# Amar G. Bose: 6.312 Lecture 16

Search transcript...

DR. AMAR G. BOSE: Last time, we set up this equivalent circuit. And we talked about what each of the elements were representing-- the electrical side, the mechanical side, the acoustical. Then we came to the box that's behind the speaker. And the reason that the box encloses the loudspeaker is to prevent the back wave from getting to the front wave.

If you go home and you have a hi-fi set, and you pull the woofer out and listen to it, it's lost all its woof. It's no low end. And that's because it's a dipole source instead of a monopole source. So you make it a monopole source by putting it in a box. But the box becomes a spring. and one end of the spring is attached to the cone, which is this moving mass here. And another is not attached, but one end moves, one end of the box. The backside of the diaphragm is what the compliance moves with.

In other words, when this goes in, it squeezes the spring, the air spring. And of course, the other side of the air spring is to Earth. And so the compliance comes right across here, as a compliance of the box, right in parallel with that of the spider and that of the surround. And you can combine them into one spring.

Force is the through variable in here. And therefore, cone velocity, or velocity of the moving mass, is the across variable. And we have chosen the radiation impedance to have the same variable. So we didn't need a transformer.

If you chose the radiation impedance to be a pressure-- well, you wouldn't choose it as a pressure over a velocity. You'd do the reciprocal of that, because you have to have something proportional to velocity across here. So if you might have chosen it as volume velocity over pressure, well, then you could put a transformer between here and here, which had the right area factor in it. Or you can just use the area factor and get out of the transformer.

Now, the only thing I wanted to add to this today is a kind of a box that you often see-- and you certainly see it in textbooks-- which has a tube like that in there. And you want to take a real look at what this does, because the way you will see it in all the texts that, at least, I know of, gives a way that it has been used for decades and decades, but I think isn't the way that it ought to be used.

Now, let's look at this structure. We want to find out what do we have to add to this schematic to handle this. Well, let's just look at this by itself. When you're trying to add something to this model, variables, or whatever they are, velocity, and force, and you're looking at a structure here, don't go back and try to say, well, OK, my variables here are force and velocity. Use the variables here that you feel most at home with. Then convert the resulting network or impedance, whatever, to match the variables that you have.

Now, perhaps your thinking is like mine on this, that when you're looking at a physical structure, it's easiest to think-- an acoustical structure-- of volume velocity as a through variable and pressure as an across variable. If you do that, what you see here is if you push this cone inward, let's say, you squeeze the air, but some of it gets out over here. In other words, there's a volume velocity feeding in here. And most of that-- or at least some of it, but it turns out to be most-- most of it goes into squeezing the air in the box. But some of it goes out.

So you look at that structure. And you say, a-ha, I have a compliance. And a compliance is if I choose my through variable to be volume velocity and my across variable to be pressure. Then a compliance goes into a capacitance, C sub acoustic. I want it to be a compliance, and I want it to be acoustic.

But I don't want to confuse it with the C sub m up there. That M is meaning mechanical compliance, because the variables are velocity and force. You'll know what it means. It's a compliance, acoustic.

Now, this would be the model if that didn't have the port there. In other words, VVS would be out here. The volume velocity that I would feed in could be out here. And then all that volume velocity would go to squeezing this.

But that's not the case. Some of it comes out. And so it comes out through something that has, essentially, 0 pressure out here. And the mass of air in here, as we know, behaves just like a little piston. And that comes as a mass in our circuit, rho 0. l over the A, where A is the area of the port.

And this would be V0, the volume of the box, minus the port, actually. But the port usually doesn't take a big proportion of the whole volume. So doesn't matter. You can subtract it if you wish. V0 over rho 0, C squared.

So that would be a model of what actually went on. Volume velocity goes in. Some of it goes to squeezing the air in the box. Some of it goes out the port. Now, we don't want that model. We want one in which the across variable is proportional to velocity, because I want to hook it up into this system. And the through variable is proportional to force. Let me just invert this thing first. And the easiest way to do that is take the dual. Then I'll get pressure as the through variable and volume velocity as the across variable so that then I can scale that to velocity and force.

So if I go for the dual of this, if we do it the way that we established before, we could just write three nodes, connect all the nodes through one and only one element. And make sure you've gone through every element. Then those nodes become like this. The source here that was a VVS, let's say, that was a current source. So it goes into a voltage source of VVS. Don't need to put a source on there, actually. I can think of this port as just having a velocity and a pressure. And it doesn't have to be a source. But I stuck one on, so I'll take the dual that way.

This element goes between the two nodes. And it becomes an inductor, again, of the same value as the capacitor, V0 over rho 0 C squared. And this one that goes between this node, the right-hand node and ground, becomes a capacitor of value rho 0 l over A.

And now our through variable in here is the same as what the across variable was before pressure. And the across variable is now what the through variable was, volume velocity. So we're almost there.

We have now an impedance, which is defined, in terms of the cross variable, VV over P. But I want U sub cone over force. So VV over P. What do I have to do if I divide this by A? Then VV over A is U. And if I multiply this pressure by A, I get force. So 1 over A squared times this impedance is equal to that.

