[User] Please transcribe this text

[Assistant] So in this case, the way I've drawn it, it means that the p-type is slightly less, the concentration of the


[User] Please transcribe this text

[Assistant] Acceptors here is slightly lower than the concentration of the donors here.


[User] Please transcribe this text

[Assistant] Okay, we'll see that this is, this is important when we design our device. We'll talk about it. Not, not this exact arrangement, but this knowledge and understanding of the depletion region with.


[User] Please transcribe this text

[Assistant] Okay, so, and then what's, how about the currents? So what is the current? So there's some current I, that for a given cross section A, corresponds to some current density J. Right? So there's a J total.


[User] Please transcribe this text

[Assistant] Going through the diode. Now this, obviously, this current, this total current, has to be constant, right? This total current, this J total, if the cross section of the diode doesn't change, the total current has to be constant.


[User] Please transcribe this text

[Assistant] Right, otherwise you're violating conservation, at least in steady state. Otherwise you're violating conservation of charge.


[User] Please transcribe this text

[Assistant] Why do you have electrons here? Where are these, if you have electrons carrying current on this side, where are they coming from?


[User] Please transcribe this text

[Assistant] Exactly, they are coming from the N side, right? They are the, these carriers that are majority, but they were injected into this thing, so they became minorities on this side. So they are minority carriers. And as they, so at the edge of the depletion region, you have a certain concentration, you have a JN here, right?


[User] Please transcribe this text

[Assistant] What would happen to this chance? The more you go, the more chances you will get, right? So as you transition, you expect the number of these electrons that are injected in this region to drop. And you can actually show it pretty easily that it's an exponential drop.


[User] Please transcribe this text

[Assistant] And how do you know it's an exponential? Because the chance of recombination is proportional to the number of the electrons to begin with. If you have a process where the rate is proportional to the number, you get an exponential because you get a first order differential equation.


[User] Please transcribe this text

[Assistant] Any process, it's called a relaxation process sometimes. Any process where the rate of something is proportional to its current value gives you an exponential. Uh, radioactive decay is like that.


[User] Please transcribe this text

[Assistant] You know, the RC time constants are like that. Everything is, a lot of things are like that. Okay.


[User] Please transcribe this text

[Assistant] So, these guys will exponentially drop. And of course, once they get to this junction, they will have to recombine. Anyway, because now you have a 3D, what we haven't shown is that there's some sort of another contact here and there.


[User] Please transcribe this text

[Assistant] Now how about the holes on this side? If you think about the holes, well, I think I messed up my color code.


[User] Please transcribe this text

[Assistant] Now, what can you say about the total net electron current on this side? So where is the rest of the current coming from? This J total has to be constant. So whatever is not carried by JP has to be carried by JN. So why is there a JN? Well, the electrons still have to be injected into this thing and going through.


[User] Please transcribe this text

[Assistant] Makeup for the electrons that are lost to these recombinations and makeup for the electrons that are injected.


[User] Please transcribe this text

[Assistant] Right? There are electrons that are going to the other side, and there are electrons that are lost in this recombination. So you have to-


[User] Please transcribe this text

[Assistant] To the shape of what you already have here. So this is gonna be the Jn. But well, matching this of course here. Something like that.


[User] Please transcribe this text

[Assistant] Of these carriers, right? So whatever makes it to this edge, edge is gonna make it to the other edge.


[User] Please transcribe this text

[Assistant] And then same, in a similar fashion, you have this happening here.


[User] Please transcribe this text

[Assistant] So this picture is important because this allows us to think about what we need to do to control and play and manipulate discharge carriers. Now,


[User] Please transcribe this text

[Assistant] The shape, the distribution of the current by the majority carriers.


[User] Please transcribe this text

[Assistant] You're right about that. But there's something that's first order, before that happened. What is dictating this shape? Why this blue line is like that? We argued why the red one is like that, right? We argued why this one is an exponential. We said this is like a recombination process. You lose more. The minority carrier distribution is this way. How about the majority carrier distribution, which is these two? Why does it have the shape it does? What dictates that shape?


[User] Please transcribe this text

[Assistant] Which is interesting to think about, right? So it is really the minority carriers.


[User] Please transcribe this text

[Assistant] And this is the basis of how you basically design a bipolar transistor, or in general a lot of devices. It is the minority charge carriers that dictate the behavior here. So, now,


[User] Please transcribe this text

[Assistant] If you wait till here, if you, so how can I harness this thing? Because if I let the minority carriers to basically just like go and propagate, eventually most of them and eventually all of them will recombine. So you have lost that leverage really over them. You want them to be


[User] Please transcribe this text

[Assistant] So you want to get them into a region where there are majority carriers again.


[User] Please transcribe this text

[Assistant] So how do we do that? We have to create a region where these ends are, for example, majority carriers, right? What kind of region would do that?


[User] Please transcribe this text

[Assistant] Let's say this is an NPN, and you can have a PNP. So let's say, we'll talk about an NPN. So you have the N and the P. Right? So that's the diode we had already.


[User] Please transcribe this text

[Assistant] And that's why you introduce another region, N, here, where they can be absorbed.


[User] Please transcribe this text

[Assistant] Before they are completely recombined. So, and how do you, of course, this one is your controlling voltage, right? So now this is a voltage.


[User] Please transcribe this text

[Assistant] For this first junction, you're lowering your forward biasing. So you have lowered the energy.


[User] Please transcribe this text

[Assistant] You've lowered it from where it would have been to here. So it would have been like that. And by applying this V1,


[User] Please transcribe this text

[Assistant] You've lowered that potential barrier, so now as a result, a bunch of your electrons that could not pass before


[User] Please transcribe this text

[Assistant] So these are the ones that could pass before they were hot enough to pass that potential barrier. Now you have a much lower potential barrier, so lower potential barrier, so these guys can pass, and these guys are bounced. So, here's the thing. You've applied this, and they are injected into the


[User] Please transcribe this text

[Assistant] This region. So you're emitting electrons from here into this region. And then you want to collect them here.


[User] Please transcribe this text

[Assistant] And you want to collect them. Now if you want to facilitate the collection,


[User] Please transcribe this text

[Assistant] What kind of electric field would facilitate, if you, if I said, okay, well I want to make an electric field that would facilitate this collection process.


[User] Please transcribe this text

[Assistant] Which way should that electric field be pointing? To facilitate collection of electrons into here. So you want electrons to go this way. Which means that you want the electric field to be pointing this way. Right? So you want an electric field here


[User] Please transcribe this text

[Assistant] We know the electric field in this region, in this part of the depletion region is pointing this way.


[User] Please transcribe this text

[Assistant] And to produce that, of course, this, this kind of p-n junction does produce that naturally because you have positively charged, negatively charged ions here and positively charged ions there.


[User] Please transcribe this text

[Assistant] So that's basically what generates the field. So what you need to do, and if you want to increase this field, what kind of bias should I apply to this PN junction? Reverse or forward? Reverse, right? So the reverse would look like this. And this should really be connected this way.


[User] Please transcribe this text

[Assistant] So this is the essence. So what happens is that by applying a reverse electric potential, I am actually making this


[User] Please transcribe this text

[Assistant] It's a steeper thing. So the electrons, once they make it past this barrier, it's kind of going to be downhill for them. So because they are going to go here, and they're going to be absorbed in the collector.


[User] Please transcribe this text

[Assistant] Now what makes them go from here to there? What makes them traverse?


[User] Please transcribe this text

[Assistant] There are several different things, but there's the first order thing, by far first order, stronger than the other.


[User] Please transcribe this text

[Assistant] Effects, the other major two second order effects.


[User] Please transcribe this text

[Assistant] What is that number one effect that makes the electron get from here to there?


[User] Please transcribe this text

[Assistant] Right? Exactly. Because remember, they're still pretty darn hot and they're pretty moving pretty darn fast. So what's happening is that once they're in the base, they're not making a lot of, I mean, for the most part, there's nothing special happening. They're just moving randomly around so fast and every so often one of them ends up in this region and gets sucked in. Goes downhill.


[User] Please transcribe this text

[Assistant] And this is important to understanding some of the behavior of the device later on when we talk about it. So,


[User] Please transcribe this text

[Assistant] Okay, fine, so this is happening, and that's good. Now, a few questions about this device.


[User] Please transcribe this text

[Assistant] The most basic question, so we are designing a device, we have some thought about this process.


[User] Please transcribe this text

[Assistant] So this region, by the way, is called what? I mean, it's obviously based on the argument that we had. This is called the emitter. Where the charge carriers are emitted from.


[User] Please transcribe this text

[Assistant] And this is where they're collected, so it's called a collector. And this is called a base for a historical reason. And the historical reason is this. If you look at the first transist- well, the first demonstrated transistor.


