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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From the course by Georgia Institute of Technology

Introduction to Electronics

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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.

From the lesson

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.

In this lesson we will look at active filters.

In our previous lesson, we introduced capacitors into our op amp circuit so

that we could do differentiators and integrators.

In this case, we will introduce the capacitors into the circuit, but

to build active filters.

Now active filters is, is a huge and, and very important application of, of op amps.

And in this particular lesson, we will introduce the active filter and

we will give you some background for

some analog filters just some of their characteristics.

And then we will look at passive filters and show their limitations and

why it is that we want to use active filters, as opposed to passive filters.

Let's take a look at an analog filter.

An analog filter is just a circuit and we look at is a way of processing signals.

So, the signal that's input to this circuit, this filter, is a voltage signal.

And then we're measuring some sort of voltage output.

And we show H of omega, it's a transfer function.

It tells me how I am going to process signals of different frequencies.

So, for example, suppose this is my input signal, and

I've got two frequencies in there.

A low frequency.

Where I can look at it here.

It's kind of the average value of that.

And that low frequency has an amplitude of 1.

Well, then I have a high frequency superimposed on it

with an amplitude also of 1.

And if I write this out.

I've got a cosine of 50t, that's the low frequency and a cosine of 800t.

That's the high frequency.

Now if I had a single frequency of a amplitude

of Ai, Ai cosine of omega 1t

into my circuit and I got an output corresponding A out

at that same frequency plus maybe some phase lag.

I can relate the input and

output amplitudes and they're related through this transfer function.

So the output amplitude is equal to the input amplitude times the magnitude of

the transfer function at that frequency.

Now, it helps to look at a numerical example.

In this example, I'm plotting a,

a magnitude of the transfer function that looks like this.

And, what I do is I look at this frequency at 50, I look at 50 right here, and

I go up and I find what this magnitude is, that's about 0.9.

So I take 0.9, multiply it by the input frequent, input amplitude,

which is 1, to give me the output amplitude.

So, that would be 0.9, cosine, 50t.

And I do the same thing for this other, component of the signal.

At 800, I look at 800, that's about say, 0.13.

And so I have 0.13 times the input amplitude, which is 1,

gives me the output amplitude.

And then I might have some phase lag in there.

So if I look at the output signal it's going to look like this.

Where I got rid of a great deal of the output sig, of the high frequency signal.

You can see this is, this has equal amount of low frequency and

high frequency signal.

They both have an amplitude of 1.

Here the amplitude of the high frequency signal is greatly attenuated Now,

I'm going to set you up to do a quiz here.

And this quiz actually is to do the same thing we just did, but

with the high pass signal.

The high pass filter.

The previous one we passed through low frequencies and

we attenuated high frequency.

This is going to be doing the opposite.

In this particular case, I'm going to set it up for you before we start.

This is my input.

I've got three frequencies here.

A DC, and DC corresponds to

a frequency of 0, no, notice that my frequency is in hertz here.

Hertz is related to radians per seconds through a factor or

2 pi, 2 pi f is equal to omega.

And so before when we were looking at cosine of omega t,

we have to relate that back to f by this factor of 2 pi.

So in fact, this is my f, and that's my f.

So if I wanted to look at the magnitude, this is the magnitude of the,

the transfer function, I would be looking at, at 10 hertz and 100 hertz.

And, what I want you to do is figure out, if this is my input, which of these,

is the corresponding output of a filter with this transfer function?

Okay, so let's look at the result here, let's look at the solution.

I see that a DC signal gets multiplied by 0, so my DC part is completely attenuated.

My frequency at 10 hertz is multiplied by about 0.45,

and at 100 hertz I get multiplied by about 0.97.

So my output looks like this and

notice I threw it in the frequency, the phase shift.

And the frequencies stay the same, and these are the amplitude changes.

So which of these signals matches this?

Well, notice that there's no DC component here.

This one right here, has a DC component.

You can see that it, it oscillates around the value of about 1.

So, that one's, we can throw out.

This one also has a DC component.

And I'm going to throw that one out as well.

And notice, this one, the amplitude, I can actually show that the amplitude of

the low frequency is about the same as the amplitude of the input frequency.

In fact, this signal is what my input signal looks like.

So, this is V in.

It's without having any sort of attenuation at all.

So that's wrong.

So I look at these two signals.

In this one, my low frequency dominates, and in this one,

my high frequency dominates.

Well, if you look at these results here, the high frequency should dominate.

So, this is the answer right here and this one is wrong.

So, passive filters are pretty easy to design and build.

Why don't we use them?

Well, they've got a couple of limitations.

If I'm going to ever build a filter on an actual circuit,

I'm going to use an active filter not a passive one.

And the reason is, these two limitations, one, it depletes power.

Because I typically will take a measurement, and

it has very low power, maybe it's from a sensor, and it's low power.

And if I build an RLC circuit,

I have to be able to drive that circuit, and it uses power to drive that circuit.

And I might not have enough power in my measurement,

my measured signal to drive that circuit.

So it depletes power.

And the other thing it doesn't provide isa, isolation.

What do I mean by that?

Well, if I have, if I want to add something to this, you know,

maybe say a load resistor, I've immediately changed V0.

And you can see that by going up here, if I add R sub L right here,

I have to reevaluate the circuit because I have changed what V0 is.

And the same thing, so

it's not isolated here because V0 depends on what I hook up to it.

And the same thing as the input.

So, suppose I measure something from a sensor and then I put my filter on it,

I can change that signal.

It's sort of like the Heisenberg uncertainty principle,

when you try to measure something, you've changed it.

I don't want to change it and that's the problem with the passive filter.

I don't have this isolation there, I change it.

So, we introduce active filters.

It has its own power supply so it doesn't deplete the power.

And, we, most common ones are made out of op amps.

And it also provides isolation.

So I don't change, so I don't change my voltage when I add another circuit on it.

I can cascade these filters together without changing these

intermediate voltages, the input and output for any given circuit.

So in summary, we've introduced an analog filter.

It's just a, a circuit that has specific shaped frequency response.

An active filter is made of op amps and it has its own power supply.

The advantages over an RLC passive filter?

Well, it provides isolation, so we can cascade filters.

It boosts the power.

And as a result of boosting the power, we can provide sharper roll-off.

To provide sharp roll-off with the passive filter,

I have to keep adding components, and I'm going to be using more power.

Well, I can be adding components to an active filter, and

because it has its own power supply,

I'm not going to be depleting the power as much, and so I can get a sharper roll-off.

So in our next lesson, we will look at lowpass filters.

Thank you.

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