Amar Bose: 6.312 Lecture 14
DR. AMAR G. BOSE: Today will be the first time that we link two of the disciplines together. We'll link the mechanical. Well, in fact, if we get to it, we'll link three disciplines together: electrical, mechanical, and acoustical. And we'll do all of this through a model of a device. Not because we want to learn about that particular device, but because it is an example of modeling.
A very interesting thing happened in the '50s. In the late '40s, the transistor was invented. And in the '50s, it became employed in a lot of circuits. And it turned out that event separated the graduates of different schools into two categories, almost black and white in their differentiation. The people who had been taught the vacuum tube and all its properties as just the vacuum tube, here is how this thing works, we're left in the dust, literally. The people who were taught about vacuum tubes as an example of modeling of a physical device just went on to the transistor as another device, which had to be physically modeled. Used the same principles that they used to get the models in the vacuum tubes, and went sailing right on. So there was actually a total separation like that of the engineers depending upon how they were trained. So don't get attached to any device, especially in this subject.
The device we're going to use is going to be this thing. That is not significant at all. It is only significant as an example of the modeling process. And hopefully, there will be many devices to replace things like that during your careers. And you'll be able to go on and model those.
Well, to link the electrical and mechanical disciplines, there is a very simple structure which we can look at first.
Just think of a permanent magnet. In other words, two ends of it. This could be the North coming here and then we'll make the other one like this. South Pole here, flux going through this, B. A wire that we insert-- by the way, everything I'm going to do here is a review of what you've already seen in I think sophomore physics, or maybe freshman.
Have a wire here. Might have a current here flowing in the wire, i. A positive direction for our measurement of displacement or velocity we could take as x. And as we know from our physics, a force is generated on this wire if it's carrying a current in this magnetic field, B.
We also know that if the wire is moved through magnetic field, it generates a voltage on it. And you've seen from your physics that you have force is the current i across-- these are, in general they're vectors. B times l, where l is the dimension here of the length of the wire in the gap. This is a gap in which sort of ideal we're just assuming that the flux goes straight here. There's no fringing flux coming like that. This is just the model that we're making.
So there is a length of wire l, which is inside the magnetic field. And then we have that the voltage is the velocity of the wire cross, the cross product, with Bl. Two relations that you got from early physics.
Now, in the particular case that we've drawn here, the motion u, is orthogonal to the B. So this is just Bl times u if you wish. When they're right angles, there's no sign term. And similarly, the current is at right angles to B. Velocity was this way. Current is this way. Both of them are orthogonal to B. And so when they're perpendicular, I'll just say perpendicular. That means everything I've said. It means i is perpendicular to B, u is perpendicular to B. Then you have force. The form that you're maybe used to seeing Bl times i. And velocity, I mean voltage, is velocity Bl times-- whoops. 1 over Bl times-- let's see. Hold it a second.
Force is-- yeah. Velocity is Bl times u. Voltage, I keep looking at a symbol and saying the wrong thing. Voltage is Bl times u. Force is Bl times i.
Now, these set of equations here really tie together the mechanical and the electrical systems. Force is the mechanical variable, current is the electrical. Voltage is the electrical and velocity is the mechanical.
Now if you put those relations in the following form where you have-- let's say I'll put the electrical on one side and mechanical on the other side. Voltage-- let's see. Hold on a second. Yeah. Voltage is Bl times u and current, the other electrical variable, is 1 over Bl times f. These two equations satisfy our definition that we had originally of the transformer.
If we take-- if we take, and we don't have to. If we take voltage as the across variable in electrical, then this side of the equation would have voltage this way, current that way.
If we do that, if we happen to make that choice, then we have no choice over here for the across variable. Because the transformer, the general symbol of the transformer defines a cross variable to a cross variable, through variable to through variable. So over here, you must have that. And of course, then force is the through variable. And the relationships then, dictate that this is a transformer of ratio Bl to 1. Because the across variable over here is Bl times the across variable here. The through variable over here is 1 over Bl. That's exactly what how our transformer was defined.
Now if you wanted to take current as the across variable here, then you would be forced to take the other one here. So I want to make it very clear that what you choose on one end of it is up to you. What you get on the other end is up to the laws of the physics.
So now suddenly, we are able to take a discipline, which is electrical here, mechanical here, and has some relationships between force and voltage or current that govern the two disciplines, and all of a sudden we're going to be able to make an equivalent circuit for a device now that spans the two electrical and mechanical domains. Any questions here? OK. Yeah.
AUDIENCE: On the top board, I was wondering where you get that voltage equation from, u cross Bl. What's the velocity there? You don't have it on there?
DR. AMAR G. BOSE: Oh, sorry, sorry, sorry. I put x up here, but I didn't put x. The velocity is the same direction as your x. Velocity is dx dt. Or u in the same upward direction. Orthogonal to the magnetic field.
DR. AMAR G. BOSE: The current, I haven't said how the currents got there. I haven't said where the rest of this wire goes. These are just relationships if you have a current flowing. And this wire is in a magnetic field, it will have a force on it. Ah, I haven't-- I should say more. I understand your question. I should say more about science.
