We'll get into a Mustang Classic Minute. Paul Taylor now joined by Middlebury head coach Dave Campbell. Coach, it was a game that was a tale of two halves. First half, you guys just couldn't get on track. Second half, you guys got it to within four or five. And it just was almost like two different games today. Yep. Yeah, our guys played with more composure in the second half. They competed. We just weren't mentally ready to play when we showed up today for whatever reason. And we need to address that and get better at that. And we will. You know, it's early in the season. But yeah, we sure wish we could have shown up the way we did in the second half when we were getting things going today. You guys have the quick turnaround. What's going to be the message overnight tonight? How much we can learn from today in every aspect. We had a tough day clearing, tough day facing off. Ground balls got better in the second half. Clearing got better in the second half. But we need to just play better as a team, be coachable, and execute. All things that we're perfectly capable of doing. You guys had a lot of man down opportunities, the wrong word, man down situations to defend. What's the biggest key to trying to working on that? I mean, obviously, you want it to be a physical game. But is it the situation where you're in a different region with different officials? Or what were your thoughts on that? I think some of those plays, honestly, probably came out of frustration, some of those penalties. So that's just our guys need to keep their heads better and not put ourselves in the situation where we're in the box that much. And then we need to execute better when we are man down. And we're capable of doing that. We just have to run our schemes better and be more dialed in mentally, which, again, we just weren't in that state of mind for much of the day today. All right, coach, will you get one more this weekend tomorrow morning? Go get them again tomorrow. Thanks so much.


So lesson one, the fundamental counting principle. Jesse, can I suggest to you that anything in math that has the title with the word fundamental in it means it's really important. It means what we're gonna do is we're gonna take today's lesson and for the next three weeks or so, we're just gonna build on this. In fact, when in doubt, I fall back to today's lesson. This is the one that I'm gonna learn inside and out and then I'm gonna show you shortcuts along the way but if I'm not sure whether the shortcut will work, I fall back to today's lesson. The fundamental counting principle or objective today is to develop the fundamental counting principle. It's really counting without counting. So here's our first example. It says a cafe has a lunch special consisting of an egg or a ham sandwich, E or H, milk, juice or coffee, M, J or C and yogurt or pie for dessert, Y or P. It says one item is chosen from each category, list all the possible meals. Well, listing them would be tough but if I wanted to do it in an organized manner, I would probably use a tree diagram. I would probably go something like this. I would say, okay, I can have an egg or a ham sandwich and after I choose my sandwich, I can have one, two, three beverages, one, two, three beverages, M, J, C, M, J, C and after I've had a beverage, how many choices do I have for a dessert? Two, I could have Y or P, Y or P, Y or P, Y or P, Y or P. What was that? Oh, I thought someone said so much writing. Oh, because that actually is part of the point here. I'm not gonna list them very often because it's too much work. What I'd like to know is how many meals are there but I'd like to try and figure out how many possible meals there are without counting them. Now I can count them right now because if I look at the end of my tree, I have one, two, three, four, five, six, seven, eight, nine, 10, 11, I have 12 meals. There's 12 possibilities. Can anybody see a way we could get the number 12 without listing them based on the information that we were given? Sorry, what? Two sandwiches times three beverages times two desserts is 12 and that's what the fundamental counting principle is. The fundamental counting principle says if you're making choices and you wanna know the total number of choices, multiply how many choices you have for each step. Oh, sorry. As long as none of the choices overlap from category to category. If I had milk up here as a choice, it's a completely different issue as well. Milk, if we're talking cards, for example, that's where sometimes choices overlap, so we'll have to deal with that later. So the cafe features ice cream in 24 flavors. You can order regular sugar or waffle cones. Suppose you order a double cone with two scoops of ice cream. Okay, how many choices of cone do we have? Read the question carefully. How many choices for the type of cone are there? Three, how many choices of flavor for the first scoop? Two, no, not two, 24, right? How many choices for the second scoop? Well, now it depends. It depends on whether I'm allowed to get the same flavor twice or not. And the phrase we're going to use is this, whether repetition is allowed or not. So I'm gonna make a little heading here with repetition, I have 24. If I'm allowed to repeat, if I'm allowed to get chocolate and chocolate, and why wouldn't I ever want to get chocolate and chocolate? Especially if it's the dark chocolate gelato, if you haven't had that, you need to do yourself favor and try that sometime. Or without repetition, only have 23 choices. How do I know which is which? This question is not clear. The questions will be clear.  They'll just know. How many different cones are possible? Well, with repetition, without repetition. When I'm during this unit area, I'm gonna give you certain tools, certain tricks of the trade. And one of them is simply drawing blanks. Miguel says, I'm good at drawing a blank. No, no, drawing blank lines. Drawing blank lines, I mean here. What I mean is if they have three choices, I'll draw a blank line for each choice. And then I'll ask how many choices do I have for the first option, three of them? How many choices do I have for the first flavor, 24? How many choices do I have for the second flavor, 24? How many choices do I have grand total, three times 24 times 24, which is what? 1,728. And hopefully you can already see Miguel, we're well beyond wanting to list all those. Like the tree works as an initial step. No, way too, 1,728 branches? No. What about if I'm not allowed to repeat? 1,656, is he right? Yeah. Which is the correct answer? Well, when you order ice cream, is there a rule that says that you can't order the same flavor twice? So I would argue that this is the correct answer, but sometimes the question will clearly suggest that you're not allowed to have a repetition. A computer store sells five different computers, three different monitors, five different printers and two different multimedia packages. How many different computer systems are available? Well, there's four different steps for making choices. I have five choices for the first option, three choices for the monitor, five choices for the printer, two choices for the multimedia package. And I'm pretty sure this is really 15 times 10. See it? I think it's 150. And this gives us the fundamental counting principle. It says this. If one item can be selected in M ways, and for each way a second item can be selected in N ways, then the total, the two items can be selected in M times N ways. The fundamental counting principle says when you're making choices, you multiply. Example, how many different two-digit numbers are there? Well, now I need to think a little bit. How many choices do I have for the first digit? It can start with a one, or a two, or a three, or a four, or a five, or a six, or a seven, or an eight, or a nine. But it can't start with a zero. That's not a two-digit number. So how many choices do I have for the first digit? Nine. How many choices do I have for the last digit? 