[User] Please transcribe this text

[Assistant] Uhm, I don't play golf but a very, very popular activity all over Canada and the United States in the summer.


[User] Please transcribe this text

[Assistant] For people who have a little bit of extra money is the game of golf.


[User] Please transcribe this text

[Assistant] It is an expensive game to play. Uhm, I have played golf before. Jen's dad actually likes to go golfing and sometimes he takes us golfing and he pays for it. So that's a nice way to go golfing. Uhm, but golf is an extremely popular sport. There are many, many golf courses in Canada and many, many golf courses in the United States.


[User] Please transcribe this text

[Assistant] Uh, for sure. Uhm, and then of course, uh, one of the things that people definitely do a lot of in the summer is they just spend time in the shade. So shade is something that you have under a tree. Uh, it's a challenging word because not every language has a word for shade.


[User] Please transcribe this text

[Assistant] When I walk, the sun makes it so I have a shadow.


[User] Please transcribe this text

[Assistant] But when I go under a tree I am in the shade. And it's usually cooler in the shade. Uhm, so people spend a lot of time in the shade in the summer for sure.


[User] Please transcribe this text

[Assistant] And then my last slide here is the slide family reunion. I don't know who these people are. This is just a picture from the internet.


[User] Please transcribe this text

[Assistant] But it is very common, uh, in my part of Canada and I think in most parts of Canada and the United States.


[User] Please transcribe this text

[Assistant] For people to have a family reunion in the summer. So this is when you meet with your uncles and aunts, your grandparents if they are still around, uhm, people who you only see once a year, cousins, distant cousins, those kinds of things. So you might have a family reunion.


[User] Please transcribe this text

[Assistant] That is a very common thing for people to have in the summer.


[User] Please transcribe this text

[Assistant] Do you think I should send him a card? Beethoven is dead. He is? Hm. I just saved a stamp. Hm. He never told me that Beethoven was dead.


[User] Please transcribe this text

[Assistant] Why should I tell you something that everyone already knows?


[User] Please transcribe this text

[Assistant] Come out, one-horned buffalo! You are under arrest!


[User] Please transcribe this text

[Assistant] I wasn't able to arrest him. We need more help.


[User] Please transcribe this text

[Assistant] Tire Town School, Episode 10, Safety Class.


[User] Please transcribe this text

[Assistant] Obey the sign, said Miss Fender. Jeb stopped at the sign. He looked both ways. Then he drove some more.


[User] Please transcribe this text

[Assistant] A green light turned yellow. Jeb slowed down. The light turned red. Jeb stopped. Zippy whispered to Eddie. You're good at slowing down. Because you're slow. I'm not slow, said Eddie. The light turned green.


[User] Please transcribe this text

[Assistant] I want to talk to you about how to build and rebuild trust, because it's my belief that trust is the foundation for everything we do.


[User] Please transcribe this text

[Assistant] And that if we can learn to trust one another more, we can have unprecedented human progress.


[User] Please transcribe this text

[Assistant] What if your employees experience a culture of bias, exclusion and worse?


[User] Please transcribe this text

[Assistant] What if there's a data breach and it feels an awful lot like a cover-up than seriously addressing it? And most tragically, what if a technological fail leads to the loss of human life?


[User] Please transcribe this text

[Assistant] But I was super attracted to going to an organization that was


[User] Please transcribe this text

[Assistant] This was an organization that had lost trust with every constituent that mattered. But there's a word about me that I should share. My favorite trait is redemption.


[User] Please transcribe this text

[Assistant] I went to Uber with the hopes of a turnaround there could give license to the rest of us who might have narrower versions of their challenges.


[User] Please transcribe this text

[Assistant] I publicly committed to wearing an Uber T-shirt every day until every other employee was wearing an Uber T-shirt. I had clearly not thought that through.


[User] Please transcribe this text

[Assistant] It was 250 days of wearing an Uber T-shirt. Now I am liberated from that commitment, as I am back at HBS, and what I'd like to do is share with you how far I have taken that liberty. Which, it's baby steps. But I would just say I'm on my way. Now,


[User] Please transcribe this text

[Assistant] Trust, if we're going to rebuild it, we have to understand its component parts.


[User] Please transcribe this text

[Assistant] The component parts of trust are super well understood. There's three things about trust.


[User] Please transcribe this text

[Assistant] If you sense that I am being authentic, you are much more likely to trust me.


[User] Please transcribe this text

[Assistant] If you sense that I have real rigor in my logic, you are far more likely to trust me. And if you believe


[User] Please transcribe this text

[Assistant] That my empathy is directed towards you, you are far more likely to trust me.


[User] Please transcribe this text

[Assistant] When all three of these things are working, we have great trust. But if any one of these three gets shaky, if any one of these three wobbles, trust is threatened. Now here's what I'd like to do.


[User] Please transcribe this text

[Assistant] And the way to do that is to understand where trust wobbles for ourselves and have a ready-made prescription to overcome it. So that's what I would like to do together.


[User] Please transcribe this text

[Assistant] The most common wobble is that people just don't believe that we're mostly in it for them, and they believe that we're too self-distracted. And it's no wonder.


[User] Please transcribe this text

[Assistant] We are all so busy with so many demands on our time, it's easy to crowd out the time and space that empathy requires. For Dylan to be Dylan, that takes real time.


[User] Please transcribe this text

[Assistant] And for us, if we have too much to do, we may not have that time. But that puts us into a vicious cycle.


[User] Please transcribe this text

[Assistant] Revealing empathy, it makes everything harder. Without the benefit of the doubt of trust, makes everything harder and then we have less and less time for empathy and so it goes. So here's the prescription.


[User] Please transcribe this text

[Assistant] Identify where, when and to whom you are likely to offer your distraction.


[User] Please transcribe this text

[Assistant] That should trace pretty perfectly to when, where and to whom you are likely to withhold your empathy. And if in those instances we can come up with a trigger that gets us to look up, look at the people right in front of us, listen to them,


[User] Please transcribe this text

[Assistant] Deeply immerse ourselves in their perspectives.


[User] Please transcribe this text

[Assistant] Then we have a chance of having a sturdy leg of empathy. And if you do nothing else, please put away your cell phone.


[User] Please transcribe this text

[Assistant] And it is super difficult to create empathy and trust in its presence.


[User] Please transcribe this text

[Assistant] Logic wobbles can come in two forms. It's either the quality of your logic, or it's your ability to communicate the logic. Now if the quality of your logic is at risk, I can't really help you with that. It's like not in this much time.


[User] Please transcribe this text

[Assistant] Super fortunately, there's a very easy fix to this. If we consider that there are two ways to communicate in the world,


[User] Please transcribe this text

[Assistant] If we consider that there are two ways to communicate in the world, and the first one is when you take us on a journey.


[User] Please transcribe this text

[Assistant] And some of the best communicators in the world communicate just like this. But if you have a logic wobble, this can be super dangerous. So instead, I implore you, start with your point in a crisp half-sentence, and then give your supporting evidence.


[User] Please transcribe this text

[Assistant] But if you represent any sort of difference, the prescription to be you can be super challenging.


[User] Please transcribe this text

[Assistant] Tempted personally and tempted by coaching of others to mute who I am in the world. I'm a woman of super strong opinion. With really deep convictions. Direct speech. I have a magnificent wife.


[User] Please transcribe this text

[Assistant] And together we have such crazy ambition. I prefer men's clothes and comfortable shoes. Thank you, all birds.


[User] Please transcribe this text

[Assistant] I hope that each person here has the beautiful luxury of representing difference in some context in your life. But with that privilege comes a very sincere temptation to hold back who we are. And if we hold back who we are, we're less likely to be trusted.


[User] Please transcribe this text

[Assistant] And without those stretch assignments, we're less likely to get promoted. And so on and so on until we are super depressed by the demographic tendencies of our senior leadership.


[User] Please transcribe this text

[Assistant] To our being, our authentic selves. So here's my advice. Wear whatever makes you feel fabulous.


[User] Please transcribe this text

[Assistant] To set the conditions that not only makes it safe for us to be authentic, but makes it welcome. Makes it celebrated. Cherishes it for exactly what it is.


[User] Please transcribe this text

[Assistant] So let's go back to Uber. What happened at Uber? When I got there, Uber was wobbling all over the place.


[User] Please transcribe this text

[Assistant] Empathy, logic, authenticity were all wobbling like crazy. But we were able to find


[User] Please transcribe this text

[Assistant] Super effective, super quick fixes for two of the wobbles. I'll give you an illustration of empathy.


[User] Please transcribe this text

[Assistant] In the meetings at Uber, it was not uncommon for people to be texting one another about the meeting.


[User] Please transcribe this text

[Assistant] It may have done many things, but it did not create a safe, empathetic environment. The solution, though, super clear. Technology, off and away. And that forced people to look up.


[User] Please transcribe this text

[Assistant] To look at the people in front of them. To listen to them.


[User] Please transcribe this text

[Assistant] Their positions outstripped their capability, and it was not their fault. The solution, a massive influx of executive education that focused specifically on logic, on strategy and leadership, gave people the rigor of the quality of their logic,


[User] Please transcribe this text

[Assistant] And it turned a whole lot of triangles.


[User] Please transcribe this text

[Assistant] Right side up, so people were able to communicate effectively with one another. The last one, authenticity, I'll say it's still mighty wobbly.


[User] Please transcribe this text

[Assistant] It is still much easier to coach people to fit in.


[User] Please transcribe this text

[Assistant] As opposed to rewarding people when they say something entirely different than what you were going to say. But when we figure out this,


[User] Please transcribe this text

[Assistant] When we figure out how to celebrate difference and how to let people bring the best version of themselves forward, well, holy cow, is that the world I want my sons to grow up in.


[User] Please transcribe this text

[Assistant] And with the collection of people here, it would be a privilege to lock arms with you and go ahead and rebuild trust in every corner of the globe. Thank you very much.


[User] Please transcribe this text

[Assistant] Alright, so by the way, we've talked about all sorts of transistors and you're doing vacuum tubes and all those things. I thought I'd show you another thing that may have some interesting implications for the way we thought about it. So this is a little kind of really fun


[User] Please transcribe this text

[Assistant] Nugget, it's really not central to the course, but I thought it's good to talk about this. So, have you heard of Kelvin generator, electrostatic generator? Has anyone heard of that?


