# Amar G. Bose: 6.312 Lecture 17

SPEAKER 1: Hi. I have an announcement to make for the Aero-Astro department. We're seeking [INAUDIBLE] stuff you guys have learned here, even plain acoustic waves to calibrate a [UNINTELLIGIBLE] stress sensor. It's a micro-machine, solid-state device, so you get to learn something about memspace electronics as well. [UNINTELLIGIBLE]

It's more geared towards electricals or people who have had acoustics. If you're interested, there's a few handouts here. They're also posted outside. You can contact myself or Professor Brewer over at the Aero-Astro department. If you have any questions, you can see me after class. Thank you.

AMAR G. BOSE: Last time, after class, someone came up and asked a very good question. And it's going to cause me to go back over something I did relative to the port model because, not only is that question important for all of you, but I made a mistake on the board which the question pointed out. And we'll correct it.

This was the case of the ported speaker, mechanical mass, moving mass, Mm, the resistance of Bl:1. And this now, what goes on here, had to do with a port. And the port was looking like this.

This is the cone here. And this was the area of the port. Let's let the area of the cone be Ac.

We made a model of this, acoustic model, of this ported structure which looked like this, where volume velocity was the through variable. Across variable then was pressure. Compliance in here was V0 rho zero C squared.

And not all the volume velocity of the cone went into the compliance because some of it came out there. And that's the part that came here. And it goes to ground through the port because we're assuming that the pressure is zero out here. So that was rho 0l/A.

Now in order to take this model and get it into this picture, here we have U sub cone as the across variable. And the force is the through variable. I call it U sub cone here, so it's simply because this moving mass is attached to the cone. And that just tells you that the velocity is the velocity of the cone.

This was your voltage across your whole thing, or whatever we called it, E electrical and current I. If we choose voltage here to be the across variable-- which is a natural in circuits-- and the current as the through variable, then we have no choice on the other side of the transformer. We have no choice because force is Bl times the current.

So the through variable must be force. And the across variable is velocity. So we don't have something here that we like that we can put on. So we can take the dual of that, which would just reverse the roles of the volume velocity, and the pressure would now be the through variable.

If we take the dual of that, you can probably see byt this time, you know that parallel elements go into series elements. But if you have any problem with that at all, just put dots in here and dot in the datum node and go through everything. This is the source, this value and this value, and put the dual of each element in.

And from that, you'll get this, in which this is V zero, the volume of this chamber here over rho zero C squared. And this is rho 0 l/A port. Now I did it again. This is the mistake I made last time, namely this, as we know, rho 0l/A port.

Now this is getting in the right direction to connect across here, but it's not quite there. It's getting in the right direction in the sense that it has something proportional to force as the through variable and something proportional to a velocity as the across variable. We have to get the proportionality constants straight.

And to do that, by the way, the best thing you can do is you write down what you want. You really want across variable as velocity of the cone. In other words, I can't hook this volume velocity here of the cone to this point because I have to have something that has the same variable.

So I want U as the across variable. U of the cone, I want not pressure, but force. That's what I had in the model. Force is the across variable.

Now what I have is volume velocity as the across variable and pressure as the through variable. And I have to get it over here. Well in order to make this an equal sign, if I put area down here, this will become force. If I put another area here, this will become velocity. Now, "what velocity," is the question. And what force?

Well in this circuit, we're talking about the force that's associated with the through variables in here, in this circuit here. And we're talking about the velocity being the velocity of the cone. So the force that would go down in here would be the same kind of force, if you wish, that goes down through the mass and through the resistance. And the across variable would be the across variable associated with all of them.

So what I have to do is-- this is a cone. And I just left it as A going zipping through last time. So what it says is each impedance which is in this circuit, a volume velocity over a pressure, each impedance must be multiplied by 1/Ac squared. So to get this into something in which I had a cone velocity here, U cone, instead of the volume velocity of the source, we would have the following circuit.

And by the way, you can choose to go from here and to a circuit that connects here by a transformer too, having the ratio in the right direction of the area of the cone. So then you could take this and stick it right on here and the variables over here would still be pressure and volume velocity of the cone. But you have a transformer in between, which gets you scaled over there.

But I'm just transforming it, so I don't see the transformer there. So this would become V0 over rho 0 CA cone squared. I think I just put A down. And this would be rho zero l.

I have to multiply the impedance by 1 over the Ac squared. I must multiply the capacitance, which is 1 over J omega C by Ac squared. So that is then the circuit that you put down here with that c Ac squared over A port and whatever this was, V0 over rho 0 C squared A cone squared. OK, and then, of course, your radiation impedance connects on the end of that simple model, being that.