[User] Please transcribe this text

[Assistant] Which was by, you know, people at Bell Labs. The transistor was invented 20 years before that. By Julius Lilienfeld. He has three


[User] Please transcribe this text

[Assistant] Issued U.S. patents that you can go and read, and they describe the transistor. So, that's one of my things. But anyway, so if you look at that transistor, which actually is interesting, I have a picture of, that I took myself.


[User] Please transcribe this text

[Assistant] of some non-conductive material. And then they had a coating of conductive material


[User] Please transcribe this text

[Assistant] And they were trying to make these two points pretty close to each other, so they made a little cut.


[User] Please transcribe this text

[Assistant] Okay, so there were two regions formed, two kind of semiconductor regions, so they had a tiny base, tiny emitter, tiny emitter, tiny collector, and gigantic base, which was that middle terminal. That's why it's called base. Now this makes sense why it's called base.


[User] Please transcribe this text

[Assistant] But anyway, so that's what it's called. So this is called the collector, the emitter, the base, and the collector. These are the names, so we will be referring to them as such. Now, so, and obviously these voltages, we call them appropriately, so this is VBE, the base emitter voltage, and this is VCB, right? The voltage between collector and the base.


[User] Please transcribe this text

[Assistant] What do we need to do about the space? What are the things that we have to do?


[User] Please transcribe this text

[Assistant] Make, make sure it's thin enough, right? You don't want it to take too long for them to go through this region. So you want to make it as thin as possible, so that you can make it across as quickly as possible. So you want to make it thin.


[User] Please transcribe this text

[Assistant] So that's the first order thing. You want to make this base as thin as you can possibly can. Of course, there are some other considerations that will come in when you do that.


[User] Please transcribe this text

[Assistant] Let's say in this case the minority carriers would be electrons, if it was a PMP it would be holes. But what would they, what would they look like? What do you think the distribution looks like in the base?


[User] Please transcribe this text

[Assistant] Straight line, okay, so that's, that's how it looks like a straight, like, like that. You mean, oh, you mean, oh, you mean flat line. Okay. So you meant, you meant a straight inclined line. Okay. Which one do you like? This one or this one? Flat. So how many people like this flat?


[User] Please transcribe this text

[Assistant] How many people like the inclined line?


[User] Please transcribe this text

[Assistant] So remember, these electrons are going through this region. It's just a possibility of finding this electron. This is basically what the homework problem will show you, and you will see it in the homework problem. So now assume that each one of these guys is, each one of these electrons are moving randomly, right, thermally. If you have random thermal movement,


[User] Please transcribe this text

[Assistant] The probability of finding an electron here at this edge is higher or is it higher to find it here? Where they were just injected.


[User] Please transcribe this text

[Assistant] So the closer you get to this region, the more likely they are that they have passed through. And this is something that you will show through a simulation.


[User] Please transcribe this text

[Assistant] Because there's a higher concentration of them. See, this is basically a variation on this exponential. Think about this exponential. Right?


[User] Please transcribe this text

[Assistant] But imagine that this was very short. Now if this was very short, this would drop very rapidly, because eventually they have to get recombined once they get to the other edge, or they have to be absorbed, and in this case they are not recombined, they are absorbed into the collector.


[User] Please transcribe this text

[Assistant] What happens is that in your homework problem, what you're doing is that you're taking some number of particles, you're assuming that they have equal probability of one half to go this way, and one half to go that way.


[User] Please transcribe this text

[Assistant] And you will see that it forms a distributionist. So this is basically sometimes called Q-base.


[User] Please transcribe this text

[Assistant] This is the minority charge carrier in the base. Sometimes we refer to the total charge as QF later on, but we'll talk about that. But this distribution, this is what controls the behavior of the device.


[User] Please transcribe this text

[Assistant] Now, the narrower you make this, the more of them make it to the other side.


[User] Please transcribe this text

[Assistant] And few of them get recombined, because there's always a chance of recombination in the base.


[User] Please transcribe this text

[Assistant] So, that's the number one thing that you want to do, if you make this transistor, is to basically making sure that this concentration is, that this width is small. So.


[User] Please transcribe this text

[Assistant] What is the current? What is the collector current? So if I look at this collector current,


[User] Please transcribe this text

[Assistant] I see. What should I, what kind of behavior would it have? To the first order. We will refine this equation. But to the first order, what do you expect it to be?


[User] Please transcribe this text

[Assistant] Well, what determines this current? What is the primary determinant of this current? This injects a charge, right?


[User] Please transcribe this text

[Assistant] Now, what determines how many of the electrons make it across this potential barrier?


[User] Please transcribe this text

[Assistant] You're just lowering the potential barrier and they go, right? So whatever electronic current you got for the diode, so we had an exponential dependence, so it was e to the VBE


[User] Please transcribe this text

[Assistant] And I actually like the K2 over Q because it reminds you that this is a thermal device.


[User] Please transcribe this text

[Assistant] Sometimes we summarize, write it in shorthand as VT, uhm, the thermal voltage. Now,


[User] Please transcribe this text

[Assistant] So times some IS, some proportionality constant, which has dependence on a whole bunch of other parameters. Among them is the area of this thing. And then it's basically the IC. So this is the collector current.


[User] Please transcribe this text

[Assistant] Except for the fact that now you have the current of a third terminal controlled by the voltage between the other two terminals.


[User] Please transcribe this text

[Assistant] So it's really to the first order a voltage controlled current source.


[User] Please transcribe this text

[Assistant] And that is very critical because what it does, this exponential is pretty powerful. That's why I say it's a powerful device. It's an exponential dependence.


[User] Please transcribe this text

[Assistant] So a small fluctuation here can lead to large fluctuations there. Which is a good thing if you're trying to amplify something.


[User] Please transcribe this text

[Assistant] Or you are trying to have, being able to switch something by having a small change in the input inducing a large change in the output. Right?


[User] Please transcribe this text

[Assistant] So, that's the essence of the transistor. That's basically the essence of what the IS, and if you plot this quantity, if you plot this, I mean it's an exponential obviously.


[User] Please transcribe this text

[Assistant] And you will see that it's kind of a exponential, of course. If you plot an exponential on a linear plot, on a linear linear plot, what you will see is that you will see a region basically on any linear plot where this is pretty much close to zero.


[User] Please transcribe this text

[Assistant] Of course, if you do it on a log log plot, you will see something else, but uhm,


[User] Please transcribe this text

[Assistant] Because another way to see this, which is the inverse of this equation, is VBE.


[User] Please transcribe this text

[Assistant] So it means that it has a weak dependence on the current, what the VBE is. So above a certain point you will have a large amount of current.


[User] Please transcribe this text

[Assistant] For the scale that you're interested in. Now if you change the scale, if you go from milliamps to microamps, then this transition will appear somewhere else. But still, it will look pretty steep.


[User] Please transcribe this text

[Assistant] This appears to be good, right? I mean, are we happy with this transistor?


[User] Please transcribe this text

[Assistant] Is there something that you're missing? Is there any problems that you can identify?


[User] Please transcribe this text

[Assistant] Okay, so that's, that's a good point. So she said the depletion regions combine. If you start, what happens if the depletion regions start hitting each other? Start getting into each other. What happens? What happens to your energy band diagram?


[User] Please transcribe this text

[Assistant] Well, you don't, if the sub-basics just drop, it starts becoming flatter and flatter, right?


[User] Please transcribe this text

[Assistant] Then you don't have this behavior, then you lose your control and you have, you have removed, reduced and eventually removed the, the insolation. The separation of the output port from the input port. From the, the control from the, the controlling. Right? So, you can't afford doing that, right?


[User] Please transcribe this text

[Assistant] So now let's say, so, so that brings us to an interesting related question, which is


[User] Please transcribe this text

[Assistant] We'll go into it. It's called punch through. Uhm, that has a name specifically. But uhm, what do we do? What do you think we can do?


[User] Please transcribe this text

[Assistant] Yes, increase the doping concentration, right? This is, you mean here, right? Yeah. So if you increase the doping concentration, then most of the depletion region according, if, if you follow this, will become shrink, shrinks on this side, shrinks on that side.


[User] Please transcribe this text

[Assistant] The relative size of the, sorry, not the doping. If I wanted to keep the relative size,


[User] Please transcribe this text

[Assistant] Of the depletion region on the either sides the same. I just want to shrink both of them by the same ratio. What would I need to do?


[User] Please transcribe this text

[Assistant] Let's, you know what I'm asking, right? I'm saying that I want to shrink both sides of the evolution by the same factor. What would you do?


[User] Please transcribe this text

[Assistant] On all of them. Increase the doping on all sides, right? So that's one way. And that's the basis for one of the things that happens in the scaling. If you want to make the transistor small, you increase all the doping content because you need to shrink all of these so you can bring these closer to each other.


[User] Please transcribe this text

[Assistant] So these are not gonna hit each other. So that's a good point, right? So that's one thing.