10, because I think I can end in a zero. How many two-digit numbers are there? We're gonna add a part B here. How many of those are odd? Let's find out. I would say I have two choices. Now, whenever they give me some kind of odd, one, three, five, seven, nine. Yeah, sorry, should have explained that for my ESL student. Even two, four, six, eight, zero. So odd numbers, the restriction to me, what makes an odd number an odd number? How can you glance at a number and tell me right away whether it's odd or even? What do you look at? If you see this, how do you know it's odd? Does that tell you it's odd? No, does that tell you it's odd? What tells you it's odd? Okay, the restriction, or the key, I'll call it the restriction, the condition, it's the last number. I always do the restriction first. So even though I've written it like this, I'm gonna deal with this. I'm gonna say, how many choices do I have for odd numbers there? Five, one, three, five, seven, and nine. I have five choices. How many choices do I have for the first number? Can I have a one there? Can I have a two there? Yes, can I have a three there? Yeah, in fact, you know how many choices I have there? Nine, now somebody said 45, it's half of that, which it is, but I've just confirmed it or proved it, and they're gonna get more difficult than this. So the first tool, the first skill, the first little trick of the trade is drawing blanks. Yes, Miguel, you're good at drawing blanks. The second trick of the trade that I often use is the idea of a grab bag or a little scrabble bag with scrabble tiles, like example three. Example three says, in each case, how many two-digit numbers can be formed using the digits zero, one, three, five, seven, and nine if repetitions are allowed and be if repetitions are not allowed? I honestly very often draw a little grab bag like that and I'm gonna put the number zero, one, three, five, seven, and nine like that, so I can kind of visualize until I get good at this. And then I'm actually going to imagine reaching in and pulling out a number. So how many choices do I have for this first number? One, two, three, four, five, six, I disagree. What can't I pick? Okay, you have to look at this very carefully. What can't I pick, Miguel? Okay, so how many choices do I have for the first number? Five.  I would cross that out. And then I would say, now how many choices do I have? Tyler, still five, because there's no other restriction, right? Doesn't say the last number has to be odd or even. So I have five. How many different two-digit numbers can I form if repetitions are allowed? Ah, it's not five. If repetitions are allowed, I'm still allowed to pick that nine, aren't I? So how many choices do I have? How many choices do I have for the second number? Six, there's 30. Compare that if repetitions are not allowed, if I'm not allowed to pick the same number twice. I would start out by saying, how many choices do I have for the first number? Five, how many choices do I have once I've picked, for example, that nine? Five, so here's my question. Many numbers repeat. How many numbers have a repetition? Pat, five. If there's 30 grand total, 25 without repetitions, the remaining five must have repetitions, from page. A true-false test has seven questions. Suppose students answer each question by guessing randomly. How many possible answers are there for each question? On a true-false test, how many answers are there for each question? Two, how many different patterns are possible for the answers? Well, there's one, two, three, four, five, six, seven different questions. Amy, how many choices are there for the first question? What about for the second one? What about for the third one? For the fourth one? Amy, can you find a shorter way to write that instead of writing two times two times two times two times two times two times two? Ah, very nice. Two to the seventh, which is what, 128? Now, if this was an actual test and there's 128 possible ways of making up answer keys, how many of those are the correct answer key? Only one. What's the probability of getting perfect? One out of 128. Come on, you kinda wondered, haven't you? Or better yet, instead of true and false, multiple choice. A multiple choice test has seven questions with four possible answers for each question. Suppose students answer each question by guessing randomly. How many possible answers are there for each question? Four. How many different patterns are possible for the seven questions on the test?  It's gonna be four to the seventh, which is what? The odds of getting perfect on a multiple choice test by guessing one in 16,384. And of course, the moral of the story is, study. Now, a better question, which we can't answer yet, is Mr. Dewick, what are the odds of passing of getting four or five or six or seven? That's a much more difficult question, tuned in about a month from here. And we will figure that out. The fundamental counting principle, now we just need to look at some weird ones. Can you go to your workbook, please, to page, I believe, 378. Page 378. Some of these get nasty pretty fast. The most common error is you get an answer twice as big as you're supposed to. It means you're double counting without realizing that you're double counting. And I still do that once in a while too. Those of you that purchased the answer keys, of which there's still six left by the way, this is where you'll find, I think, it very helpful because the answer keys, he does show you his steps. In fact, this is where I learned how to do most of it because I had only done about three weeks of this in college and here we're trying to teach it to the grade 12 when this came out in 2001. So determine the number of distinguishable four-letter arrangements that can be formed from the word English if.  Right now, I'm picking my words very, very carefully. I'm making sure that none of them have the same letter twice because that's gonna be tomorrow's lesson, next class's lesson. What if you got two E's, we'll deal with that later. How many letters are we picking grand total, Justine? We've got four, okay? So if letters can be repeated, one, two, three, four blanks. If letters can be repeated, what that's saying is you pick a letter, look at it, and then you put it back in the bag and shake things up again. I think it's gonna be seven choices, seven choices, seven choices, seven choices. It's gonna be seven to the fourth, which is 49 times 49, 2,500, 2,398. No, what's, what, 2,401? What if no letters are repeated? So we're not allowed to have repetition, which means when I pick a letter, it stays in my hand and there are no further restrictions. So I'm gonna draw one, two, three, four blanks. Irwin, how many choices do I have for the first letter? Seven, now I hold that in my hand. How many choices do I have for the next letter? Six, then, then, seven times six times five times four. Five times four is 20, seven times six is 42, 42 times 20 is gonna be 840. Someone double-check me, 840, yes? Okay. Now they're gonna give me a condition. Dylan, can you read to me part two what's the condition they gave us here? Now, I will always do the condition first, even if it wasn't the first letter, if they said the second letter or the fourth letter, I always deal with that first. I'm still gonna draw one, two, three, four blanks. What's the condition? The first letter has to be what, Dylan? So I'm gonna show that by drawing an E underneath the first blank and I'm gonna look at my grab bag and I'm gonna say, how many choices do I have for a letter E? It's a trick question. How many ways are there to get an E? Only one. So I've picked an E, it's in my hand. Now how many choices do I have left for the next letter? Six, then, then, five, then, four. If the first letter has to begin with an E, it looks like I have a total of 20 times six, 120 choices. Okay. We're gonna come back to the third one in a second, we're gonna do the fourth one. The first and last letters must be vowels for my ESL students and actually for all of you. Vowels are A, E, I, O, and U and never Y, not for the purposes of the provincial exam and not for the purposes of my test. It's gonna be five vowels, A, E, O, I, and U. Those are the vowels. What do we call the non-vowels? Consonants, okay? So one, two, three, four. It says the first and the last letters must be vowels. How many choices do I have for the vowel? What vowels are there here? How many choices do I have? Two. Now the last letter also has to be an E or an I. If I put an E or an I at the front, how many vowels are now left for me to pick from? How many letters am I holding in my hand right now? Two grand total, the E and the I. How many letters are left for this choice? Five, then four, three. The word must contain a G. Does it say it has to be at the front or at the end? It just has to be there somewhere. So this one I'm gonna do, I'm gonna draw one blank separately and then three blanks here and this is gonna be the letter G. How many choices do I have for the letter G? One. How many choices are remaining for the rest of them? So I've got a G in my hand. How many? Six, then five, then four. Except this would be if G was in the front. In fact, I can put the G right there, right there, right there, right there. How many choices do I have for a location for G? 20 times 24. 