And again, just in review, when you want to calculate, the only time at which this thing really resonates and puts out energy, and the pressure that it puts out is related to the voltage across the capacitor, which the voltage across the capacitor is the volume coming out of here, to get volume velocity out here, you just multiply by A port. Multiply the voltage you had across here by A port.

So volume velocity out of here plus the cone volume velocity gives you the radiation at whatever frequency you're interested in. This thing only radiates near its resonance, which is very low-frequency. The device is very small compared to the wavelength.

But the only thing you have to remember is that it's the volume velocity, which is U cone times area of the cone minus this voltage times the area of the port because remember that, when the cone was defined to go positive-- and just look at the circuits at low frequencies. At very low frequencies, when the cone is going this way, volume velocity is going in this way. And this circuit doesn't show that.

This circuit shows that, at low frequencies, this is very high impedance. This is very low impedance. This velocity is in-phase with that.

So I could choose to put a transformer in here with a minus 1 turns ratio and whatnot. But the whole object in making a model is so that it can be simple, and you can look at it. So it's easier to remember to subtract the volume velocity here from the volume velocity over there.

Oh, while we're on this, I asked last time or one of the times, or I was talking about putting a wall up behind the loudspeaker, and we got mirror images. And the way you could get rid of the wall is put another loudspeaker here and take away the wall because the boundary conditions were met. No normal velocity.

And then I asked the question about what happens to the pressure out here. The pressure out here doubles. And if the pressure out here doubles, when you put the wall up here, how could that possibly happen?

And then we got an answer to it, that what happens is the mirror image causes a pressure on here. And the pressure on here then causes more power to come out of the loudspeaker. And I saw a lot of looks like, wait a minute, how can this be? It's worth just talking about a little bit.

When you put a wall here, which is equivalent out here to putting another speaker back here and taking away the wall, the mirror image, when you do that, the first effect of that is really to increase this fellow and at the low frequencies. At the high frequencies, this thing radiates this way. This one radiates this way.

And the wall doesn't make any difference at all. So at the high frequencies, nothing happens. At low frequencies, when they each radiate omnidirectional, effectively, you've increased this capacitor.

Now when you've increased that capacitor, we saw that-- take the idealized speaker, transfer a function from here to any way you want. If you want to take it just to volume velocity, let's see, it goes like this. Take it all the way out, 12 DB, 6 DB, flat, 12 DB.

Now when you increase this capacitor, what actually happens is this thing is a-- oh, I don't know. It depends on the type of speaker and the frequency range. but this is smaller than this capacitor, but not 1/20 of it. Maybe it might be 1/3 of it or so. So at the low frequencies, by increasing the capacitor, you just decrease the resonance point by a little bit. And you lower the gain in the mid-band, as we saw when we made different equivalent circuits.

So now what happens is you can get all this extra power out of the speaker without getting more in because the impedance seen in here is, essentially, this at the low frequencies. This is the dominant one, if you look at it. When you're down in this region of frequencies down in here, you basically are seeing this impedance and a very low inductance which is coming from-- well it depends on whether you're ported or not at this point.

But this is about the dominant one. So the amplifier is feeding 99 watts into this to give you one watt out. And now you, all of a sudden, are getting two watts out. Fundamentally, what you've done is you've loaded this thing in a way such that the efficiency of the system increases.

Depending on where you are in frequency, you might have a tiny bit more power going into the thing. And actually, it's possible, I think-- let me see if it's possible to even have a load change out here that increases the efficiency so much that you've got less power in there and more power out. Yeah, no, I don't think that's possible in this model. But I don't think it's impossible either in a general situation that you could load something here, cause this thing to be so much more efficient now that you've got more power out for the given power in when there's a lot of loss in between.

But in general, what would happen in a model like this is the power in would be, essentially, the same. And you doubled the power out just because the system got more efficient. Any questions on that side? I want to leave that at the moment, if we can. Yeah.

OK, let's see now. Tell me. I need to know how far I went last time. Anybody have any notes? Did I get to the transformer?

SPEAKER 2: No.

AMAR G. BOSE: No, OK, all right. Ah, let's see. Did I get as far as a core which had two coils on it? Ah, good, OK.

V1, I1, V2, I2, flux going in this direction, let's see, yes, we got the resistive model of this thing which related flux and current. The resistive model was the reluctance, two sources, N1, I1 into I2.