[User] Please transcribe this text

[Assistant] And you usually kind of like add something like salt to it to make it conduct.


[User] Please transcribe this text

[Assistant] And these are conductive containers. So take these conductive containers, contain some water, already, obviously, because they've been filling up.


[User] Please transcribe this text

[Assistant] That's all. That's all there is. And you can actually make one. It's fun to make. So what is happening? What's happening with this?


[User] Please transcribe this text

[Assistant] Propose why there is some arcing and electrostatic buildup. Obviously one side becomes charged and the other side becomes charged in the opposite direction. And what is happening? Any thoughts?


[User] Please transcribe this text

[Assistant] Why do we have electrostatic charge buildup when something like this happens?


[User] Please transcribe this text

[Assistant] Yes, but what is the reaction? So it's a purely electrostatic behavior, right? You're right, in the sense that, but what, how does it start? How does it begin? Why is there some charge buildup?


[User] Please transcribe this text

[Assistant] So think about it this way. So let's say there's some fluctuation in charge, some minuscule noise fluctuation, instantaneous fluctuation in charge. So let's say this side gets slightly positively charged. And this side gets slightly negatively charged. Right? What's gonna happen?


[User] Please transcribe this text

[Assistant] Through this tank of water, these are connected through the water, right? So this will attract negative charge here, and I guess positive charge here, right?


[User] Please transcribe this text

[Assistant] As the water is going through this tube, it will attract the positive charge here and negative charge here. Now, what happens?


[User] Please transcribe this text

[Assistant] Part of this positive charge will end up in these droplets that have detached and end up in this bucket. And part of this negative charge


[User] Please transcribe this text

[Assistant] This one becomes more positively charged, which will


[User] Please transcribe this text

[Assistant] Enhance this process, because now this is more positively charged, so attract more negative charge, and likewise here.


[User] Please transcribe this text

[Assistant] Until you build up so much arch that you basically just create an arch.


[User] Please transcribe this text

[Assistant] Obviously this is an example of positive feedback when we talk about feedback, but there's something else here that there's a reason I mention this to you. What do these look like to you?


[User] Please transcribe this text

[Assistant] FETs, yea, FET transistors, exactly right. They're field effect transistors too.


[User] Please transcribe this text

[Assistant] Because you're applying electric field, you're controlling the channel charge, and then you have a little bit of a pinch off here, because your water droplets have separated.


[User] Please transcribe this text

[Assistant] So you've basically made some sort of a positive feedback amplification system, and there's the energy, there's the bias for it, there's the energy coming from.


[User] Please transcribe this text

[Assistant] In a regular circuit, the energy is coming from the C terminal, the supplies, right? Here, where is it coming from? Where is the battery?


[User] Please transcribe this text

[Assistant] Which you can see shows a negative resistance. So this looks like that.


[User] Please transcribe this text

[Assistant] It is hot. Maybe we should take a break.


[User] Please transcribe this text

[Assistant] The mountain of flames is near here. It's always burning.


[User] Please transcribe this text

[Assistant] It's impossible to pass! We'll go around it.


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[Assistant] All the land around the mountain is covered in flames! The land stretches far to the north and the south! It's impossible to go around!


[User] Please transcribe this text

[Assistant] The flames are magical. There is just one spirit who can put them out. Who? Asked Wuxing.


[User] Please transcribe this text

[Assistant] She has a special fan that can put out the flames. It's made from a plantain leaf.


[User] Please transcribe this text

[Assistant] Chapter 88, The Priest Plan.


[User] Please transcribe this text

[Assistant] The monkey is a powerful spirit, cried another. We must leave, cried the third. Poof.


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[Assistant] The women vanished. Wukong ran into the house.


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[Assistant] Those women are gone now, said Wukong. Then he smirked. Bajie scared them away. The pig was a real hero.


[User] Please transcribe this text

[Assistant] Bajie was on the ground, still struggling to get out of the webs. He pretended to laugh.


[User] Please transcribe this text

[Assistant] The road left the forest and headed up a mountain. The travelers arrived at a Daoist abbey.


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[Assistant] Said Wu Jing. The group entered the abbey. An old priest was inside mixing a potion.


[User] Please transcribe this text

[Assistant] Your Buddhists? Where are you going?


[User] Please transcribe this text

[Assistant] The Tang monk opened his mouth to answer, but Wu Kong spoke first. We're on our way to visit friends, said the monkey.


[User] Please transcribe this text

[Assistant] Said Bajay. The priest stood up and left the room.


[User] Please transcribe this text

[Assistant] He was looking for rice when he realized he wasn't alone.


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[Assistant] The three young women stood nearby and they looked angry.


[User] Please transcribe this text

[Assistant] But we had to flee our house said another woman. A monkey and a pig attacked us said the third.


[User] Please transcribe this text

[Assistant] The monk Wu Jing and Bajie were already sipping their tea.


[User] Please transcribe this text

[Assistant] In fact, these grades would indicate that President Obama is the best friend the NRA has ever had in elected office.


[User] Please transcribe this text

[Assistant] I don't know why, but some days, everyone's in a bad mood.


[User] Please transcribe this text

[Assistant] Maybe it's these bells. I'll take them off while I sleep.


[User] Please transcribe this text

[Assistant] Maybe he can help me deal with the Iron Immortal. He was able to stop the Great King Yellow Wind's wind magic.


[User] Please transcribe this text

[Assistant] No thank you. I must get back to my companions.


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[Assistant] It was me! Shouted Wukong. I'm in your stomach now! And I won't stop punching and kicking until you give me that fan! Woof! Pow! Please stop!


[User] Please transcribe this text

[Assistant] Journey to the West, Chapter 78, tricked again and again.


[User] Please transcribe this text

[Assistant] Bajay threw his hands up.


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[Assistant] Said the demon. He was very rude. Said the woman. He demanded I let him borrow my fan.


[User] Please transcribe this text

[Assistant] Suddenly, light sparkled and Wukong stood in front of her.


[User] Please transcribe this text

[Assistant] Said the bold demon king with a big smile. But then his smile faded.


[User] Please transcribe this text

[Assistant] Your friend son Wukong was here. She snapped.


[User] Please transcribe this text

[Assistant] He tricked me into giving it to him. He made himself look like you!


[User] Please transcribe this text

[Assistant] Wukong walked along the road carrying the plantain leaf fan.


[User] Please transcribe this text

[Assistant] A second later, the bold demon king stood next to Wukong. The demon laughed.


[User] Please transcribe this text

[Assistant] There was another flash, and the demon was gone.


[User] Please transcribe this text

[Assistant] Journey to the West, Chapter 82, The Magic Cloth.


[User] Please transcribe this text

[Assistant] The monkey jumped back to his feet and ran fast out of the monastery. When he got outside, everyone was waiting for him.


[User] Please transcribe this text

[Assistant] We can't worry about the bags now. Whispered Wukong. Yellow Browse is coming! We have to go! Quickly!


[User] Please transcribe this text

[Assistant] He helped the Tang monk up onto the horse. The group hurried down the road.


[User] Please transcribe this text

[Assistant] Run faster, Bajie! Hissed Wujing. Zhu Bajie breathed heavily.


[User] Please transcribe this text

[Assistant] They hadn't gotten far before yellow brows flew over them.


[User] Please transcribe this text

[Assistant] Wukong, Bajie, and Wuxing pulled out their weapons. The gold dragon bared its claws.


[User] Please transcribe this text

[Assistant] He reached into his robe and pulled out the cloth. Bage lowered his rake and laughed.


[User] Please transcribe this text

[Assistant] The monkey leaped up to the sky just before the cloth wrapped around his companions. Quickly the demon grabbed the cloth with the group in it.


[User] Please transcribe this text

[Assistant] As Wukong was wondering what to do, two spirits approached him. One was a turtle and the other was a serpent.


[User] Please transcribe this text

[Assistant] The two spirits each carried a sword and a shield.


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[Assistant] Said Wukong. He quickly told them what had happened.


[User] Please transcribe this text

[Assistant] We are much too powerful for him.


[User] Please transcribe this text

[Assistant] The door flew open. Yellow Brow stepped out and threw the cloth.


[User] Please transcribe this text

[Assistant] Laughing, the demon grabbed the bundle and went back into the monastery. Wukong flew back up into the sky. He sat on a cloud and moaned.


[User] Please transcribe this text

[Assistant] Wukong looked toward the voice and saw a man with a big belly. The man was smiling.


[User] Please transcribe this text

[Assistant] Wukong smiled. You're the Buddha Maitreya!


[User] Please transcribe this text

[Assistant] Okay, so before the break, what we did, we said that we are going to do a parallel treatment of the BJT MOSFET and ATD, arbitrary three terminal device.


[User] Please transcribe this text

[Assistant] And we did the development of the BJT, we looked at the IV characteristic, we looked at the large signal and small signal, and we are going to do a similar thing for MOSFET.


[User] Please transcribe this text

[Assistant] Now, since we've done some of these things already, uh, in a certain way, you will see there are similarities, and there are places that are similar, and we are doing the similar thing, we may not discuss it in as much detail as possible.


[User] Please transcribe this text

[Assistant] Anyway, so let's go to the MOSFET and see what happens in the MOSFET. We did develop a current voltage characteristic for MOSFET, so we set ID.


[User] Please transcribe this text

[Assistant] Right? The drain current was mu N, we developed this equation, C ox, if it's a PFET it would be mu PC ox, W over L, which is the width over length ratio for a transistor,


[User] Please transcribe this text

[Assistant] Square. So VGS minus VT, square. And then we have


[User] Please transcribe this text

[Assistant] The equivalent of early voltage, which was the channel length modulation, which you can write it either as lambda or a VA, uh, you can write it as V...


[User] Please transcribe this text

[Assistant] The DS, drain source, over VA or lambda VDS. You can do it in either way.


[User] Please transcribe this text

[Assistant] Uhm, now, that was the expression we had. So you can see the similarity with this. You can actually, if you write it in more mass-centric language, you can write as lambda v ds. Let's just do it that way.


[User] Please transcribe this text

[Assistant] Keep it consistent, so you can see lambda is 1 over VA essentially.