This a cross product. Yes, it boils down to this. But which way are the signs? In fact, I should say a lot about that because I may have mentioned on the first day or so, one of the reasons I like to give quizzes in the evening is because I can remember pulling on one side of the paper and the instructor on the other. Well, the issue was exactly over this. Namely, it was a physics quiz, second year. And had a lot of the i cross Bl's in it. And I'm a left-hander. And I did the whole thing, the cross product this way. And I was reviewing the quiz in the last five minutes and I realized the whole thing was based on this. And so I had to do it all over and get all the signs right. And I was attempting to do that while the instructor was nagging on the other end on the paper. So I should never have forgotten to talk about signs here.
Let's see. i cross B. Cross product. By the way, there is a book, it's an old, old book but it's the best book that I personally have seen in waves and in things like vector products. It really gives you insight. It was written by a Stanford University professor and it's called Fundamentals of Electric Waves. And it's an old book. It's a Wiley book. Fundamentals of Electric Waves 1948. And if you really want to have a feel for electric-- well, for E&M, for E&M fields, and for all things like this even, I don't think there's a better place to get it than that book. He has pictures. You come out of that thing just believing that you can see fields, and you can see cross products, and you can see divergence. You can see curl. So that is about the best reference I can give you.
Since that time, there have been many professors who have been promoted writing about the same thing-- textbooks, new books on the same fundamentals, but I think that one stands out as, in my opinion, the best.
OK, yeah. i cross B. Let's see. I've taken i this way. B is this way. If you rotate i into B, you'll get up this direction a force. So the force on this wire here would be in the positive direction of x for a positive current that's flowing this way.
This one is a cross product, which reduces to this without the sign term when they're orthogonal. But it's u cross B. u taken in the positive direction here, cross that into B this way. And you find then that this end of the wire becomes positive with respect to that end. And so the cross products totally determine the sign that you're involved with. Thank you very much for asking. I don't know why I completely skipped it.
Now, as to where this wire goes, it doesn't even make any difference. I mean this can be closed. It can be hooked to a battery to get the current going. However it's there, however the current gets going, these are the relations that govern. These relations care nothing about or say nothing about what happens to both ends of the wire.
OK, now I'd like to start with a model. It won't be the only one, but this one we will go into more carefully than we've gone into any other model. Again, with the emphasis strictly upon modeling.
This thing, as you might know-- this is a big one-- is called a loudspeaker. How many of you have ever seen the innards of a loudspeaker?
OK, well for those that have not, I better attempt to draw some of that. Here's a little one, a different kind of a magnet. A totally different structure. We'll get to talk about that at some point. And this magnet. You notice this magnet here is surrounded by an assembly, the flux flows through the magnet and around here. And there's no such thing at all in this magnet. And the difference happens to be-- we'll get to that when we get to magnetic circuits or permanent magnets. The difference happens to be that the magnetic materials are very different. And one requires this structure and the other that structure.
But for the moment, I'm not going to worry about that. I'll try to tell you what these parts are, and then give you a drawing of it.
This is what the thing looks like from the front. This is called the dust cap. It does much more than prevent the dust. This little felt one in the middle here is also called a dust cap. It does perform that function because if you didn't have that, as you'll see in a moment, you could look right into the magnetic circuit, and it would attract all metal filings. And the first time you get a metal filing in there, boy, the sound changes dramatically. So it prevents dust, but it has another function, which I hope we'll talk about.
Dust cap. Cone. This part. And this thing is called the surround-- sometimes called the annulus. It's what enables the cone to move in and out. Not all speakers have that. This kind of a speaker, which you'll see in cheaper kinds of units doesn't have that. But it has a corrugation at the end of the cone, which you can see when you come down here afterwards if you wish. And that corrugation at the end of the cone serves much like this role in the annulus here to enable you to travel in this direction, but make it hard to travel this way. Because you've got to keep this thing aligned as you'll see, again, soon.
This thing doesn't look like one, but it's called a spider. And it's a very interesting mechanical device because try and move this center horizontally. You can't do it because if you try to move it, let's say horizontally, you'd have to be stretching the material that was here and here. No go. Try to move it this way and it's extremely easy. So I can move it in and out, actually. But I can't move it horizontally. The whole thing is you have to keep this thing aligned horizontally, but give it freedom to move vertically. So this is really a spring for moving vertically. And a very strong restriction to any other movements.
That's supposed to be the purpose of this thing as well. This one is usually a lot stiffer. This fellow you could move horizontally because you could stretch this material. But it's not as crucial in the horizontal dimension as the one that's very close down here where your tolerances are small. So spider, cone, surround, or annulus.
Then comes the voice coil. Now this is a rather husky loudspeaker. In fact, I think it's-- I believe it's a speaker out of what the company calls its acoustic cannon. You'll get to see that and hear it. That's a wave guide speaker, which is used in big theaters, in public places, Olympics, things like that. It's the only one I know of that really goes down to 25 Hertz beside all the advertisements. This really does.
This is the voice coil. Now, it's wound in this case on an aluminum bobbin. All the turns here have to be insulated from each other, otherwise you have a big short circuit. And they all have to be insulated from the aluminum. And as we'll see when we get into magnetic circuits, you want to get as much wire as you can in the gap. That's the name of the game. Your efficiency is related to that.
So all of the insulation must be very, very thin. The thinner you can make that insulation, the better off you are because the insulation is air in the magnetic field gap. And you don't want air, you want copper in there. Or aluminum as the case may be.
In fact, if you look at this voice coil, you-- well, maybe I can draw it. Let's see. Let me draw it here.