[User] Please transcribe this text

[Assistant] Well, if you make it very thin, that's true. The electrons will tunnel over. But yes, and that's a good, good, good thing. Uh, we are not there yet.


[User] Please transcribe this text

[Assistant] But if you make it very small, that can happen. We are not at that level yet, but it's a good anticipatory comment.


[User] Please transcribe this text

[Assistant] Alright, any other thoughts? What would happen? Isn't it a more basic problem? I mean, the problem I'm talking about is very basic.


[User] Please transcribe this text

[Assistant] Think about where these currents, there's some base current, right? Why is there base current? Where is this base current going by the way? We'll talk about it in a second, but there will be some current in the base. Now this current will be traveling in this direction, right?


[User] Please transcribe this text

[Assistant] And if you make this thinner and thinner, you remember that slab that we drew on the first lecture? What happens to the resistance of that slab?


[User] Please transcribe this text

[Assistant] If you make it thinner and thinner, it can reduce its cross section area. If I make this thin,


[User] Please transcribe this text

[Assistant] The resist, the series resistance will go up. So you will have a series resistance here that will, it will keep increasing.


[User] Please transcribe this text

[Assistant] And that can get in the way, because if you have a series resistance where there's current flowing, then you will have voltage drop, so some of your voltage drop is not really dropping across where it should be, it's dropping across some, pitting some resistor, which is not going to be used for controlling the device. So you lose some of the control, some of the sensitivity of the device. So that's another problem.


[User] Please transcribe this text

[Assistant] And one way to deal with that is to reduce the resistivity, of course, as you scale it down. Which happens by what? How do you reduce the resistivity? Or increase the conductivity? You remember the expression we had for conductivity? We derived it from microscopic level calculations. So what were the parameters in there? There was mu. There was, so there was q electronic charge. It's kind of hard to change that. There's the mobility. Just again, not an easy parameter to change.


[User] Please transcribe this text

[Assistant] Uh, I mean, there are parameters, but increasing it beyond certain result. And then there's n.


[User] Please transcribe this text

[Assistant] The concentration of the charge carriers, right? In this case it would be P, because it's P type. Uhm, so, how do you increase P? How do you increase the number of holes in the design?


[User] Please transcribe this text

[Assistant] For a piece of semiconductor. If I want to get more peas at room temperature, what do I do? More holes per cubic centimeter, per cubic meter of it.


[User] Please transcribe this text

[Assistant] More dopants, right? Which basically goes back to what we did already.


[User] Please transcribe this text

[Assistant] So as you increase it, that's helpful. That goes back to, that's another reason you increase the doping concentration when you do scaling.


[User] Please transcribe this text

[Assistant] Because as dimensions become smaller, you want to make the resistance the same.


[User] Please transcribe this text

[Assistant] The total current in this junction, let's say this emitter current in that case, would be carried by what?


[User] Please transcribe this text

[Assistant] Aren't holes going backwards? As we lower the potential barrier.


[User] Please transcribe this text

[Assistant] Aren't holes being injected back because now we have shifted the potential barrier so this thing has moved up. So now there's a whole bunch of new ones that can pass, right?


[User] Please transcribe this text

[Assistant] So as we lower the potential barrier, not only electrons get injected from here to get into the base,


[User] Please transcribe this text

[Assistant] So, good thing or bad thing first? How many people say it's a bad thing? How many people say it's a good thing?


[User] Please transcribe this text

[Assistant] Right? I mean it's gonna get recombined eventually here because there's no way to collect them. Right? This is the bridge to nowhere. Right? It's just like they're gonna go and get recombined eventually.


[User] Please transcribe this text

[Assistant] Unless you put another P here, which would kind of like be kind of a funny thing. I mean, it would be an interesting thing to discuss, but that's a separate discussion.


[User] Please transcribe this text

[Assistant] But if you want to keep it this way, you need to do something about this.


[User] Please transcribe this text

[Assistant] There is an efficiency associated with this process. We call it emitter injection efficiency. Which should make sense. The name should make sense.


[User] Please transcribe this text

[Assistant] It's the efficiency of how many of the, what percentage of these carriers are useful carriers, right? Essentially, it's defined, shown in gamma, it's defined as I n of, uh,


[User] Please transcribe this text

[Assistant] In the emitter, the emitter current, part of the emitter current that's carried by electrons divided by the total, which is the sum of the part that's carried by electrons and part that's carried by the whole. Because if you remember this picture,


[User] Please transcribe this text

[Assistant] There's some current carried by the electrons and some carried by the holes. So we want to make sure that this current


[User] Please transcribe this text

[Assistant] This blue one is the component that's useful because that's the one we are absorbing afterwards. We are collecting.


[User] Please transcribe this text

[Assistant] What do you want it to be, ideally? Like most efficiencies, right? If you talk about an efficiency, you want it to be 1. 100%. Some people aim for 110%.


[User] Please transcribe this text

[Assistant] I don't know how they do it. Uhm, anyhow, so, I, so, this is what we are trying to maximize. How do we do that? Proposed solutions. And there are more than one solution.


[User] Please transcribe this text

[Assistant] So, okay, so that's a good suggestion. He says, look, you know, you want to start with fewer holes on this side to begin with.


[User] Please transcribe this text

[Assistant] If you have fewer holes, so they have equal probability of going both ways, right? The Boltzmann distribution tells you what the probability is. But this probability is multiplied by the initial number, right?


[User] Please transcribe this text

[Assistant] If the probability of crossing is, let's say, I don't know, 10%, if you start with 1000 people, you will get 100 people that cross. If you start with


[User] Please transcribe this text

[Assistant] A million people, you will get 100,000 people that cross. Right?


[User] Please transcribe this text

[Assistant] So it's not just the probability of crossing, but it's also the number that is multiplied by.


[User] Please transcribe this text

[Assistant] So it says, okay, control the number that's multiplied by. So in other words, have a lot more electrons here than you have holes here.


[User] Please transcribe this text

[Assistant] How do you have, how can you make this device such that you have more electrons here than you have holes here? The density, yea. You suggested it, so why didn't you?


[User] Please transcribe this text

[Assistant] So increase the doping concentration of emitter with respect to what? The base. So we actually, we dope the heck out of it.


[User] Please transcribe this text

[Assistant] Sometimes you show that's N plus. Sometimes you show that's like N plus plus.


[User] Please transcribe this text

[Assistant] And meaning that you have a high concentration. You basically are looking at 10 to the 18, 10 to the 19, 10 to the 20 per cubic centimeter of dopants. To the point that actually, at some point, it becomes degenerate.


[User] Please transcribe this text

[Assistant] You have so many of them yet you don't get more because of the interactions among them and then you get ironized and all those things. So you, but you want to have a lot more. And if you make this ratio for example, the ratio of dopant concentration here, the dopant concentration here. So let's say this is 10 to the 18 and this is 10 to the 16.


[User] Please transcribe this text

[Assistant] What can you roughly say your emitter injection efficiency is? Roughly.


[User] Please transcribe this text

[Assistant] So if I make this 10 to 1, it would be like about 90% efficiency. If I make it 1000 to 1, it would be like...


[User] Please transcribe this text

[Assistant] 99.9% efficiency. For emitter injection efficiency. But it is what it is, right?


[User] Please transcribe this text

[Assistant] So that's one way to do it, and that's a traditional, classical way of doing it. Uhm, can somebody suggest a second way of improving? This is by the way done in almost all transistors.


[User] Please transcribe this text

[Assistant] Can you suggest another way? By the way, is there a problem with doing this, first of all? There's a little bit of a problem, because it means that your base needs to have a lower diluting concentration, which was in conflict with a couple of things. One was this resistance.


[User] Please transcribe this text

[Assistant] Two is this depletion region, because now the depletion region, if I have that kind of concentration, the depletion region will be mostly where? On the base side. Which, yep, it will happen. It does happen. And you live with it.


[User] Please transcribe this text

[Assistant] What are the things that are controlling this current? We said there were two different things I mentioned, right? One is the probability of crossing. The other one is the density of charge carriers you begin, you start with, right? So we played with the density of charge carriers you start with. That was the first solution.


[User] Please transcribe this text

[Assistant] It's used in almost all, I wouldn't say even one, I shouldn't even say in almost all, in all bipolar junction transistors. That technique, the doping, the doping concentration of emitter is higher than the doping concentration of base in all bipolar transistors that I know of. Yes?


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[Assistant] So the suggestion is interesting. So he says, if you lower the doping concentration on this side, then you can get these to touch, and then you will get rid of it. Which is okay for the problem you're trying to solve, but it actually introduces a bigger problem, which is the lowering of this potential barrier, and you lose control. So it's a problem to a,


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[Assistant] I guess first order problem, which is causing a zeroth order problem.


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[Assistant] So you want it such that you want to make it uphill for the holes. That's what you're trying to do, right?