[User] Please transcribe this text

[Assistant] Now, and similarly you have, we had a treat, a, a, an IC characteristic, right? I mean we had the ID now, the drain current is a function quadratic or linear depending on what you have. So we saw also for a velo, deeply velocity saturated device, this was, if this velocity saturated,


[User] Please transcribe this text

[Assistant] This was above threshold, so the assumption here, there are two things. First of all, the transistor needs to be on.


[User] Please transcribe this text

[Assistant] Now, with the strong inversion assumption, what is the requirement for the transistor to be on? This is simplification. What do we need to happen for the transistor to be on? Vgs needs to be greater than


[User] Please transcribe this text

[Assistant] The threshold voltage. Now, by the way, this VT has nothing to do with this VT.


[User] Please transcribe this text

[Assistant] In the travesty of terminology, uh, let's go up. So this Vt is the threshold voltage.


[User] Please transcribe this text

[Assistant] Which is determined by the flat band voltage, the work functions, the implant, the charge interface, all of that good stuff. So,


[User] Please transcribe this text

[Assistant] Anyway, so regardless of that, so basically what it means is that at some up to a point, VT, this transistor based on the simple model, it's off, we know that there's current flowing below that because of the weak inversion, because of the subthreshold


[User] Please transcribe this text

[Assistant] Or weak inversion, which translates to sub-threshold current. But in this simplistic model, you just ignore that, and after that you have a quadratic dependence.


[User] Please transcribe this text

[Assistant] And again, if we ask ourselves, what is the slope? The same thing happens. The slope is the definition of the transconductance, or GM.


[User] Please transcribe this text

[Assistant] Depending on which one of these expressions you use, you will get some expressions. So let's use this and ignore this for the time being. You will get mu and c ox w over l.


[User] Please transcribe this text

[Assistant] Vgs minus vt. Which you could also write if you want to write it as a current, you could write it as 2 mu n cox w over l id.


[User] Please transcribe this text

[Assistant] Alright, so, so those are the expressions for this thing. And of course, all of this also, there's a second condition for this to happen. The same way that this was for forward active region.


[User] Please transcribe this text

[Assistant] This holds for what kind of region of operation? Pinch off, right? So this is only valid in pinch off. And what is the requirement for pinch off?


[User] Please transcribe this text

[Assistant] Now it's gonna be this. So we know this, but we also knew something else. We knew that this lambda was proportional to L effective. So what you have, we ended up having this being L effective, uhm, over ID.


[User] Please transcribe this text

[Assistant] Times dxd, and we did derive this before, dvds.


[User] Please transcribe this text

[Assistant] And we saw that this, we treat this like some sort of a fixed parameter. Although it's not exactly a fixed parameter, we treat it like a fixed parameter. The key point, the key takeaway from this is that basically you can just approximate as L, is that in the


[User] Please transcribe this text

[Assistant] MOSFET, you actually, since you're controlling the L yourself as a design parameter, unlike bipolar, which is a fabrication parameter,


[User] Please transcribe this text

[Assistant] You control L, so you have actually control over the RO on resistance. But, okay, so that's what it is. And,


[User] Please transcribe this text

[Assistant] So that was the current, and then we had a sim- we can have a similar I-V characteristic as the bipolar. So if you draw the I-D versus V-C-E, or sorry, V-D-S, you have similar curves.


[User] Please transcribe this text

[Assistant] And then there are two regions, this is where you are in triode, and this is where you are in pinch off.


[User] Please transcribe this text

[Assistant] To the left of that right, you're in trial, and to the right, you're in pinch off. And then, again,


[User] Please transcribe this text

[Assistant] The question is, what are we gonna do with this? How can we make an amplifier with this? So the next step is to apply a load. And again, by the same arguments that we used in the bipolar, the case of bipolar, this load needs to be referenced to a supply voltage to provide power and make sure also that you're sufficiently high enough voltage that allows you to operate this properly. So when, when you do that, you will end up with a stage


[User] Please transcribe this text

[Assistant] So the simplest stage, by the way, this, this stage was called common emitter.


[User] Please transcribe this text

[Assistant] Because the emitter is a common terminal between the input and output, and this would be called a common source. So, the common source stage would be something like this.


[User] Please transcribe this text

[Assistant] So you have some VDD, you have some RD, and then you have a VIN.


[User] Please transcribe this text

[Assistant] That consists of two parts, and you have a V out. That consists of two parts. The two parts again, the AC and the DC. So this is the V in, and


[User] Please transcribe this text

[Assistant] So you have the relationship between Vout and Vc, ehm, and Id, let's say this current, is the same. So you have Vout is Vdd minus Rd Id.


[User] Please transcribe this text

[Assistant] And we have an expression for id, we can use either one of those, let's say use the, if you use the quadratic one, what we are gonna end up with is vdd, minus mu n, c ox over 2, w over l, rd, uh, v in, minus vt squared.


[User] Please transcribe this text

[Assistant] And for, for this one we can actually for now ignore the same way that we ignored the, uh,


[User] Please transcribe this text

[Assistant] We can ignore the channel length modulation. If we don't ignore it, we can end up with a more fun quadratic equation to solve, uh, which would end up in the same result at the end as what we get otherwise.


[User] Please transcribe this text

[Assistant] So this is the equivalent of that. Now what you see is that these equations are different, right? One has an exponential in it, and one has a quadratic.


[User] Please transcribe this text

[Assistant] So let's say we can't try to calculate the gain using the same procedure that we used last time, namely defining it as the derivative of the v out versus v in. And if you try to plot this by the way,


[User] Please transcribe this text

[Assistant] So, you need to build up a minimum amount of VIN before the transistor turns on.


[User] Please transcribe this text

[Assistant] Right? Right? Based on our model. Because we need to be above the threshold.


[User] Please transcribe this text

[Assistant] So according to the simplistic model, if your below the threshold voltage transistor is off, so what would the output be?


[User] Please transcribe this text

[Assistant] So it would be a VDD, so you start off at VDD, and you end up here. And then after that you go down quadratically, up to when? Would you continue going quadratically forever? No. What happens?


[User] Please transcribe this text

[Assistant] What is the thing, what is the change that would produce a different kind of behavior?


[User] Please transcribe this text

[Assistant] Triad region, yes, you can enter the triad region. So what is the condition for the triad region?


[User] Please transcribe this text

[Assistant] The condition for being in trial, or being in pinch-off, is that, being in pinch-off, the condition is that VDDS needs to be greater than VGS minus VT. So if we now draw a line,


[User] Please transcribe this text

[Assistant] The line of v out equals v in minus v t. If you are to the left of this line, you are in pinch off. So you are in pinch off here. Well, you are in pinch off here. You are off here. And you are in triad here. To the right of this line.


[User] Please transcribe this text

[Assistant] So once you pass this line, you're getting to try it. And you get another equation, if you apply the tried equation, and you can solve that too, if you really want to. But it behaves like that. And there's more of that in your handout. But,


[User] Please transcribe this text

[Assistant] Inverting gain stage. Now, the exact dependence of course is different.


[User] Please transcribe this text

[Assistant] Because you had an exponential, and you had a narrow, you know, the saturation region, you can have a wider triode region, so when you go here and the slope reduces, but essentially, you have a region where the slope increases, and then you have a region where the slope decreases. The absolute slope. Now, what is exactly the slope? One can calculate it. Right? So that's the gain.


[User] Please transcribe this text

[Assistant] Which is D V capital V out to D capital V in. And you look at all of that.


[User] Please transcribe this text

[Assistant] And the derivative of this thing is basically just, you get two times this. So you basically get negative rd mu n cox w over l vgs minus vt.


[User] Please transcribe this text

[Assistant] Now again, if you look at this and compare it with what we had before, here we recognize something that this term is simply what?


[User] Please transcribe this text

[Assistant] So what you have, really, is again, negative GMRD.


[User] Please transcribe this text

[Assistant] So what you get is negative GMRD, which is similar to negative GMRC.


[User] Please transcribe this text

[Assistant] We call it RD because it's connected to drain, and we called it RC because it was connected to collector. But it's the same resistor essentially, right? So fundamentally they are, from a small signal perspective, they are the same. In that regard.


[User] Please transcribe this text

[Assistant] So basically that means that the input resistance is open.


[User] Please transcribe this text

[Assistant] Input is open, and then you have a current source, gmvgs. You had the ro, and then you have


[User] Please transcribe this text

[Assistant] The RD. And this was the drain, this is the gate, and the source is connected here. And this is VGS. Lowercase VGS. Which would be VIN.


[User] Please transcribe this text

[Assistant] So again, if you do look at these two, almost by inspection you can see that you can expect the exact same result.


[User] Please transcribe this text

[Assistant] Then it produces a voltage at the output. This is your V out. That would be what? Would be


[User] Please transcribe this text

[Assistant] Almost exactly the same thing. So what you can see is that AV, the current, the voltage gain,


[User] Please transcribe this text

[Assistant] Which is defined as v out over v in. You can calculate, now this is a linear circuit.


[User] Please transcribe this text

[Assistant] So we can apply all the linear circuit techniques and what you get is negative gm rd parallel with rr.


[User] Please transcribe this text

[Assistant] Well, when you get the AV max, if you don't resort to negative resistances, it's negative GMRO again, right? So it's negative GMRO. Now, parametrically, then you have to see what these things are.


[User] Please transcribe this text

[Assistant] And we can just basically say for this kind of transistor, what is our gm?


[User] Please transcribe this text

[Assistant] Our gm was, for example, so now we can calculate it, right? We can say it's mu n, c ox, w over l, v gs minus v t, that's one way to write it. And what is r o?


[User] Please transcribe this text

[Assistant] So that for a long channel for a non-velocity saturated quadratic device, that's the thing. We can actually simplify it a little bit more, right? We can simplify this and see what we get. So, what do you get? Let's look at this.


[User] Please transcribe this text

[Assistant] This L and that L, so you have these W over L's, then you have the IDs.


[User] Please transcribe this text

[Assistant] The rest of them are constant, so the things that are non-constant, you have square root of w over l, so you have a square root of w.


[User] Please transcribe this text

[Assistant] Oh, let's say, let's try it this way. You have W to the power of negative, to the power of one half. You have L to the power of negative three thirds.