This relationship came from integrating the curl of H around here. And that was equal to the MMF, the NI cutting that area. And we established the fact that you can get a relationship between currents and fluxes. We got that far, no? Yeah, OK.

Now that gave us, if you know the current, gave us the flux which was the through variable here, flux being related to MMF. And then F is HI, OK? HL, sorry is flux times the reluctance, just like voltage is current times resistance.

All right, now I think we also got a little bit further, namely, that the voltage-- yes we did. The voltage induced by a change in flux, the voltage induced here was in d phi, dt. And we went through the signs in this. And that came from the curl of E. Curl of E integrated around this core was integral E sub s ds.

And we found that, if the flux increased in this direction, it caused an E field, like it was going this way. But that E field caused the electrons-- this is open circuit-- to be pushed the other direction down here. And it caused this to be plus and minus.

And when that is plus and minus, of course, there is another E field which travels exactly opposite to that. Plus to minus, there's an E field here. And the summation of those two E fields had to be zero on the wire because there's no current flowing through it.

And sigma E, if there were a current flowing through this thing, then you would have that related to the E field by the conductivity. But there is no current because it's open-circuit. So that's how we established the sign of this voltage.

And that resulted in the complex ampli-- E equals nd phi dt. Again, if you plug in exponentials, E to the J omega t representation to go to complex amplitude, you immediately get V1 is J omega N1 I1. I think I got that far, no? Yeah, OK.

And V2 is J omega N2 I2. OK-- sorry. J omega flux, J omega flux, the N d phi dt. And this is, in complex amplitudes, d phi dt become J omega.

OK, now if we go to the goal of the whole business at the moment, the goal is to get at a relationship, a two-port relationship, between everything that's in this box and which you, as a user, don't care about at the moment what's in the box, if you have a set of VI equations that describe it. From a network theory point of view, that's where you want to go. And I'll come back to that later.

From a design point of view, you're interested in what's in the box and how do you design what's in the box to have minimum size, minimum cost, whatever your objective is. But we're now looking at the terminal. What I want to get is how does this box look from the terminals. So I want to get a relationship not between V1 and V2 and phi, but between and V2 and I1 and I2, two-port network.

So I already have a relationship for between I1 and phi right here. Just from this circuit, phi is equal to N1, I1 plus N2 I2 over reluctance. That's the current net network is the sum of the voltage, which happened to add up in the right sign here, in phase, divided by the resistance or the reluctance.

So all I have to do now is take these two equations with this and get rid of phi. And if I do that, I get V1. And then I have my two-port relationships. V1 is equal to J omega N1 times phi.

Well, let's see. That's going to be J omega N1 times something with I1, something with I2. I'll separate out the I1 and I2 because I'll get my two-port relationships like I like them then. So I will get J omega for the first term, J omega N1 squared over reluctance times I1 plus J omega N1 times N2 times I2 over the reluctance, again. That's the first equation, substituting phi in for here.

Second equation is I'll get the I1 on this side and the I2 on this side, so it's N1 N2. OK, omega N1 N2 over reluctance times I1 plus J omega N2 squared, N2 squared, I2 over reluctance. OK, that is a two-port set of equations.

Now I'd like to get a network. We go back to what we built as the background some weeks ago. We want to get a network that represents this.

Did I go this far? No. OK, now it's new territory. I can ask some questions. Here's my box. I want to fill in the elements such that I have V1 here, I1 here, V2 here, I2 here. Give me-- yeah?

SPEAKER 3: Just one question. Lenz's law, is the direction of phi related to voltage the way they did it correct?

AMAR G. BOSE: Yeah, well I forgot what Lenz's law is. But I went to Maxwell's equations and got it. So if Lenz's law says some-- Lenz's law has to agree with that. What's Lenz's law?

SPEAKER 3: Well, if we have a divided T possibly through that, shouldn't we be generating the one, the other polarity?

AMAR G. BOSE: Not for the way I wound the coil around there. And the reason I'm so confident is I went to Maxwell's equations. So if you tell me what Lenz's law is, either now or afterwards, I'll be happy to--

SPEAKER 3: We should generate a voltage which would oppose the change in flux through the coil.

AMAR G. BOSE: Oh, let's see if that works. We should generate a-- this is what?

SPEAKER 3: [INAUDIBLE] a current.

AMAR G. BOSE: Yeah, but the voltage that I generate here-- in other words, you're saying, even on an open circuit, I can't-- this voltage doesn't oppose anything if I don't have any current.