So in order to get that, I simply take this impedance, defined as VV over P, in this case, divide this by A squared, multiply this by A squared to change its impedance, because upside-down, 1 over j omega, the capacitance. So I multiply this by A squared, divide that by A squared.

I have my new network that represents the port in terms of the variables that I want to see, namely, force as the through variable and velocity of the cone or velocity as the across variable. So I'd have V0 over rho 0 C squared A squared. And I'd have over here rho 0 l times A.

Now, where do I connect these two terminals to the-- all I've done here is give a velocity force relationship for this thing here. How do I connect this into the existing circuit? If you're having a problem there, let me back up here and separate out these compliances that I put together before. One was the mechanical compliance of the suspension. The other was the mechanical compliance of the box for just a box speaker, a speaker without the port.

Now I want to change this model so that instead of having just a sealed box, I have the ported box. Anybody have any idea what kind of a change I'm going to make here?

MALE SPEAKER: The capacitor in a series with--

DR. AMAR G. BOSE: The capacitor in series with this--

MALE SPEAKER: The box.

DR. AMAR G. BOSE: Exactly. Now, that makes the assumption that when I stuck the port in the box, I had the same compliance I had with the sealed box. That's OK. I've taken up a tiny bit of volume. So in theory, I should reduce this compliance by the amount of the volume taken up by the port. But that's negligible. I won't bother.

So that comes in there why? Because this now, as seen from here and here, this is the impedance that this speaker sees looking back into here. Now a question about the radiation. Something gets out. That's why it's here. Something gets out of here. And something's coming out of here. So when you're standing in front of this box, you now have radiation not just from the cone, but you have radiation from that.

Where is that radiation, the port radiation, represented in this thing? I know where the business end is for the cone. It's right here. There's a cone velocity working into a radiation impedance.

By the way, if I just know this-- if I calculate, put a source on there, calculate the velocity across here, the voltage across here, that gives me the velocity of the cone. And in fact, at low frequencies, that's all I need to find out what the radiation is, the volume velocity of the cone at the low end.

And by the way, I'll just say another thing, which isn't obvious from the beginning here. But this structure, as we will see in a moment, only radiates at the very low frequencies. If you took a woofer, for example, made from a decent speaker, that would be designed to radiate around 50 hertz or so. So at that point, the wavelength is huge. And all you need to do to find the radiation is find the volume velocity from here and add it to the volume velocity from here. That'll give you the total volume velocity. And that's all you need to do the radiation, to find what's radiated.

So where is the point that I would go to in this network to find out what the volume velocity is from here? I know the final velocity from here, that's this voltage. Where's the volume velocity from the port? Yeah?

MALE SPEAKER: When you did this, and you used an impedance model, didn't you invariably assume that the pressure at the output of that tube was 0, and therefore, the radiation impedance was 0?

DR. AMAR G. BOSE: No. Actually, remember when we look at opened tubes, we said, look, as far as what goes on in the tube is concerned, we can consider this pressure as 0. Because it's not going to be big enough to affect anything that's going on in the tube. So we find out what's going on in the tube. And under that assumption, we find out that this part here is acting like a little mass. And that's a volume velocity. So we can calculate what the volume velocity is under the assumption that the thing is 0, because remember, those radiation impedances went down to 0 as you got down to low frequencies. So it's OK to calculate the volume velocities if the pressure were 0.

Now we have the volume velocity. And what I'm really looking for is where could I get-- well, in this network, clearly you have two variables, velocity and force. So I won't go to a point in that where I'll get volume velocity. But give me a point where I get the velocity of this piston. And then I'll just multiply by the area of the piston. Then I have the volume velocity.

MALE SPEAKER: [INAUDIBLE]

DR. AMAR G. BOSE: Voltage across the inductor, here to here. Hm?

MALE SPEAKER: Capacitor.

DR. AMAR G. BOSE: Capacitor?

MALE SPEAKER: Yeah.

DR. AMAR G. BOSE: How do you know?

MALE SPEAKER: It's the velocity of the mass, right?

DR. AMAR G. BOSE: Right. It's the velocity of the mass. And the mass is the capacitor. In this-- it wasn't a few minutes ago. It was an inductor. Now when we took the [UNINTELLIGIBLE] of the thing, it's a capacitor. So yes, this point here is the velocity of the port, U sub port, right there, measured across that capacitor.

When you look at sines, you can decide to trace all these things through carefully from the beginning. The way I always tend to do is put the thing up. And then I look at what happens physically. And then I know what the sine is. Now, physically, you know very well that if you push this cone in nice and slowly, the air is going to come out of the port. So as you get down to near DC, when the cone is moving-- we've defined this as the positive direction for the cone. So when the cone moves in its positive direction here, the velocity here is going back in.

Well, go down in frequency in your model. Go down in frequency in the model. And what happens here, in this, is that this velocity becomes equal to this, because there's no drop across this inductor. The inductor gets to be very small. So the velocities are equal. So in terms of getting the total volume velocity that's actually coming out of the physical structure, what you have to do is you take this thing times the cone area minus this thing, the U sub P times the port area. Because you know, in the physical structure, that at low frequencies, they're going out of phase.