This is the bobbin that it is wound on, the aluminum. In this case, aluminum.
If you look at this coil, this one has a multi-- yes. It's wound around this. And the wires then come up and out here. Well, I'll draw it one from each end. But you can see they're pretty close together here. Wires come out. Now this thing is moving back and forth, of course, to drive the cone. And the cone is fastened on the top here. So what happens is that you-- normally you will see in a speaker like this, there will be wires. Normal wires. Wires like this, copper wires, that come out and connect to the cone here. They're actually connected right through the cone. And then you have things like this, which are called in the industry, tinsel, which is very, very flexible wire. Which go to the fixed structure, to the terminals that are on here. And so all the motion of the cone takes place through these flexible tinsel, if you wish.
If you tried to use this wire to connect to the fixed, your speaker would last a day. If that. Because the wire would just break right off.
So you have wires that come through, connect to the cone-- the stiff wires. And then the flexible wires will connect from the cone to the fixed part.
Oh, I might tell you that way back-- and I guess it was when I first got interested in this in the late '50s. I looked at loudspeaker design and said, if we apply some physics to this, we ought to be able to optimize it. It turns out that most of the field that doesn't have engineers, even in speaker companies. And the ones who make these things, not at all-- there's no engineering involved. And they've been made the same way for year after year.
So you think if you'd think about and you'd applied physics to it, you ought to be able to come up with an optimum structure. So we tried to do that. And we mailed off what we wanted to the company that actually made this thing. And didn't know that there was no engineering there. And got back a letter. They looked at it and first of all, they didn't know what B in webers was. There was a complete disconnect of communication.
But they spotted something that made them totally convinced that these academic people are 30,000 feet up and don't have their feet on the ground. So the letter came back, I can remember it so well. I wish I had saved it. And I had designed a one layer coil because I didn't want the extra spaces again that you're going to have with more wires. And it said, "Dear Professor, we regret to inform you that you cannot make a coil with an odd number of layers." Because if you do the other end of the wire comes out the bottom, and there's no way of bringing it out. And if, of course, you bring it back up here, you have a space for the whole thing. And they knew that you wanted to get a lot of copper in the gap anyway. So we got that letter.
It actually took until 1973 when one of the people who was a student also here, finally made a one-layer voice coil for the most efficient loudspeaker that you could build. And he did it by making-- he did a number of good things. He made rectangular wire, so that you didn't have these big spaces in between the turns. When you have two round wires, you have a heck of a lot of air. So he made rectangular wire and things like a micron spacing here. And if you can imagine, let me see. Give me something like that.
This rectangular wire was worse proportion than I have shown. It was four times in width. It was about like that. And it's easy for me to bend this this way. Try and bend it this way.
That copper wire that's going around there is bent that way. And there are only five machines in the world that can actually build that. And two MIT fellows designed all five of them.
And by the way, just as a matter of interest to show you again how you can cross disciplines. They were both electrical engineers. However, they did work with another, a third one who was a mechanical engineer. But the three of them together. No coil winding machine companies, the people who build coil winding machines. There are a number of companies that specialize. Their whole business is in that. None of them would even consider it.
And a lot of electron microscope work had to be done because when you tried to bend the wire that way that it doesn't want to go, it cracks, of course. And you're trying to get an insulation on there that's a micron thick. You have to study and have to develop new insulation materials. You have to look at what happens when you blow this thing up 50,000 times. A tremendous amount of work. And it turns out that there are five machines now and they will turn these things out every eight seconds, one of them. And perfectly wound. And they do the job, insulated, everything. But they finally came up with this and, believe it or not, it's one layer. I'll let you think about how they took the other end of the wire and got it out the top.
And if you can't think about it, if it doesn't come to you at some point by the end of the term, I'll answer it.
OK, so voice coil bobbin, voice coil. These wires running all around here like so.
Now, the spider then is attached to the neck of the voice coil. Where is it? The spider now comes like so. And it's glued on like this. And it goes into this structure here. There is the slot.
The magnetic field is-- it's radial here because the-- I'll draw a cross section now. What I'm going to do on the board is, if you took a saw and went right through here, I will draw the cross section of that as near as I can. Let's see. Yeah.
Cross section looks like-- here's the coil. Here is what we call a pole piece. This is the magnet. This is strontium ferrite that's in here.
Now, this is a cross section through the middle. Question? This is a cross section down through the middle. This is the magnet that you see right here. The big, thick ceramic. Pole piece. I'm sorry, the front plate this is. Front plate. This is the so-called pole piece. This comes down and it's connected to, or an integral part of the back plate.
This is all steel. And this is the magnetic material. Magnetic flux flows like this. The B field if you wish. And similarly, by symmetry, it's going like this. So we have the orthogonal condition that led to these relationships in the loudspeaker because the coil is always orthogonal to the magnetic field. And of course, the velocity of this voice coil that's wound on the pole piece. I should make this a little bit clearer.
If this thing is the bobbin, maybe I'll make the bobbin in orange. That's the aluminum that the voice coil's wound on. The pole piece then comes inside with a tiny bit of clearance, a few thousands clearance to the back plate. So steel, magnet, and that's the structure. Any questions on that part of it?
