This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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來自 Georgia Institute of Technology 的課程

电子学基础

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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Op Amps Part 1

Learning Objectives: 1. Develop an understanding of the operational amplifier and its applications. 2. Develop an ability to analyze op amp circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to electronics.

This is Dr. Robinson.

In this lesson we are going to design a balanced output amplifier to meet a set of

given specifications.

But let me begin by explaining to you what a balanced output amplifier is and

why we would want to use one.

Let me draw a block that represents the balanced amp.

Now, a balanced amplifier has a single input we'll call it, Vin and

produces two outputs.

That are related to the input.

One output equal to AVin, and a second output equal to minus AVin.

Where A is the gain of the balanced amplifier.

Now you can see how the outputs are related to the input and to each other.

The outputs are the negatives of each other.

Or you can think of them as two signals that are 180 degrees out of

phase from each other.

Now, why would we want to use a balanced amplifier?

Say for example, we have an electrical signal that represents an audio source.

So for example, this audio signal [NOISE] could be

the output of an MP3 player, possibly a phonograph.

And we want to transmit it through a cable or

a transmission line to the next stage in our audio system.

So for example, a preamp.

Pre amp.

Well, during this transmission.

Along this cable electrical noise can be introduced by external electrical

components.

So that the input of the preamp,

we not only have the signal that we're interested in.

But we have, in addition to that, this added noise component that I'm calling N.

So during the transmission we have distorted our signal that represents

the audio by this external noise.

But what if instead we transmit the signal using a balanced amplifier.

So I make the input of our balanced amplifier, the audio signal, or

the electrical signal that represents the audio signal.

We then transmit the signal using two transmission lines or two cables.

And we make the input stage of our preamplifier, a differential amplifier.

A diff amp.

With a gain equal to one-half.

Then, any noise that's introduced due to external devices.

The noise.

Is applied to both the AVin signal and the minus AVin signal equally.

So at this input to the diff amp, we have AVin plus the noise.

And at this differential amplifier input, we have minus AVin plus the noise.

So then at the output of the differential amplifier Vout,

we can write than Vout is equal to one of the inputs.

AVin plus the noise minus, because it's a diff amp,

the other input voltage, which is minus AVin plus the noise.

All times the gain of the diff amp, one-half.

So Vout is equal to one-half times

AVin plus AVin plus the noise minus the noise.

Or the output voltage is equal to AVin.

So this combination of a balanced output amplifier and

an input differential amplifier tends to cancel any common mode noise.

That's introduced into the signal during it's transmission.

Now, let's look at an a op amp circuit that can be used to implement a balanced

output amplifier.

Here is our input voltage Vin applied to the non inverting terminal of an op amp.

We have a feedback resistor, RF1.

We have another resistor, R1,

that's applied to the inverting terminal of a second op amp.

And we have another feedback resistor, RF2.

Here is one output of the circuit and here is the second output of the circuit.

Now we consider this circuit to be composed of two simpler op amp amplifiers,

a non-inverting amplifier and an inverting amplifier.

Remember, these two voltages must be equal to each other.

The voltage here is ground,

which means the voltage at this node is zero volts or ground.

So we can consider this portion of the circuit

to be a non-inverting op amp amplifier with

a gain of Vo1 equal to one plus RF1 over R1.

And we multiply that by Vin to get the output voltage Vo1.

Now, this voltage here must also be equal to the voltage here.

So the node voltage here is equal to the input voltage Vin.

We can consider this circuit, or this portion of this circuit,

to be an inverting op amp amplifier with an input voltage Vin.

So Vo2 is equal to negative RF2 over R1 times the Vin.

Now what I want to do is design this circuit so

that it meets a couple of conditions.

One, I want the magnitude.

The right design.

I want the magnitude of the gain A of this balanced output amplifier circuit

to be equal to 6.

And I want to impose a second condition, and that is that

Want the maximum

current through RF1 and

RF2 to be one milliamp.

When the magnitude of Vo1 is equal to

the magnitude of Vo2, is equal to 12 volts.

So, you can see that from these two equations we can impose

this gain condition.

We can choose RF1, RF2, and R1.

So that the gain here is equal to 6 and the gain here is equal to negative 6.

But to implement this second condition we need another equation.

An equation for

the current through RF1 and RF2 in terms of the circuit component values.

Now, remember there's no current into the input terminals of an op amp.

So the current into this inverting terminal is zero.

And the current into this inverting terminal is zero.

So any current that flows through our RF1 must also flow through R1 and

flow through RF2.

Because there is no way out at this node.

Because of the infinite impedance looking into the op amp here.

And there's no way out at this node.

Because of this infinite impedance.

So we have, flowing from here.

A current that flows through RF1, through R1 and through RF2.

So I can use Ohm's law to calculate that current.

The difference in voltage across this series combination of three resistors.

Must be equal to the current through the resistors times the resistance.

Or in other words, I can write that Vo1 minus Vo2, the total

voltage across the three resistors divided by the sum of the three resistors.

The series resistors are RF1 plus RF2 plus R1 must be

equal to the current I through the three resistors.

So to impose our second design condition I can

write that Vo1 when it is equal to 12 minus Vo2.

Which is equal to minus 12 divided by RF1

plus RF2 plus R1 must be equal to 1 milliamp.

Then we know that 1 plus RF1

over R1 must be equal to 6,

and we know that, minus RF2 over

R1 must be equal to minus 6.

So this equation implies that RF1

is equal to 5 times R1.

This equation implies that RF2 is equal to 6 R1.

So I can use these two equations to eliminate RF1 and RF2 from this equation.

So I can write that 24

divided by 5 R1 plus 6 R1,

plus R1 is equal to 1

milliamp is equal to 24

divided by 12 R1.

Which implies that R1

is equal to 2k ohms.

Then, I can use RF1 is equal to

5 R1 is equal to 10k ohms.

And.

RF2 is equal to 6R1 is equal to 12k ohms.

So if we construct a balanced output amplifier using

these resistor values we'll meet our design specifications.