So the total volume velocity, VV total, would be the cone, the velocity of the cone, times the area of the cone. Let's just-- don't really need-- times the area of the cone minus the velocity of-- I'm just going to call that U sub P here-- times the area of the port.

Now, interesting. Take a look at this. Oh, let me first tell you what the textbooks tell you to do. This is a, of course, highly resonant circuit. So this is going to go like heck near the resonant frequency, because all the-- near the resonant frequency-- resonant frequency between two elements is defined when their impedances are equal and opposite. This is equal and opposite of that. This is a short circuit, therefore, across here. So all the force that comes out of this transformer says, to heck with all these elements. They have a finite impedance.

This fellow has 0. All of it's going to go down here. Tremendous amount of current going down here. Big voltage across here, but of course, an equal and opposite voltage across here. So there's no voltage here. But there's a heck of a lot of voltage here. The amount of voltage here is the force that's coming out of this transformer times the impedance of either one of these elements. Those are equal. So that's how you'll get a huge voltage here and no voltage here. That's a normal situation of resonant circuit.

So what happens is that's a highly tuned resonant circuit. So you get a radiation versus frequency that's big right at the resonance frequency of this and very small outside. And so what they do and what you'll see in texts is, here's the idealized response without a port. And this thing here is the fundamental resonance. They'll say, now, tune the port to a frequency just below the fundamental frequency.

And the net result of adding these volume velocities-- I'll tell you the result first, then we'll go back and look a little bit more-- is something like this. This is the volume velocity without the port. The volume velocity contributed by the port only has real action around here. So you get a thing that looks like this.

And then, of course, it goes down like heck. Let me write that in. It goes down here much faster than the other one. Why? Because now, at low frequencies, instead of just having a spring back in here, when the cone moves forward, the port sucks in the air, and vice versa. So it goes down much faster here. And you get a little bit of a boost here.

Now, actually, there's been a lot in the folklore literature of the audio magazines and, to some extent, Audio Engineering Society. A lot of good articles have appeared in there on many subjects. But an enormous amount of folklore has appeared in that magazine also. And they'll talk about port speakers not sounding so good, and this and that, and somehow, distortion. Well, it's easy to see where the distortion could come from or does come from.

What happens is, when you put the port in here and you're below the port frequency, there is nothing, there's no longer a compression back in here, because all the air goes out here. And so when you put a voltage across the speaker terminals, the speaker tends to go like heck. There's no radiation. But the speaker is traveling a long distance, because there's nothing, except its own suspension compliance, to stop it from moving. There's no longer the spring back here that you had when the box was sealed.

And so the cone tends to move a lot. And when it moves a lot, the coil goes out of the magnetic gap. And when the coil goes out of there, you may be feeding other frequencies in, of course, as you would be normally up here. But the low frequencies that you feed in are causing the coil to move out of the gap. And that's causing the other stuff to get modulated by this.

In other words, take the extreme that it moves all the way out of the gap. There's no sound during the time it's out. And there is sound when it's in. It doesn't go that extreme, but that's a way to see it. And so the frequencies that are down here that aren't radiating at all because they're heavily attenuated, but they're in the source that goes to the thing, that goes to the loudspeaker.

An organ is playing down here. And a flute-- well, a woofer wouldn't be used for a flute. But maybe a cello or something is playing here. And every time this goes out of the gap, that's affecting the sound that you're getting. It's modulating one frequency with the other. The frequency that you have down here gets modulated on this. That's why the ported speaker is thought by many, and for real reasons, to have distortion, normally.

And you get such a little gain in here that it's a question whether it's worth it. And you saw manufactures, some sticking to one philosophy, some sticking to the other, changing with time depending upon the reviewing process or the trends of the day.

Now, it turns out there's another way to use the ports entirely. But you can only do this when you take a system point of view to your engineering and acoustics, which, in the audio field, is saliently missing, or has been saliently missing.

Namely, a system point of view would say, there's an orchestra playing here, and there's you in your home. And everything between there is the system. And you design each component to go with every other component, much as you would do in any other field of engineering.

You'll never find-- at least, I don't want to go up in it if you find it-- a shuttle going up into space in which the ground control computer was designed not knowing which shuttle it was going to control. Well, this is exactly what happens in the audio field where system engineering is somewhere else. They design the amplifiers to have a nice, flat frequency response. And you can supposedly connect them to any loudspeaker. And you're supposed to get something that's good.

Well, you have many deficiencies in one component that you can make up for in another. And when you realize that, you can do all sorts of interesting things.

For example, you take a look at this port. And you could do something very interesting. It turns out that if you look at what kind of excursion you need to get this volume velocity at the low end-- and that's easy for you to do now; you have a model-- the excursion goes up as the frequency goes down. And that's what limits your distortion, of course, if the coil moves out of the gap. So this thing kills the excursion.

In other words, where the port is radiating, it turns out that the cone is hardly moving at all. It can't move, because there's a 0 impedance across here. So a-ha, that gives you an idea. Now, why don't you take that down.