SPEAKER 3: If a current could flow, it would--

AMAR G. BOSE: If a current would flow. The only way a current would flow is if I had an external load on this, OK? So let's see what would happen. If I generated a voltage that's this polarity and I had an external load, current would flow out, OK? And so current flowing out would, in fact, generate a flux going down.

So Lenz and Maxwell agree. Yeah. Good. OK. Now two-port relationship, ha, ha, ha. Anybody want to give me-- well how many elements are going to be this thing? In the box?

SPEAKER 4: Three.

AMAR G. BOSE: Three? OK, yeah. This is a set of loop equations. In the normal two-port, this is a set that looks like this.

We have 1 equals Z1 1, I1 plus Z1 2 I2 of V2 equals Z2 1. But we know that these things are symmetrical, OK? That's why I chose the current going in, in this case. Plus Z2 2 I2, so there are three impedances.

I shouldn't have said how many elements. It's true. But in general, it's three impedances. You're going to make a network that looks like this, where Z1 1 looks like this.

Now, by the way, let's look at Z2 1. It represents the voltage that would be generated at terminal two on open-circuit. In other words, kill this.

This is Z2 1. So with no current coming out of terminals two, it's the ratio of the voltage at terminal two to the current at terminal one. The voltage at terminal two, with this thing equal to zero, as a result of the current at terminal one, I1.

Well let's see, if this current is zero, I2 equals 0. In other words, to make this term stand out, kill this term. How do I kill that term? Set I2 to zero. And it's the ratio of voltage at terminal two to current at terminal one.

Well current at terminal one, if this current is zero, all goes down here. And the voltage across here is the voltage V2. Let me say it again. There's the physical network realization.

Here's the equations. I want this thing to stand out. So the best way I can do it?

I say OK, let's set I2 to zero. Well that's easy to do. I open-circuit this thing. If I set I2 to zero, I want to compute V2 as a result of I1. So V2/I1, is Z V2/I1 V2/I1 with this thing zero, because I made that zero, is Z2 1.

I1 goes along like this, right down through here. It generates, as a result, a voltage across here. But that voltage across here is equal to V2 because there's no current flowing there.

So the Z2 1 is this impedance. It's the ratio of V2 divided by I1 when this is zero. So that is this impedance here.

So the impedance I have going down here then is an inductor, J omega, N1 N2 over the reluctance. Now see if you can fill in Z1 1. Ah, sorry, see if you can fill in this element. Anybody can give me that element.

We know there are going to be three impedances in this thing. We nailed one of them, this one. Now there's going to be an impedance here and there's going to be an impedance here. I want this impedance. Yeah?

SPEAKER 5: N1 squared minus N1 N2/R?

AMAR G. BOSE: Exactly. I don't know if I have room here. N1 squared minus N1 N2 all divided by reluctance, now what was your reasoning just to get that? That is right.

SPEAKER 5: Well to find what V2 was, you set I2 equal to zero. So that if I2 is equal to zero, just going back to V1, you know that it sees current I going through a total impedance of Z1 1, which has to be the sum of that impedance and the impedance there.

AMAR G. BOSE: Exact, namely, you wanted to make this thing stand out and you set I2 equal to zero. And then-- I'm just repeating what you said-- you set I2 equal to zero. And you want to get V1 over I1. It's the impedance that you see looking in here when I2 is zero. If I2 were zero, the impedance you see looking in there is the sum of these two blocks. And the sum of these two blocks must be an inductance of value N1 squared over R.

So when you look into this network, you already have this fellow in there. And the sum must be N1 squared over R. So this fellow must be N1 squared over R minus whatever this one is. And if you do exactly the same thing on the other side, you'll get here N2 squared minus N1 N2 over reluctance for this inductance that's here.

And now you have the whole thing. Any questions about that? Doesn't the minus sign bother you? Or can you see that that will always be a positive element?

Think N1 N2. You're free to just suppose. Can you all see that this will always be a positive element? How many can see that? Nobody.

Now let me go the other way around. Can you all see that it'll be negative? Somebody can see that it might be negative. How did you come-- what? One of these has to be negative, yeah.

Let's pick an N1 equal to 1, one turn, and let's say an N2 is equal to 2. So this fellow is 1. This is minus 2. This fellow is 4. And this fellow is 2.

So this is positive, that's negative. Ah! So we can have a negative element.

Now this is a very good illustration of something, that when you see a negative element in something like this, it does not mean that this thing is not realizable with positive elements in a different structure. It means that, in that structure which we have forced this two-port into, what's actually-- well, there are many different things that could be in this box. And they don't all have to obey this thing, which was the simplest realization of a two-port network.