[User] Please transcribe this text

[Assistant] And then you have an i to the power of three-thirds. Three-halves. Three-halves. Right, so this is a proportionality. I mean, I've dropped the other parameters.


[User] Please transcribe this text

[Assistant] So this is times that. So this is, uhm, L effective divided by ID. Times DXD. DVDS. I was looking at it and I said, well this doesn't make sense.


[User] Please transcribe this text

[Assistant] And it's always important to look at your equations and see if they make sense.


[User] Please transcribe this text

[Assistant] Uhm, okay, so what do we have? So we have now w, square root of w and l, so if the proportionality becomes square root of w and l, and then you have over square root of


[User] Please transcribe this text

[Assistant] Do we agree with this? Let's double check, make sure we haven't made any mistakes. Or I haven't made any mistakes. So it's good now?


[User] Please transcribe this text

[Assistant] It tells us a couple of interesting things, and you have to really pay attention to.


[User] Please transcribe this text

[Assistant] So, one is that, what, how can I maximize the intrinsic gain of my transistor with my, through my design, according to this expression? What do I need to do? I need larger width of the transistor, right?


[User] Please transcribe this text

[Assistant] So when you make a larger width, basically you're making a larger transistor. It's like putting multiple transistors in parallel. That's not that surprising. Okay?


[User] Please transcribe this text

[Assistant] It sounds like it's weakening the transistor, right? When you make the transistor channel length longer, you're making it weaker because you're reducing the transconductance. But somehow, it's saying that increase the channel length.


[User] Please transcribe this text

[Assistant] And why is that? Because this dependence is square root, and this dependence is linear. So what that tells you is that for a square law device,


[User] Please transcribe this text

[Assistant] This dependence dominates the amount, the amount you gain by increasing the channel length, which basically translates to an increase in the output resistance, is stronger than the amount you lose by lowering your trans conductors for a given current.


[User] Please transcribe this text

[Assistant] Modulation, right? Channel length modulation means that what fraction of the channel length changes when you apply the change, the drain voltage. And if you make it a long channel, the fractional change would be smaller.


[User] Please transcribe this text

[Assistant] So of course, we didn't say anything about speed. This has significant implications on speed. All three of them. Now,


[User] Please transcribe this text

[Assistant] The most interesting one perhaps is this denominator.


[User] Please transcribe this text

[Assistant] So every time we write a bunch of equations, we have to stop and say, what does that tell you? What does that teach you? What do we learn from this equation? Or why does it tell us what it's telling us?


[User] Please transcribe this text

[Assistant] Make that transit, make the current smaller, right?


[User] Please transcribe this text

[Assistant] So use smaller and smaller currents. This sounds kind of like a little bit counterintuitive at first in me. Well, it's not totally counterintuitive again because


[User] Please transcribe this text

[Assistant] It tells you what you can do to make the whole thing blow up. What if my current is zero? It seems to indicate that for a current of zero, there is no limit on maximum gain I can get.


[User] Please transcribe this text

[Assistant] That's what it says, right? And that's what this equation says. Now whether or not that's true, that's a different question. And if it isn't, you need to understand why.


[User] Please transcribe this text

[Assistant] But according to this, the gain, the maximum gain of this MOS transistor


[User] Please transcribe this text

[Assistant] It's doing something out of nothing and say, well, the less current, the less power you consume, the larger you make your transistor, the more gain you get. The larger you make the transistor, the more gain you get. That's true.


[User] Please transcribe this text

[Assistant] So when you see something like this, what does that mean? In your equations. It means that we've made certain simplifying assumptions that were apparently too simplifying for what we are considering, right? That's what they mean. Or you have discovered something really cool and new and all those things. But it's more often the former, not the latter.


[User] Please transcribe this text

[Assistant] What happens to operation point of the transistor from a device physics perspective?


[User] Please transcribe this text

[Assistant] In terms of Vgs- let's go back to the current voltage dependence, right? Let's go back to, well, this equation. Let's just forget about this part, just to keep this part.


[User] Please transcribe this text

[Assistant] What would happen if your current is becoming smaller and smaller? What does it say about the gate overdrive? VGS minus VT. It's called gate overdrive. It becomes smaller and smaller. So in the limit, when the current is almost zero, what is VGS minus VT? Zero. So,


[User] Please transcribe this text

[Assistant] Now we said, if this looks like this, and it comes, it's zero up to here, and then it keeps going up.


[User] Please transcribe this text

[Assistant] But this was an area where we had already made an approximation in this, in this region.


[User] Please transcribe this text

[Assistant] We had made an approximation, we said we are ignoring something in the simplified model. What was that thing?


[User] Please transcribe this text

[Assistant] Subthreshold current, right? Weak inversion.


[User] Please transcribe this text

[Assistant] Because we assume that there is no current below, we can, below threshold. And now that assumption is coming back to haunt us.


[User] Please transcribe this text

[Assistant] That's why we needed to understand that assumption. So when it comes back to haunt us, we can go back and fix it.


[User] Please transcribe this text

[Assistant] Well, you remember we also did a little derivation of what the-


[User] Please transcribe this text

[Assistant] What was the, well not the collector current, the drain current versus VGS. We did a little derivation of, it depended, it was proportional to what? In the subterrential. Exponential.


[User] Please transcribe this text

[Assistant] Or nkT, and this was Q something, right? So this was, this was, the way, the way we wrote it was we wrote that Q vgs over n.


[User] Please transcribe this text

[Assistant] Okay, let's write it that way because the two VTs involved here, so that's really a mess.


[User] Please transcribe this text

[Assistant] Right? You remember that. Right? Hopefully, you remember that. Now, if you have a current,


[User] Please transcribe this text

[Assistant] It was exponentially dependent on this. We have already, we know what the transconductance is. So this is not the right expression for transconductance.


[User] Please transcribe this text

[Assistant] It's DID-DVGS, which we already calculated.


[User] Please transcribe this text

[Assistant] It's like GM of a bipolar transistor, right? So GM here, which would be DID, DVGS,


[User] Please transcribe this text

[Assistant] Right? And if you have that expression, and if you plug it into here, what happens to the ID dependence? If this was the expression we use for GM.


[User] Please transcribe this text

[Assistant] You can see that now what happens, the ID dependence disappears.


[User] Please transcribe this text

[Assistant] So at small currents, the ID dependence, is it clear that the ID dependence disappears? So what happens in reality is that, yeaah, this is true.


[User] Please transcribe this text

[Assistant] And at some point, you will just get a constant, because of subthreshold.


[User] Please transcribe this text

[Assistant] You can actually do it at very low currents, which also had the added benefit of being very low current, or low power.


[User] Please transcribe this text

[Assistant] So, if you have something that doesn't have to operate fast, and needs gain, and it can be low, once you want it to be preferably low power, then sub-threshold operation is preferable in terms of what you're trying to achieve.


[User] Please transcribe this text

[Assistant] Because it would allow you to get the maximum gain that you can out of a device.


[User] Please transcribe this text

[Assistant] And if you had a velocity saturated one, then what happens is another interesting thing happens. So if you had a velocity saturated one, let's say deep velocity saturated, then your GM would become, so let's show it with a different color. So velocity saturated.


[User] Please transcribe this text

[Assistant] GM now becomes what? The derivative of this, so it basically becomes mu n C ox over 2 W e sat. And if I plug that in here,


[User] Please transcribe this text

[Assistant] Instead of that, you will see that the L dependence is still there.


[User] Please transcribe this text

[Assistant] You can, and then what, the other thing that you have is that, yes, your w depend, the w dependence becomes linear, so in velocity saturated region, you have a w here, and then you have an l here. Okay? So we come proposed, so this is velocity saturated.


[User] Please transcribe this text

[Assistant] So the game, maximum game becomes proportional to these two. And then, what else do you have? You don't have that ID dependence, right?


[User] Please transcribe this text

[Assistant] So you cannot gain that much by playing with the ID. In, in there.


[User] Please transcribe this text

[Assistant] So it's basically you still can play this game, same game, until you get to the subthreshold.


[User] Please transcribe this text

[Assistant] Okay, so basically you have that dependence, but again, you will be limited by the sub-threshold, the same way that you were limited here.


[User] Please transcribe this text

[Assistant] Now why do we do these things? Because as we design circuits, we will see that we need to make decisions and choices in terms of operation of things.


[User] Please transcribe this text

[Assistant] So, so that's for that. So now we sit at the very beginning of this discussion that we'll talk about this


[User] Please transcribe this text

[Assistant] ATD. The arbitrary three terminal device.


[User] Please transcribe this text

[Assistant] So let's look at that. Let's just say what, what is the commonality and what would happen in general in any three terminal device that has some of these general problems. So what we're assuming is that there's some level of monotonicity.


[User] Please transcribe this text

[Assistant] So this would be I2, terminal 2, versus V21 for different values of V31.


[User] Please transcribe this text

[Assistant] So you can see, one is like the source, or like the emitter. Two is like the drain, or the collector. And three is the controlling terminal. The third terminal, the controlling terminal. Which is the gate, or the base, or whatever you want.


[User] Please transcribe this text

[Assistant] So you have those kind of things. So now what let's say, let's see what we can say. Well we don't know for an arbitrary transistor what we know is that ID in general is a function of V1, V31 and V21.


[User] Please transcribe this text

[Assistant] Now, and we know that it's a stronger function of V31 than V21, ideally. It doesn't have to be. So if it is, it's a good device. And if it isn't,


[User] Please transcribe this text

[Assistant] It's a problematic device, and well, it would be a less useful device, it's not a completely useless device, but it would be more challenging. So, we say, if we could assume that it's only a function, primarily a function of v3,1,


[User] Please transcribe this text

[Assistant] It would have certain properties, but we don't have. So, for this thing, we can always define a GM. GM is defined as DID.


[User] Please transcribe this text

[Assistant] Or awesome, not ID. So this is I2.


[User] Please transcribe this text

[Assistant] Of the terminal 2 with respect to variations of terminal 3 with respect to the common terminal, terminal 1. Which is a control. So the rate, the fluctuations of the controlled current to the controlling voltage. Right? That's GM. You can define. Now, what are, what is its exact, exact parameters? Well, it would depend on this functionality. Whatever it is.


[User] Please transcribe this text

[Assistant] So you have a new functionality, you have a new device, you can still define a GF.