This is the clearance that you want to keep down as much as you can because this is where all the energy of your magnetic curves right in the gap of the whole magnetic circuit. In very good designs, this back plate will be an integral part of this pole piece. That didn't happen until, I would say, maybe a decade or so ago.
Basically, what used to happen was that the-- by the way, this particular pole piece is hollowed out because-- as you can see, a whole right through the thing. Because if it's a big enough one and you have flux that can go without saturating this iron, then you don't need the centerpiece. And then, that is a saving in weight, and a saving in money, and no change in performance. So sometime you'll see that in some speakers.
But in many speakers, you will see-- in, for example, ones that are in things like a 901. You'll notice in many-- well, in many commercial speakers, you will find that this piece is fastened onto this piece. It's staked on there. But there's a slight gap always. You can't get those things exactly even. And you lose magnetic energy when you do that. So you can make all of this by cold forging out of one piece.
In fact, the design is tapered like that. And how that's made is by a machine that's about as big as from here to the wall. And it takes in a one inch or a 3/4 inch rod into the machine and one big-- it cuts it off and then whacks it on the tail end. And it becomes flat like yes. It was all a rod of this diameter and one big smack, and that's why the machine is so big. You get something like that out. And now there is no gap at all, no loss of energy in here. So that's the device from the bottom end.
Then what happens up here to the bobbin is attached the spider. The spider, we'll picture it like this in cross section. And the purpose of the spider again, is to keep the voice coil centered, so that the coil doesn't hit the front plate or the bobbin doesn't scrape the pole piece. And you want that extremely stiff in this dimension.
Then to this, the cone is attached. And the cone finally goes to another what we call hard part. Namely attached to the front plate is this thing called the basket. This part here. And that comes up and forms the seat upon which the annulus or the surround goes in.
So when you're actually assembling this, you'd have all of this thing like so. Well, you'd glue this in first. Then you'd put this on, glue it to the top of the bobbin, glue it to the outside. So this goes into some rigid structure, which is connected to-- this is the basket. So that is the whole construction.
Then you finally put on your dust cap across here.
I'll tell you, maybe it's a little bit premature. But I'll tell you since so many of you have seen the inside of one of these things, you may know what that thing does besides keep out dust. See, the dust, by the way, now you can see.
If dust gets in here-- metal, and metal filings in here, there's a strong magnetic field. They go zip right in down in here and right in here. And then it's all over. So dust cap, yes. But what do you think it does acoustically? Anybody have any idea at all? Yeah.
AUDIENCE: Keep air going. Make sure that the air moves with the cone?
DR. AMAR G. BOSE: It's going to move it-- you mean in here?
DR. AMAR G. BOSE: Yeah, it does do that. It makes a whole surface that moves and doesn't leave a hole here that's not moving. In other words, doesn't leave the front of your pole piece there. But actually, that's a small area. That's an unusual dust cap. The more common one in size would be something like that. But that's not a big area compared to the other, so it wouldn't make too much difference in a total volume velocity. But there's another very important reason. And that reason has to do with the behavior of this fellow, this cones.
You look at this thing when I move it in my hand and you think, it's a big plate that just moves back and forth. If you look at this thing at higher frequencies. In fact, you don't even have to look at a cone. Just look in any book on vibrations and you will see a circular diaphragm, or any shape diaphragm. But this happens to be circular. Breaks into all sorts of modes. Standing waves on the cone in the material, not standing waves in air now. And it turns out that the cone will do things like this.
A diaphragm, never mind this actual cone. You'll get to one frequency where this part's coming forward. This is going backward. This is going forward. This is going backward at the same time.
You go to another frequency and you'll find a ring like this in which this is superimposed with this. Plus, minus, plus. So all these different regions of the cone are going different directions.
When you get to frequencies where you have standing waves running out the cone and going back, not acoustic waves in the sense we're thinking of them, but standing waves in the material. So what happens is you can't count on this thing radiating like a piston anymore.
And if you put a smaller surface on there that's connected more to the center, you will get an extended region of frequencies in which, at least this thing moves like a piston because it's smaller diameter. And so these always affect the high frequency radiation of the device.
Now a device like this isn't meant to go too high. Meant to go to maybe 100, 150 Hertz. But the high-end of that is controlled by the dust cap. And so the material of this thing is very, very important. You change the material of this, the stiffness, for example. You change that stiffness and you change the whole characteristic of the unit, of the loudspeaker.
So that's the structure now and we want to analyze it. OK. Let's see. Maybe I can make the equivalent circuit right here. Or you can make the equivalent circuit.
I've shown you all the parts-- the electrical, mechanical. Acoustical? Yeah, there are no acoustical parts of this so to speak. But there is a huge acoustical part when you think of what happens when this moves through the air. So we want all this together in one equivalent circuit that we can look at, that we can analyze the frequency response.
AUDIENCE: I have a question. You said that all the energy, all the magnetic energy gets stored in that air gap, so it's critical to keep it small. But what about all the steel there? It seems like you should make that out of, instead of steel, something that has a lower reluctance?
DR. AMAR G. BOSE: Yeah.
AUDIENCE: Wouldn't you get a lot of loss in the steel?
DR. AMAR G. BOSE: Yeah. We would love to be able to have a material that could carry more flux before it gets saturated. When we get into magnetic circuits, what you'll see is that when you try to increase the flux through a rod, let's say a steel rod. You get to a point where the BH curve becomes nonlinear and, all of a sudden, that rod behaves just like air. Has a mu of mu 0 instead of a mu of 1,000 or something. And so yeah, we would love to get a material that had a higher saturation level than steel, so that we could carry more flux through a smaller area. But this is still the best overall choice. Just cold rolled steel.
