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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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 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.

This is Dr. Robinson.

In this lesson we're going to work out a lowpass filter design example.

Let's look at how we would design a Butterworth 2nd order low-pass filter.

Now, once I've told you it's a Butterworth filter

we know that that implies that the quality factor

of the secondary transfer function is equal to 1 over the square root of 2.

And I want to design this filter so that we have an f knot of 1 kHz.

And I've chosen to use the special case 1 equations for

the low-pass filter And I've also chosen to

simplify the equations further by making this assumption that R1 is equal to R2.

Now we can write that C1 is equal to q of 1 over R2.

Times omega naught in terms of f node is

two pi f node times 1/R1+1/R2 but

these are equal to each other so

I can write that as 2/R1, then C2 = 1/-

2 pi f naught.

Times 2 R1 times the square root of 2.

Now knowing that it's harder to find capacitor values because they

have 20% tolerances than it is resistor values.

The way I solved for C1 and C2 was to look at the available resistors that I had

in the resister bins in the laboratory, and choose different values of R1

until C1 and C2 were close to values of capacitors that I had available to me.

So trying out different Rs, I reached the conclusion that by letting R1 = 1000 ohm.

A resistor value that's found in every resistor bin in existence.

I get C1 is equal to approximately 0.11 microfarads and

C2 is approximately equal to 0.22 or so microfarads.

Well in my 20% capacitor bin, I can choose a 0.1 microfarad capacitor,

and I have available a 0.22 microfarad capacitor.

So if I implement the circuit using these values R equals 1k, both R1 and R2 = 1k.

C1 = 0.1 microfarads, and C2 = 0.22 microfarads.

I get a circuit topology

that looks like this, both 1k's, a 0.22, and a 0.1.

And here you can see that I've set the gain equal to 1 by

making this resistor a short circuit.

And the resistor that was here, an open circuit.

The simplest way of making the gain one.

Once this is a short circuit, the value of this resistor doesn't matter.

But, the best solution is just leave it out

because that makes the circuit a little less complicated.

Now if we plot the Bode magnitude plot, for this circuit,

we get this red curve here on log log scales.

You can see that, as expected, it is low-pass filter.

It has an F naught or caught off frequency of 1.12 kHz.

And the reason this isn't 1 kHz is because my approximations

in choosing the capacitor values.

Now, we would expect this slope because it's a second order transfer function or

a second order filter to have a slope of minus 2 decades per decade.

And we can verify that, let's look at this decade increase in frequency.

So in moving from 1 kHz to 10 kHz we increase by a factor of 10 and

in doing that we decrease the magnitude of the transfer function from 1 to 10 milli.

A change in two decades in magnitude, so

from 10 milli to 100 milli to one is a factor of 100 or two decades.

So we went down 2 decades for this 1 decade increase in frequency.

So in summary during this lesson, we've designed a second order low-pass

filter using the design equations that we've talked about in previous lessons.

We looked at some practical design considerations.

And we examine the behavior of a simulated version of our designed filter.

In our next lesson we will look at a filtering demonstration where I

demonstrate for you using active filters the extraction of particular

frequency components from a signal.

So thank you and until next time.