Suppose you wanted to do this with a system design now. You could say, I want to extend my frequency response way down here. So I'll put my port at the place where the biggest excursion's likely to be. And I know that at the port resonance, everything's going to come out of the port, nothing out of the cone. And you park it down here somewhere, where you have a big radiation. And then you don't have radiation in this region much. It comes like this. And you equalize that electronically to come up.

So you've taken care of your excursion problem at the worst place where it can exist. And you've got a funny looking frequency response. But so what? You equalize that electronically in the system approach.

This is why this notion of going to the store and buying a loudspeaker from one manufacturer, an amplifier from another is as archaic as the field of audio, 1948 or '49. And it's because the companies that went into audio didn't have any engineering. Had RCA gone into it with the stuff that Olson had done at the time, it would've been a different world. They would have seen things like this. They didn't at the time. Olson was writing his books in the '40s. But they would have been the first to realize something like that.

So with more knowledge and less reliance on the correctness of what's in the book, but a willingness to go step outside of what's in the book and say, wait a minute, what's going on, you can use a structure like this very, very effectively. It cuts your excursion. That's all it does. I mean, that's what you use it for.

OK. I want to shift gears here. Any questions on-- yes?

MALE SPEAKER: Can you talk about those electrostatic speakers?

DR. AMAR G. BOSE: The which ones?

MALE SPEAKER: Electrostatic.

DR. AMAR G. BOSE: Oh, yeah, electrostatic speakers. I mentioned anything about that. Very good. Before we go off with loudspeakers. Again, with the electrostatic speaker, you can make a whole model. I chose this one because, as I'll mention, for a couple of reasons. This one is a very interesting one to make models of and to understand the models. And that's our object here, not to particularly learn anything about a loudspeaker.

The electrostatic speaker is a speaker in which it has great interest from the modeling from one point of view, that when you're all said and done and you linearize it, your force is proportional to the voltage. Here, force was proportional to current from the magnetic transducer. From the electric one, f equals qe, the charge.

So electrostatic speaker is basically a charged diaphragm, maybe a piece of Mylar, that you place a charge on. And you do that with that a very, very, very, very, very big resister to a voltage supply, to a battery. And then you have plates here that you can apply a signal to and attract this. Electrically, discharge gets attracted if you make this plus or minus, for example. And you can move the thing this way.

Well, it turns out that-- If you do that, by the way, you have to go in your model now, since force is proportional to voltage, you wind up with the electric circuit on the other side of this. If you use voltage as the across variable in electrical, you will now come with the dual of this circuit on the other side because of the constraint that the new transformer, voltage proportional to force, gives you.

So what happens is you attract the thing with a field that's going through here. Now, the problem is that the force, fundamentally, is proportional to the square of the voltage. But then what you do is you bias it, and you work over a small region. This is very obvious when you get an the electrostatic speaker.

For example, if you take a sine wave generator and go through it, from low frequencies to high, the thing goes, boop, like this. Well, what you'll hear, if you do it at the magnitude that's within the range of operation of the device, at the maximum range, as the sine wave generator goes, boop, you'll hear, beep, beep, beep, beep, beep, coming down, the modulation of it, because the fundamental nonlinearity is there. And you can't move it much. You can't get enough of a field in here, because you have 30,000 volts per centimeter breakdown voltage of air.

And so this is why the electrostatics that play well, that have any base, are always big. There like as big as a door, because in order to move that air with a little displacement-- and you've got to have little displacement-- you have to have-- there literally are seven-foot ones that you have to have a big diaphragm.

Now, the folklore part of the electrostatic speaker is that it produces much better high frequencies. And the folklore argument is because the mass is so light. Now, that sounds good at first. It's very light mass. Here, we were straining to get-- if we want better, high frequencies, we found out we needed a smaller mass here.

So anything wrong with that? Sounds like a good argument. The smaller the mass here, the more you got out at high frequencies. So now you say, ah, electrostatic is much than a cone. It has this Mylar thing, which is very, very light. That's true. It's a better unit to reproduce high frequencies with. Can you find any flaw in that logic?

Well, let me suggest an analogy. If I told you that a Volkswagon was much better to have fast acceleration, height frequency, than a Mercedes, would you believe me that a Volkswagon accelerates faster than a Mercedes? It's lighter. Ah, depends on the engine. A-ha. Depends on the force. F equals MA, for God's sake.

So the force that you can get out of the electrostatic speaker is tremendously limited by the 30,000 volts per centimeter. You can't get a big enough field. In fact, the force is so limited on electrostatic that the electrostatic, if you place it against the wall, it's done for, because the wave reflecting back has such an influence on the cone motion.

So that's why you always have to play them a considerable bit out in front, because they are not a good velocity source. They are a force source. And a force source, if you have any force reflecting from the wall-- the best way to play them is, in fact at a diagonal-- if you have any force reflecting from the wall, it'll kill the motion of the device.

So you need them large so you can have small excursion. And that small excursion is what you can get from a small force. You have to linearize the thing. And you can do that at low volume levels only. But it gets highly nonlinear at high volume nodes.