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[Assistant] Same thing is true for people. Right? So, it's a filter.


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[Assistant] That's a very good suggestion. So the suggestion is to change the semiconductor material in the metal. So use a slightly different semiconductor material in the base.


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[Assistant] So instead of this, you will get that. So this whole thing gets shifted up, which means that you have a larger potential barrier.


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[Assistant] And this is the basic of what we call heterojunction bipolar transistors, or HBTs, that are used extensively, uh, for very high frequency applications.


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[Assistant] So for example, silicon germanium transistors are like this. You have silicon, but then you have silicon and germanium alloy in the base.


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[Assistant] In germanium, having smaller band gap, the alloy will have a band gap that's determined by the ratio of the alloy. It will have a smaller band gap, and what happens is that you will get fewer holes.


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[Assistant] Now what's the advantage of that? The advantage is that now you can have, actually you can live with the smaller ratios of the N2P. And you can have higher concentration of doping in the base.


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[Assistant] And as a result, your resistance is not going to be as large. Your serious resistance. Right?


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[Assistant] And the other interesting thing about it is that when you do this,


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[Assistant] But regardless, what is happening is that you can actually control that and by controlling this you can have a different set of parameters for this device.


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[Assistant] So this is called HPT, heterojunction, as opposed to homojunction, which is basically made all of the semiconductor. It's the same semiconductor. You have a heterojunction, which are made out of different junctions, and, uh,


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[Assistant] You have basically something like that, and it, it, it, it's very common. The other, the other combinations that they have, make it gallium arsenide, gallium all, gallium, uh, gallium, aluminum arsenide, gallium indium, uh, arsenide or indium phosphide transistors, they actually can engineer the composition.


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[Assistant] So we can see we lose some of our electrons due to this back injection. You have some emitter injection efficiency loss. Because some of these electrons, some of these holes that are back injected. So these holes that are back injected are lost because they are going to be recombining here. Is that the only mechanism for losing some of our current?


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[Assistant] So what happens to some of them along the way?


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[Assistant] They recombine, right? They are in a p-type region. There's a lot of holes there. So there's a chance for recombination. So you lose some of them through recombination. So there's another loss mechanism.


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[Assistant] It's expressed as alpha base, which is basically called base transport factor.


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[Assistant] So there's a second efficiency, if you will. How many of them make it across? There's no man's land in the middle, right? Go across and without getting recombined. So you have this and you have these two.


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[Assistant] Mechanism. And there's a third one in the collector side, but since the collector is typically reverse biased, that's a small one, so it doesn't matter to worry. We don't worry about some textbooks put it in there, and then eventually they kind of, when they cut to the numbers, they say, oh yea, this is almost one, so. So let's not worry about that.


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[Assistant] If you were doing hardcore device physics, we would worry about it, but not here. But anyway, so.


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[Assistant] There's this alpha, we call it alpha zero, or alpha DC.


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[Assistant] Because the collectors that make it to the, the electrons that make it to the collector are


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[Assistant] Constitute this current, and the electrons that are coming out of the emitter constitute this current. The current that you have to basically put in here, well, you're really putting, taking out of it because of negative charge. So this current is IE, and that current is IC. And what you have here is that essentially IE over IC is gonna be alpha 0.


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[Assistant] I'm sorry, I see, uh, I see over. Alright.


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[Assistant] And where's that current going? You have KCL for this device, right? For this thing, it has a, it's a three terminal device.


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[Assistant] If it's a circuit, then this third current has to be the base current, right?


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[Assistant] So that's the current going through the base. So there's another way to see this, which is this picture. I'm gonna overlay it on this with a different color, hopefully it wouldn't make it too busy.


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[Assistant] So, the vast, in a properly designed transistor, the vast majority of the current is carried by these electrons, right? Going across, so there is a big flow of electrons. Now, some of them are, the discharge, what happens?


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[Assistant] In the emitter. And where does it go? It's balanced by the hole currents, right? So there's some holes that are coming in and then there's a recombination happening. So this is recombination.


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[Assistant] You have also some of these holes that are back injected into the collector, right? Into the emitter, sorry.


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[Assistant] Yeah, they would recombine with an electron there. So that would be contributing some of the current that needs to come in and balance that. So your total IE coming in is gonna have two components. So this is the external current. So if you have an external current coming in,


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[Assistant] There would be this segment that's going there to make up for that. And this big chunk, that's the word that ends up in the collector. So there's some loss here, and some loss there.


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[Assistant] This is accounted for by gamma, this is accounted for by alpha t.


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[Assistant] Based on the alpha t, we can actually, and the other thing that we know is that basically the beta of the transistor, we can define another parameter, which is basically the ratio of the base current to the


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[Assistant] Collector current. And that's pretty straightforward to calculate. I'm just running out of space a little bit here. So, let's try to open up some space here.


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[Assistant] Now we call this quantity beta, beta zero. So this is alpha zero for DC.


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[Assistant] And this is a famous parameter for bipolar transistors, it's called beta. So IC is beta IB.


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[Assistant] So the collector current is beta times IB. Now, you remember, if it is alpha, we want it to be close to 1.


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[Assistant] This is a high efficient, if, if, in a properly designed transistor. Let's say if alpha was 0.99, 99%, what would your beta roughly be?


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[Assistant] But then the other thing that you see is that this is quite sensitive to alpha. So if alpha changes a little bit, beta changes a lot when alpha is close to one. Which is where you want it to be if you're properly designing.


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[Assistant] And that's why beta of a transistor is a pretty sensitive parameter.


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[Assistant] Because if a design depends on beta, it means that the slightest change in the temperature or parameters or from one transistor to the next one will result in large changes in your design. Which is not what you want in a robust design.


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[Assistant] The robust design, you want something that is independent of these parameters, to the extent that's possible. Okay, uhm, any questions on this?


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[Assistant] I doubt she would have married you. Why not?


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[Assistant] Master? Said Wukong. While I was at Mount Podolatka, I thought a lot.


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[Assistant] I realize now that it's wrong to harm people. I won't ever do it again.


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[Assistant] Wukong laughed. Not much has changed since I left. Bajie still complains all the time. Wu Jing wiped his brow.


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[Assistant] Where does the Iron Immortal live? I'll demand that she put out the flames!


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[Assistant] I am as powerful as a god. Just tell me where she lives.


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[Assistant] I am Sun Wukong. I need to borrow your fan. Sun Wukong?


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[Assistant] Well then, can I speak with the bull demon king? asked the monkey. He's at a banquet, said the iron immortal.


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[Assistant] I'm not leaving until I borrow your-


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[Assistant] Lousy doesn't like me much. Thought the monkey. He won't be happy to see me.


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[Assistant] So far we've done a study of the device physics. We understood the, how the devices, uh, work, and we've developed some model. Of course, what we've done is not a thorough study of the device physics, but we've taken, we've gone from the physics of it to an equivalent circuit. And what, this is what we did.


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[Assistant] So now, the question is that once you have this equivalent circuit, what do you do? How do you analyze it? I mean, it's really an abstraction, right? We've gone from


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[Assistant] Well essentially all of our sciences are abstractions. But it's another level of abstraction where you actually create something that's defined in terms of currents and voltages and we saw that for example we had the large signal model for the transistors and we had the small signal model which was a linearized perturbation model. The question is how do we analyze these things?


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[Assistant] And what I'm going to do quickly right now, we are going to quickly review.


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[Assistant] Some of the underlying, uh, methods for circuit analysis. This is again, this is not a class on circuit analysis. So this would be a very brief review of what we cover, for example, at Caltech in W44 or things like that.


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[Assistant] There are different methods of analysis. So the most general method of analysis for a circuit is essentially what we call the nodal analysis. So very quickly reviewing nodal analysis.


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[Assistant] So let's say you have something like this.


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[Assistant] Now, let's say you have a network like this, you can give them names, you can call them G1 or R1, uh, C1, C2, G2, uhm,


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[Assistant] You can call this G3, C3, L3, and then this can be GM. Let's say this is GM related to this V pi. And you can see this is kind of starting to resemble a model of a transistor, of course with certain kind of load, so we don't make an arbitrary circuit, we make something that's relevant. So,


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[Assistant] This is, let's say you're given this circuit, so this is your I in. Let's say I in. Now if you're given this,


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[Assistant] How do we analyze this? There are different ways to analyze it. The most general method of analysis is called nodal analysis.


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[Assistant] And the method basically has the following steps and procedures. So first you define one of your nodes as a reference potential.


