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

In this video, I'm going to solve for the transfer function for a filter formed by

Â cascading a first order low pass filter, and a first order high pass filter.

Â We're going to put the transfer function in standard form, and

Â look at the limitations of a filter formed in this way.

Â Let's say we form a filter, by cascading.

Â Two first-order filters.

Â So we have a first-order, low-pass filter that has some transfer function,

Â Hlp of omega.

Â We cascade that with another first-order filter.

Â A first-order high-pass filter, that has a transfer.

Â Function H, HP of omega and this is the output voltage.

Â So the overall filter is formed by cascading these two first water filters.

Â And the overall function of this cascade is the ratio of this VO to this VI.

Â A first order low pass filter has a transfer function.

Â Standard form for the denominator would be s over omega u plus 1.

Â It's a low pass filter, so we have a 1 in the numerator.

Â And I'll call the gain constant k1.

Â This high pass filter has, again.

Â Same form of denominator.

Â But I'll call its minus 3 db frequency omega l.

Â It's a high pass filter so we have the highest order term in the numerator s over

Â omega l, and it'll have some gain constant, say k2.

Â So the overall transfer function V O over V I would

Â be equal to the product of the two transfer functions,

Â H H P of omega time H L P of omega is equal to.

Â K1 K2 1 over s over omega u plus 1 times

Â s over omega l over s over omega l plus 1.

Â And remember that in writing a transfer function like this s is equal to j omega.

Â Now what I want to do is write the transfer function for

Â this cascade in standard form of a second-order filter, and

Â you can see that when we multiply these two transfer functions out, we're

Â going to have a second-order denominator, an S squared term in the denominator.

Â So by mani, manipulating this and

Â putting it into standard form for a second-order filter, remember.

Â For a second-order filter, we can write the denominator like this.

Â S over omega naught squared, plus one over q, s over omega naught plus one.

Â Some gang constant K out front, and

Â then the numerator is composed of one of the terms of the denominator, or

Â some combination of these terms [NOISE]

Â So, let, let me just multiply out this transfer function.

Â So, I can write that Vo over Vi is equal to K1.

Â K 2 [NOISE] and the numerator we still have our S over omega L.

Â And I'm just going to distribute these terms so I have S squared,

Â omega U omega O plus S times one over omega U.

Â Plus 1 over omega L plus 1.

Â And I can write this, this 1 over omega U plus 1 over omega L,

Â I could write that as the sum over the product.

Â So I can write this as K1 K2 s over omega l,

Â here we have an s squared, omega u, omega l,

Â plus s times omega u plus omega l,

Â over omega u omega l plus one.

Â Now remember we want this to be in standard form given by this.

Â So we want the numerator term, this is an S to the first power term, we

Â want this term to be the same as the S to the first power term in the denominator.

Â Which means then I want, instead of a one over omega l here,

Â I want it to be equal to this.

Â So I can re-write the transfer function like this.

Â K one, K two.

Â I'm going to bring the one over omega l out front,

Â and I'm going to replace it by.

Â And S times omega U plus omega L

Â over omega U omega L, and then in the denominator I can write this as

Â S over the squa.re root of omega U.

Â Omega L all squared plus

Â s omega u plus omega L over

Â omega u omega L plus 1.

Â Now, I have this in standard form because these two terms are the same.

Â The problem here is, of course,

Â is that these two transfer functions are not the same.

Â To make it the same I have to cancel out this term by multiplying this by an omega

Â u omega L over omega u plus omega L.

Â So now this transfer function is the same as this transfer function.

Â But it's now written in standard form where we can identify k, q,

Â and omega naught.

Â So let me make just one more adjustment to

Â this I'm just going to bring this term to this side.

Â So I'm going to have, write this as K1.

Â K2, omega U over omega U plus omega L,

Â and then we have our transfer function in standard form.

Â S, omega U plus omega L over omega U.

Â Omega l, s over the square root of omega u and omega l, squared,

Â plus s omega u omega l plus,

Â over omega u times omega l plus 1.

Â And now that it's in standard form, we can identify by inspection a mega nought.

Â Remember a mega nought is what is, what is sitting right here in this position So

Â I can write that a mega nought is equal to the square root, or the geometric mean of.

Â Omega U, the upper cutoff frequency, and the lower cutoff frequency.

Â And then we know that the term sitting here,

Â must be equal to one over Q omega knot.

Â So I can write that one over Q omega know is equal to mega U,

Â plus omega L over omega U omega L.

Â And then I can substitute this value for omega naught into here, so

Â I have that Q is equal

Â to, it would be omega U omega L.

Â Over omega u plus omega L

Â times 1 over the square root of omega u, omega L, is equal to.

Â So one of these square roots would cancel one of these square roots so

Â I can write it as the square root.

Â I'm going to make a U, I make it O, over, I make it U plus I make it O.

Â SO, here is one of our equations, I make a note in terms of I make a U and

Â I make a O, here is another equation, the quality factors in terms of the tape, and

Â then by inspection we can see that the game constant k must be equal to k one,

Â k two times omega u over omega u plus omega l.

Â [NOISE] And the form of this transfer function

Â it's a second order denominator and

Â the numerator consists of the middle term, the s to the first power term,

Â which indicates that this is a bat, band pass filter.

Â So by cascading a first order low pass filter and

Â a first order high pass filter we obtain a second order band pass filter.

Â But there's some restrictions to.

Â Creating a filter in this way.

Â And we can determine that with these equations.

Â Now remember, the overall magnitude of the transfer function would,

Â would look something like this.

Â A low pass filter and a low pass filter cascaded.

Â Where this is omega L, and this is omega U.

Â Now we can see that, if omega U and

Â omega L are far apart, in other words, omega U is much much greater than omega L,

Â then because omega U dominates omega L here, we can neglect this, this becomes 1,

Â and this overall gain here, k, is equal to k1 k2.

Â So if we pick K one and K two equal to each other,

Â we could have a flat band here equal to the product of K one and K two.

Â But, as omega U gets closer to omega L, you can see that the gain decreases.

Â So as omega L moves in this direction, and

Â omega U moves in this direction, we'll start to get filters that look like.

Â Like this, and then as we, the two frequencies get closer together,

Â you can see that the gain, from my picture here, the gain K, overall K drops.

Â Because as, say omega,

Â you were equal to omega at all, then this would give us a factor of one half.

Â K1, k2. Now look at

Â how the quality factor changes as a mega u and mega l become closer together.

Â So if a mega u were exactly equal to a mega l here, so

Â say a mega u equals a mega l, then q would be equal to.

Â Omega U over two omega U is equal to a quality factor of one half.

Â And remember the quality factor for

Â ban pass filter Q is equal to F not over the band width.

Â So if I keep our F not here a constant as I decrease the band width I

Â move omega U and omega L closer and closer together.

Â You can see that the bandwidth decreases, so the quality factor increases as,

Â as omega u moves in this direction and omega l moves in this direction.

Â So it turns out that this is the maximum quality factor that you

Â can get from forming a second order bandpass in this way.

Â So the maximum value of Q is one half.

Â And, as a mega U and a mega L move further apart, the Q decreases from this value,

Â approaching a Q of zero.

Â Right, a center frequency with an infinite bandwidth.

Â So you can form a band pass filter in this way but

Â there are some limitations on the transfer functions that you can implement

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