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Sample Problem: DC Op Amp 2

The module describes some basic principles used in circuit analysis.

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.

Impartido por:

Dr. Bonnie H. Ferri
Professor
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
Dr Mary Ann Weitnauer

Transcripción

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.

So we have a circuit that has one voltage source in it, it has a 5 volt source.

And Iout 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 0,

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

And that the voltages at the inputs are equal.

V- = V+.

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 Iout is a current which is confined to this loop that we have at

the bottom of our circuit.

So if we want to find Iout, 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, Iout is measured in a counterclockwise direction.

So we're going to sum our voltages up counterclockwise around this circuit.

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

we first encounter the negative polarity of the -5 volt source, so it's -5 volts.

And then we have the voltage drop across the 20 kiloohm resistor,

which is 20k times

Iout plus the voltage drop across the 10k as we continue around this loop.

And the voltage across the 10k is going to be 10k times Iout.

And then we return back to the starting point.

And so we know through Kirchhoff's voltage law that

the sum of those voltages is equal to 0.

So if we use that equation and we solve for Iout,

then we end up with an Iout which is equal to one-sixth milliamps.

Now that we know Iout, we can go back and perhaps find Vout.

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 Iout.

So the voltage at the non-inverting

input is equal to 10 kiloohms times Iout,

and Iout is one-sixth milliamp.

So our voltage at the non-inverting input is equal to ten-sixths of a volt.

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

So the voltage at this node, at the non-inverting input, I'm sorry,

at the inverting input, has a voltage level of ten-sixths 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 Vout as the unknown.

So let's sum the currents into that node.

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

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

which is ten-sixths volts, divided by 10k.

0- ten-sixths.

Divided by 10k.

We also have the current which is flowing from

right to left through the 20 kiloohm resistor.

It's going to be Vout- ten-sixths volts divided by 20k.

So that's the current flowing right to left,

into node 1 through the 20 kiloohm 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 0.

We're going to add it in for completeness.

And that's all of our currents that are summing into node 1.

So this is an equation which just has Vout as the unknown,

it's the only unknown in the equation.

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

So now, we have Iout, which we've been looking for, and also Vout.

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