480? Yeah. These are already considered fairly tough. These are completely fair game. These are fair game as a curve ball, but not 15 of these on your test. We've already done the multiple choice. Let's do example four. So as the telephone number is allocated to subscribers in a rural area, it consists of one of the following. The digits 3, 4, 5, followed by any three further digits or the digit two, followed by one of the digits one to five, followed by any three further digits. Kinda weird, but we'll live with it. Oh, do you? Looking for me? You're just the one for the ride. Okay, we'll be done in about five minutes. Like you to draw a little arrow pointing to the word or and in combinatorics, the word or really is the same as adding a plus sign. In other words, when I read this question and I see the word or in my mind, actually on my paper, I'm doing that already. Says telephone numbers consist of one of the following. The digits 3, 4, 5, followed by any three further digits or plus the digit two, followed by one of the digits one to five, followed by any three further digits. Yeah, let's walk through this. What has to go right here? 3, 4, 5, because it has to go there, what it's really saying is it's only got one choice. You gotta put it there. How many choices do I have here? Well, it says any three further digits and it doesn't say that we're not allowed to repeat. It doesn't say we can't use zeros. So I think we have 10 there, 10 there and 10 there or the digit two has to go there. How many choices do I have for the two? Only one choice, followed by one of the digits one to five. How many choices do I have if I have to put one of the digits from one to five there? Five, followed by any three further digits. I got 10 to choose from, 10 to choose from, 10 to choose from. This one is going to be 1,000 plus 5,000. I think if I've done the math right in my head, it's a 10 times table, that I can do 6,000. Classic question that we love to throw at you is license plate questions. So car plates in an African country consist of a letter other than I or O, followed by three digits, the first of which cannot be zero, followed by any two letters which are not repeated. So it looks like we have six choices to make along the way here.  How many choices do I have then? 24, because there's 26 letters in the alphabet. Followed by three digits, the first of which cannot be zero. So this can't be a zero. How many choices do I have? Is there any restriction here? 10, any restriction here? 10, followed by any two letters, but you can't repeat. So how many choices do I have for the first letter? 26, then 25. Did say any three letters, I can bring that I and O back, apparently. How many license plates are possible in this African country? I don't think so. 900 times, ah, read me the digits, one, four, one, four, zero, four, 14 million, 40,000. Okay, yep, louder please. The first letter, like here, it doesn't say that I can't use I or O over here, so I assume it says any letters, any two letters, so I got 26 to pick from again. Yeah, I think it means that I can't use the second over here, I can't repeat this letter. I don't think it's saying that I can't use this one, and if it did mean that, I think they'd phrase it that way, okay? Two more, we're done. It says, consider a six-digit numeral note. If you start with a zero, that's a five-digit number, so no beginning with zero, and I'm not gonna tell you that from now on, I'm gonna say to you, when I ask for a three-digit number, don't start with zero, please. How many odd six-digit numerals have no repeating digits? So we're gonna have one, two, three, four, five, six. And just so I can visualize what's going on, I'm gonna do my little scrabble bag here. Zero, one, two, three, four, five, six, seven, eight, nine. There's my digits. What's the restriction here? Odd, what makes an odd number an odd number? So it has to be a one, or a three, or a five, or a seven, or a nine. How many choices do I have for this last one then? Let's assume, because it says we're not allowed to repeat, so now we're gonna cross off numbers along the way. Let's assume we got a three. How many choices do I have for the first number? No, I don't have nine. Why eight, why not nine? Okay, I can't start with a zero. Can't start with a zero. So I look at what's left, but I can't start with a zero. I have eight choices. Let's suppose, as an example, Alex, that it was a four. How many choices do I have left now? Still eight, because I can have a zero there. Let's suppose it was a nine. How many choices are left now? Seven, then, six, then, five. Eight times eight times seven times six times five times five. Doin' my head? That's a 40, that's a 30, so that's 1200. 1200 times 56, yeah, I can do this in my head. Double 56 is gonna be 112, 112,000? No, really? No, no, no, no, no, Mr. Dewick, you can't be off that much. I doubled it, I got an extra two in there. Like I said, my problem is double counting sometimes. I hope I don't trust you, Miguel, but, you know, 67,200, good gosh, Mr. Dewick, way off. It is, if I had any kind of an ego. Oh, sure, go ahead. Oh, no, I never make mistakes, believe me. How many even six-digit numbers? This question is tougher, and the reason this question is tougher is, what makes an even number? What can we end with? Rattle him off to me, Pat? He said, zero, two, four, six, or eight. What can't I start with? Zero, the fact that zero is in both restrictions is gonna be a problem. I can't start with a zero, and I have to end possibly with a zero. In fact, the only way to do this is to look at two cases, end in zero, or end in two, four, six, four, six, or eight. And remember earlier I said the problem arises when one of our categories, one of our scrabble tiles happens to be in more than one condition. And the best way to do that then is to say, well, let's look at what happens if it's in one, or what happens if it's in the other. And what does the word or mean? And once again, I'm gonna do my scrabble tile here. Zero, one, two, three, four, five, six, seven, eight, nine. We're doing six digits, so one, two, three, four, five, six. One, two, three, four, five, six. And in this first one, we're gonna end in a zero. How many choices do I have to end in a zero? Only one. And the reason that's important is, okay, I just picked a zero, I'll hover my little mouse above it. How many choices are left for the first number now? Not eight, I don't think, why? Why nine this time? Because I've already removed the zero from my scrabble tile. I now have nine choices. Let's suppose I got a two. So I've used the first two. How many choices are left? Eight, then, seven, then, six, then, five. That's if I end in a zero. Or I could end in a two, four, six, or eight. And the reason we're having to separate these again is because zero happens to be in both restrictions, in both conditions. How many choices do I have for a two, four, six, or eight? Four. Let's suppose I picked a two, this guy. How many choices do I have for the first digit? Careful. Eight, because the zero is still in the scrabble tile bag.  