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[Assistant] So you define one of the nodes. Usually you want it to be one that has the most connections to it. So the idea here is that assumption is that all of them are actually physically connected. And we call that ground. Now, this by itself, so this is a mathematical abstraction, right? We can say this is a reference potential. This is zero potential. Now, this is not necessarily true


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[Assistant] In f- from a physical perspective in the sense that if you have a large physical circuit, different parts


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[Assistant] May be connected through a piece of metal, but that connecting metal may have resistances, so there may be resistance here, there may be inductance here, there's capacitance to ground, there are all sorts of other parameters, and that means that you have to differentiate between a mathematical abstraction and a physical abstraction, and a physical model.


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[Assistant] So, if you are, for example, in the lab, you may not, you may have everything connected to some big sheet of metal, so that's my grout.


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[Assistant] So you have to think about it. Or if you're making a chip, you may have a wire bond connecting this path to the, to a larger piece of metal, and that path to a larger piece of metal, and you say all of them are ground. Now if you assume that they are identical potential,


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[Assistant] We've made an assumption here that can lead to errors, especially at high frequencies and things of that sort that we will see. But now, from a mathematical perspective, this is what it is. Now in those cases, in those scenarios, if you for example have something like this, if let's say you have a resistor here and a resistor there, then this becomes a separate node.


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[Assistant] So you really treat this as a separate node, and if there's a resistance intervening between these guys, this becomes a separate node.


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[Assistant] So let's go back to the mathematical model, the simplified mathematical model. So we define one potential as ground, or zero. So let's say your circuit has n plus 1 nodes.


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[Assistant] Or, this is a node, and this whole thing was a node. Right? We defined. Now,


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[Assistant] We define a voltage for each one of the nodes, so we call this for example V1, the voltage here, V2, and V3.


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[Assistant] Now, if you have something like this, and what you see here is that we have only linear


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[Assistant] Elements right now, right? The way we've defined it is linear. So, and then neural analysis really by itself is only applicable to linear circuits.


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[Assistant] So for example, if you have a small signal model, you can apply the nodal analysis. If you have a large signal model, you cannot apply it in its current form. So, we'll talk about that in a second.


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[Assistant] Now let's say you have these things, so what you do, you write basically KCL, Kirchhoff current law, at each one of the nodes in terms of the voltages of the nodes.


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[Assistant] So we can do it different ways. We can actually write them in terms of currents, so we can give these currents names. We can call this I1, I2, I3, and this would be I3.


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[Assistant] Conservation of energy, and, I'm sorry, conservation of charge, not conservation of energy, conservation of charge, and the fact that the dimensions are much smaller than the wavelength.


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[Assistant] At least temporarily or, or, or charge being drawn out of the node. Net, there's no net charge accumulation on any of the nodes. And that comes from Maxwell equations if you essentially assume that the speed of light is infinity.


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[Assistant] Which means that dimensions, which is another way of saying that the dimensions are much smaller than the wavelength.


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[Assistant] So for us, from a circuit perspective, the assumption is that the speed of light is infinity.


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[Assistant] The propagation happens instantaneously, and that simplifies the Maxwell equations to what turns out to be the circuits equation.


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[Assistant] So that's one. And the other one is basically you can write it as negative i3, so all the currents leaving that node, i4 plus i5 plus i6 plus i7 equals zero. So you have


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[Assistant] Two sets, two, uh, two, so two, and then there's a third one, well, we define this as a node, so, so this wasn't, you should not have, I should not have called this i3. I should have called this i3, and this i, whatever, 8.


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[Assistant] This would be i8, if I wanted to deal with that as a node. So this becomes i8, and then I will have another equation from here that says negative i3 plus i8 equals 0, which would then equate i3 and i8. So you can see that I have three equations here.


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[Assistant] Ohm's law here. Or for example, I2 is going to be C1S, uh, V1. And so on and so forth.


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[Assistant] And you can write all these things. So when you write all of those, what you will see, for example, is that you will come up with a, so we can actually try writing these in this form. You can actually write it as a matrix, what we call a Y matrix, multiplied by V1, V2, and V3.


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[Assistant] Times C2S. C2S times V1 minus V2. That is the current across this one, because it's this voltage minus that voltage times the admittance of that element, which is C2S. Equals IN.


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[Assistant] And then the second term would be negative c2s, and the third term would be zero, because there's no v3 in that equation. And then on the right hand side, you will have the i in.


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[Assistant] And similarly you can write the second equation, which is this one really, I just wrote them in the wrong order, which basically says that this current, which we calculated to be negative C2S V1 minus V2,


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[Assistant] So that would produce a term for C1, for V1, that's negative C2S.


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[Assistant] And then you get a term for v3, which is negative g2. From that one. This is basically really this equation. I've reordered them.


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[Assistant] So that's that current, negative i8, plus i4, which is gonna be now gm v-pi. Now what is v-pi?


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[Assistant] In terms of the node voltages, can we express it in terms of node voltages we have?


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[Assistant] Vpi is just V1, right? Because it's this voltage. So gmV1.


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[Assistant] And then for v2, you have the, so the term you have for v2 is negative g2, right? That's the only one that you see.


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[Assistant] And for V3 you have a whole bunch of stuff. So you get negative G2 from this one. Uh, you get the...


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[Assistant] I'm sorry, you get positive G2 because this is negative times negative. G2 plus G3 plus C3S plus 1 over L3S. So this is what we call the Y-Matrix.


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[Assistant] Right? And you're familiar with that. This is called the nodal analysis. This is the admittance matrix. Now, it has certain interesting properties. Now, if you didn't have any dependent sources,


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[Assistant] This Y matrix would be symmetrical. In other words, if your GM was zero, you can notice, you notice that it's symmetrical.


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[Assistant] Now the other thing that you'll find interesting is that the diagonal elements are all the conductances attached.


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[Assistant] Now, or for example, for this node, you have G2 and C2 attached to it. Or for this node, you have G2, G3, C3S, and 1 over L3S.


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[Assistant] So the diagonal elements are essentially the elements attached to those nodes. And the off-diagonal elements, again, not counting the dependent sources, are basically the negative of the elements directly connecting those two nodes.


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[Assistant] So for example, nodes 1 and 2 are connected through C2S. So you get negative C2S.


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[Assistant] 1, 2 and 2, 1. Or elements 2 and 3, notes 2 and 3 are connected through G2, so you get negative G2 here.


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[Assistant] And that's basically, these are the things that account for the action of the transistor and those are the things that make the network what we call


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[Assistant] And you will see how that behaves. Now, the nodal analysis, and you could do something else if you wanted to simplify it. So this is kind of like a little bit of a larger form. We could have combined these two and treated them as one thing, one element.


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[Assistant] With the value of g2 plus c2s. I'm sorry, no, g2 really times c2s divided by g2.


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[Assistant] Plus C2S, because these are two conductances in series that behave like two resistors in parallel.


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[Assistant] So, you could take that and then combine them and behave, treat them like that and instead of a 3 by 3 matrix you would have gotten a 2 by 2 matrix. But then you wouldn't get the information directly for node v2.


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[Assistant] If you're not interested in what's happening in that node, that's fine.


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[Assistant] Because it's a voltage divider between that conductor and the capacitor. So, so that's fine. So this is kind of like a very standard way of analyzing circuits, right? I mean, this is like, this is the fallback. This is what I call a fallback in the sense that it solves any network you throw at it.


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[Assistant] The key to it is that it's actually, and if you want to solve this, how do you solve this? So basically this is the Y matrix, and then this is the, what we call the voltage vector, the vector, so this is a matrix, I'm going to put a tilde on top of it to emphasize the matrix, and this is the vector of a stimuli.


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[Assistant] Is. So the compact form of this expression is something like this. So these are the unknowns, and these are the stimuli. These are the inputs of the circuit.


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[Assistant] So, how do you solve this? Of course, you, the easiest way is to, you've reduced it, I mean, like most 80% of math, once you reduce it to linear algebra,


[User] Please transcribe this text

[Assistant] And you apply it to the vector stimuli, and that basically produces all the voltages. Now if you're interested in one particular one, you can derive that. But now, the key point about this is that if you look at the inverse of y,


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[Assistant] What, how do you define it? It's the cofactor matrix divided by the determinant of Y.


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[Assistant] Now, the key point is that, first of all, for this inverse to exist, the determinant of this thing needs to be what? Non-zero.


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[Assistant] Right, so this matrix cannot be singular, as we know from linear algebra. And, in other words, this inversion blows up when this determinant is zero.


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[Assistant] So when they are zero, the frequencies at which they are zero, define the poles of the transfer function.


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[Assistant] And we will see how we derive them and we find them and we'll use them in the circuit design and we take advantage of it.


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[Assistant] Now, and if you, of course, want to find the, and you can also see that if you're trying to find the relationship between an input and an output, you of course depend on both of these parameters, so it not only depends, but, and that would be the definition, the frequencies at which those become zero, or what we call zeros,


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[Assistant] And that would depend not only on the Y matrix by itself, but the definition of the input and output.


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[Assistant] And you will see when we get, for example, the generalized time and transfer constants method, when we talk about that.