However, many people who knew music well would correctly observe that there was a nice, high frequency sort of sound to this. This is what gave rise to all the folklore artists who were saying it's because of light mass and all. No, it isn't. It's because-- and we'll get to that a little later-- it's because 50% of their sound didn't come to you directly. The diaphragm's vibrating, it hits the wall, and it goes out around the room, because it has reflected sound, which you can do, of course, with a magnetic speaker. But it normally hadn't been done. And people always made these boxes that point straight at you, without any thought of how music arrives at your ear when you are in a live performance, unamplified live performance. And we'll talk about that when we get to room acoustics.

But in essence, I felt that because it didn't have to go through all the linearization and what-not, this is a much more interesting model to illustrate the modeling process, which is my real objective. And that's why we haven't, again, gone in-- I will make a model of one thing that is electrostatic, and that is the microphone, at some point. We'll do that later. OK? Thank you.

OK, now, oh, boy. I don't even know if it's-- well, I hate to start some area and then get up to the point where it just looks interesting and drop it. How are we going to do it? Let's see.

We're going to talk about a subject now which, I have to tell you, is not only absent from normal acoustics books, but probably should be. What I tend to do in the subject is I drag in everything I can that will cause the result that I want. And that is that you can understand modeling and that you can think about models in different fields, and hopefully, one day, in fields we don't know anything about today. And I use a very loose excuse to do all this, that they're somehow related. But you can relate anything almost to any other field.

And this field that I want to talk about is magnetics. I can say, ah, this is obviously related, because you saw a big magnet already in this hull attached to a speaker. That's my excuse. But the real thing is to let us get another example in another field of modeling and lump parameters.

Suppose I take a piece of steel here. A lot of this is review for you because you've seen it in physics or some course in fields. And suppose this thing has a length, l, an area, cross-section area, A, a magnetic permeability, B equals mu H-- magnetic permeability, mu, and has an H field going through here, H.

Now, the integral in magnetics, integral of Hdl is called-- from one point to another. I'll take it from here to here in the integral I'm going to do. And the integral is going to be trivial because I'm assuming for the moment, for this little block, that H is constant through it. So that integral, Hdl, from this point to this point, is just H times l. But that's called magneto motive force, MMF. That's the name that they give from A to B, let's say.

And you can see the analogy beginning to come up. The E field, integrated from here to here, is voltage between the two points. This is like a voltage, if you wish. The integral of H over that l is like a voltage.

OK, now, in our case then, this becomes H, that's in this bar, times l is equal to the MMF of that piece. Now, H, I can express as B over A-- sorry, H is B over mu times l. Flux we have as B times the area. Magnetic flux. B is flux density. B times A is area. So I can express this, then, as B is-- I'll write it this way-- B mu A times flux, substituting for B flux over area.

And this fellow here, this coefficient, we call reluctance. It's like resistance, and so we give it a symbol that looks like that. But it's a script one. So this, then, becomes equal to reluctance times flux. So you have something, which we're calling at the moment an MMF, which-- magnetics, that's what they call it-- is equal to reluctance times flux.

Now, if you had a model in which flux was the through variable and MMF was the across variable-- MMF from here to here, flux going through here-- what would this element be represented as? Flux is the through variable. MMF is the across variable. What the heck is the element? A resistance. There it is. It would be a resistance of script R. OK.

And if I had a magnetic circuit with a lot of little arms like that, the whole thing is steel-- in another words, let's say I had something like this. I could, if I wanted to, break this thing up into four sections and have a reluctance in this section, a reluctance one, a reluctance two, reluctance three, et cetera, reluctance four, all like a resistive circuit. And those drops would add up to something. And what they add up to we now would have to talk about starting from the usual place, whenever you have magnetics or electrical stuff, Maxwell's equations. Let's take a look. Heck with it.

OK. You saw the beginnings of an element. You're not sure what this MMF is yet. Now we'll get to it. I'm not sure which order is better. But let me first-- you know what's going to come. You have a lump parameter element. You have to connect elements together. You connect them together. You have a KCL and a KVL. So let me do the KCL and KVL for magnetics first.

The through variable we'll take as flux. Flux one, flux two. Let's say this one's going this way, maybe this one's going this way. Now, there are pieces of material with a high mu, let's say, or with a mu that's different than the air, much higher than the air, because I like the flux to stay here and not go outside. This. We look at this little connection point here.

Now, from Maxwell's equations, we have divergence of B is equal to 0. But the divergence integrated over the volume of this-- this is some little volume here-- the divergence integrated over the volume-- integrated over the volume, dv-- is equal to B, the normal component integrated over the area-- divergence theorem-- equal to B integrated over the area, ds-- B dot ds, integrated over the area. And that, of course, if this is 0, this is 0, this is 0, is zero.

Now, B integrated over the area, we would have this area, we'd have this area-- let's take it like this. B times area is flux. So in the discrete case, this is B times summation Bj Aj. Well, Bj Aj-- in other words, this is the B coming in here times the area, which happens to be the flux. I'll write this as summation phi j, add a node, must then be equal to 0.

In other words, that's all I've done. I've taken this integral and performed it over this surface, this surface, and this surface. And it's all equal to 0. So that is our KCL of magnetics. It all starts, this is what gives me the 0. Divergence theorem gives me this. This is fluxes in the discrete case. B dot ds over each area is B A. B A is flux. Flux equal to 0 KCL for the unit.