How many choices do I have left now? Eight again, then, seven, then, six, then, five. And I get, what was that, 15, 120? Plus eight times eight times seven times six times five times four. 53, seven, 60, and I get 68, 800, and 80. Well beyond our ability to list them and count them. I skipped example six, because I figured example seven kind of covered it. Just really quickly, if they say that you have to be less than 400, how many choices would you have for your first digit here if we're doing three digit numbers and you have to be less than 400? Two, and then whatever and whatever. What's your homework? Yep, the fact that zero is part of the restriction when it says even, zero is included in there, and anytime I'm listing numbers, I can't begin with a zero. Zero appears in two restrictions. You can't start with a zero, make sure you take that into account. Oh, and this time, since it said even numbers, oh, one of your options is you have to be able to end at a zero. There's also a two, four, six, or eight, but I have to look at what if I end at a zero? If I end at a zero, can I possibly start with a zero? Nope, because I've used it up already. That's why I had to list them separately. Look at two, four, six, or eight at the end where you can start with a zero, or end at a zero, which means you can't start with a zero, because you nine choices there. That's the big issue. This is already considered borderline too tough for the provincial exam. In other words, these are, they stay fairly basic for a while, okay? What's your homework? Number one, number two. Take a look at number two. Gonna give you a hint, the answer to number two when it says how many possibilities are there for the score at the end of the first period. The answer is not 18. It's not six times three. I'll let you think about it. Number four, five A, five B, I, five B, three, five B, part four, seven. Three letters followed by three digits. Try eight. Nine A is good, nine B is good, nine C is good, nine, 12 A, 12 B, and we'll pause there, okay? And then next class remind me to show you the ABC menu, restaurant menu that I thought initially was mathematically incorrect but turned out to be right.


In the last lecture, we had looked at the fundamental problem of the delays introduced by long interconnect. We had seen that while scaling improves the speed of active electronic circuitry. The interconnects in case of local interconnects retain their old speed, whereas long interconnects actually become much slower. This puts an inherent limitation on the maximum speed at which the system can operate. We had looked at the existing methods of improving interconnect speed in case of long wires and buffer insertion is the dominant technique which is used for this. However, buffer insertion is far from perfect. It takes power, it makes the line unidirectional and ways of making the line bidirectional are not easy. In order to reduce the total amount of power consumed by the long interconnect, we had looked at low swing signaling. Low swing signaling in voltage mode is already in use. Essentially, what it means is that over the bus, we do not use the same 0 and 1 levels as we do in the main logic. When the signal is to be conveyed over long wires, the line swings by a small amount of swing. This reduces the dynamic power considerably. At the receiving end, we have an amplifier and this amplifier will then restore the logic levels so that the rest of the logic does not know the difference. This technique is quite successful and we had seen a variant of this which is called the current mode logic. The current mode logic differs from the voltage mode logic because it terminates the line in a low impedance. We had also seen a active circuit which can be used for producing a very low and controllable input impedance. There are many other ways as well for loading the line with a low impedance so that it operates in the current mode. The current mode essentially makes the impedance of the line low and therefore it is able to charge and discharge capacitors to the small voltage levels much faster. As a result, it has inherent advantages over the voltage mode of signaling. Having established that there is a lot of promise in low swing signaling over the traditional buffer inserted lines, we now see how we can optimize this mode of signaling so that we can get the best possible performance from this. One of the methods for this we had briefly discussed yesterday. We shall take it up again in detail today and that is called inductive termination. There are two other methods also which we shall see today and then see what are the kind of optimizations which are possible in this technique. Let us look at the basic current mode signaling technique. We need special drivers which are called low swing drivers. These ensure that the voltage at the line does not go from rail to rail but swings around a small common mode voltage here. This small swing is then conveyed here and terminated into this load resistance. Notice that the common mode voltage means that there is a steady current flowing through this and therefore there is static power consumption. This small fluctuation at this point is then amplified by this receiver to a level where the output now again swings from rail to rail. So this is the basic model of the current mode signaling which we have been discussing and now we want to see what are the possible improvements which are possible. Notice that this wire is an RC line and therefore has a low pass characteristic. It will therefore selectively attenuate high frequency components of the signal and therefore distort the wave shape. When we want to carry out very high speed communication over these lines then we need to worry about this low pass nature and we are always looking for ways which will improve the bandwidth of this link. There are various ways which have been suggested in the literature for this. There are three dominant ways which we shall discuss. One of them is called inductive peaking. The idea is that rather than terminating the line in a pure resistance we terminate it in a load which has an inductive component. Because of this the termination now has a high pass characteristic which to some extent mitigates the low pass characteristic of the line. As a result we can enhance the bandwidth and get higher speed transmission on the same infrastructure. I will just mention two other techniques here. We shall see the details of these when we come to it. Both of these techniques do the correction at the transmitter end rather than at the receiver end. At the transmitter end in dynamic over driving what happens is that whenever there is a transition on the signal that means it is not resting steady at 1 or 0. Then you employ additional drive and this additional drive enhances the high frequency components because it is used only during transitions. That then combats the attenuation of high frequency signals down the line. Capacitive peaking is a very simple implementation of dynamic over driving and it essentially puts a series capacitor in series with the line here and that has a high pass characteristic and that can compensate to some extent the low pass behavior of this line. There are side effects to this which we will worry about but we shall come to these details when we look at these two techniques a little later in this lecture. But to begin with let us look at inductive peaking. We had seen this in the last lecture and we will quickly go over this again. Essentially the termination that we had seen here now has an inductive component and using ideal components first we find out what is the effect of placing this inductance here. We will worry later on how to implement this inductance. These three curves correspond to no inductance at all this lower curve putting 100 nano henries which is this middle curve and putting a micro henry is this last curve. I would like to remind you again that these curves are on a log scale and therefore this small extension of the bandwidth is not negligible. In fact you can increase the bandwidth at the 3 dB point by about 500 megahertz in this. From this it appears by the way the values of R and C have been taken from current technologies and it is of the order of 10s to 100s of ohms per unit length here and femto farads for the capacitor. Using those values of R and C we find that typically we are looking for inductances which are in the range of let us say 100 nano henries. Now a 100 nano henry inductor is not easy to put on chip. If you were to make a coil whose inductance is 100 nano henry it will actually consume just too much area to be practical. Therefore we need to look at electronic implementations which will give us the equivalent of this kind of termination. The amount of inductance that we need also depends on the line length. So here we have taken a 4 millimeter line and these 3 curves correspond to different values of the terminating resistance and these are 500 ohms 1k and 1.5k. We now choose various values of inductance and see the effect this amount of inductance will have on the bandwidth. So you notice that essentially these curves for a wide choice of resistors actually