[User] Please transcribe this text

[Assistant] The same way that you did it for JFET and you did it for, you know, vacuum tube and things of that sort, right? So, and then, there will be an RO.


[User] Please transcribe this text

[Assistant] Similarly, right, some output resistance, which is a measure of the slope of these lines.


[User] Please transcribe this text

[Assistant] So that would be by definition what the RO is. Again, it would depend on the parameters or functionality. You may not even know the parameters. You may just have a curve.


[User] Please transcribe this text

[Assistant] You may not have an expression. You may not have an analytical expression for it.


[User] Please transcribe this text

[Assistant] So there's nothing fundamental about any of these things, which is good or bad, right? I mean, which, which, the more details of this thing you know, the more meaningful statements you can say, make about what design parameters will affect what. If you treat everything as a black box,


[User] Please transcribe this text

[Assistant] Yes, you can still do these things with just partial derivatives. And you can always analyze things. You may not be able to design very well.


[User] Please transcribe this text

[Assistant] So this is what you get in parallel. I mean, we can't go beyond that, right? I mean, it may have different regions of operation, et cetera, et cetera, but it is what it is. And now if you come to this picture,


[User] Please transcribe this text

[Assistant] So this is, now we're still doing this for ADT, ATT, sorry, uh, arbitrary three terminal device.


[User] Please transcribe this text

[Assistant] You take terminal 1, connect it to ground, and you apply a voltage consisting of an AC and a DC port. Uhm, so you have a V in, and then you have a V in.


[User] Please transcribe this text

[Assistant] The DC part and AC part apply to this device and you have a V out.


[User] Please transcribe this text

[Assistant] And if you want to find the gain of this thing, so this would be V out versus V in. Because the load line characteristic is still the same, right? Why? Because if you write this expression that we had before, you still can write this. V out is V, whatever, let's call this V22.


[User] Please transcribe this text

[Assistant] And whatever that functionality is from that function above, so you can actually write this as v22 minus r2 f of


[User] Please transcribe this text

[Assistant] And if you just ignore that first part, if you could, and if you don't, if you can't even, you can solve it for both ways. So let's even write it the most general way. V2-3, V2-1, V3-1, and some functionality on V2-1.


[User] Please transcribe this text

[Assistant] And this is also V2,1. So V2,1 becomes V out. Right? And V3, what is V3,1?


[User] Please transcribe this text

[Assistant] In this picture. It's V in. Right?


[User] Please transcribe this text

[Assistant] Either graphically or numerically or, you know, there are all sorts of methods you can solve this with. Or if you know the analytical expression, analytically.


[User] Please transcribe this text

[Assistant] And if you want to find the gain of this thing, by definition, the AV,


[User] Please transcribe this text

[Assistant] I mean, if you ignore that Vout part, it becomes just a GM.


[User] Please transcribe this text

[Assistant] This is RO, this is RO, and likewise, looking down into here, you see the RO. And looking up here, of course, you see R2.


[User] Please transcribe this text

[Assistant] So it's a parallel combination of those two resistances, a small signal resistances that appear on that node. Here you see Rd.


[User] Please transcribe this text

[Assistant] Here you see RC looking up from a small signal perspective. Now, and again, if you take all of this, and try to bring it to this, with this picture,


[User] Please transcribe this text

[Assistant] Again, exactly the same model. I mean, so you may have some input dependence, some input resistance, depending on the current nature, the nature of the current that goes into terminal 3. So you have v, you have some sort of an r-pi, or you can call it r31, and there is a v31, there is a transconductance here, gmv31,


[User] Please transcribe this text

[Assistant] And then there's some RO. So this is terminal 3. This is term- I'm sorry, this is...


[User] Please transcribe this text

[Assistant] Terminal 2, this is terminal 3, and this is terminal 1.


[User] Please transcribe this text

[Assistant] Obviously from this, you arrive at the same conclusion.


[User] Please transcribe this text

[Assistant] Skirmish, skirmish, across the border. Uhm, so I have a a v max.


[User] Please transcribe this text

[Assistant] And then you have, so basically when the load is maximum, when you basically have a current source as the load, and that's what you get. GMRO.


[User] Please transcribe this text

[Assistant] Which would be, I mean, if you know these two derivatives.


[User] Please transcribe this text

[Assistant] divided by that derivative. So it's basically, you can write that as


[User] Please transcribe this text

[Assistant] So it's maximum, so we don't need the negative sign. We need, we need the D, I, or consider that even F. So it's D, F, D, V,


[User] Please transcribe this text

[Assistant] Right? So, in general, the maximum intrinsic gain of a transistor


[User] Please transcribe this text

[Assistant] And now how it behaves and what it depends on and the parameters that depends on the physics. This is just a mathematical abstraction.


[User] Please transcribe this text

[Assistant] So, a lot of this analysis that will follow, a lot of the small signal models that we will use later on, are applicable to essentially all devices. And it should not be too concerning or too limiting to you.


[User] Please transcribe this text

[Assistant] If you see, we use examples, we use bipolar examples, mosfet examples, this and that.


[User] Please transcribe this text

[Assistant] You should be able, your expectation is that you will need the people who can...


[User] Please transcribe this text

[Assistant] Transcend the specific of the device as well. You understand it deeply and always work.


[User] Please transcribe this text

[Assistant] And the specifics would depend on the physics and the design. And it's important to understand the specifics because if you want to make something useful out of it.


[User] Please transcribe this text

[Assistant] That's what will come into play in the trade-offs and decisions that you will make.


[User] Please transcribe this text

[Assistant] So, we did basically cover multiple things really here, but I think it's important for you to see the similarities and the differences among these devices. Any questions?


[User] Please transcribe this text

[Assistant] I don't want to delay our journey, but you three can defeat demons. We must report to the palace.


[User] Please transcribe this text

[Assistant] Talk a little bit about an example of a potential problem or typical scenario. So let's say you're trying to make an integrated radio receiver.


[User] Please transcribe this text

[Assistant] You're trying to make a chip that goes into your phone or your device and then use it. It's a receiver that receives signal from the communication signals from the cell towers or radio or something, right? What is the characteristic of these signals, the radio signals that you receive?


[User] Please transcribe this text

[Assistant] They are very small, right? Why they are very small? Because they are going over a large distance, they experience r over, you know, one over r squared losses and things like that. So you have a very small signal coming in, let's say arriving at your antenna.


[User] Please transcribe this text

[Assistant] And then you are going through some traces, you are going through your chip, and then you are having an amplifier. So first thing you do is an amplifier, a low noise amplifier, LNA.


[User] Please transcribe this text

[Assistant] And we talk about these later, uh, in terms of RF. But then, let's say on the same chip you have some digital circuitry, which is very common nowadays. We do a lot of things with digital. And then there's a digital line passing by here. So this is a clock, let's say. A digital clock passing by. So you have a weak signal arriving in your antenna, and you have this big digital clock next to it.


[User] Please transcribe this text

[Assistant] And it somehow couples to it. Somehow, maybe it electrostatically couples to it. Maybe it magnetically couples to it.


[User] Please transcribe this text

[Assistant] Or maybe it couples through it through the substrate, the silicon substrate or something like that. So it couples through it somehow. Right?


[User] Please transcribe this text

[Assistant] What's gonna happen to the signal? So you have this weak signal coming in. Let's say, let's say it looks like that. And then you have this big signal next to it. What's gonna, what is that going to do to your signal?


[User] Please transcribe this text

[Assistant] You will see coupling, right? For example, with this thing, if it's capacitive coupling, you will see spikes like that.


[User] Please transcribe this text

[Assistant] Coming every time the clock edge switches, you will see something like that. And of course, this is not a highly desirable thing.


[User] Please transcribe this text

[Assistant] Because now your signal has this undesirable element on it. As you amplify it, this will get amplified and, you know, it goes through the system and it can saturate your system. It can create all sorts of noise and errors. If it's digital, things of that sort, right?


[User] Please transcribe this text

[Assistant] So this is obviously not a good scenario. So what do we do? What can we do when you have a problem of this sort? And this is a general statement. When you have coupling problems, there are essentially three things you can do. So what are those three things? If you have undesirable signal bothering you. Right? If there's some annoying person


[User] Please transcribe this text

[Assistant] What do you mean by filter? So let's talk about filter. So I think you showed a different picture, maybe you can start with that, we'll come back to filter in a second. So one of you guys did this. So one is that make yourself less sensitive to it. Right? Reduce your sensitivity. Put earmuffs on, right? Noise cancelling headsets. Annoying friends cancelling headsets.


[User] Please transcribe this text

[Assistant] So, so that's one thing you can do. You can reduce your sensitivity. The other, what is the other thing that you can do? You can tell the person, say, well,


[User] Please transcribe this text

[Assistant] Reduce your sensitivity, reduce the source, or the third thing, which may have been essentially the kind of combination of other, reduce the coupling. Go sit somewhere else. Put a wall between you. Right? So reduce the coupling, reduce the signal source, or reduce your sensitivity to it.


[User] Please transcribe this text

[Assistant] I can't think of anything fundamentally different. Now, when you say filtering, it's an interesting statement. You mentioned something about filtering earlier. Now, it's interesting because filtering implies that the signal that's bothering you is at a different frequency.


[User] Please transcribe this text

[Assistant] Then, then, what, what, what, what you're trying to operate. Now, if it's out of band, if it's in a different frequency, there are filtering techniques that you can play. But if it's inside your band, then the filtering tricks do not work as well. You need to do something more sophisticated.


[User] Please transcribe this text

[Assistant] There are other techniques, sophisticated techniques, filtering techniques that can help. But essentially this boils down to this. So now, think about it. How can we reduce our sensitivity to this signal?


[User] Please transcribe this text

[Assistant] And how can we reduce the impact of the source? So one of the things you can say, well don't draw this line here, draw it farther away. Sure. I mean, it would reduce the coupling a little bit.


[User] Please transcribe this text

[Assistant] But it won't eliminate it. And if you have something that's quite sensitive, yes, these spikes may not be this large, they may become like half as large, but still, they're large.


[User] Please transcribe this text

[Assistant] What do we do? How do we reduce our sensitivity? How can we think, can you think of ways of reducing sensitivity and reducing the source? If they are located in a given location.


