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

326 ratings

Georgia Institute of Technology

326 ratings

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 2

Learning Objectives: 1. Examine additional operational amplifier applications. 2. Examine filter transfer functions.

- 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 This lesson is on First-Order Lowpass Filters.

Â In our previous lesson, we introduced active filters.

Â In this lesson, we want to introduce active lowpass filters.

Â We'll go through the characteristics and then show you how to design them.

Â So, starting with the characteristics.

Â A Lowpass Filter passes low frequency components and filters out, or

Â we say attenuates, high frequency components.

Â They're characterized by a transfer function H of omega.

Â So if I plot on a linear plot,

Â we have certain characteristics that we define for this.

Â Two particular things that we use in design actually.

Â One is the bandwidth.

Â And the bandwidth is the value at which it's 0.707 of what we call the DC value.

Â First is the frequency plot.

Â DC value is at omega equals zero.

Â And so we look at the value of the plot right there,

Â and I'll call that the DC gain.

Â So the DC gain is that, and

Â then the bandwidth is the frequency at which we're at 0.707 of the DC gain.

Â Now this is a linear plot where I'm plotting the magnitude H of h.

Â The other way we often plot this is a Bode plot, where I'm actually plotting

Â 20 times the log of h.

Â So, then, this plot, this value is going to be 20 times the log, and

Â this is base 10 of the DC gain.

Â And then, at the point where I'm at 3 dB below the DC gain, that is my bandwidth.

Â Now, in this particular class, we're really going to be concentrating a little

Â bit more on the linear plots here rather than the Bode plots.

Â But Bode plots is a common way of seeing this.

Â It's the same plot, except on different scales.

Â This being a log scale here.

Â Log in, in terms of omega.

Â So, this is on a log scale.

Â And this is on a linear scale where the units are decibels.

Â Now, let's look at the characteristics of a First-Order Filter.

Â A first-order filter has this generic frequency response.

Â This, this transfer function right here.

Â And we showed the same plot as what we, what we had before.

Â And the bandwidth is this value that is 1 over tau,

Â where tau is a multiplier of j omega.

Â The way we get that is if I look at the magnitude of H.

Â It's equal to K of DC, 1 over, and in the magnitude of the denominator.

Â Well the magnitude of the denominator is the square root of this sum that

Â squares of the real part plus the imaginary part.

Â So, the real part is 1,

Â squared is 1 plus, the imaginary part squared, which is Tau Omega squared.

Â And at 0.707, this is equal to 0.707 when Tau is equal to 1 over Omega.

Â So, that's the bandwidth.

Â So for that value, this whole thing is equal to 0.707.

Â Now, the DC gain is a value of H when I let it equal to zero.

Â That's because, again, I want to emphasize again, that's the DC,

Â is at Omega equals zero.

Â So, that's just a generic First-Order Filter.

Â So, going from a passive to an active filter is probably the easiest way to

Â design your first filter.

Â It is to look at a passive filter, which is an RC circuit.

Â This is a, a lowpass filter where my input is right here, and

Â my output is measured across this capacitor.

Â And so this is the generic form for this circuit.

Â If I want to make this into an active filter and I want to isolate at the input,

Â then all I have to do is add a buffer circuit at the input.

Â So the buffer circuit doesn't change this waveform.

Â So this V in signal is the same here, V in.

Â The waveform looks the same,

Â doesn't change in amplitude, but it increases the power here.

Â And it isolates it from the imput.

Â So, if I'm measuring a, a measurement device and I want to, I don't want

Â to change my measured signal, then I just add in the buffer circuit before this.

Â And then I've got the characteristics of a lowpass filter.

Â If I want isolation at the output, then I take my circuit and

Â I add my buffer circuit at the output.

Â In this case, if I were to add, say a resister right here,

Â I'm not going to change the out.

Â So, I can add a whole circuit on here,

Â cascade something else to it, and I won't be changing the out.

Â So, this is just taking a simple passive filter and creat, and

Â making it in to active by adding an up amp.

Â Now an alternate way of creating a lowpass filter

Â is to put the filter into the feedback loop.

Â So putting it in the feedback loop means that I can do two things at once.

Â I can isolate at the input and

Â at the output both with this one, with this one circuit, and the other

Â thing is because I've got two resistors in there, I can change the gain, the DC gain.

Â With the other two, I had a DC gain of one.

Â I might not want a DC gain of one.

Â I might want something else.

Â So, this particular circuit has this transfer function,

Â this part right here is my H.

Â Let's go ahead and, and derive where we get this, this transfer function from.

Â I want to start with this circuit right here.

Â It's a little bit simpler.

Â And we're going to use this, because later on when we do high pass filters,

Â we'll start with this configuration, as well.