peak at a value of inductance which is of the order of 100 or a few 100s of nano henries. And once you choose that value this value will remain optimum or close to optimum whatever be the termination resistance. Therefore in short irrespective of the terminating resistance that we actually use we somehow require inductors of a few 100 nano henries. And clearly passive inductors of this value are not practical and we therefore look for active inductors which will provide this much of inductance. In fact the beta multiplier circuit which we had seen as the termination resistor earlier itself can provide inductive component if it is properly designed. The reason for this is that the beta multiplier essentially forms a generator circuit which has 2 gm elements connected back to back. So these 2 have positive feedback and they are connected back to back. When you have 2 gm elements connected back to back they do have the capability of converting a capacitor into an inductor. And therefore a basic beta multiplier circuit which we were anyway going to use as an inductor in case of current mode can exhibit inductive input impedance for some frequency range if properly designed. So our next task becomes how to design the beta multiplier so that we will have a substantial and a required inductance which should be in the order of hundreds of nano henries. We can actually put up an equivalent circuit for the beta multiplier. These are essentially the 2 active parts of this and the active circuit can be represented with this with the 2 gate capacitances CG1 and CG3 coming here and CG2 coming here and this is the gmp2 of transistor 2 just to refresh your memories. This is mp1 and mn1 and mp2 and mn2 these are referred to as 1, 2, 3 and 4 in this circuit. So this is mn1 and this is the capacitance across that and these capacitors play an important role in determining the equivalent inductives. This network is a little tedious to solve but it can be solved given enough patients and we have in fact done this and eventually you get the input impedance of this at this point where we connect the interconnect. The total input impedance is given by this somewhat long expression. The component of this expression involve various time constants and that old gamma factor that we had defined which is a ratio of ratios here. So we have essentially gm by c behavior here. So the time constants are the c by gms of various components here. This is for transistor 2, this is transistor 3, this is transistor 1. The various resistors are the 1 by gms or the output resistances of various transistors and then we have defined two additional parameters. The k is r1 by r3 which is essentially this divider and gamma which is that ratio of ratios which makes the beta multiplier so robust. Notice that when we had done the earlier analysis we had ignored the output resistance of all the current mirrors and as a result these two terms were missing and the effective input resistance that we had gone and got at that time was 1 minus gamma upon 1 by gmp1 into 1 by gmp1. Now we have made a more careful analysis compared to the previous one and included the output resistances of all the current sources and when we do that then this is the equivalent circuit we get, this is the input impedance we get and once we define these intermediate values this can be put down in terms of these. Now this has various poles and zeros and the relative location of poles and zeros determine the nature of the impedance which can be inductive or capacitive. If the first zero occurs a decade prior to the first pole then the input impedance is indeed inductive. This means that this term gamma minus 1 upon gmp1 into r output 1 should be greater than 0.9. And if any two time constants are equal that will ensure that a zero occurs a decade or so prior to the first pole this gives us the design equation. Once we do that then we can further simplify the input impedance equation and the effective L comes out this value. It also has a resistive component and it is a capacitive component and the actual equivalent circuit is given by this where we have this r equivalent, L equivalent and C equivalent. Recall that what we want is only the r equivalent and L equivalent. But our advantage is that this C equivalent comes in parallel with the line capacitance. The line capacitance is in any case large and therefore even though we did not wanted and got this capacitance this capacitance will not make too much difference. In short by proper design of a beta multiplier we can get a circuit which terminates this RC line so that this C becomes part of that line RC and which terminates this line therefore in the kind of configuration which we originally wanted and this will then compensate for the low pass filter of the line all the way up to this point. Obviously these terms will then become frequency dependent and we are interested in a frequency range where which is which corresponds to our data rate and which is of the order of gigabits per second and essentially that is this decade this is 1 giga this is 10. So essentially we are interested in this range and this is the Z and this peaking that you see is the inductive peaking and the inductive peaking depends on the amount of current that we give to the transistors which in turn determines their gm and as we have seen before the gm determines the impedance of these circuits. So we can summarize that algebra as saying that the beta multiplier shows an effective inductance of hundreds of nanohenry's for a practical range of input current and transistor geometries these currents are not too shocking they are like of the order of 25 micro amps to 35 micro amps. So it will not cause too much power consumption and by using appropriate geometries and currents of the order of tens of micro amperes in the range of 30 micro amp we can indeed get the amount of inductance that we need in order to improve the bandwidth. It is also possible to control the input resistance and the effective L with very little interaction between the two and the inductance changes from about 100 nanohenry to 980 nanohenry's while the effective resistance remains within 12 percent of its nominal value for a 20 micro amp change in the bias currents. So this gives us a very good handle on getting the inductance that we need this is indeed the range of inductances which we want without changing the terminating resistance too much. Once we have done that we can actually come with the actual design notice that we are using actually two beta multipliers one for positive going incursions the other for negative going incursions. This shorted inverter essentially provides the reference voltage at which this line will be maintained by the beta multiplier. So this provides the DC reference for these beta multipliers and indeed for the line voltage. So this line will be maintained at a common mode voltage which is determined by the by this reference voltage generated by this shorted inverter. The output then is amplified by this in this these inverting amplifiers and the low swing voltage at this point will then be converted to a rail to rail signal here. Notice that apart from the power generation in this circuit this shorted inverter also will consume static power and the V ref generation circuit it does consume static power. However when we report various performance parameters of this we will include the power consumed by this and this circuit in the driver power that we claim. So here we present a comparison of three cases. One is the buffer inserted voltage mode the other is the beta multiplier termination and third if we just had a diode connected current mode system that means the terminating resistance is just provided by diode connected transistors. These are taken from these references and reference one is in fact our own work. As you can see the voltage mode has a very large delay about 1000 picoseconds. Just going to the plane current mode reduces this delay to about 500 picoseconds. This is a plane classical current mode line in which the termination resistance is provided by a diode connected MOS transistor. And finally the delay of the beta multiplier case is 420 picoseconds. The throughput actually goes in the opposite direction the throughput of the diode converted case the diode terminated case is the lowest at 2.45 gigabits per second. The throughput of the