[User] Please transcribe this text

[Assistant] So the amount of coupling, let's say you've done all you could have done about reducing the coupling. Directly by placement and things of that. So now it's still not enough. What do we do? Can you think of a way? Yes.


[User] Please transcribe this text

[Assistant] You can subtract a copy of the clock. So the idea here is that you're generating this clock. You might as well generate the opposite phase of this clock.


[User] Please transcribe this text

[Assistant] And, well, I should probably use a different color, but let's use a different one.


[User] Please transcribe this text

[Assistant] And pass it exactly the same way around this thing so I have the same sensitivity. Alright.


[User] Please transcribe this text

[Assistant] So you have two of these that are coupling the opposite amounts in the opposite direction into the signal. And you have to make sure that these are well balanced, right? This coupling. So you have to make this clockwise. Yeah, this could work.


[User] Please transcribe this text

[Assistant] This could help to some extent. But you can imagine, you can start seeing that it's a little bit kind of a challenging thing to do in a systematic way. First of all, it means that any time you're routing a clock, you're routing two lines instead of one. So your routing becomes more complicated. You have to drive two lines. And then you have to make sure that it remains symmetrical. So it's a good idea.


[User] Please transcribe this text

[Assistant] And it does some improve, gives you some improvement, but let's say for some reason you don't want to carry two lines of clock, or you don't want to do things of that sort. So, but, but what else can we do? Yes?


[User] Please transcribe this text

[Assistant] So you can create two couplings that cancel each other, right? That's what you're saying. You're saying you're using except second coupling to cancel the first coupling.


[User] Please transcribe this text

[Assistant] In principle, that's possible. In practice, it requires a lot of detailed work. Because then, you're not talking about one or two lines.


[User] Please transcribe this text

[Assistant] You're talking about a big chip. There are lots of digital lines going everywhere. You know, it's not just one line, there are lots of data lines going across that are not even periodic, so everything needs to be modified that way. So it's a major overhead and also...


[User] Please transcribe this text

[Assistant] Difficult analytical thing to do on a large number of things. Maybe if it's one line, yes. So what else can we do?


[User] Please transcribe this text

[Assistant] Have a separate ground plane with analog and digital. That helps, but that doesn't address this problem.


[User] Please transcribe this text

[Assistant] Right? The ground planes would be basically, if the coupling is happening through the ground planes and all those things, that's correct. But I'm saying that this coupling, I'm just saying that it's more direct. That's a good...


[User] Please transcribe this text

[Assistant] That's a good design practice, that's a good practice in design, but it's to address a slightly different problem. So, what else? Can we do anything else?


[User] Please transcribe this text

[Assistant] Shield it. Well, which is basically reduce the coupling, right? So you put another metal layer here.


[User] Please transcribe this text

[Assistant] Measure against the copy of the coupling, so active cancellation.


[User] Please transcribe this text

[Assistant] Right? So these are all good ideas. So, you know what, if you know what you're coupling into it, you can try to actively subtract it till it cancels, right?


[User] Please transcribe this text

[Assistant] Take it differentially as, as two opposite versions. So instead of, so one way, for example, let's say if you have a patch antenna, you can take the output from both ends of this thing. Right?


[User] Please transcribe this text

[Assistant] You can, there are many different ways. If you have a dipole antenna, you can take these two inputs.


[User] Please transcribe this text

[Assistant] And these two inputs will be in opposite phase. Now, in general, take the signal and break it into two equivalent parts that are in opposite phases. And carry those.


[User] Please transcribe this text

[Assistant] So let's see what happens if you do that. So let's say, now instead of doing this, you're gonna have two lines. So my antenna is a differential one.


[User] Please transcribe this text

[Assistant] So I take my signal, let's say vn, and divide it into


[User] Please transcribe this text

[Assistant] And I have two signals, let's say V1 and V2.


[User] Please transcribe this text

[Assistant] Whose difference is what I'm looking at. So what if I look at the difference of the signal? In other words, define Vd as V1 minus V2.


[User] Please transcribe this text

[Assistant] What would that do in this scenario? In a coupling scenario?


[User] Please transcribe this text

[Assistant] Well, you have this line passing by, and it couples to both of them.


[User] Please transcribe this text

[Assistant] And it would induce the same kind of spikes on both of them. So what you have, let me just do a click better.


[User] Please transcribe this text

[Assistant] Drawing. So let's say this is your one of the signals and this is the other one. So let's say you have


[User] Please transcribe this text

[Assistant] Two signals that are going out of phase, but then you have a coupling signal, right? So you have spikes here, and spikes there, you have spikes here, and spikes there, and spike here, and spike there, so on and so forth, right?


[User] Please transcribe this text

[Assistant] I'll see something like that. With very little interference.


[User] Please transcribe this text

[Assistant] So, what if I could carry all of my signals in differential form, or basically a difference of two different things? For that to happen, I need to have an amplifier that can amplify these things in a pair.


[User] Please transcribe this text

[Assistant] We can do a variety of different versions, so we'll.


[User] Please transcribe this text

[Assistant] Let's say I have two signals coming in that are going in opposite phases and I want to amplify them and keep them differential. So what do I do? Well, you can make an amplifier here, right? You can make an amplifier like this. Sorry.


[User] Please transcribe this text

[Assistant] Make, let's make a common base, common emitter amplifier, or common source amplifier in the MOSFET. Let's say we had something like this. So this is RD, so this is V1, uhm, or VI, and then this is V, uh, VIN1, I1, and VO1. Right?


[User] Please transcribe this text

[Assistant] And then I can actually apply the next one, this one to another amplifier, just like that.


[User] Please transcribe this text

[Assistant] Now, sure, I can do that. Did I solve the problem by doing this? Did we solve the problem? In other words, if there were spikes here, let's say,


[User] Please transcribe this text

[Assistant] No, right? Because the spikes will go through the amplifiers, probably they will saturate the amplifier now, because this spike is now going big here, and this spike is going big there, and then this spike is just messing this up here, and this is gonna mess it up here.


[User] Please transcribe this text

[Assistant] Right? Intuitively, kind of like just qualitatively looking at.


[User] Please transcribe this text

[Assistant] So we haven't solved this problem, we just amplified both of them, and yes, you can say, oh, you can look at the difference of these two still would be good. It would be good if these were amplifying linearly.


[User] Please transcribe this text

[Assistant] Right? If there's some nonlinearity, then the subtraction is not going to work anymore.


[User] Please transcribe this text

[Assistant] So I haven't really improved the situation. I have created two lines, but I don't have a good way of amplifying. So what are we looking for? What is the quality of behavior of the stage we need to do this thing? We need something that amplifies this, I don't know, blue thing, the signal, and actually suppresses the red one.


[User] Please transcribe this text

[Assistant] That would amplify the blue and attenuate the red.


[User] Please transcribe this text

[Assistant] So, what is different between the blue and the red? What is, what, how, so if this was V1 and V2, right? Blue is captured as V1 minus V2. We show it as VD. V differential.


[User] Please transcribe this text

[Assistant] There's another component, we can call it VC, the common mode, which is V1 plus V2 over 2.


[User] Please transcribe this text

[Assistant] So, if I just look at v1-v2, what do I see? Do I see less of the red or more of the red? Less of the red, right? I see more of the blue. So this one is gonna generate more of the blue. And this one is gonna be more of the red.


[User] Please transcribe this text

[Assistant] So we want to make an amplifier that amplifies this and suppresses that.


[User] Please transcribe this text

[Assistant] Heights. Let's say the two ends of this, this thing represents the, they represent the voltages, V1 and V2, right? So this is V1, let's say this is V2. Right? For the blue stuff, what's happening? One side goes up, the other side goes down, right? You see a seesaw operation. For the red stuff, what do you have, what do you see? You see a common mode operation.


[User] Please transcribe this text

[Assistant] So, what happens is that when these two inputs go up, you get gain. Why do you get gain? What is a gain operable mechanism?


[User] Please transcribe this text

[Assistant] As this voltage goes up, your VBE is going up, your current is going up, you draw more current through the resistor, right, and therefore you produce, you lower the voltage, you get amplification through the exponential.


[User] Please transcribe this text

[Assistant] How can we enforce that? How can we enforce that if both sides are going up together, the current is not gonna increase. The current of either one of them is not gonna increase.


[User] Please transcribe this text

[Assistant] Have a common current source, right? So if they share the current source, if I take these two, and put them side by side,


[User] Please transcribe this text

[Assistant] Then what's gonna happen if both inputs go up? To the total current in the collector. So, let's say I take these two and push them up together. If this is an ideal current source. Is, are these two, the sum of these two currents have to be IEEE.


[User] Please transcribe this text

[Assistant] This voltage goes up, but this current does not increase, so this current is gonna be half of that current, the emitter currents are gonna be half of that, and this is gonna be alpha times half of that.


[User] Please transcribe this text

[Assistant] So that current is not going to change. So is there going to be a change on this voltage? No. Right? Because this terminal goes up


[User] Please transcribe this text

[Assistant] The total current remains the same, therefore the current through the resistors are the same.


[User] Please transcribe this text

[Assistant] And this voltage is simply, what is this voltage? It's basically VCC minus RC times IC. If IC doesn't change, that voltage is not gonna change. So what does it do to the red stuff? It doesn't amplify it, right? It actually attenuates it. How about the blue stuff? Does it do anything to the blue stuff?


[User] Please transcribe this text

[Assistant] Well, if this side goes up and this side goes down, right?


[User] Please transcribe this text

[Assistant] So the current is steered in this direction, and as a result, this voltage will go down and this voltage will go up.


[User] Please transcribe this text

[Assistant] Differential gain and common mode attenuation.


[User] Please transcribe this text

[Assistant] That's why this stage is made instead of this stage. This is the fundamental understanding of it. Basically, this current source is critical.


[User] Please transcribe this text

[Assistant] If you make this an ideal current source, this would be completely suppressing.


[User] Please transcribe this text

[Assistant] You get a little bit more current because there will be some extra current going through that resistor. So you will get some amplification of the red stuff.


[User] Please transcribe this text

[Assistant] So the quality of that current source would determine your common mode rejection ratio.


[User] Please transcribe this text

[Assistant] And these are the insights you need to have to develop and expand on before you write equations.