When you force it into three impedances, you may get negative elements that might not have to be negative in another embodiment. And let's see. Look at this, for example.

Suppose I tell you the following thing. What I'm going to do right now wouldn't be something that you would jump to all by yourself the first time around. If you have more experience with circuits or transformers, you might come across something like this and you would be tempted to do it.

But suppose I tell you that you're allowed to have a transformer in that box, an ideal transformer. This is an ideal one. And let's just look at this circuit and see first. Let's say here are the terminals of my box like here and here. And I wanted to make a network in here which was equivalent to this, i.e. to this set of equations, which is what I get from the physics.

If I wanted to make just the driving point, suppose I just said, well gee, if it's going to be equivalent, it's got to be equivalent for the driving point impedance. What element or elements would you add here to make just the driving point, if that's your only concern, make the driving point impedance the same for this thing?

The box is here. What would you do over here? This is I1. This is V1.

SPEAKER 6: [INAUDIBLE]?

AMAR G. BOSE: Series inductance? Let's see, if I have a series inductance in here, the driving point impedance is what I would get. V1/Z1, this is Z1 1 with I2 equal to zero. Now if I put I2 equal to zero here, what's the current over here?

SPEAKER 7: Zero.

AMAR G. BOSE: Zero, so that won't work. Any other ideas?

SPEAKER 3: You could branch on a magnetizing branch so it's shunting you into series?

AMAR G. BOSE: Yeah, but [UNINTELLIGIBLE] magnetizing are things we don't know about yet.

[LAUGHTER]

AMAR G. BOSE: Nobody knows what they are, I think, except you, maybe, or one or two. But your idea is OK. Let's try an inductance in parallel with the transformer.

If I kill I2, set it to zero, then the current here is zero. And everything I see looking in here is this inductance. So this must be N1 squared over reluctance.

I killed I2. I look in V1/I1, I see an inductance of this value. Now that would make the driving point impedance the same.

Now suppose I took a transformer and I just put a turns ratio on it like this, N1 to N2, ideal transformer. What would be the impedance that I see in here? The transformer steps up the voltage by the ratio steps down the current by the same ratio.

So if I have N2/N1, looking in here, I would see an inductance of value N1 squared over reluctance times N2 squared over N1 squared. Aha! That's exactly what I would see for this circuit if I kill I1. And I'm looking for this parameter, Z2 2.

So far, doing this and this got us the two-- by the way, there are only three things you have to worry about. If you have them, you've nailed it, all right?

Where are they? Yeah, Z1 1, Z2 2 and one these, they're equal. So you'll have the network. So we now have this fellow, m this fellow, OK. Wouldn't it be nice, wouldn't it be lucky, if we had this one OK? We'd be all done.

So let's check this one. With I2 equal to zero, this Z2 1 is the ratio of V2:I1. So I stick a 1-amp in here, if you wish, and V2 would be Z2 1.

OK, let's see. If I take a look at V2 over here, this is V2, V2 is whatever this voltage is times N2/N1. Well put an I2 in here and you'll get this voltage to be J omega N1 squared over reluctance times I1 is the voltage across here.

But that voltage times N2/N1 is the voltage across here. So now this is V2 with I2 equal to zero. So that all comes out to be J omega N1 N2 over reluctance times I1 is equal V2 with open-circuit I2.

Ha, ha, ha, everything checks. So here is a simple device which represents that transformer up to the point that we have modeled it. This is equivalent to this. And this fellow has positive elements and an ideal transformer, which is one of our elements, if you wish, so something like that.

I want to give you an example so that, one day when you model something that's totally new and you get a negative, you go, oh my God. I've got to have power sources and this and that in there. I have to synthesize it actively. Not necessarily.

And there are some things which we won't go into in network theory now. But it's interesting, when you get a matrix like this, to know whether you actually can realize it with passive elements or not. That's more advanced. In network theory, you can find it out. But in these simple ones you can move right along without knowing that.

OK, now one thing that we modeled-- let's see where I modeled the thing, there it is-- wasn't really-- well, a number of things aren't representative of the real world. But what happens in cores like this? The reason the flux runs around here in the first place and doesn't go all over-- why does it follow this path? It's only simple.

It follows the path because the reluctance of the path is very small compared to that of air. Remember reluctance? How did it go now? It was the length over the area and over the mu, yeah?

So now for-- whoop, over the mu. Is that right? The higher the mu, the less the reluctance, yeah. OK. And this was just like the resistance l sigma A for resistance.