OK. Yeah, equivalent circuit.
Here are electrical terminals, input. I'm going to choose the voltage as the across variable, the current as the through variable. But leave a little more space here.
OK, there's the pile of parts. Now I want an equivalent circuit.
If you want to look at anything else in making it, go ahead. Where do I start?
AUDIENCE: Voice coil?
DR. AMAR G. BOSE: Voice coil. That's this fellow right here. How would you like to start?
DR. AMAR G. BOSE: Inductance? It's reasonable. It's a coil. In fact, it's even more reasonable because the coil sits down on this pole piece, which means there's iron in the middle of the thing. So the inductance of this coil is one thing out here. It's another thing when I put a slug of iron in the middle. So there really is an inductance in there. How would you like that? Across from here to here?
Before I ask you where I put it, what else is associated with the coil?
AUDIENCE: Parasitic resistance.
DR. AMAR G. BOSE: Resistance? I don't know about parasitic, but it's sure there. The resistance is there.
OK, so here's one end of the coil. Here's the other end of the coil. If I just we're looking not at the loudspeaker model now, but just in these two terminals with this thing in air, it's some inductance and some resistance. What kind of model would you make?
Inductance in series with the resistance. Sure. The same current's going through the coil that's going through the resistance. Current is the through variable as we have chosen it. So the two things are in series. But now the only question is where to put them. From here to the ground? Or from here to going to something?
AUDIENCE: To the ground.
DR. AMAR G. BOSE: You want to put them to ground? All right, let's try that. That's the inductance. This is the resistance of the coil. We'll call it electrical just to put a little e on it to remind us.
AUDIENCE: [INAUDIBLE] --going through your [INAUDIBLE] there, so they should be in series no matter.
DR. AMAR G. BOSE: A-ha. So suggestion is I can't make it go to ground because the current that goes through this coil is, in fact, the current that's going through the transformer. So I can't make that thing go to ground then if that's the case.
So we have electrical resistance, electrical inductance. Or the inductance of the electrical circuit. And that current is the very current in the transformer that we have up here. So transformer, transformer. As we saw up there, Bl to 1.
What's the across variable here? Velocity. No choice, dictated by the fact that we chose the across variable to be voltage on the left side over here.
OK, so look at that. Right down the line here we have made the transition electrical to mechanical. And by the way, just so that you can feel that this isn't so specific. This thing is nothing more than a linear motor. That's all it is. Absolute linear motor. So you get exactly the same equivalent circuit, the same analysis kind of things that come up in the linear motor case, will come up in this. Yes?
AUDIENCE: From that point of view, shouldn't there be-- in addition to the back EMF from the coil, shouldn't there be another one from you moving it on and off that base post because you're-- it's IDLDT.
DR. AMAR G. BOSE: When the voice coil moves up and down?
AUDIENCE: Right. Then you're changing the reluctance.
DR. AMAR G. BOSE: Changing the reluctance.
AUDIENCE: I guess that's from the coil side. You're going to have a different amount of magnetic material in the coil. So the l that you're modeling actually changes, correct?
DR. AMAR G. BOSE: Oh, no. Yes, the answer is yes, but in a well designed system very, very little change. Because that turns out to be a nonlinearity. In other words, if any circuit element changes with position, with voltage, or with current in any system, it's nonlinear. Because then the element isn't-- it's no more representing constant coefficient linear differential equations.
So what you try to do is you try to make your coil and you design your pole piece such that, for example, the pole piece will always come up a little bit further than the coil. Such that when it moves it sees about the same flux field. And the reluctance of anything in here won't change. Of course, the reluctance there, this stuff is non-magnetic. And so the reluctance as seen around this path won't change.
But the inductance as you say, could change if I pulled that coil a little bit up out here. And then you have nonlinearity, and then you have distortion. And so that's what you design around. And poor designs really do have distortion. You'll find plenty of woofers that are used in pop concerts and whatnot where it seems like volume is more important than quality. In fact, they even have-- literally, you come buy what they call distortion boxes for these things. And so when they don't worry about that, then they just bump the thing back and forth gaily. And as long as it doesn't pop out. And that does happen sometimes.
Once it pops out, once this thing pops out of the gap, it's extremely unlikely that it's going to go back in again. So that's a one-time event.
So let's see. Where were we? OK. Any other questions up till now? Yeah.
AUDIENCE: I was wondering where the-- you said the wire's attached from the voice coil to the terminal on there. But it's not [INAUDIBLE].
DR. AMAR G. BOSE: The wires that are coming out of the voice coil. Now, question where are they? Yeah, very good point. Why aren't they here and here?
If I give you this thing in air, that's exactly the model you're going to make.
AUDIENCE: Well, that's just one point in the circuit. I was wondering where the rest of it was, the rest of the electrical circuit.
DR. AMAR G. BOSE: I'm sorry.
AUDIENCE: Where's the rest of the electrical circuit?
DR. AMAR G. BOSE: There isn't any. The rest of the electrical circuit is-- here it is, the whole darn thing. Just this. And this is actually going to be part of the mechanical circuit as you'll see in a minute too. Because this thing is moving. And it's a mass, and the mass is moving. So it's going to be part of the electrical and part of the mechanical.
