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.

Â In this lesson we will look at active filters.

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

Â that we could do differentiators and integrators.

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

Â to build active filters.

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

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

Â we will give you some background for

Â some analog filters just some of their characteristics.

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

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

Â Let's take a look at an analog filter.

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

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

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

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

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

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

Â I've got two frequencies in there.

Â A low frequency.

Â Where I can look at it here.

Â It's kind of the average value of that.

Â And that low frequency has an amplitude of 1.

Â Well, then I have a high frequency superimposed on it

Â with an amplitude also of 1.

Â And if I write this out.

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

Â That's the high frequency.

Â Now if I had a single frequency of a amplitude

Â of Ai, Ai cosine of omega 1t

Â into my circuit and I got an output corresponding A out

Â at that same frequency plus maybe some phase lag.

Â I can relate the input and

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

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

Â the transfer function at that frequency.

Â Now, it helps to look at a numerical example.

Â In this example, I'm plotting a,

Â a magnitude of the transfer function that looks like this.

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

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

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

Â which is 1, to give me the output amplitude.

Â So, that would be 0.9, cosine, 50t.

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

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

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

Â gives me the output amplitude.

Â And then I might have some phase lag in there.

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

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

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

Â high frequency signal.

Â They both have an amplitude of 1.

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

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

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

Â with the high pass signal.

Â The high pass filter.

Â The previous one we passed through low frequencies and

Â we attenuated high frequency.

Â This is going to be doing the opposite.

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

Â This is my input.

Â I've got three frequencies here.

Â A DC, and DC corresponds to

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

Â Hertz is related to radians per seconds through a factor or

Â 2 pi, 2 pi f is equal to omega.

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

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

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

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

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

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

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

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

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

Â My frequency at 10 hertz is multiplied by about 0.45,

Â and at 100 hertz I get multiplied by about 0.97.

Â So my output looks like this and

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

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

Â So which of these signals matches this?

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

Â This one right here, has a DC component.

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

Â So, that one's, we can throw out.

Â This one also has a DC component.

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

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

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

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

Â So, this is V in.

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

Â So that's wrong.

Â So I look at these two signals.

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

Â my high frequency dominates.

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

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

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

Â Why don't we use them?

Â Well, they've got a couple of limitations.

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

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

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

Â Because I typically will take a measurement, and

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

Â And if I build an RLC circuit,

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

Â And I might not have enough power in my measurement,

Â my measured signal to drive that circuit.

Â So it depletes power.

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

Â What do I mean by that?

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

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

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

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

Â And the same thing, so

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

Â And the same thing as the input.

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

Â I can change that signal.

Â It's sort of like the Heisenberg uncertainty principle,

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

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

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

Â So, we introduce active filters.

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

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

Â And it also provides isolation.

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

Â I can cascade these filters together without changing these

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

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

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

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

Â The advantages over an RLC passive filter?

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

Â It boosts the power.

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

Â To provide sharp roll-off with the passive filter,

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

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

Â because it has its own power supply,

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

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

Â Thank you.

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