Now if something has a large mu, like the iron, if it has 1,000 times that of air, the reluctance is very small. So the resistance becomes very, very small. And all the flux goes around there.

This is exactly analogous to if you had a conductive blackboard, if you wish, and you painted on this part of the blackboard. Let's say you had a poorly conductive blackboard, and you painted on the blackboard, this part here, with a highly conductive material. And you put a voltage source in here, right in series with it.

We would expect that all the current would go this way. Why would it go to a higher resistance path? But some would. Some, however small, would.

And that's exactly what happens in the magnetics. The calculations are just the same. Some of the flux goes like this. Some of the flux goes like that.

In other words, its flux, how it's very small compared to what's going in here because the mu of this thing is very small compared to the mu of this. Just think of it as the resistive case and you can see with the voltage source in here.

So we have a thing called leakage flux. And the leakage flux is defined as the flux that is caused by a current in this coil that does not link the other coil. Similarly, the leakage flux for this fellow here is flux caused by the current in here that links only this coil, does not link the other coil.

That is called leakage flux. And now it turns out that that is an extremely important element, as you'll see in a few minutes, when you want to embed a transformer in a circuit. And so what we need to do is find out what does that leakage flux do to the model that we just constructed?

Let's see now. I think I have most of the equations up here for the model that we just built. Good. Now we have only one thing new in this model, that the orange flux exists now.

Let's say we'll call this-- I'll represent it all on one side here. The actual flux clearly goes around both ways. And its strongest lines are right down here.

It has weaker lines that go around here through bigger paths, et cetera. I'll just call this all phi leakage 1. And I'll call this whole thing-- I've already made it on the complicated side here-- I'll call this phi leakage 2.

OK, so the only thing that's going to change in this whole story now is a result of these two leakage fluxes. And we want to find out what happens to the model. Or it's easier to think, at least for me, in terms of that model, so we'll alter that model.

OK, let's see. Well the model for this with the leakage inductance in it, for the relationship between current and flux, is pretty simply related because you just have the same voltage source, the NI. And you have flux leakage 1 going down through here. And this is a R leakage 1.

You similarly have another one out here, R reluctance leakage 2. And the flux going this way because the source is this way, flux leakage 2. I want to make sure that you see this. In other words, this is really a voltage source across two paths. It's across this path and it's across the other leakage path, so the resistive circuit changes just like that.

Now there's a slight error in what I told you, and I don't want to tell you what the error is now. Just if you accepted what I just said, please keep it that way for a little while, and we'll come back and talk about what the error is. If I tell you what the error is now, the whole thing will look complicated.

OK, so I have a way of calculating the total flux that's flowing, let's say, through this path. The flux that's flowing through this path is this flux plus the flux that went over here. If I can find the flux going through that path, that's all I need to get the voltage in here because we have the relationships already that the voltage now would be J omega N1 times the flux that went through the path plus the flux leakage 1 that went through the path. And the voltage, V2, would be J omega phi times the leakage flux, OK?

Now everything that's in orange just adds the leakage. It just adds the effect of having leakage flux. So now we have to get down to altering these equations.

So to get from these equations to these, the only thing we did was substitute. To get rid of flux we substituted this and the terms of I1 and I2. Well similarly, there's a relationship between flux 1 and the I1 and I2.

In fact, it's only I1 because this is a voltage source across here, N1 I1. That voltage source is across here, this path, and it's across this path. So to find the additional flux that's in here, I only have to take N1 I1 and divide by the reluctance 1 to get the additional flux that's in there.

Maybe I'll write it all out once, instead of just using the changes. Here we go. Voltage 1 is J omega N1 times the flux. And the flux was this part here, N1 I1 over reluctance 1 plus N2 I2 over reluctance.

Plus I now have this fellow to contend with. I just put this in here and got what we got before. Now I have to contend with this term, J omega N1 times this. But this is N1 I1 over reluctance 1, so N1 I1 over reluctance leakage 1.

OK, so let me write this as V1 now is J omega N1. I'll gather all the I1 terms. J omega N1 squared over reluctance. That'll take care of this term, plus J omega N1 squared over leakage reluctance 1 times I1, OK? And then, of course, I have J omega N1 N2 I2. So much for that equation.

Now the second equation and the last one, V2, I have to take into consideration this only involves I2. So the terms for I1 are the same as they were before here. I'll put them down.

J omega N1 N2 I1 plus the terms J omega N1 squared over reluctance-- this, N2 squared over reluctance, N2 squared over reluctance times I2-- but I'm going to save that for a moment-- plus the term due to this. And this is nothing more than the flux that's going down here is N2 I2 divided by the reluctance leakage 2, J omega N2 over reluctance leakage 2, and now times I2.