[User] Please transcribe this text

[Assistant] And also you need to understand the thought process of how you arrive at this stage. This stage was first invented in vacuum tubes actually. Was used in vacuum tubes, the differential pair.


[User] Please transcribe this text

[Assistant] And of course it has been reused, and it's perhaps the best, the greatest of all, uh, of the two transistor stages because it has a lot of interesting and good properties. But this is the most essential property that you're talking about. So, something like this,


[User] Please transcribe this text

[Assistant] Would be very interesting to work with. But let's now analyze it. Let's actually write equations and see how it behaves. We can do it for bipolar and then we'll do it for MOSFET next time.


[User] Please transcribe this text

[Assistant] So you have something like this. Let's, let's understand how it does. So there are really two things in this. There is a differential operation and common mode. Let's start with the differential operation because that's really the desired operation. So let's say you have an input that's V1. So you have two voltages, V1 and V2, right?


[User] Please transcribe this text

[Assistant] You can always decompose them into VD and VC, right? Because you can write these two things.


[User] Please transcribe this text

[Assistant] For any arbitrary two voltages, you can write a common mode part and a differential part, defined as these.


[User] Please transcribe this text

[Assistant] And therefore you can actually express them in those forms. So this is actually equivalent, and we'll see it in a second.


[User] Please transcribe this text

[Assistant] What does it generate as, so look, let's look at what you get on this terminal. You get VC plus VD over 2, right?


[User] Please transcribe this text

[Assistant] If you calculate VC plus VD over 2, what do you get?


[User] Please transcribe this text

[Assistant] Well, vd over 2 is v1 minus v2 over 2. And vc is v1 plus v2 over 2. So you add them, you get just v1 over 2 plus v1 over 2. So this becomes v1.


[User] Please transcribe this text

[Assistant] And the same way, vc minus vd over 2 is gonna be v2. Very simple, you can solve for them. In terms of vc and vd. Right?


[User] Please transcribe this text

[Assistant] Uhm, so you can think of any two arbitrary sources, really, as a combination of a differential part and a common part.


[User] Please transcribe this text

[Assistant] And somehow you maintain the base bias. If this bothers you, you can keep it this way.


[User] Please transcribe this text

[Assistant] So, I'm trying to find what's the relationship of the output currents, let's call this IC1 and IC2, to the VID.


[User] Please transcribe this text

[Assistant] Even more interestingly, I'm interested really in Delta IC.


[User] Please transcribe this text

[Assistant] So what we are trying to do is to define, find the functionality of, we want to show this as a function of VID.


[User] Please transcribe this text

[Assistant] And we'll do the large signal one first, and then we'll get to the small signal later. But let's do the large signal. Let's do the nonlinear analysis first, because it's worth doing it here to see how it behaves.


[User] Please transcribe this text

[Assistant] So, what do we get? Well, I need to understand what IC1 and IC2 are, right? So, what do I know about this bipolar transistor? I know that this is a VBE1 and this is VBE2.


[User] Please transcribe this text

[Assistant] So I know, we know that VID is VBE1 minus VBE2, right? KCL around this loop.


[User] Please transcribe this text

[Assistant] This voltage minus that voltage is VID, the difference between these two. Okay? Alright, so that's VID.


[User] Please transcribe this text

[Assistant] And then, but what do I know about VBE1 and VBE2? How are they related to IC1 and IC2? Well, the exponential transistor behavior, right? It's the relationship between the IC and VBE.


[User] Please transcribe this text

[Assistant] Isn't, isn't it true that VBE1 is VT natural log of IC over IS?


[User] Please transcribe this text

[Assistant] This is another way of writing this. IC is ISE to the VBE over VT. I'm ignoring, we are ignoring the, uhm, early effect for now.


[User] Please transcribe this text

[Assistant] Right? So that's the exponential, this is the inverse of the exponential. So this is IC1.


[User] Please transcribe this text

[Assistant] Over that. And then VBE2 is gonna be VT natural log of IC2 over IS.


[User] Please transcribe this text

[Assistant] So VID is gonna be VT natural log of this divided by that, the argument.


[User] Please transcribe this text

[Assistant] Which would be basically IC1 over IC2. Now there's an assumption that we've made here, when we wrote this. Essentially when I cancelled the ISS. What assumption did I make?


[User] Please transcribe this text

[Assistant] Right? That these two transistors have the same IS. Which means that they have to behave very exactly the same. So the matching of these two transistors matters. If they're mismatched.


[User] Please transcribe this text

[Assistant] It would be a problem, and we'll talk about this later in more details. The matching of these two is important, because otherwise this could not be done.


[User] Please transcribe this text

[Assistant] So of course you can take this and you can invert it so you can write it as IC1 over IC2 equals


[User] Please transcribe this text

[Assistant] Is this the only thing I know, IC1 over IC2 is exponential? So I know that the ratio of IC1 to IC2 is an exponential. Right? What else do we know about IC1 and IC2? Do you know another relationship between IC1 and IC2? Well let me ask you a question. Go ahead. Go ahead. There was somebody.


[User] Please transcribe this text

[Assistant] But we know the relationship between IE1 and IC1 if the transistor is in forward active region. Right? Which is what? How are they related?


[User] Please transcribe this text

[Assistant] Alpha, exactly. So this is basically just divided by alpha or in another way basically saying IC1 plus IC2 is alpha IEE. So now you have two equations for these two.


[User] Please transcribe this text

[Assistant] Right? So you can actually find what IC1 and IC2 are in terms of EID. So, for example, if I factor IC2 from this thing, so I get IC2 times 1 plus IC1 over IC2.


[User] Please transcribe this text

[Assistant] Right? And I know this is alpha IEE.


[User] Please transcribe this text

[Assistant] Is going to be, and this term, we already know it's exponential, so it becomes IC2, 1 plus,


[User] Please transcribe this text

[Assistant] So from this we can calculate IC1 and IC2. So IC2 from this is simply alpha IEE divided by 1 plus X from E to the VID over VT and likewise you will get an IC1


[User] Please transcribe this text

[Assistant] And if you plot these, what you will see, if you actually look at it versus VID,


[User] Please transcribe this text

[Assistant] It's relatively straightforward to plot this, right? Because what you have


[User] Please transcribe this text

[Assistant] So when VID is zero, there's no differential voltage, what do you get? This exponential is one, this is one, so you get one half, one half. So you're gonna be at half of alpha IEE.


[User] Please transcribe this text

[Assistant] Now as you increase VID, let's say you make it positive, this denominator becomes larger, right? And therefore IC2 becomes smaller.


[User] Please transcribe this text

[Assistant] And IC1 becomes larger. So this is IC2, let's say, and then let's say this one is


[User] Please transcribe this text

[Assistant] If the VID is 2, let's say 3 VTs, you have e to the power of 3. Right? It's a pretty large number, it's greater than 10.


[User] Please transcribe this text

[Assistant] So 3T is about 25 mV, right? In bipolar. By the time you're at 75 mV, maybe 100 mV, you'll almost completely switch to one side or the other. So it switches quite rapidly. If you think about it. Now,


[User] Please transcribe this text

[Assistant] These are good, these are useful, but we still haven't calculated the key, one of the key things. Which is, what we wanted to calculate was the difference between the ICs. We wanted to calculate the delta IC. Which was IC1 minus IC2. Right?


[User] Please transcribe this text

[Assistant] So we need to do a little bit of an algebra, right? So what we need to do, we take these two. So we have a common part, alpha IEE.


[User] Please transcribe this text

[Assistant] And then what is that? Well, there's a little bit of manipulation we need to do. I will factor half of this and half of that from the denominator. So when we do that, you get alpha IEE, okay, and then half of that would be


[User] Please transcribe this text

[Assistant] Plus e to the positive vid over 2vt. So this was the worst part.


[User] Please transcribe this text

[Assistant] Right? So factor that out. So we are going to have alpha IEE divided by


[User] Please transcribe this text

[Assistant] e to the positive plus e to the negative.


[User] Please transcribe this text

[Assistant] of that thing. So what we are left with, this is gonna go here, this is gonna go here, so it's gonna basically be


[User] Please transcribe this text

[Assistant] I'm gonna put them in the denominator. So you get e to the positive here, minus e to the negative here, times alpha IEE. When I say negative, I mean negative VID over 2VT. I'm not just showing it. Uhm, and this has a name. This function, right? What is it? Yea, hyperbolic tangent.


[User] Please transcribe this text

[Assistant] Hyperbolic tan of VID over 2VT.


[User] Please transcribe this text

[Assistant] Large signal describing, the function describing this, is going to be a hyperbolic tank. So if I want to see actually what is delta V out, the differential voltage, Vo2 minus, well Vo1 minus Vo2,


[User] Please transcribe this text

[Assistant] I have to write each one of them in terms of ICs, right? So I can write this as VCC, the first one, is VCC minus, uhm,


[User] Please transcribe this text

[Assistant] And then what you have, what you're left with is RC. So you get negative RC, delta IC.


[User] Please transcribe this text

[Assistant] Which is gonna be given by that, so it becomes negative alpha R C I E E.


[User] Please transcribe this text

[Assistant] The relationship between the differential output voltage and the differential input voltage is given by that.


[User] Please transcribe this text

[Assistant] Now, that you can actually show, again, hyperbolic tan behaves like that.


[User] Please transcribe this text

[Assistant] So if you look at the differential input, you can call it VOD, it behaves like this.


[User] Please transcribe this text

[Assistant] And at some point, basically after 304KT, it basically saturates.


[User] Please transcribe this text

[Assistant] Now, what we'll do next is to look at this large signal behavior and look at the small signal equivalent of this circuit. And we'll see different ways to analyze it and simplify it to think about the differential stages. These differential stages are very critical to a lot of design.


[User] Please transcribe this text

[Assistant] For a variety of reasons. One is that because you have the differential signaling and suppression of the common mode signals so you can actually get a lot less sensitivity and this is important if you are putting a lot of stuff very close to each other on a chip. The other thing that's very interesting is that these stages are very conducive to being cascaded without having to have DC level shifters.


[User] Please transcribe this text

[Assistant] So there's another strength of this differential signalling. And that enables you to basically make a cascade of these differential signals and have them do certain kinds of operations.