Now, let's see. Suppose we had a magnetic circuit now, some steel here, and we had a coil here, and we had some flux. How it got here, I don't know. We have some at the moment. We'll see how it gets there later.

Now, I want a relationship between the flux and the current that's flowing here. How do we get that relationship between flux and current? Maxwell have anything to say about it?

MALE SPEAKER: The curl of H dot current is--

DR. AMAR G. BOSE: Curl of H is equal to conduction current plus the displacement current. We're going to integrate this. This one, we're going to neglect on the same basis that we did in-- we neglected things in the electric circuits, that if the thing is small enough, the contribution to curl of H, relative to that of the conduction current, is going to be very, very small. If we're not, have this is [? related ?] [? D ?] because epsilon E.

So the radio fields that are floating around and whatnot, are they fields from the-- it's not always true for the fields from the 60 hertz, depending on where you are. But we're going to say that the fields that are generating this, what's going to be MMF, are negligible compared to the conduction current.

Now, curl of H is equal to J sub c, the current density. If I integrate the curl over an area-- curl of H over an area-- that is equal to the line integral of H dot dl around the contour of the area. So if I took an area that is, let's say, something like this, I integrate that-- and this, by the way, of course, from here is equal to the integral of J sub c-- I integrated this thing over area-- ds.

So it says that the line integral of H around a path here is equal to the conduction current. Conduction current density times area is equal to the conduction current flowing through the contour that I took the line integral over. That's what we are told from Maxwell.

So Hdl all around here, if H is the same all around here, it's H times l. If you want to make this thing up of a lot of little blocks like we had up there, that's fine too. The sum of those H's in the discrete case-- Hj Lj, summation around a closed path-- is equal to the conduction current, I sub c.

Now, the conduction current here, through this particular thing in this model, would be N times I, where N is the number of terms, because let's say you have an I coming whichever way goes-- it's in or out of the page at the moment, I don't care-- I goes down there, and then comes and goes down again through there. So it's N times I is the total MMF.

These terms, as we've seen, are called MMFs. We call the MMF, which is NI, the MMF of the source. The source is giving rise to the H. And of course, the H in here is then equivalent to a flux, in the sense of B equals Mu H, and B times area is flux. So this is Ic. And in our case, this equals N times I, which equals the MMF of the source.

OK, so this is KVL for magnetics. OK. So we have a KCL for magnetics. We have a KVL. The through variable is phi. The across variable is MMF. And we have the elements, the individual elements, which cross variable is the MMF. And it's equal to this.

By the way, does anyone happen to remember-- this is a block of steel, in any case. I told you that. But this block has a certain electrical resistivity, resistance. Well, the material has a resistivity. Does anybody remember, in the relationship v equals RI, for a block like that, how I would express R in terms of the materials? Just wondering.

MALE SPEAKER: Rho l over A?

MALE SPEAKER: Rho 0, l over A?

DR. AMAR G. BOSE: Rho 0? What's rho--

MALE SPEAKER: Where resistivity times the length of [UNINTELLIGIBLE]

DR. AMAR G. BOSE: OK. Let me give you a little bit hard time here and say I don't like resistivity. I like conductivity. How would you express it?

MALE SPEAKER: Rho 0 over [UNINTELLIGIBLE].

DR. AMAR G. BOSE: Yeah. Conductivity's one of our easiest-- [INAUDIBLE] You'll see why I want that that way. So the R was equal to, if you remember from this, sigma A. The sigma's one of your rho. So this is just the same kind of a form here. The only thing is that sigma is replaced by R in the case that you're looking at the electrical resistance. So it just brings the analogy closer and closer and closer.

OK, so now let's suppose I had a magnetic circuit, and it was like this. Some steel around here. I'm going to make up a flukey one here. You wouldn't ever make one like this. But I want to illustrate a point. We have I here. We have N turns here. OK.

Now, I could break this thing up and call this thing, from here to here-- I don't care how you do it, diagonal things or square. The end parts aren't going to be that important for me at the moment. But I'm going to call this thing maybe a length l1-- I'll let the cross section area of these steel pieces all be the same so we have to fool around with that-- l2, l3, l4, l5, l6, l7. And I have N turns on that thing. And I want to find out the flux that's going everywhere. I'd like to find out the flux here. I'd like to find out the flux here, here, wherever.

So what I would do is simply take each of my legs there-- this is l1 over mu times A for this one. And then I have one down here. I have one over here. I have one down here. All resistors, very nice circuit to analyze. One here, one here. And all of this, the sum of the drops around here, is equal to a rise, the MMF of the source, which is NI.

So I have a nice DC battery here-- well, it might not be DC. The current that you're feeding in the coil might be AC. N times I. A voltage source of value, N times the current that you have in the coil.

So anybody can solve this resistive circuit. The through variable is flux. And so the flux that flows down here, or the flux that flows down here-- whatever you want to call that, 2 and 1-- is the current flowing down there. So the current, in this simple resistive circuit, then gives you the flux at each point here.