scheme with the beta multiplier which we have suggested comes out marginally higher than this it is about 2.56 gigabits per second. The voltage mode is actually capable of very high throughput. The throughput possible in this case is 2.85 gigabits per second, but this comes at the cost of immensely high values of power. That means it will require 300 3000 micro watts of power in order to get this throughput. The power consumed in case of the two current mode options one with the diode termination the other with the beta multiplier termination is much smaller. It is between 300 to 400 micro watts. The diode termination circuit takes 380 micro watts and the use of the beta multiplier actually reduces the power to 310 micro watts. So, there is almost an order of magnitude of power between this and should we want to match the throughput offered by this voltage mode it should be possible using the beta multiplier configuration also. However, in that case the power will go up slightly it will still remain much lower than the 3000 micro watts required by the voltage mode option. The area also shows much better figures here the two current mode circuits consume hardly any area remember no buffers are required down the line whereas, the voltage mode requires buffer insertion along this line the line is 6 millimeters long and many many buffers will have to be inserted and therefore, the silicon area taken up by that is quite considerable it is about a factor of 6.5 higher than the area taken by the current mode. So, in short inductive termination gives 16 percent improvement in delay and about 18 percent improvement in power also and this is compared to the ordinary current mode. So, more than 50 percent improvement in delay at the same time and order of magnitude lower power compared to the voltage mode having looked at inductive termination we now look at other techniques of improving the current mode transmission. And one of these is that of dynamic over driving or pre emphasis after all we have a low pass line and what we are looking for is a high pass characteristic in order to combat the attenuation of high frequency signals by the line. One possible way was what we have seen that is inductive termination, but now we look at yet another way which has been suggested. Now, if a line is sitting steady at 1 or 0, obviously the signal has very little high frequency component. High frequency components come in because the line switches at a high rate from 0 to 1 or 1 to 0. Therefore, one way of boosting the high frequency component in advance before it can be attenuated by the line is actually to pre distort the signal in such a way that the distortion is in the opposite direction to what the line will bring about. We do this by a pre-emphasis circuit and the pre-emphasis circuit essentially dumps a lot of extra power only when there is a transition from 0 to 1 or 1 to 0. This actually makes sense if the line is sitting steady at 0 or 1 then it does not require too much power. It is only when you have to change the state of the line with its large associated capacitance from 0 to 1 or 1 to 0 that you have to either put in or remove large quantities of current. Therefore, if we have a smart driver which puts in high amount of power only selectively only when it is needed namely when there is a 0 to 1 or 1 to 0 transition. In that case the overall power consumption will be kept in check at the same time we will be able to send data much faster. It is this idea which is used by the dynamic over driving techniques. So, this is what we are saying that when the line remains at 1 or 0 from 1 bit to the next we just use a small drive current to maintain the line at the required voltage. However, when there is a change from 0 to 1 or 1 to 0 we either dump or remove current and large quantities of it at that time. That means, we now we need two different kinds of drivers. One is the weak static driver the other is the strong dynamic driver. The strong driver will come on only when there is a transition. The weak driver will make sure that adequate power is supplied even when there is no transition at the input. Let us look at the weak driver first. This weak driver is taken from the work done by Atul Katoch et al from the Philips laboratories and this paper was originally published in the European Solid State Circuits Conference in 2005. The full scale digital input is applied here. So, the traditional rail to rail swinging input appears at the input of that buffer. We now need a special driver which will not force the output all the way to VDD or all the way to ground. Remember this is the weak driver and this is largely used when the input remains steady at 0 or 1 for multiple bits. If we did not do a careful design what will happen is that this line even though weak will continuously dump current or continuously sink current till the line is taken to a voltage which is close to VDD or close to ground and we do not want that. We want this drive to stop as we start approaching VDD or ground and that is ensured by the circuit configuration suggested in this paper. If you look at this circuit these two transistors in short form a traditional inverter this is a P channel transistor and this is an N channel transistor. So, therefore, there is not much to understand in this part of the circuit. The other two transistors act as switches controlled by the bus voltage itself. Now, consider the case that we are trying to drive the line high that means the input is at 1. When the input is at 1 this node because of the inversion will go down to 0 and it will go all the way down to power supply because this is a traditional inverter. Now if this voltage goes down to 0 then this transistor will turn off. Because this transistor is off there is no current taken from the line towards ground. On the other hand because the output is at 0 this P channel transistor turns on it starts pouring current into the line which is largely RC type which starts charging up. As the line charges up this particular switch remember the line was originally low and that is why this switch was also on. As the line charges up as we approach VDD minus the absolute value of VTP the effective negative bias on this becomes less than VTP. As a result this transistor now turns off and even though this is an inverter the power supply from the inverter has been taken away. And therefore, this driver stops driving the line towards VDD. In other words for voltages from the resting voltages close to VDD by 2 to VDD minus VTP this upper transistor is on and therefore, this acts like an inverter and dumps current into the line. This continues till this voltage reaches VDD minus VTP once we reach this voltage the upper transistor cuts off. And now there is no drive available to take the voltage any higher at least the output is automatically limited to going up to VDD minus VTP and no higher exactly the same thing happens when you want to drive this line low. Suppose the input is 0 this is a traditional inverter and therefore, this output will go all the way to VDD. Because this point is at VDD both the P channel transistor will be off. Recall we are trying to drive this line down to low values now and therefore, we want to discharge the line down to low values which is done by the N transistors. Because the input is 0 and this node is at 1 which is close to VDD this N channel transistor will now turn on. The line is presumably high otherwise why are we trying to discharge it. So, because the line is high this transistor is also on as a result this series combination is now on and it starts discharging this line towards ground. This discharging will continue as long as the line remains above VTN. As soon as the voltage goes to VTN and drops below this this transistor will now turn off and this will discontinue the discharging of the line. As a result we have seen that this weak driver accepts an input which is rail to rail the standard logic kind of input. It drives the output like an inverter, but make sure that the output high value remains a VTP below VDD when charging and remains a VTN above ground in case of discharging. Therefore, full charge and discharge is not possible and this line has limited swing from VDD minus VTN a VDD minus VTP to VTN. Should we want to control the swing even further we could actually use two transistors in series here and then we will