[User] Please transcribe this text

[Assistant] So we'll use them a lot, we'll deal with them a lot, we'll have a lot of variations of that, but that's the core essential behavior of the bipolar one. We'll do a similar thing for MOSFET.


[User] Please transcribe this text

[Assistant] Journey to the West, Chapter 95, The Raptor Demon.


[User] Please transcribe this text

[Assistant] The Tang Monk and Wu Xing were tied up inside the Demon's Cave.


[User] Please transcribe this text

[Assistant] Bajie was also tied up, floating in the tub of salt water again. The three demons talked in another room.


[User] Please transcribe this text

[Assistant] Said the lion. He felt his broken tooth.


[User] Please transcribe this text

[Assistant] He saw two fiends coming down the road.


[User] Please transcribe this text

[Assistant] He hid behind a rock and waited for the fiends to pass by.


[User] Please transcribe this text

[Assistant] They ate the Tang Monk, and now they'll live forever, said the first fiend. He sighed. It would be nice to live forever.


[User] Please transcribe this text

[Assistant] Said the other fiend. The two fiends continued along the road. Wukong was heartbroken, tears poured from his eyes.


[User] Please transcribe this text

[Assistant] Finally, he stood up and wiped away the tears.


[User] Please transcribe this text

[Assistant] The monkey burst into tears again.


[User] Please transcribe this text

[Assistant] Wukong looked up. Are you sure? He asked hopefully.


[User] Please transcribe this text

[Assistant] Wukong swung his iron bar at the cave entrance.


[User] Please transcribe this text

[Assistant] Suddenly he looked up and saw an enormous figure.


[User] Please transcribe this text

[Assistant] Light flashed as Buddha turned the two demons into ordinary animals. Only the raptor demon wasn't scared.


[User] Please transcribe this text

[Assistant] His eyes burning with fury. The bird shot toward Buddha.


[User] Please transcribe this text

[Assistant] But Buddha raised a finger, and the bird was paralyzed in mid-air.


[User] Please transcribe this text

[Assistant] Journey to the West, Chapter 89, Many Eyes.


[User] Please transcribe this text

[Assistant] Wukong jumped to his feet and pulled out his iron bar. The priest backed away.


[User] Please transcribe this text

[Assistant] Just then, a door burst open, and the three women ran into the room.


[User] Please transcribe this text

[Assistant] He jumped up, did a flip in the air, and landed directly behind the women.


[User] Please transcribe this text

[Assistant] Both women were gone. In their places, lay dead spiders.


[User] Please transcribe this text

[Assistant] He covered his own eyes and ran all over the room. But he couldn't escape the blinding light.


[User] Please transcribe this text

[Assistant] Finally, he slammed his iron bar into the appy stone floor. A huge hole opened up. The monkey jumped in and turned into a mole.


[User] Please transcribe this text

[Assistant] Using his claws, he tunneled until he was sure he was far from the abbey.


[User] Please transcribe this text

[Assistant] Even from far away, the light burned the monkey's eyes. Wukong turned away.


[User] Please transcribe this text

[Assistant] Prolombra pointed her finger at the priest and he turned into a giant centipede. Prolombra took a bottle out of her sleeve.


[User] Please transcribe this text

[Assistant] Said the Tang Monk one day. Bajie snorted.


[User] Please transcribe this text

[Assistant] Before the break we talked briefly about the differential to single ended conversion. And we talked about different ways of converting it. And one of the stages that we talked about was using a, uhm, current meter.


[User] Please transcribe this text

[Assistant] To achieve that. So one of the things that we can do, we are going to now look at this a little bit more briefly, uh, but in the context of a MOSFET stage.


[User] Please transcribe this text

[Assistant] So if you have a MOSFET stage, the same deal, right? I mean you can actually implement a current mirror to reflect the current from the input to the output. And the idea here is again, the way to think about it is that if you're applying a differential input, so this is VID over 2 and this is negative VID over 2,


[User] Please transcribe this text

[Assistant] If you apply VID over 2 here, the drive current here is gonna be GMVID over 2. Let's call it GMN for a MOSFET. Because each one of them will have its own W over L. And they don't necessarily have the same, uhm, GM. This is one of the differences. In a bipolar,


[User] Please transcribe this text

[Assistant] If you have two transistors, the NPN and PNP that are carrying substantially the same current, they will have the same...


[User] Please transcribe this text

[Assistant] GM, right? Because GM is IC over VT.


[User] Please transcribe this text

[Assistant] So we have to choose a w over l for each one of these things. But before we do that, this current is gonna be now reflected down here. So this current is gonna be gmvid over 2. Now this current going in, so this again is the important part, this going in is gonna be negative gmvid over 2, which you can actually flip the polarity, so you can make this plus and make it point into the node again.


[User] Please transcribe this text

[Assistant] You can see that they were additive all the time, but this way you can see that they were additive. They're not subtractive. Because they're both now injecting into that node, and of course the total resistance on this node


[User] Please transcribe this text

[Assistant] Is the R-O-N, I'm sorry, R-O-P in parallel with R-O-N. Here.


[User] Please transcribe this text

[Assistant] Why? Because it's just reflecting this current here, which was determined by GMN. This is just a mirror.


[User] Please transcribe this text

[Assistant] And high output resistance. High trans conductance and high output resistance.


[User] Please transcribe this text

[Assistant] Well, it appears that you want to get high g, high gain, right? High gm. So if you want to high gm, both expressions, both in terms of current and voltage of gm, have a direct relationship with W over L.


[User] Please transcribe this text

[Assistant] So you obviously don't want, it appears that you don't want a very small W over L.


[User] Please transcribe this text

[Assistant] A beefier transistor, it appears. So let's pick something. I'm gonna make some numbers. 10 microns over, let's make them long. 1 micron.


[User] Please transcribe this text

[Assistant] 10th of a micron, 100 nanometers or 65 nanometers or whatever. Let's just pick some numbers. I mean, the ratios are really not what that matters. So, if you pick something like this. Now, how about this guy? From a gain perspective, do I need my MOSFETs to be large or small? Just purely from the gain perspective.


[User] Please transcribe this text

[Assistant] What determines my output resistance? Let's, let's write the expressions, right? What is GM? GM...


[User] Please transcribe this text

[Assistant] In terms of, if your current is fixed, it's gonna be 2 mu n, or mu p, w, c ox, w over l, i d. And r o is basically, uhm, l over i d, times d x d, d v d s, or 1 over d x d d v d s, inverse. So these are the things that will affect me. Now,


[User] Please transcribe this text

[Assistant] So, let's give some numbers. Let me just say, mu n Cox, let's say of 100 microamps per volt squared, mu p Cox, these are like,


[User] Please transcribe this text

[Assistant] They stand at classic numbers, just long channel numbers, just to get a feel. Uhm, okay.


[User] Please transcribe this text

[Assistant] So, what do we get? Let's do the calculations. Let's say we bias this at, I don't know, 200 microamps. So each stage gets, each side gets one micro, let's do that actually at 1 milliamp because we did 1 milliamp for the bipolar just to get a feel.


[User] Please transcribe this text

[Assistant] Of comparison, right? So if you have 1 milliamp, what are the Gm's?


[User] Please transcribe this text

[Assistant] Well, GM is a function of W over L, so for this one we've picked a GMN is going to be 2 times 100 microamps, 200, 200 microamps, times 10, 2 milliamps per volt squared, times 100 microamps. Right? So, did you see what did I say? Sorry?


[User] Please transcribe this text

[Assistant] So that's one, oh nice, one millisiemens.


[User] Please transcribe this text

[Assistant] That's what you really need. I mean, you don't need a GMP. It doesn't appear in our calculations. But if you need it, we can calculate. Now, how about RON? Let's calculate RON.


[User] Please transcribe this text

[Assistant] RON is L, it's at 1 micron.


[User] Please transcribe this text

[Assistant] Every, these are in microns too, so I would just like, this would be a one. Divided by.


[User] Please transcribe this text

[Assistant] And then we have to calculate ROP. To calculate ROP you need to know what L is.


[User] Please transcribe this text

[Assistant] How do we choose that L? How do we choose the channel length of the p-fet? What are the considerations that you have to think about?


[User] Please transcribe this text

[Assistant] Now, what, what do they, what does that trade with? What are the trade-offs with, of using a larger L? Speed is one.


[User] Please transcribe this text

[Assistant] Because you're making your device larger, it would have more capacitance, therefore, uhm, you're limited by that. What else?


[User] Please transcribe this text

[Assistant] Yes, but yes, theoretically yes, the size, you're concerned about the size, right? But these transistors are usually much smaller than the actual thing, because you're limited by other things typically when you're doing analog, the pads and connections and all those things. So that would not be a primary consideration, although it could be if it becomes very large. What else? If I make a large W over L,


[User] Please transcribe this text

[Assistant] Is mu n C ox over 2 W over L Vgs minus Vt squared, or delta Vgs, the gate overdrive. Alright? If I make this small, by making L large, what does it imply for this guy, for a fixed current? This needs to be large.


[User] Please transcribe this text

[Assistant] Which means that I need to maintain a large voltage, which means that I can't really crunch this too much. I mean, this will go into triode region quickly at a higher voltage. Essentially my headroom will be limited. So there's another trade off in terms of how much swing I can have.


[User] Please transcribe this text

[Assistant] Okay, at five micrometer, what do I get? What do we get? So we get five, right? So we get five divided by


[User] Please transcribe this text

[Assistant] No. Because now the parallel combination is dominated by RON.


[User] Please transcribe this text

[Assistant] So, yeaah, I've done it, we've done it already, but just like, let's, let's do it right. Okay, so let's, let's say, what would you put it at? I mean I put, let's put it at 2 microns.


[User] Please transcribe this text

[Assistant] If I put it at 2 micro, then what you get instead is gonna be 80 kilo.


[User] Please transcribe this text

[Assistant] Alright, let's keep it that way. So what is the game?


[User] Please transcribe this text

[Assistant] Well, it's 20 in parallel with 80, whatever that is. So it's 2 times 8 divided by...


[User] Please transcribe this text

[Assistant] Not incredibly high, right? So what do we do? To improve it. Let's say if you want to get something higher than that.


[User] Please transcribe this text

[Assistant] Cascade for example, right? How do we cascade this? So, first of all, I can make this a cascade.