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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En provenance du cours de 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.

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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.

In this lesson we will look at first-order highpass filters.

In our previous lesson we introduced active lowpass filters.

In this particular lesson our objective is to look at highpass filters.

In particular we'll start with the characteristics of them and

then we'll show you how to design them.

So a highpass filter has this basic frequency response characteristic.

It passes high frequency components and attenuates low frequency components.

So, it's, the magnitude of the transfer function, this being.

H of omega plotted here is low at low frequencies and high at high frequencies.

And the Bode plot looks like this.

Bode Plot is 20 times the log base 10 of the magnitude.

So let's particular start with a first-order filter.

This is a generic first-order filter.

Now let's look at, at its characteristics.

There are two things that we normally look at.

One is the corner frequency.

And in a particular a highpass filter we talk about the,

the region where it passes the signals as being the passband.

So the start of the passband is this corner frequency, omega sub c.

And we define it as being equal to 0.707 of the past band gain.

The passband gain we call it case of PB, is a gain at high frequencies.

So we'll define it this way.

So if I have a generic first order transfer function that looks like this,

my corner frequency is 1 over tao.

It is the whatever multiplies by the j-omega, and then I invert it.

And the passband gained is actually the limit,

as omega goes to infinity of h of omega.

And that's the, the passband gain.

[SOUND] A look at an inverting highpass filter.

One, it gives me isolation like we talked about in the last lesson.

Isolation's a good thing.

Let's look at the, the highpass the passband gain.

Again that's, I'm going to abuse my notation a bit.

If I were to look at the limit as H goes to infinity.

And that limit as H goes to infinity, this being H right here,

then my 1 becomes negligible, and I've got minus RfC omega over R1C omega.

So, reducing that down, I get minus Rf over R1.

And then, my corner frequency is, well, I look at whatever multiplies,

whatever constant multiplies by, by j omega and I invert that, 1 over R1C.

Now we want to derive this transfer function.

Let's go back to our, thinking about our last lesson.

In our last lesson we looked at a configuration like this, and

we said that the transfer function for this is equal to minus Zf over Z1.

We derived it in the last lesson.

So Zf in this case is Rf and Z1 is equal to the series

combination of this, the impedance Z sub C plus R1.

And Z sub C is j1 over j omega C plus R1.

So if I substitute that into this formula,

I will be able to derive this transfer function.

Let's look at the frequency characteristics of the highpass filter.

So it has this transfer function right here.

[SOUND] If I were to look at the magnitude of the transfer function, it's right here.

Remember that that's the magnitude of the numerator, which is 1, or

which is this, over the magnitude of the denominator,

which is the real part squared plus the imaginary part squared, square root.

Okay, then I take the angle.

Well i've got a minus 90 because I've got a minus sign and then I've got a J.

So the angle of minus J is minus 90.

And then this is minus the angle of the denominator.

That means in the passband we're going to have a gain of this which is

found by taking the limit of this as omega gets very large.

Because it's a, it's a highpass filter, the passband is in the high frequency.

So as omega gets large, I take the limit and I get minus Rf over R1.

And then the corner frequency is a frequency at which I get

a value that were this imaginary part magnitude is equal to the real part.

So that is 1 over R1C.

If I were to plot this, this magnitude has to be positive.

So, this is on a linear scale, so I'm looking at positive values,

that's why I have a value of Rf over R1 being positive.

And plotting the angle, I get this right here.

Now, why does this start out as minus 90?

Because it starts out at minus 90 there.

It ends up at minus 180, because at high frequency this has a, a minus value.

A minus and the minus corresponds to-

Minus 180.

So that's the angle.

And the magnitude, we have a, the 0.707 value is about right here.

So the passband.

[SOUND] Is considered here, where for

the most part we are passing through signals without attenuating them.

Thi, this is a stop band over here.

This is where we are attenuating our signals.

The corner frequency again is, is this point right here.

That's at the point where we get the 0.707 times this and the K being the.

The gain of the passband.

So now we want to design a highpass filter to have a passband gain of say minus 2 and

a corner frequency of 1000 radiance per second.

Remember our transfer function for

this filter is minus RfCj omega over R1Cj omega plus 1.

And my passband gain is H of omega.

When omega goes to infinity,

we found that to be equal to minus R sub f over R1.

And we want that to equal to minus 2.

Now, the other equation we have is our corner frequency, omega C.

That's equal to 1 over whatever multiplies this J omega.

So that's R1C and that should be equal to a 1000.

Okay, if I let.

I've got three parameters to chose and only two equations.

So can let R1 equal to 1000, because it's a very common resistor value.

So I'll just pick that value and then I can solve for Rf.

Would be equal to 2,000 ohms.

And C would be equal to 1 microfarad.

Now one thing that I want to comment on is if my corner frequency is given in hertz,

then I have to make sure that I have this 2 pi factor.

Remember that omega is equal to 2pi f, so

omega C is equal to 1 over R1C.

So I would have 2pi f sub C is equal to 1 over R21C.

So if my corner frequency is given to hertz,

I have to convert it to radians per second first before I do this.

So in summary, a highpass filter passes high frequency components and signals and

attenuates or filters out low frequency components.

We looked at one basic design, this is inverting highpass filter.

And it provides isolation at the input and the output.

It has this transfer function right here.

And we showed how to do a design.

We based the design on two factors that are given.

If we're given the corner frequency of the passband and the passband gain.

And then we showed how we were able to pick these parameters.

In our next lesson, we will look at more complicated filters,

ones where we can build cascaded filters together, put them together and

create more interesting filters like passband or bandpass and notch filters.

Thank you.

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