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[Assistant] That the poles of the system, the poles of any


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[Assistant] Transfer function are functions of the, of the network itself.


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[Assistant] But the zeros, deterred, depend on the choice of the input and output. So if you say, if you choose a different output, you get a different set of zeros, but not a different set of poles.


[User] Please transcribe this text

[Assistant] Then you would have, if you take it out from a different note.


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[Assistant] And we'll see this in detail and we'll see that exhibits itself. But really it's coming from the fact that the poles are really determined by the Y matrix. Not the choice of which one of these voltages is the one that you're interested in.


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[Assistant] But if you want to do the zeros, you will need to take that into account. That choice would matter. Okay. Any questions on that?


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[Assistant] I mean this is a very powerful analysis, but it has a major limitation still. What is that major limitation?


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[Assistant] Well, there are several different limitations from different perspectives, but what is the major, the most glaring limitation of this thing, if you wanted to apply to, apply this to the devices that you've developed so far.


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[Assistant] And the other question, so that's correct, that's the most glaring limitation, and we'll talk about that in a second. How we deal with that.


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[Assistant] The other limitation of this thing is that this is not particularly conducive to design. This is not particularly useful for coming up with new design. It's good if you want to analyze something that you already have.


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[Assistant] And even in that case, it's not particularly good for human beings. It's very good for computers.


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[Assistant] Because computers are really good at dealing with large matrices of numbers and inverting matrices and all those things. And we are not.


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[Assistant] Okay, so, the key question is, but what is it good for then? Well, it's good for writing computer simulators.


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[Assistant] And that's what your, the computer simulators that you use, you will use, uh, in the class, are essentially using variations of this thing. And we'll talk about how they use these variations in different cases, depending on what kind of analysis you do, they use it, use it differently. But they use this essentially, this nodal analysis as the basis for the subsequent analysis they do.


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[Assistant] So now the question is that, let's say you are interested in a behavior of a circuit, but you are trying to kind of model something with a non-linear device. Let's say you have a transistor or something like that. And then the question is, how would you use this? Can you possibly use this? Let's say you have a transistor and it's being driven by a, I don't know, a square wave. And you're trying to find out what the output of that thing looks like.


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[Assistant] Can we think of a way to have our cake and eat it too?


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[Assistant] Okay, so replace the transistor with some equivalent linear components, and then do what?


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[Assistant] Okay, so if you do the linear analysis, then you have linearized it, right? So you've suppressed all the nonlinear behavior it sounds. Right? Okay, but that's a good starting point.


[User] Please transcribe this text

[Assistant] So let's say you have a transistor, we talked about small signal model, right? So let's say you have a transistor and it's operating at some, so,


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[Assistant] So let's say, very simple model. Just like let's say you have a bipolar transistor or MOSFET, it doesn't matter.


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[Assistant] And then let's, we will talk about these stages in detail soon. But let's say you have something connected with a resistor to some supply. Some VDD.


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[Assistant] And then you have some input and an output. So this is V in and this is V out.


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[Assistant] Now, this is obviously a nonlinear behavior transistor. Let's say the model, you want to use the quadratic model, or a more complex model, there are more complex models that the computer simulator is used, but they're all nonlinear.


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[Assistant] Right, so you have this non-linear behavior between the V, the ID and VGS and it also depends on the VDS and all of that stuff. So how do you do that? So, one of the things is that let's say you have some applied some input DC voltage to it.


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[Assistant] And then you have an AC voltage on top of that. So let's call this VGS. And then this is VGS lower case. And you are trying to find, but this is not necessarily small.


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[Assistant] So, so let's say this, or this thing, let's say it's a square wave.


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[Assistant] How would you deal with that? So let's say you start with a point, say well at this point, the beginning of this thing.


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[Assistant] This is the voltage, so you can actually say, okay, I am thinking about this as a DC operation point that I've been at. Let's say I start my simulation at this time.


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[Assistant] This is a simple set of equations, just like to solve it locally and statically. So you basically solve it, you find what the ID, the DC current would be, or the total current, and the V voltages are.


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[Assistant] So once you define that what ID and VGS are, so you can actually define things like GM. So you can say this is ID versus VGS. And say, okay, at that point I'm here. Right?


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[Assistant] And then what do you do with that? Say, well okay, I'm here, so now I linearize it around this operation point. I do a perturbation. I say, okay, this is my GM. The slope of this line. Or this is my RO.


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[Assistant] For the VD, you know, if you have the slope of the, uhm, or you can basically have the other curves, let me just draw them.


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[Assistant] You can say this is basically ID versus VDS for that operation point and you have this slope and then this slope is of course 1 over RO.


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[Assistant] Right? So you find these parameters. So from that you develop what? You can calculate a small signal model.


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[Assistant] Right? And your small signal model is not limited to the static, the low frequency parameters, right? You can also calculate the capacitors.


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[Assistant] So you develop a small signal model at this operation point, at this particular operation point that you found. Now, you use that small signal model


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[Assistant] Use a little analysis to estimate what the next step of the circuit is. So in time domain


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[Assistant] You have found one operation, let's say this is one of the voltages you're interested in, node voltages, right, or two voltages. So you have defined V in and V out.


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[Assistant] And then you say, okay, I'm gonna take the nodal analysis, so now I use the nodal analysis to solve


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[Assistant] The circuit, linearized circuit at this point. And say, okay, the linearized circuit is going to behave this way and this way, for example, I don't know. You can calculate, right?


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[Assistant] But you won't take it all the way down. You say, okay, let me take a small step.


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[Assistant] You say, okay, I'm taking a small step, I'm only going this far, so now my operation points have changed a little bit.


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[Assistant] And as a result, my GM has changed to this new value, and my RO has changed to this other new value. So I update the parameters of my small signal model.


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[Assistant] And that sets the tolerance, one of the tolerances. So you take this stepwise iterative process and you constantly solve the nodal analysis. You do a nodal analysis repeatedly on this modified


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[Assistant] There are lots of analysis, there are a lot of details in there. But that's the essence of it. So this analysis is useful for even the nonlinear system because you constantly recalculate the operation point from this.


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[Assistant] And evaluate it based on the previous thing. And every so often, if you see a big jump, what happens is that you take a step and say, oh, okay, well, a big thing happened. Big change happened. Bigger than my maximum range of variation. Well, you take a step back, you take a smaller step.


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[Assistant] And if that's still too big, take a step back, take a smaller step.


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[Assistant] Till you get to, the simulator gets to a point that it reaches that level of change that you, maximum level of change that you're allowed, you're permitting, and then it maintains a certain level of accuracy. So that's what we call a transient analysis.


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[Assistant] That's the basis. But now, from a circuit design perspective, nodal analysis is not particularly useful, in my opinion. It's the last resort. We should not use it unless we absolutely, positively have to.


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[Assistant] So we make a lot of effort in not using nodal analysis. Because


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[Assistant] There are other things that will allow us to see what's going on. So what are those things? So some of those things are the classic theorems of the network theory. So what are some of the classic theorems of the network and circuit theory? So you have the most prominent one, of course, for linear circuit is superposition. So it basically says in a linear circuit,


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[Assistant] Effect of sum of stimuli is the sum of the effects of an individual stimuli.


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[Assistant] And the necessary and sufficient condition for superposition to hold is what? Linearity, right?


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[Assistant] So if you have something that's linear, superposition holds. If you have something where superposition holds, linearity, it's linear.


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[Assistant] Current sources and voltage sources and dependent sources.


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[Assistant] And, uh, of varying kinds and elements, capacitors, inductors, um, resistors, what have you, whatever. And they're all connected to each other somehow, in some way.


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[Assistant] And we are interested in certain behavior of certain parts of the circuit. Let's say we are interested in certain voltages or currents.


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[Assistant] What it says is that if I am for example interested in this current, Ix,


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[Assistant] I can calculate Ix by doing what? By setting all, now, and this is the key part, again I like to emphasize this because this can lead to a lot of problems if not paid attention to.


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[Assistant] You null with zero all independent sources. Only the independent sources.


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[Assistant] Individual elements, conductances, resistances and all those things. So, you take the independent sources, you null them. What does it mean to null a source? Again, to refresh everyone's memory.


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[Assistant] Nulling a voltage source, what does it mean? A voltage source with zero voltage is what? A short circuit. A current source with zero current is what?


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[Assistant] So we open all the current sources, we short all the voltage sources, and we keep one at a time. You look at what this current is, so you get IX1. And then you do it for the second one, you get IX2. So then total IX is gonna be sigma IX.


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[Assistant] So you calculate this i, and sometimes it may be positive, for some of them it may be negative, you do an algebraic sum. That's the definition of superposition. How superposition works, right?


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[Assistant] You do not do this to the dependent sources, obviously. Just to emphasize.