get a 2 VTN row. That was a weak driver. So, the job of the driver was essentially to ensure that the input is buffered to the wire. The only difference being that while the input will swing from rail to rail the wire will swing over as low swing region. The dynamic driver needs an arrangement so that it turns on only when there is a transition. So, let us look at this circuit. This driver is a traditional inverter driver. These transistors are made strong so that it can dump or withdraw large amounts of current from the wire. We use a feedback inverter here and this feedback inverter apart from inverting the logic converts this low swing value on the wire to a full swing output. That means, this buffer is actually required to amplify as well and therefore, does consume power unlike standard CMOS gates which do not consume static power. This one however, is biased to be near VDD by 2 because this wire rests near VDD by 2 and therefore, it continuously draws current. However, there is sufficient amplification here so that a relatively small swing on the wire results in a full digital swing at the output. This feedback value and the input are combined with this NANDOR logic which ensures that drive is applied only when there is a transition. Let us analyze the functioning of this strong driver a little better. Consider the case when the input is 1 and the wire voltage is low that means, there is a transition the wire is resting at 0 and the input is 1 because the wire is resting at 0 the inverting output is 1. So, the digital input is 1 and this feedback value is also 1 because both of these are 1 this NAND goes to 0 also because both are 1 the NOR also goes to 0. That means, both these inputs are at 0 and this circuit acts like an inverter with both gates shorted and return to 0. As a result it starts charging up the output towards 1 and because these are very strong transistors we have applied strong additional drive to the wire when there is a 0 to 1 transition. What happens when this voltage reaches 1? Remember this feedback inverter will now go to 0 that means, the input is at 1 and the feedback is at 0. The line has now reached 1 therefore, this feedback voltage will become 0. When this is 0 and the input remains at 1 this being 0 is sufficient to force this NAND to 1 and if this NAND goes to 1 the upper transistor turns off.  On the other hand if the output has now reached 1 then this point is at 0 the input is at 1 and therefore, this output remains 0. Therefore, the N channel is off in other words when there is a transition for a short time when there is a 0 to 1 transition for a short time the P channel turns on and dumps current into the wire. As the wire voltage rises and actually becomes 1 this configuration automatically turns off and now both the transistors are off. As a result this wire is now maintained at 0 or 1 by the weak driver that we had discussed earlier. Now let us consider the other case that means, the previous output was 1 and the input is now 0 because the input voltage was 1 the wire voltage is greater than the trip value or the mid value of this feedback. As a result because this voltage is actually high and we are trying to drive it to 0 this high produces a 0 here. In other words the digital input is 0 and this voltage is also 0. If both of these are 0 this ensures that the NAND is at 1 and therefore, the P channel transistor is off. But since both are at 0 the NOR output is 1 and that turns this transistor on. Once this transistor is turned on it starts discharging the bus and this is a strong transistor very wide transistor. As a result it will take away current very quickly from this wire bringing it below Vm. As soon as this wire is brought to a 0 the feedback value will become 1. Now the input is at 0 and the feedback is at 1. A 0 input means that this NAND is guaranteed highly irrespective of the feedback. However, since the output has now reached 0 the feedback value is 1 and that by itself is enough to drive this output to 0 and therefore, the drive is now removed from the N channel transistor. That means, both transistors are off and now we shall not consume any more power in keeping this wire at 0 or 1. So, in short this NAND NOR combination ensures that power is applied of the proper kind to the wire only for a short while while it remains at the wrong digital value. If both the values are equal then because of this inverter both these values are unequal. That means, once the wire has reached the desired value equal to the input then these two are guaranteed to be unequal and that means, exactly one of them is 0 the other is 1 it does not matter which is which. If the two inputs are unequal then at least one of them is 0 and that ensures that the NAND output is 1 and the P channel transistor is off and if the two are unequal it ensures that at least one of them is 1 and that is sufficient for the output of this NOR to go to 0 and that ensures that the N channel transistor is also off. So, essentially what it means is that once the wire logic value in low swing becomes equal to the input logic value then this drive is automatically removed and both the P channel transistor and the N channel transistor are off. So, therefore, the high drive provided by this strong driver is applied only as long as the wire and the input are different values. Once the wire acquires the input value this guy is turned off and now will not be turned on unless an input different from the current value namely during a transition. So, when an input like that arrives only then power will then be used in these and either a current will be dumped or a current will be sunk depending on whether the input is 0 or 1. So, this is the basic operation of this dynamic or strong driver. There are many other configurations which are possible for this dynamic or strong driver and we shall look at these configurations as well a little later. However, this one using feedback makes it clear how we continuously monitor the line and as soon as this line has reached the desired logic value no further power is consumed by the strong driver it automatically turns off. This gives us a very interesting idea remember dynamic over driving is done at the transmitter and inductive termination is used at the receiver. So, should we try to combine both and apply dynamic over driving at the transmitter and inductive termination at the receiver. We examined this suggestion and did computations to see whether additional improvement is brought about by this combination. Here is the wave form of the input digital value and here is the effective drive applied to the line. During this short time when there was a transition we have seen that the strong driver turns on and the total current dumped into the line is very large. Once the line reaches the value 1 then the strong driver turns off and now only a small static current is provided by the weak driver. When there is a 1 to 0 transition on the input then our line is actually coasting at 1 and we need to discharge it and that is done by a very large drive which is provided by the strong driver now which detects this 1 to 0 transition as we have seen in detail just sometime back. And therefore, essentially the n channel transistor is turned on very heavily. This means the large amount of current is taken out of the line for a short while. As the line this charges and just comes below the apparent 0 for the low swing inverter then the n channel transistor turns off and now there is no further drive applied. So, the shape by the way this is the plot for the current which is dumped or removed from the line a negative value means current is removed and it has this shape a positive value means current is dumped and it has this shape this dotted line represents 0 current. So, therefore, if we use a dynamic over driver then this is the shape of the current pulse which is then applied to this line. If we now integrate this into the RC which is represented by the line then you get a slight overshoot here, but essentially a flat curve and a slight undershoot, but then a flat curve maintaining the line at 0. The question that we now want to answer is whether both can be used. And to answer this question we simulated four combinations. One of these combinations the first one has the current mode scheme with dynamic overdrive on the transmitter end, but a pure resistive load. This