The question I thought you were asking was, here's a coil. If I give you this, you will give me this with this lead and this lead being here. And now what you have is these two leads coming out. And we're saying, hey, this is the equivalent circuit. That's the back EMF if you wish, induced in this coil through this arrangement here.
AUDIENCE: Where do those wires attach?
DR. AMAR G. BOSE: These wires actually attach to both ends of the coil. But you can think of this thing if you want as being inside of that whole thing. In other words, whatever current is flowing through this coil is causing a voltage drop that's in addition to these circuits. We model it like this. But that's a voltage drop that comes because of the mechanical motion in the magnetic field. And it sure as heck, appears across these two terminals as an addition. And so I can put it in the model this way. But it's actually happening inside the coil.
AUDIENCE: Tell me more where the current comes from in the first place.
DR. AMAR G. BOSE: Oh, you mean the rest of the electrics? Yeah, they have these things in hi-fi shops. Amplifiers that they pump tons of watts in here. That's what drives this thing. In other words, there's whole electronics out here that drives the loudspeaker from its terminals.
OK, now let's go. Mechanical. I haven't done a real good job here. The annulus is here. The role in what I call the annulus or the surround is here to the basket.
Now, for this model I will-- let's assume that the cone moves as a piston. It doesn't have all of this business. Where in an ideal world, the cone-- which by the way, for a good range of frequencies. For example, if this unit is used in one of these acoustic wave devices. That'll go from about 50 Hertz to 125 Hertz. And over that region, centrally over that region, this cone is a piston.
So with that, what do I do over here? Yep.
AUDIENCE: You could have a capacitor based on the mass of the bobbin and the cone and everything else.
DR. AMAR G. BOSE: Where would you like the capacitor to go?
AUDIENCE: Do they all move at the same velocity? Do all of the pieces move at the same velocity?
DR. AMAR G. BOSE: Yes. Do all the pieces move at the same velocity? We will make the assumption that the voice coil, this end of the spider, and the whole cone move as a unit.
By the way, since you're asking about it. Those are major problems. It turns out that if you apply good technology in these things, both from the point of view of measurement, if you make holographic measurements on these, you find out that these adhesive joints here are extremely critical. That for many adhesives used, this cone here and this one are not going in the same direction necessarily at the same time. Because when you get to higher frequencies, as you'll find it very easy to compute, the distances are very small.
And if that adhesive in elastic, you can actually have the voice coil coming up and the cone going down. Or you can have poor transmission from here to here, so the motion here isn't anywhere near what it is here. And then you lose your whole output because your volume velocity is driven by what happens on the cone, not on the voice coil. So for the moment, assume that they're all the same, same velocity.
OK, back to where would you like the capacitor?
DR. AMAR G. BOSE: This is going to be our reference if you want to call it that, our datum. And this capacitor value is M mechanical and it's the mechanical mass. Watch now. It pays to write this down.
This is force through here and this is velocity. So we're dealing with force and velocity. f equals ma. This is actually, the mechanical mass. The value is the mechanical mass of the moving parts. The moving parts in this thing are the bobbin, the voice coil. So it's the mass of this. It's roughly maybe half the mass. This is pretty light compared to this. Half the mass of the spider because this end of the spider's fixed. The middle is going up and down. It's the mass of the dust cap, the mass of the cone, and maybe half the mass of the surround or the annulus. Because this edge of the annulus is fixed. This end is moving. So half the mass of the annulus, half the mass of the spider roughly. Total mass of the cone, total mass of the voice coil and the bobbin. That's what gives you M sub m. Just weigh it. determine m, that's it.
DR. AMAR G. BOSE: Of the inductor because of the bobbin?
AUDIENCE: I'm sorry, not the bobbin. The spider.
DR. AMAR G. BOSE: Oh the spider? OK. Because that's a spring. Yup. Where would you like to put it?
AUDIENCE: In series with something.
DR. AMAR G. BOSE: In series with something? Ah, let's see now.
AUDIENCE: [INAUDIBLE] --associated with spider too? [UNINTELLIGIBLE PHRASE].
DR. AMAR G. BOSE: Oh, is there a loss? Yeah, we'll get to that. Yeah, exactly. Somebody anticipated that the way you saw this thing, you might use energy if you push this up and pulled it down again. That you wouldn't get all the energy back in. That there's some damping or friction in this. And to be sure, there is. And I'll ask you where that goes in a minute.
But the spring. Where would you like to put that spring now? It's an inductor in this. When force is the through variable and spring is an inductor. Don't memorize that, but just get it from f equals kx. So where would you like it?
AUDIENCE: It should actually go to ground because the spider's attached to ground.
DR. AMAR G. BOSE: Yeah, exactly. The one end of the spring is ground is the inertial system for this. Which is this. This is the inertial system for this whole thing.
If I were to hang this on a string and play it, and you measured finely enough, the speaker would be moving back and forth with the music. But you'd be hard-put to see it because that's a lot lighter than this. But this is the inertial system for this whole thing.
So one end of that spider goes to the inertial system. That's where the mass went. So we'll call this the mechanical compliance of the spider. We're getting there.
Next. One second. Think about what you haven't included. Here are all the parts on the table.