By the way, that kind of thing is very, very, very useful, because it turns out that iron-- we will talk about this, but later-- the iron has a BH curve like so. And the slope of this thing is that of mu 0, air, when all the domains are lined up. And you don't want to work up here. B over H is mu. You want to work in the region of high mu.

And so when you have a structure like this and you calculate it out and you get a flux here, let's say, maybe in this branch here, which is much too high, then you have to increase this resistor to get it down. And so you know exactly what you want to do in this. So you'd start off with this. And then you'd know what flux you wanted in the iron. You want to get the maximum up here that you can, because if you don't, you waste a lot of iron. And that's expensive. And so a little DC circuit like this gives you, right away, the first indication. And we'll talk about things like leakage in a minute.

Now, that's all interesting to find the flux. But that's not enough. In a real design, what you want to do is you might take, for example, a second coil on here, I2 and I1. And you have a voltage across these coils. Now, nothing that we've done so far tells us anything about voltage. Let's begin to ask questions about that.

What I'm interested in is getting the relationship between the voltage here and the voltage here. Let me take a simpler circuit than that. I made this thing complicated just because I wanted to show you the kind of circuits you can get into. Let's just take, again, the old story, take the simplest thing you can possibly take to illustrate your point. That's all right. I1, V1. I've used complex amplitudes here at the moment. I might talk about other things when we-- I2 V2.

What I really want is a relationship between these things. and I want to know how to design this thing. So given I1 and I2, you have no problem finding the flux. We do that as follows. We just make our little resistive network. In fact, to do that, I can take it in pieces. But why bother? This is all one element. I'll take that as one reluctance all the way around there.

There's no other branches in it or anything else. You can choose to make it up of four pieces if you want. So this is the reluctance of this whole path, which is the l over the mu A. Let's see.

Be careful now. This goes this way. This goes this way. Yeah. So N1, N1, N2. Different turns on both. N1I1, N2I2.

So the flux now runs around here. Flux is very simple to get. Flux is then the total MMF of the sources divided by the reluctance. So MMF of the sources is N1I1, just as if you're doing it as a resistive circuit, N2I2, and over the reluctance. So that's the flux, easy to do from everything we did before.

But that only gives us flux to current. In other words, I have a relationship between the current and the flux. But I want the relationship between voltage and current. Because ultimately, as a network block, this thing is going to be put in a big bag. You don't have to know what's even in the bag. You just want to know the VI relationships at the end of the terminal. So when you insert this into a circuit, it does what you hope it will do.

By the way, if I haven't said it before, it took me a long time after graduation to realize something that I don't think students still realize when you go through, that most of what you will do in engineering is synthesis, because you're trying to synthesize something. You're trying to do something that wasn't done before. So you're synthesizing. But synthesis isn't taught. And it isn't because we forgot to teach it. It's because, in most of the disciplines, synthesis is actually accomplished by assembling preanalyzed structures.

Things that you know the analysis of, you put them together cleverly to do something that hasn't been done before. But when you first get out and you go to industry, and somebody tells you to design something, you think, oh my god, there's got to be some chapter somewhere in a book that tells me how to synthesize this thing. That isn't so.

That is why it is so important to analyze simple things. Because the best engineers are the ones that can just see what these parts do and see what they do in-- parts, systems-- what they do when one is connected to another. And they can realize this function just that way. And that's the way it really is. I mean, there are literally-- as I told you-- there are books written on network synthesis. And you're not going to use any of them for the reasons that we talked about. What you're going to use is the insight you've got from the simple things you've analyzed. You'll put them together to synthesize something.

That's why the most valuable part of the subject is the homework. It's not me sitting up here chatting. It's not even your teaching assistants. It's the thinking you do to understand simple things. And that's what makes you be able to synthesize. Because I can almost guarantee you, assuming that what I'm saying doesn't have a major effect, that when you first get out and you're given a problem, your thought process will be, oh, god, course six, such and such and such. That's got to give me the answer. Well, it won't. If it did, it probably would have been done before. And so it'll be the thinking time you have spent on simple problems.

OK. Now, where in the heck were we? Let's see. Oh, yeah. Voltage. Let's see. Well, when you get stuck, you go back to the beginning. If it's not obvious how we're going to relate voltage-- well, let me give you a clue. What we're going to do, we have a relationship between I and phi. We want a relationship between I and voltage. And the clue is we're going to get it by relating voltage and flux. In other words, if I could write down another equation here which relates voltage and flux, I get rid of flux between the two of them, and I have a relationship between voltage and current at the various terminals that I'm interested in.

So together relationship between voltage and current. Seems like the safest place to start is curl of E is equal to minus partial of B with respect to t. I want flux. And this is flux within an area, within a factor of an area. Curl of E is minus-- now, if I integrate curl over an area, that's the line integral around the area. Let's take the area that I'm going to integrate over to be this thing here. In other words, we'll integrate over the area of the core. And you can see what's going to happen if we do that.

Well, let's write the integral. The integral of curl of E over the area, ds-- this is the area of the core, S core-- is equal to minus this fellow integrated over the area. I'm going to assume B, the field, is constant for my integration, that B is constant over the whole area. Not quite true, but good enough for this.