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[Assistant] But, and then you do that. So, so that's a superposition. So that's one of the theorems that's helpful in understanding the effect of different things. We will not use it that much directly, but we will use some of its implications. The other ones that are quite interesting and important to remember is Thevenin and Norton theorem. Norton equivalence.


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[Assistant] So what Thevenin and Norton say is that if you have a network like this,


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[Assistant] And it's made purely out of linear elements. They could be sources, they could be dependent, independent sources, inductors, capacitors, uh, you know, resistors, what, or what have you. Now, and it has two terminals, just two terminals, namely one port.


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[Assistant] Okay, what it says is that this whole thing can be modeled, so Thevenin equivalent says this can be modeled as a source in general.


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[Assistant] A voltage source in series with some impedance z.


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[Assistant] The equivalent, let's say. And the seven.


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[Assistant] And then, or a current source in parallel with the


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[Assistant] Same z equivalent, which would be, well, y equivalent would be 1 over z equivalent. Let's see. Equals 1 over y equivalent. And this is i Norton.


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[Assistant] So it says that if you take a box, if this is a black box,


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[Assistant] With purely electrical measurements, you cannot differentiate. And purely electrical means currents and voltages. Just currents and voltages.


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[Assistant] There are other things that you can differentiate these things, right? For example, if you put them on an x-ray, you can differentiate these things.


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[Assistant] Or if you look at the temperature, you can see that this one is warm and this one is cold.


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[Assistant] Or if you look at the magnetic field for example, you will see that this has magnetic field around it and this one doesn't have magnetic field.


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[Assistant] So when we say they are indistinguishable purely by voltage and current measurement, not by other means. So,


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[Assistant] It means that all of this thing can be reduced to something like this. Now, what are the values of these things? So there are a couple of different ways to think about it. One is that z-equip- so what is V-thevenin? V-thevenin, which is also shown sometimes better, I think, as VOC, the open circuit, is the voltage you experience here.


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[Assistant] If you don't load it, if there's no current flowing, if you keep it open circuit, what's the voltage you see here? Now, that voltage may be a constant, may be a variable voltage, so this, this V equivalent doesn't even have to be a constant. It can be something, you can say V equivalent is this. Or VOC looks like this versus time.


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[Assistant] This would be the current that would flow if you short-circuited this. And this would be this current. So ISC.


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[Assistant] And again, ISC doesn't have to be constant, it can be a function of time or a function of frequency.


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[Assistant] So in this case it can be a function of frequency. It can have certain behavior of frequency or a certain frequency in time or...


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[Assistant] Whatever, so in general it can be some voltage, but it can still be represented as such.


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[Assistant] Now, how about the z equivalent? What is z equivalent? Well, z equivalent is basically the impedance seen


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[Assistant] Looking between these two terminals when all the independent sources are nulled.


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[Assistant] Again, we know what that means, nulling the source, setting its value to zero.


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[Assistant] Meaning, shorting the voltage sources and independent voltage sources, not the dependent ones, and opening the independent current sources. Okay? So that's the T equals. So that's the Thevenin-Norton, again, review for you.


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[Assistant] So there are two other theorems that are probably, so Thevenin and Orton are useful, I mean we use them from time to time because we want to think about a part of the network as an equivalent when we talk, deal with the rest of it. Then the other ones that are quite useful, one is source transportation. So we've already used it in this class.


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[Assistant] And then there's a voltage dual to this, which basically says that if you have two nodes and a voltage source here,


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[Assistant] You can actually remove this and put this two volt, the voltage source, two voltage sources here of the same amount.


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[Assistant] I like to think about that toothpaste. You squeeze it out and it goes into this thing. Like a paste that goes into tubes. But anyway.


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[Assistant] So, so that's, that's useful. We use this where? You remember? When we were trying to convert, for example, the pi model to t model.


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[Assistant] We use that theorem. There's another theorem which is perhaps the most


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[Assistant] Useful and most commonly used theorem in electronics design, which we actually don't explicitly say it, and most people don't mention it explicitly, is the source substitution.


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[Assistant] The source substitution theorem is actually the most commonly used theorem in electronic and circuit analysis for active devices. Which basically says, if at some point in this network,


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[Assistant] You, for example, you have an element where you see a voltage. So you measure the voltage across this element. Right?


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[Assistant] Uhm, and phase up 45 degrees, and nothing will happen.


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[Assistant] The same thing is true, it has a current tool, which basically means that if you saw a current through a branch to be something, you can replace it with a current source of the same value. And nothing else changes. Right?


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[Assistant] Which makes sense, because everyone else from the outside sees the same voltage or same current, so nothing is changing. And you can actually prove it formally. None of these we are going to prove. If you're interested in the proof, you can see the 44 lectures and things like that. So, this is basic, why is this useful? We'll see this frequently.


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[Assistant] This is particularly useful if you have cascaded stages of amplification.


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[Assistant] So this is your input, and this is your overall output.


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[Assistant] And this is stage one. So now what happens of course is that each one of them represents a source impedance and has a load impedance. And we'll talk about this in more detail.


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[Assistant] And calculate the gain of this thing. What this does, it allows you to abstract away different stages.


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[Assistant] And analyze them independently. Again, we'll see examples of this. So, but when we see them, I want you to know that this is coming from the source substitution. I will point it out to, if I remember.


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[Assistant] So, those are some of the key theorems behind this. Now, there's one more thing that we generally need to talk about to make sure that we are on the same page, and that's the concept of two ports, right? So you can actually have


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[Assistant] A box that, that looks like this. And again, we don't use to, we try to actually stay away from two ports to the extent that we can in design. Some people just use it extensively, I don't think it's great, but we use it in


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[Assistant] Some of the analysis that leads to or understanding of the analysis that leads to the design. So what is a two-port? So, is it some box with four terminals? Is any box with four terminals a two-port?


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[Assistant] See, remember, this was a one-port, when we had a box with two terminals, and if this whole thing was much smaller than the wavelength, then that there was no charge accumulation, net charge accumulation inside,


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[Assistant] You may say, what's the fun about a capacitor? Capacitor doesn't accumulate net charge, right?


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[Assistant] Meaning that's that's another way of saying speed of light or propagation speed is infinity. Okay, so.


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[Assistant] Is anything a two-port? Is every arbitrary four-terminal box? No, right?


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[Assistant] What is that subset? What's the definition of that subset? What kind of subset? Well, satisfy that.


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[Assistant] This current has to be identical, identically equal to that current. And this current has to be identically equal to that current. So for, for this to be a port, this is the condition you need to satisfy.


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[Assistant] And if that condition is satisfied, then you can write certain set of equations. You can simplify this into a two by two matrix representation. So you can, for example, and there are different boxes. For example, you can start with currents as stimuli and measure the voltages. And that will produce the z parameters.


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[Assistant] What I can say is that I can write, so the inputs or stimuli are I1 and I2.


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[Assistant] And that matrix is called z11, z12, z21, z22, and this is called the z, these are called z parameters.


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[Assistant] And they're easy to- fine, I mean, and we'll talk about them in a little bit more detail later when we need them. For example, z11 is the ratio of v1 to i1 when v2 is 0. I'm sorry, when i2 is 0.


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[Assistant] You set I2 to zero, meaning, and how do you set I2 to zero? How do you set, what happens when you set a current source to zero?


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[Assistant] Open circuit. So basically means that it's the ratio of this voltage to this current when the output is open. So you open this from a practical physical perspective and you measure that. So that's how you measure z parameters.


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[Assistant] Now you can imagine there's a Y parameter, set of Y parameters, right? So what are the Y parameters like? Well, that's if you stimuli, your stimuli are voltages. So if I replace these things with voltage sources, V1 and V2, and monitor I2 and I1.


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[Assistant] And there is a relationship between these, but it's not a one-to-one relationship that y12 is equal to 1 over z12. It's a matrix relationship. Why? Because


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[Assistant] For example, if you're trying to calculate y11, what is the setting? v2 needs to be 0.


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[Assistant] When you set v2 to zero, what are you doing? A voltage source with?


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[Assistant] A voltage of zero is a short circuit. So this is a different setting. You're measuring the properties of the input when the output, for example, is shorted versus open. Or when you do the trance parameters.


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[Assistant] From V1 to I, V2 or V2 to I2, this, this diagonal, off diagonal elements, those would be affected by whether or not you're shorting or opening the output.


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[Assistant] There are all sorts of other parameters. There are the TT parameters, ABC, S parameters, things of that sort. So, eh, we will deal with them when we need them. Okay. So this is a very, very quick high level review.


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[Assistant] Of the circuit and network theory that just we want to make sure that everyone is on the same page before we move forward. Any questions?


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[Assistant] You're evil, Wukong. You can't go to the Western Paradise with me. Now leave before I recite the Tidehead Band spell.


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[Assistant] But first, I must figure out how he creates the fire, smoke and sand.