OK, so we now have this set of equations for the VI relations at the terminal versus this set. Take a good look at the difference between them. The difference between them resides in this term and in this term, that's all. Now think about that and tell me what I have to do to either of those networks to include leakage.

SPEAKER 8: [INAUDIBLE PHRASE].

AMAR G. BOSE: I left out a--

SPEAKER 8: [INAUDIBLE].

AMAR G. BOSE: Yeah, thank you.

SPEAKER 8: [INAUDIBLE].

AMAR G. BOSE: OK, but-- oh, oh, thank you. Yeah, theta, that's by definition, almost, huh?

OK, only difference between this set and that set are these two terms. They occur in the diagonal terms. That's a hint. So I ought to be able to go over to this network, either one of these networks, and change them immediately and have the whole problem leakage taken care of.

In the loop method, which is you can consider this as a loop method, what are the diagonal coefficients? The sum of all the impedances going around the loop. The off-diagonal terms, are the coefficients, are the impedances that map between the two loops, the loop you're writing the equation for and the loop for which it's a coefficient for, between one and two. Yeah?

SPEAKER 9: Shouldn't it be the off-diagonal of the different side?

AMAR G. BOSE: Ah, they would be. If I had turned the current around, they would be negative. The two off-diagonal terms would be negative. I just happened to choose the current in this direction, so I got a bunch of positives.

OK, so what I've changed from that network to this network is I've added something to the self-terms, the diagonal terms. I've done nothing to the mutual term. So if I've added something to the self and done nothing to the mutual, what have I done to the end network? We need to spend more-- yes?

SPEAKER 10: Put one inductor on this side and one inductor on the other side.

AMAR G. BOSE: Yeah, this inductor is the added term to the 1, 1, Z1 1, which is N1 squared over reluctance leakage 1. This fellow, which is added to the second term, N2 squared over reluctance leakage 2. Done. That's all that it did.

In recitation, you need more. I didn't get enough response, so you need more attention. You will need more. Can you ask any questions?

I expected more of you to be able to see what the change was. And if you can't see it and there's a problem, we haven't presented it in the best way possible or something. So any questions? No?

SPEAKER 11: I just--

AMAR G. BOSE: Yeah?

SPEAKER 11: Can you assume that the two leakage reluctances are the same?

AMAR G. BOSE: Can I assume that the two leakage reluctances are the same? In this particular problem, the way I've drawn it, yes. If the iron is a different configuration on the right side than it is on the left, which it wouldn't be true for a simple transformer like that, then it would be no.

But the leakage reluctances, for example, wouldn't be changed by the number of turns you have here, if you changed that, because the mu of the copper is the same as air. And so it's determined by the geometry of the iron and that's all. So in the case I've drawn two leakage reluctances would be exactly the same.

Now let me take a look and see what all of this means when you put this in a circuit. And that's where it gets interesting, namely, if you use the transformer. They're used in many, many applications, high-frequency, low-frequency applications. Because vacuum tubes weren't a good match to loudspeakers, so they were always used in amplifiers.

Now when you put something like this in-- and by the way now, a term that was used earlier, we can introduce the magnetizing inductance and the leakage inductance. This fellow here, in people that design transformers, is called magnetizing inductance, magnetizing l. And this fellow is called leakage, for obvious reasons, leakage inductance.

But now think of this transfer. Suppose you have just a resistive load out here at high frequencies, and you have a voltage source across here. The transfer function you'll want to this point, at high frequencies, how is that transfer function going to behave? How's it going to go as a function of frequency when you go high enough?

Think of making a Thevenin and, just mentally, looking back in here. Going to be any C's in the Thevenin? Any capacitors? No.

Going to be any R's in the Thevenin? No. There's no R's in the circuit yet.

There's going to be one L. The impedance you see, it's going to be the parallel combination of the-- to make a Thevenin, you go back and you set your sources to zero. So that nails this point to zero.

This is in parallel with this. And that just appears as an inductance from here to here because N2 over N1 squared quantity times whatever that other one was. So you have one big inductance from here to here.

And your Thevenin then has this as an impedance and has some source. And the Thevenin is quite good for determining anything outside, not inside. This thing doesn't exist, OK?

Now how does that circuit behave at high frequencies? Transfer a function from here to here. It goes as-- I see all kinds of things-- as?

SPEAKER 12: Low-pass filter.