Â I'm going to just derive this, and

Â then we'll substitute back in for these impedances up here.

Â So, if I look at this.

Â I want to do a derivation for this.

Â I want to get V out in terms of V in.

Â Okay. So I can do a KVL across here.

Â Looking at that as plus.

Â That's equivalent to having a V in here with respect to ground.

Â And remember that the voltage drop here is zero.

Â So I can get an equation that is V in is equal to Z1 times I.

Â Now, the I that's going up here is the same I that's going in here.

Â Because, remember that there's zero current going into here.

Â Zero current into there.

Â So all the I goes through this way.

Â So that is the same I.

Â So Z1i plus zero, if I do around here, because that's a, a zero full drop there.

Â And the other KVL that I can do is across this way.

Â Over to here, and then down to ground.

Â So, doing that one, I get V

Â in is equal to Z1i plus Zfi plus the out.

Â Well, these two are going to cancel.

Â And what I'm left with is V out is equal to minus Z sub f i.

Â And I can actually solve for i from this equation.

Â I get i is equal to V in over Z1.

Â So I plug that into here, I get V out

Â is equal to minus Zf over Z1 times V in.

Â And so this is, this is simple.

Â I've, I've been able to come up with the input output equation for

Â this circuit right here.

Â Now I want to use that to go up to this circuit.

Â Well, what's ZF in this circuit?

Â ZF is equal to the resistor and

Â parallel with the impedance of this capacitor.

Â So if I write that, I get 1 over, 1 over Rf,

Â and 1 over the impedance of the capacitor is j omega c.

Â So if that's Zf, and then Z1 is equal to R1.

Â So if I make these substitutions into this equation,

Â I derive the transfer function that we needed.

Â Let's look at the frequency characteristics of a lowpass filter.

Â This is a transfer function right here.

Â And if I calculate the magnitude and the angle, I get this.

Â So the magnitude is, I just take the,

Â in the denominator I'm going to take the sum of the squares of the real part and

Â the imaginary part and then take the square root of that.

Â And then the angle, notice that I've got a minus sign right there, so

Â the angle starts out a hun, at 180 degrees.

Â And then I subtract off the angle of the denominator.

Â The DC gain is found by setting the frequency equal to zero.

Â So, H of zero is that DC gain.

Â The bandwidth is the frequency at which I have 0.707 of the DC value.

Â So, this plot right here is the magnitude, and I've got 0.707 of the DC value.

Â Now as the DC value.

Â The magnitude of the DC value.

Â So, this plot,

Â because this is magnitude, this has gotta be equal to, positive value.

Â So I've stripped off the negative sign.

Â And then the frequency at which we have .707 of that DC value.

Â The magnitude of that DC value is the bandwidth.

Â So this range, right here, we call the pass band and this range, the stopband.

Â The passband is

Â frequencies at which I pass through a signal with very little attenuation, and

Â the stopband is when I start attenuating those signals.

Â So, low frequencies are passed through.

Â And high frequencies are attenuated.

Â The angle looks like this, the angle plot, so remember it starts out at 180, and then

Â at high frequencies, this will, will go to 90 degrees.

Â So, what we want to do is an example here.

Â We want to design an inverting lowpass filter to have a DC gain of -2, and

Â a bandwidth of of 500 radians per second.

Â So, let's go back to these formulas.

Â The DC gain is H of 0,

Â which is minus Rf over R1 and that is equal to -2.

Â And the bandwidth, Omega

Â sub B is equal to 1 over RfC.

Â So the bandwidth is equal to 1 over RfC.

Â And that is equal to 500 radians per second.

Â So now I have two equations and

Â I actually have three param, design parameters, Rf, R1 and C.

Â So I can just pick one.

Â So I'll say pick R1 is equal to 1000 ohms.

Â Very common resistor.

Â Let's go ahead and pick that one.

Â So if I solve for that, then Rf would be equal to 2000 ohms,

Â and C would be equal to one microfarad.

Â And that's the design of my circuit, so I would build this, and

Â I would get a filter with this bandwidth and that DC gain.

Â So in summary, a lowpass filter passes low frequency signals and

Â attenuates high frequency signals.

Â We looked at three first-order lowpass configurations.

Â Two of them were made up by just taking a passive lowpass filter and

Â putting a buffer circuit on it,

Â either at the input or at the output to provide is, isolation wherever we want.

Â Now these particular circuits, this one right here, and

Â this one right here, provided a DC gain equal to one.

Â And they were noninverting.

Â So, in this particular case we have an inverting amplifier.

Â It provides isolation at both the input and the output and

Â it also has a DC gain that is not one, so it can actually amplify our signal.

Â Minus Rf over R1.

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

Â Thank you.

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