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[Assistant] You wait here, your majesty. Tonight I'll steal those bells from the demon. Once I defeat him, I'll come back and rescue you.


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[Assistant] Journey to the West Chapter 90 The Demon's Plan


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[Assistant] Trembling with fear, the Tang monk fell from his horse. Wu Jing rushed over to help him stand up.


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[Assistant] Looking closely, Wukong could tell that he was not an ordinary man.


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[Assistant] Hello Venus, said Wukong with a smile. The man smiled back.


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[Assistant] Several demons were talking in front of a cave. They were dressed like military officers. Wukong spied on them from behind a rock.


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[Assistant] That was very kind of our leaders, said a captain.


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[Assistant] Then he stepped out from behind the rock and hurried over to the officers.


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[Assistant] I have to read this book and do a report on it for school by tomorrow.


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[Assistant] I bet all of you are familiar with this view of the ocean.


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[Assistant] But the thing is, most of the ocean looks nothing like this.


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[Assistant] Below the sunlit surface waters, there's an otherworldly realm known as the Twilight Zone.


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[Assistant] At 200 to 1000 meters below the surface, sunlight is barely a glimmer.


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[Assistant] Tiny particles swirl down through the darkness, while flashes of bioluminescence give us a clue that these waters team with life.


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[Assistant] Microbes, plankton, fish, everything that lives here has amazing adaptations for the challenges of such an extreme environment.


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[Assistant] There could be ten times more fish biomass here than previously thought.


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[Assistant] And life in the Twilight Zone is intertwined with Earth's climate. Yet the Twilight Zone is virtually unexplored. There are so many things we still don't know about it. I think we can change that.


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[Assistant] I was drawn to oceanography by just this kind of challenge. To me, it represents the perfect intersection of science, technology and the unknown.


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[Assistant] As a college student, I went on an expedition across the Atlantic with a team of scientists using a high-powered laser to measure microscopic algae.


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[Assistant] This amazing discovery happened because we used new technology to see life in the ocean in a new way.


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[Assistant] We know so little about the Twilight Zone because it's difficult to study. It's exceedingly large, spanning from the Arctic to the Southern Ocean and around the globe. It's different from place to place.


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[Assistant] It changes quickly as the water and animals move and it's deep and dark and cold and the pressures there are enormous. What we do know is fascinating.


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[Assistant] So much sound bounces off this layer. It's been mistaken for the ocean bottom. But if we look, it can't be, because the layer is deep during the day.


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[Assistant] It rises up at night, and the pattern repeats day after day.


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[Assistant] These animals and their movements help connect the surface and deep ocean in important ways. The animals feed near the surface,


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[Assistant] The migration may help keep carbon dioxide out of our atmosphere and limit the effects of global warming on our climate.


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[Assistant] Who is trying to eat them, or how much carbon they're able to transport?


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[Assistant] For me, curiosity about these things is a powerful driver. But there's more to the motivation here.


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[Assistant] We need to answer these questions and answer them quickly, because the Twilight Zone is under threat.


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[Assistant] The animals are ground into fish meal to support increasing demands for aquaculture and for nutraceuticals such as krill oil.


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[Assistant] Industry is on the brink of deepening fisheries such as these into the mid-water.


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[Assistant] Operating outside the reach of national fishing regulations.


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[Assistant] This could have irreversible global-scale impacts on marine life and food webs.


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[Assistant] Together, we are ready to launch a large-scale exploration of the Twilight Zone.


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[Assistant] We have a plan to begin right away with expeditions in the North Atlantic, where we'll tackle the big challenges of observing and studying the Twilight Zone's remarkable diversity.


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[Assistant] This kind of multi-scale, multi-dimensional exploration means we need to integrate new technologies.


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[Assistant] Satellite tracking devices on animals such as sharks.


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[Assistant] We find that their feeding hotspots are linked to ocean currents and other features.


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[Assistant] We used to think these animals found all of their food in surface waters.


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[Assistant] But we still need to figure out how they find the best areas to feed.


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[Assistant] What they're eating there and how much their diets depend on Twilight's own species.


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[Assistant] We will also need new technologies to explore buildings with climate. Remember these particles?


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[Assistant] Salps are like super-efficient vacuum cleaners, slurping up plankton and producing fast-sinking pellets of poop. Try saying that 10 times fast.


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[Assistant] We need to know where and when and why and whether this kind of carbon sink has a big impact on Earth's climate.


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[Assistant] To meet these challenges, we will need to push the limits of technology.


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[Assistant] We will deploy cameras and samplers on smart robots to patrol the depths and help us track the secret lives of animals like salves. We will use advanced sonar to


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[Assistant] We will sequence DNA from the environment in a kind of forensic analysis to figure out which species are there and what they are eating.


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[Assistant] Just look at these beautiful, fascinating creatures. We barely know them.


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[Assistant] And imagine how many more are just down there waiting for our new technologies to see them.


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[Assistant] The excitement level about this could not be higher on our team of ocean scientists, engineers and communicators. There is also a deep sense of urgency.


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[Assistant] We can't turn back the clock on decades of overfishing in countless regions of the ocean that once seemed inexhaustible.


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[Assistant] The Twilight Zone is truly a global commons. We need to first know and understand it before we can be responsible stewards and hope to fish it sustainably.


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[Assistant] This is not just a journey for scientists. It is for all of us.


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[Assistant] Journey to the West, Chapter 91, The Demon Officers.


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[Assistant] The Demon General looked at Wukong, who was still in disguise.


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[Assistant] That is great news, said a demon captain.


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[Assistant] Said the general. The other demons nodded in agreement.


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[Assistant] A captain looked surprised. What about eating the Tang monk? Don't you want to live forever? Wukong shook his head.


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[Assistant] Said a captain. The general bit his lip.


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[Assistant] Soon Wukong could see the demon army marching away in the valley below. The monkey smiled.


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[Assistant] Still in disguise, Wu Kang entered the cave. He came to a large room.


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[Assistant] Three huge demons, a lion, an elephant, and a raptor, sat on thrones.


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[Assistant] Said Wukong. What? shouted the raptor.


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[Assistant] As the two demons argued, Wukong began to laugh.


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[Assistant] Soon he was laughing hard. The demon stopped arguing and stared at him.


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[Assistant] Journey to the West, Chapter 84, The Sad King.


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[Assistant] The streets were lined with people selling food and other goods.


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[Assistant] He pointed to a crowd of people reading a sign on a wall.


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[Assistant] The king looked up, but didn't say anything. The guard whispered to the travelers.


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[Assistant] Wukong, hissed the Tang monk, show some respect. Wukong ignored the monk and looked at the king.


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[Assistant] Asked Wukong. The king got up and walked to a window.


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[Assistant] Several small demons were standing outside a cave entrance.


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[Assistant] A large demon with broad shoulders came out of the cave. He wore three bells


[User] Please transcribe this text

[Assistant] Alright, before the break we did cover the common emitter stage in more details, especially when we saw what happens when you have resistances in the emitter. So that's one stage which basically has an input at the base and the output taken out of the collector.


[User] Please transcribe this text

[Assistant] That's not the only thing you can make, right? I mean, you can think about taking the input, uh, you know, taking the output from other places or applying your input to other places. The transistor has three terminals. So there are different topologies and each one of them has its own...


[User] Please transcribe this text

[Assistant] So the next one we are going to talk about is basically the common collector or also known as emitter follower.


[User] Please transcribe this text

[Assistant] Before we start analyzing it, let's see what we can expect to see in something like that. What do we expect to get? From understanding of our input and output resistances, right?


[User] Please transcribe this text

[Assistant] So, the input resistance of the stage, what do you expect it to be, roughly? And I want, I encourage you to start thinking about it without having to write equations.


[User] Please transcribe this text

[Assistant] Because this would be a helpful way of dealing with these as they grow. So you have to develop some intuition about it. So the input resistance, if you have a stage like this, Rn.


[User] Please transcribe this text

[Assistant] You remember we just talked about, before the break we talked about the common emitter, right? With degeneration.


[User] Please transcribe this text

[Assistant] You can use it, you can use the reflection rule, we talked about the reflection rule, we saw that if you have some resistance in the emitter, whatever you have in the emitter, can be viewed from the base side if you multiply it by beta plus one, or roughly beta.


[User] Please transcribe this text

[Assistant] So what do we have? What do we have in the emitter? What is the total resistance in the emitter right now? You have of course the RE.


[User] Please transcribe this text

[Assistant] And then you have this RM, alpha RM sitting here. Which is implicit inside the transistor. So there's a little alpha RM here, on the emitter side.


[User] Please transcribe this text

[Assistant] So, you can see that your input resistance is still can be quite large, if you have a sufficiently large Rm, or, or, or, or pi, or Re, right, because it gets multiplied by beta. So you have an input resistance that's large, which means that it's a good voltmeter.