Oh, I forgot to-- whoop. I forgot to tell you what this is. This is just a ring whose principal purpose is that when you put the thing down like this and ship it, you don't destroy the cone. Otherwise you don't need it. Yeah?
AUDIENCE: Well, so you have a generalized element whose value is z naught to represent the complex impedance of the air on the outside?
DR. AMAR G. BOSE: I haven't gotten there yet. I want to stick to the mechanical things. We actually haven't included all the mechanical ones yet. Then we'll get to what the air does.
You're looking at all the stuff, you haven't used it yet.
DR. AMAR G. BOSE: Oh, the resistance. The laws? Yeah, OK. That's a good one. I'll get to that in a minute.
AUDIENCE: Don't you need something to represent the piston of the cone actually moving?
DR. AMAR G. BOSE: Well, we put that in the--
AUDIENCE: Is that the [INAUDIBLE]?
DR. AMAR G. BOSE: The moving mass was in here. We said that was in there. You're from Bose, I can't-- you got to hold it for a second. Yeah.
AUDIENCE: Do you need some resistance from a spider [INAUDIBLE]?
DR. AMAR G. BOSE: Yeah, the resistance we will-- that's certainly one that we haven't included that you couldn't see easily.
AUDIENCE: Spring on the edge of the cone.
DR. AMAR G. BOSE: Yep, I'm holding it in my hand. This spring. Where would you like to put it?
AUDIENCE: To ground.
DR. AMAR G. BOSE: To ground because the outside of this spring is connected to the inertial system. And the other side of the spring is connected to the moving mass. So compliance, mechanical of the [INAUDIBLE]. I won't call it surround. I'll call it annulus now because we have two words for it. So now I think you've used all of them that are physically obvious. And now, you've brought up the condition that this thing has loss in it. And for that matter, this one even, this suspension has a small amounts of loss in it. Hysteresis actually that causes this.
And so we have to represent that. How would you choose to do that? For this and this they're the same. Because look, they're both connected to the-- one end to the moving mass and the other end to the inertial system, which means that the element in here goes to ground. So we put that in as, let's say some r mechanical. I'll just put in r mechanical. You know what it is. It's to account for loss in this.
Now, that loss to be sure, if it was hysteresis, for example, the loss would go up with frequency. Because you go around a hysteresis loop every time. To the extent that it's viscous loss, it would be a force proportional to velocity. There'd be a real resistance to the sense that there was any hysteresis in it, this resistance wouldn't be a real resistance. It would have something to do with frequency. But the losses are pretty small. And if you represent them as a viscous loss, it turns out to be good enough that the analysis and the measurements do coincide.
OK, now we have included everything that is on the table. Now comes the rest of the story. How do we do it? That loudspeaker now, let's think of the following thing.
The loudspeaker, you see this basket on the back. It has holes in it. Big holes. If it didn't, what would be the result?
DR. AMAR G. BOSE: You'd have compliance. You'd have one heck of a stiff area. When that cone started to move, you'd have a very small volume. And so you would have a very, very small compliance back there. And it'd be a heck of a thing for the cone to move in and out. So let's at the moment, just for the beginning say that-- don't worry about-- I'm going to lift it this way. Don't worry about the fact that there is something going on out here. Just look at the front of the cone. I want to ask you, what does it see? What does it see and what do we do about it? Yeah.
AUDIENCE: What is the speaker connected to?
DR. AMAR G. BOSE: An amplifier.
AUDIENCE: No, no, no. The mechanical part.
DR. AMAR G. BOSE: The mechanical part. Well, I'm telling you not to worry about what's going on in the backside. So imagine that we cut a hole in this wall and we put the speaker in the hole and it's facing out here. And the rear of it goes next door. But I'm not worried about what the rear is yet.
AUDIENCE: You've got the air. It sees the air.
DR. AMAR G. BOSE: It sees the air. So now we're in a world of impedances over here with lump parameter elements. So what does it see? It's cut into here. This is a wall.
DR. AMAR G. BOSE: Z naught?
DR. AMAR G. BOSE: A z. That's safe. Here it is looking out. What kind of a z does it see? Acoustic z, I know that.
AUDIENCE: Well doesn't it see any [INAUDIBLE] volume velocity or spherical source, it sees some complex impedance that we went over last time.
DR. AMAR G. BOSE: And what did we give that name, that kind of impedance that you see looking out into the rest of the world?
DR. AMAR G. BOSE: Radiation impedance. That's exactly what it sees.
Now the simple model of the radiation impedance, let's see. For the across variable being velocity and the through variable being force, the simple model that's approximately true. And I've asked you look to look at, I think it's chapter five or something in Beranek for all of these models-- is this.
Oh, by the way, let's just see if you know this. Here's force and here's velocity. I want this element in that without you looking it up in the book. I want that element in this circuit with this as force and this as velocity. I want to get the value of that.
If I had asked you this question with this pressure and the through variable velocity, what would be this element? z0? z0? Pressure over velocity. When you go to high enough frequency, the loudspeaker goes-- radiates is if it were in a tube of infinite length. Remember, it gets directional as you go to higher and higher frequencies. It looks like a plane wave that it's radiating. So pressure over velocity is z0 or rho 0c. So now if I change it to this, we had pressure over velocity was z0. But we want pressure-- no, force in this case. Force over velocity, which is a times pressure over velocity. Which is a z0. So this is rho 0c times 4 pi r squared, where r0 is a radius of the cone. The radius of cone.
