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 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 is Dr. Robinson.

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

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

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

Â we know that that implies that the quality factor

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

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

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

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

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

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

Â Times omega naught in terms of f node is

Â two pi f node times 1/R1+1/R2 but

Â these are equal to each other so

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

Â 2 pi f naught.

Â Times 2 R1 times the square root of 2.

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

Â have 20% tolerances than it is resistor values.

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

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

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

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

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

Â I get C1 is equal to approximately 0.11 microfarads and

Â C2 is approximately equal to 0.22 or so microfarads.

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

Â and I have available a 0.22 microfarad capacitor.

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

Â C1 = 0.1 microfarads, and C2 = 0.22 microfarads.

Â I get a circuit topology

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

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

Â making this resistor a short circuit.

Â And the resistor that was here, an open circuit.

Â The simplest way of making the gain one.

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

Â But, the best solution is just leave it out

Â because that makes the circuit a little less complicated.

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

Â we get this red curve here on log log scales.

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

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

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

Â in choosing the capacitor values.

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

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

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

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

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

Â A change in two decades in magnitude, so

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

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

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

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

Â We looked at some practical design considerations.

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

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

Â demonstrate for you using active filters the extraction of particular

Â frequency components from a signal.

Â So thank you and until next time.

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