AMAR G. BOSE: Low-pass filter. How many DB per octave coming down?

[INTERPOSING VOICES]

AMAR G. BOSE: Six DB per octave. Yeah, yeah, yeah, the pole pattern of that thing, the transfer function, you should be able to construct like that. So now it turns out that most of the things you really want to drive don't exactly look like a resistor. When you go high enough in frequency, they look like a capacitor.

If you had a capacitor here instead of this thing going down at 6 DB per octave, when the capacitance became important, it goes down at 12 DB per octave. Then a very interesting question would be-- well I won't ask the question. What would happen at low frequencies to this circuit? You should think about it.

But now it turns out that it's very hard to buy wire that has no resistance and so very expensive. So basically, all of these things have a resistance in here. And when you go to high enough frequencies, all of the coils have distributed capacitance between the windings, et cetera.

To a first order, you can represent that in terms of little capacitors across here. And if you do all that, you have quite an accurate model, actually, of what goes on. And you find out that you do have resonances in transformers that you don't like, especially if they're communication-type transformers.

If you're designing a 60-Hertz transformer, it's a totally different story. You just need one frequency. But if you're designing, say, a transformer for audio bandwidth communication, you need a bandwidth. And you have to worry about how this thing goes.

Now let me see if I could get an answer from you for another question. I'll tell you a fact and then see if you could tell me why.

In the days of audio transformers, the ones that were high-quality for audio were always bigger. You could buy the low-quality ones, and they were small. Now is there anything that you think we have done that suggests why a, so-called, better performing transformer would be larger, obviously, more expensive too?

It's on the top board. This question's open to anyone even the engineers that are in here. Yeah?

SPEAKER 13: Being a larger force, you won't saturate the lower frequencies.

AMAR G. BOSE: Let's see, suppose I were going to impose a certain voltage that I want across this transformer, across this thing here. The voltage that I have across there determines, what I have picked in one, it determines the flux. Now if I have that, I want that same voltage to be applied at low frequencies as well. In order for this to be fixed, it says, holy smokes, I've got to increase the flux in the circuit or I can't do that.

It's another way you can look at the relationship between flux and current. So I can get this voltage by upping N1 or by upping the amount of flux. To up the amount of flux, flux is B times area. And I can't get the high B's because the darn thing saturates.

So I have a maximum B. And the only way I can get more flux is to increase the area of the core. And it's only a problem at low frequencies because this thing is very small. And I'd better make this very big, or this.

Now if I choose to make N1 a heck of a lot more turns, what happens is then we introduce these resistances here with the wire. In other words, I have many more turns of the wire, I've got much more resistance. And I have big losses in the transformer, let alone what it does to the frequency response. Yes?

SPEAKER 14: But if you increase the core area, then that means that you're also going to have to-- oh, no, never mind, I guess. I was thinking, if you increase the core area and you keep the length of the wire the same, then, of course, the number of turns is actually going to go down. So you get no effect. So you have to keep the-- OK.

AMAR G. BOSE: Yeah.

SPEAKER 14: But that means you have to have a longer wire. So doesn't that--

AMAR G. BOSE: Up the resistance, exactly. So what I'm showing you now is that the equations are simple. The number of variables is pretty small, the number of turns, the area of the core. But you have to play with these things to see what an optimum design means.

So there's definitely a limit on this. If you could make this big without getting resistance in there and whatnot, that's the way it would be. You might have a little problem, that you got it so big you couldn't fit the coil on the core, minor problem. But basically, there's a limit to this, you can see, and so then you have to make this bigger.

This is why, in communication transformers that are used in RF, for example, the darn things are this big. And they can handle a heck of a lot of power, big V and I, which is power. In other words, here, let's see if I have an expression, yeah. If I have big I here, I'm going to have big flux, OK? If I have big V, I have big flux. But who cares about it? It doesn't happen, if the frequency is high.

So basically, the high-frequency transformers are small. Low frequency transformers are big. And the reason that Hi-Fi transformers cost so much is that they would be rated by how far down in frequency they would go at what power level. And the further down the transformer would go, smaller omega, the bigger phi you had to have, the bigger the core that you have.

OK. Oh, let me just see, magnetic circuits, we did. Transformers, we did. Permanent magnets, we did not do.

Now that will be next time. And we'll find out something very interesting about permanent magnets which will come right out of the same basic stuff that we've been dealing with. But if you didn't know it, with almost certainty, you would fall into an enormous trap. So we'll see you this next time. Yeah?

SPEAKER 15: [INAUDIBLE PHRASE].