So that is a model and the reactive part you can get. For force and velocity, now what do I do with that model? How do I get it connected to what I have on here? Pardon?
DR. AMAR G. BOSE: No one? Yeah.
AUDIENCE: Well, your forced air is a way to-- well, pressure. You have to get one of your across variables to relate to your through variable because you've got an f there. The mechanics is [INAUDIBLE] and that's related through the area of the cone to the pressure. So you need like a gyrator type of something instead of a [INAUDIBLE].
DR. AMAR G. BOSE: Well, I could take the dual now and I could reverse those things. In fact, I could just do this if I-- if I were really sneaky about it, I could do this. That's actually right.
AUDIENCE: Wait. If you take the dual of that, how can you--
DR. AMAR G. BOSE: Same thing?
AUDIENCE: Yeah, how can you have the same thing?
DR. AMAR G. BOSE: Oh, it's special magic. I screwed up, I gave you the circuit that was actually for the across variable being u and the through variable being f. And this thing then is rho 0c times what? Or what is it?
AUDIENCE: It should be 1 over.
DR. AMAR G. BOSE: 1 over.
The other circuit that goes with that, the one I should have given you is this, which is the dual of that. All right? And this is the one, as I seem to recall, which has force and velocity this way. This is the dual of that, or vice versa as this.
So I now have this, what do I do with it?
AUDIENCE: I was putting a transformer in there.
DR. AMAR G. BOSE: Yeah, I could put it in a transformer, but the transformer isn't too interesting. It's 1:1. Because nothing wrong with that. But since I have this.
For example, if I had pressure, if I had the acoustic impedance up here for pressure, then I could make an interesting transformer with an a:1 ratio. So I can throw away the-- if I need a transformer to scale by a, I can put it in. But I don't like to put transformers into places where I don't have to because if you're used to looking at circuits and you put a transformer in there, you'll find out that it discombobulates you. You look at it and you say, oh my god, I'm not sure what's happening as you go to different frequencies. We're going to be able to look at this thing by inspection and tell what happens at low frequency, middle frequency, high frequencies. Less transformers, better.
So since it's a 1:1 transformer, I already have the through variable as force over here and the across variable as velocity, which is right across here. I simply can do this.
And this point here divides the world of mechanical, which extended from here to here, and now acoustical.
Now, if I wanted to find out-- let's say I have an amplifier driving here, and I want to find out how much power is radiated by this loudspeaker at a given frequency under the conditions that the cone is moving like a diaphragm. Under the conditions that the frequency, the wavelength of the frequency is bigger than 2 pi times-- much bigger than 2 pi times the radius. Or much bigger than the circumference.
Let's assume that with the voltage on here, you could compute this voltage across here, for example. How could I determine what the acoustical power radiated is of this loudspeaker? Acoustical power radiated. Yeah.
AUDIENCE: [INAUDIBLE] that would be the power dissipated in that resistance.
DR. AMAR G. BOSE: Exactly. Just the power dissipated in this fellow. Because, remember when you're looking out at the rest of the world, the power radiated, though it is not dissipated in the model that you make of the rest of the world, that turns out to be the real power that goes across these terminals, the average power that goes across these terminals that doesn't come back. That's the power dissipated in the model of the resistance.
In the real situation, it is not dissipated. It just goes on. But when you're looking out there and you made a model of the rest of the world, which is this. That is the power dissipated in this resistance. So you don't have to go to things like volume, velocity and whatnot to find out what the power is. From volume velocity to pressure, and from pressure over a sphere to compute power. You don't need any of that, you can get it right from here.
And by the way, it's a very interesting thing. Let me see if I can get this across. I should repeat it again. But the first time, here we go.
This voltage here in this model, in this simple model, believe it or not, this voltage is proportional, voltage across this resistance to the far field pressure.
Now, why? Because the power in this resistance is proportional to the voltage here squared. The power that you got outside of the source somewhere in a big sphere, the power out here is proportional to the pressure squared that you would measure out there. We computed that from, remember our intensity calculations and power calculations. Power out there is proportional to the pressure squared. The radiated power here is proportional to the voltage squared across this resistance, v squared over r, one of the three Ohm's laws. So the voltage here is proportional to the far field pressure, which is really interesting. You bring a whole darn thing down to looking at a voltage in an equivalent circuit and finding out a relationship between pressure.
Now, when we had expression for volume velocity and radiation, remember the pressure in the far field was something like this. The pressure of r and t was basically j omega, rho 0 over 4 pi r, e to the minus j omega t minus r over c times volume velocity.
OK, this thing, the far field pressure went up as omega. You look at this thing and that's exactly what's happening. Because when you're at low frequencies, this impedance is dominated by the very large impedance of the capacitor. And so the current that flows down here as you double the frequency, when you're at very low frequencies, this thing is-- all the current that goes down here is due to the capacitor. This resistance very small compared to that.
So as you double the frequency, twice the current goes down. Therefore, twice the voltage across that resistance. And therefore, twice the pressure. So you can look at these models. This was in terms of the volume velocity of the cone. But the volume velocity is nothing more than this across variable times area. So it all comes together. This is the first time now we've actually been able to bring three of these disciplines into one.
OK, we'll continue that next time.