source in it as a five volt source and

I out is measured as the current in the bottom

leg of our circuit in this configuration.

V out is the voltage measured at the output of our op amp, so

to solve this circuit we're going to use the properties of an ideal op amp.

First of all, we have the symbol for

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

Each one of those inputs has a voltage and a current associated with it.

We know that the ideal op amp has these properties.

First of all, the currents into the op amp are equal to zero.

There's no current flowing into the input of an ideal op amp.

And that the voltage is that the inputs are equal.

V minus is equal to V plus.

So, these are the properties that we're going to use to solve our circuit that

is given in this problem.

So, the first thing we're going to use is this first property of the circuit that

there's no current flowing into the ideal op amp.

So if there's no current flowing into this non inverting input of the op amp,

that means that I out is a current which is confined to this loop that we have

at the bottom of our circuit.

So if we want to find I out, we can use Kirchhoff's Voltage Law to do that and

we can use Kirchhoff's Voltage Law around that mesh.

And in this case I out is measured in a counter clock wise direction.

So we're going to sum our voltages up counter clock wise around this circuit.

So if we start at the bottom right hand corner and start around the circuit

we first encounter the negative polarity of the minus 5 volts source.

So it's minus 5 volts.

And then we have the volts that drop across 20 kilohm resistor which

is 20k(I out) plus the voltage drop across 10k as we continue around this loop.

And the voltage around the 10k is going to be 10k(I out),

and then we return back to the starting point.

And so we know through cross voltage law the the sum of those

voltages is equal to 0.

So if we use that equation and we solve for I out,

then we end up with an I out which is equal one-sixth milliamps.

Now that we know I out, we can go back and perhaps find V out.

And the way we're going to do that is using the second property

of the ideal op amp.

If we know what the voltage is at the non-inverting input

it's going to be equal to the voltage at the inverting input.

So can we find the voltage at the non-inverting input,

which is the voltage at this point in our lower loop.

It's going to be equal to 10k times I out so

the voltage at the noninverting input

is equal to 10 kilohms times I out and

I out is 1/6 milliamp.

So, our voltage at the non-inverting input is equal to 10 6th of a volt.

We also know that that is equal to the voltage at the inverting input.

So, the voltage at this node, it's a non-inverting input.

I'm sorry, at the inverting input, has a voltage level of 10, 6 volts.

Then, we can use Kirchhoff's Current Law at that node.

Let's call it, node 1, and we're going to sum the currents into that node,

in order to find an equation, which has the V out as the unknown.

So, let's sum the currents into that node.

We have the current through the 10 resistor flowing from left to right.

It's going to be 0 volts minus the voltage here,

which is 10 6th volts, divided by 10k.

0 minus 10 6ths.

Divided by 10k.

We also have the current which is flowing from right to left through the 20 kilohm

resistor.

It's going to be Vout minus 10/6 volts

Divided by 20 K.

So that's the current flowing right to left into node one through 20 kilom

resistor.

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

the op amp.

We know that that is equal to zero.

We're going to add it in for completeness.

That's all of our currents, they're summing into node one.

So this is an equation which just has V out as the unknown.

It's the only unknown in the equation.

So we can solve for V out and if we do, we end up with a V out equal to 5 volts.

So now we I out, which we've been looking for, and also V out.

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