is the pure CMS scheme with dynamic overdrive. We also had a control signal which uses the current mode scheme with a simple driver without any dynamic overdrive and with a resistive load. Then we designed a circuit which uses the current mode scheme. However, it uses an inductive load and finally, a CMS scheme which has both namely a dynamic overdrive on the transmitter end and inductive load on the receiver end. Now, we only want to know whether it is worthwhile and therefore, we do not use actual circuits here. We use macro models. So, the dynamic over driving driver is implemented by ideal voltage control voltage sources, voltage control current sources and with the current wave shape as we had just seen. So, the controlling voltage is the digital input. The simple driver is implemented as a voltage control current source with a square wave output shape. If the input is 1, it provides certain amount of current. If the input is 0, it provides a certain amount of negative current and therefore, the input current ranges from minus a i average to plus i average. The i average is given as the weighted sum of i peak here. i peak is the current provided by the strong driver times the time for which the strong driver is on plus i static times t minus tp. So, this weighted average over the time t that is i average and the input average current ranges from minus i average to plus i average. Once we do these simulations, this top curve is in fact, the input given as a voltage. So, this is the input voltage plotted as a function of time. Now, we have 4 curves here. The red curve is the line voltage at the receiver with inductive loading. The yellow curve is the line curve at the receiver, but with just resistive loading. Notice that these 2 curves are very similar. However, the inductive loading does improve the speed. So, that comes out quicker. At the same time, the 2 dynamic over driving options are these 2 and while this is plain dynamic over drive without inductive, this is the inductive. However, notice that in the 2 cases, the rise time remains roughly the same and this is as a function of i peak, the voltage fluctuation on the line caused because of switching. Now, we operate these under 2 separate conditions. Once when the over drive is very large like 500 micro amps for a short time, the other when there is a small over drive 50 micro amps. Our results are that if you are using strong over drive which is what you will do if you want very high speed. In that case, the inductive peaking does not offer substantial additional advantage when combined with dynamic over drive. As you can see here, the 2 show practically the same rise time and therefore, the advantage of inductive termination is not that very high. On the other hand, if the over drive is small, in that case, the improvement is better and then it is worthwhile having inductive peaking. We will stop this lecture here. What we have done is, we have seen the effect of inductive loading and the effect of dynamic over drive. These 2 combined can be used one at the transmitter end and the other at the receiver end. However, our studies show that if you want a high speed link which will require substantial over drive at the transmitter end, then the incremental advantage offered by inductive loading is not very much and then it is not worthwhile adding inductive loading. At the low power spectrum, however, you would benefit by having an inductive loading. So, in short, whether to use both or not depends on whether you are at the low power, low speed end of the applications or high power, high speed end of the applications. We shall look at a few other dynamic over driving solutions and then go over to the capacitive driving solution in our next lecture. We stop this lecture here.


Hi, my name is Marc and in the last videos we talked about the DNA of your organization and the trends that are going to have an impact on your company. Now it's time to paint a very clear picture of where we want to be in a couple of years time And it's the most fun part of the entire output process. Let's start by determining how far we want to be looking into that future. For some companies the world moves rather slow and the company themselves don't move that fast either. In these cases it's possible to look maybe five or even seven years ahead. For other companies like Startup for example the world moves really fast. So fast even that it might be hard to look two years or maybe one year into the future of your company. We grow about 20% year-on-year and we expect this to accelerate in the upcoming period. The communication market is a really fast-moving market for us. So looking three years ahead is a solid ground to start from. Now that we have defined the period which we're going to look at it's time to fine-line the details. To do this we split the three-year picture into four main segments. What are we going to do? Where are we going to do it? How are we going to do it? And what are the conditions that apply? It's something we call the why. We often call the three-year question a dream question. In a perfect world even if you close your eyes, where will the company be in three years time? What will it be doing? What has changed? And how is it interacting with the world? The what question is about your product or your service. Imagine it being perfect. What does it do but also what doesn't it do? That's what the what question is about and the more concrete you can make it the better it's going to be. For us the where question often contains new countries we want to operate in since we're expanding to international markets. It also contains the type of markets we are addressing. Do you work for consumers, corporates or SMEs? And if you're working for SMEs, how big are they? What are the characteristics they share and the more concrete you can make this the better it will be? The third part of the three-year picture is the how. The how has multiple components to it. How are you delivering your product or service? Do you have a central office? Do you work completely remote? Do you support your customers by a telephone or do you have physical appointments? All these questions are going to be in the how. Again, the more detail in there, the more concrete it becomes, the more visible it becomes and the more easier it's going to become to get there. The final segment of the three-year picture is the while part. The while contains the conditions and the constraints. An example for our organization is while being part of society and not causing any unnecessary harm. When we are working on a three-year picture, we ask about eight people to write it down in as much detail as possible. You'll often find a huge overlap and this overlap makes it into one story. The more detailed the story, the better. The reasoning behind that is very, very simple. If you know where you want to go, the only thing you have to do is plan a route to get there. If you get stuck while writing this story, look at your purpose, your DNA, but also at the trends. What also helps is focusing on what you don't want to become. You'll see that some of your colleagues only write down one sentence while others write an entire story. Both are completely fine because both will help you get closer to the core. I recently looked at the picture we had written down as an organization three years ago. And it's incredible how close we have come while making small adjustments along the way. Now that you have a very clear picture on where you want to go, you can start looking at actions or drivers that give you a really strong push forward. We call these actions key thrusts. To get your key thrusts, you can ask six to eight people to see where they see really big steps you could take that could really help. Examples of key thrusts are launch in X countries or build this new product or service or set up a new office in country X. They are very big projects, which might take years to finish, but will help you get closer to your three year goal or five year picture. Now that we know exactly where we want to be in three years time and we know which things we can do that give us a really big push forward, it's time to make this measurable. But metrics and KPIs are a completely different ballgame, so we'll talk about that in the next video. I wish you good luck with figuring this all out and see you next time. Cheers.