Integral db dt, partial of B, respect to t, dot ds. And B times area is d phi. So it's integral of partial respect to t-- there's a minus sign still hanging around-- of phi. B dot ds is phi. OK. Now, curl of E is also equal to-- oh, boy. That's equal to line integral of E dot dl around the closed path, in other words, around the core. E got dl, then, is minus respect to t of phi. Now, E dot dl. When you integrate E, as we had mentioned earlier today, when you integrate E over a length, you get a voltage drop, right? So going around this core, you integrate E over the wire now. And you get the voltage that's going to be from here to here.

Now, actually, I don't know if you've thought about it or not. But the voltage developed in a coil, when a field's going through it, isn't as trivial as it looks, because you can get the E. And we'll have to worry about signs in a minute. But turns out there is no E field on the coil. Yet we can get the E by integrating this curl of E. And it's the line integral around the thing. But there is no E field on that coil, because if there were an E field traveling tangential to the coil, the thing that I'll integrate to get the voltage, if that really were there, the coil has conductivity, and I is equal to sigma E.

So there would be current. But there's no current. There can't be, because the darn thing's open. So what actually happens is it's minus d phi dt as the curl of E. d phi dt, if I take it this way, I will take it negative for the moment because it's a minus sign. The curl of E is going this way. So E, the actual curl of E, before the coil got there, was in that direction.

Now, in this direction, force on the charge, f equals qe. But q is negative-- electrons. So it pushes the electrons down here. When the E is pointing this direction, the electrons go down here. That make this thing positive.

But when this is positive relative to this, that is the same-- for example, let's suppose I had a d phi dt, which came out with a V0, a voltage V0 from here to here. I could put a voltage source on there of the V0. And it wouldn't make any difference to the terminals, because V0 was already induced there. It doesn't make any difference. I hope this doesn't complicate it. I'll do it without it also at the end.

But if I had V0 here, then I would have a field-- and I have a wire connected to it-- I would have an E field going down, that way. Well, what happens is the induced field-- the curl of E, which resulted in this E going around the closed loop, was in this direction, caused the electrons to pile up down here, the voltage here to be positive relative to that, that creates a field that's equal and opposite to what the inducing field was. And there's no field in the wire.

So you've probably just thought, d phi dt is E, and it's all done. But it's a little bit more complicated when you stop and say, hey, wait a minute. There can't be any field on that wire, because it's open circuit-- any field tangential, going along the wire. So that's a sort of aside.

The book that I gave you reference to, Fundamentals of Electric Waves by this Stanford fellow, this Skilling, again, I reference it to you because simple things like this that you take for granted, he explains so beautifully. And he has a filament of an electric light bulb and shows you some really interesting things about electric light bulbs. Of course, when you put a current across it, there's charge all along this thing. Well, the charge means that there's a normal field, normal E field, out of that electric light bulb. And yet, inside the conductor, there isn't such a field. The charge is residing on the surface. So just putting a voltage, plugging a light bulb into a socket, is a real interesting field problem. And it's explained about the most beautiful way I have ever seen it explained in that little, thin book.

OK. So what we have, then, is that E equals the d phi dt. Edl, which is, eventually, the voltage across the coil-- and it'll be in this direction-- is equal to this times the N. If you just integrate it around that loop once, you get this. But the coil goes around it a number of times, goes around it N times. So in our case, we would have the V1 is equal to-- and I've already taken care of the signs up here, so I'm not going to use this-- it would be N1 d dt of phi. N1 d dt of phi.

Now, I should use, for the moment, let me use instantaneous quantities here, because I have a d dt. And that's-- phi. In terms of complex amplitudes, if I let this equal cap V to the j omega t, this equal kappa phi E to j omega t, the j omega t's go out. But when I differentiate, I drop one down. So that immediately goes over to j omega N1 times phi, the complex amplitudes, phi.

So now I have a relationship between voltage and phi. I have a relationship between current and phi. I can get rid of phi.

And I have a relationship-- whoop, I need one more thing. I obviously need V2. j omega N2 times phi. That's the other one, because I had a second coil on here. And if you watch this-- and I'll let you do that-- this sign is also positive. The flux is going around this way. Look at which wavy the curl of E goes. And you'll find out it's the same direction as for this.

So taking that relationship-- and we'll do this-- taking that structure, you make a DC circuit, you find out what the flux is right away. And then you have this relationship. And from the relationship here, you get immediately that this gives a relationship between flux and voltage. We're going to eliminate this. And we will then have left a two-port network, a black box like this. I1, V1, I2, V2. And it'll be representing exactly what that core is. Only restrictions at the moment are linear.

But I also will talk a little bit-- since we're so close to it-- about the fact that if you really try to do this, you did just this, would find that you got a representation-- this is what you know as a transformer, or going to be-- you would find that you got a representation that would be OK, but it would only work at very low frequencies. And when you got to a little bit higher, maybe a few hundred hertz, let alone 1,000 hertz, you wouldn't get the output that this would predict. And so there's something that's missing from this circuit that we haven't talked about yet. And we'll put that in. And then as a side benefit, we'll get a complete model of a transformer. OK.