[User] Please transcribe this text

[Assistant] But you can also convey more information. Uhm, you're not limited to 140 characters. Uhm, and then a blog or a, there's a new blogging platform many people will have heard of called Medium. Uhm,


[User] Please transcribe this text

[Assistant] And those are places where you could talk at length about science, explain anything you want. Uhm, but you're gonna have a much smaller readership, so you can give a lot more information to a much smaller number of people.


[User] Please transcribe this text

[Assistant] And navigating which of these is best depends on what you're trying to achieve. Um, so it really depends on the kind of fruit salad you want. Um, and understanding these different tools can help you figure out what it is that you want to use to achieve whatever your goal may be.


[User] Please transcribe this text

[Assistant] Okay, so the first question you want to ask yourself before diving into social media is, is why are you engaging there? What is your point? So back to the fruit salad analogy, how are you going to try to figure out which fruit you want in your salad if you haven't thought first about what kind of salad you're trying to make? So what are you trying to convey to people?


[User] Please transcribe this text

[Assistant] So there's a number of different reasons why people within science, um, engage in social media. So at an academic level, you know, we're at Stanford.


[User] Please transcribe this text

[Assistant] People would engage in social media to, uhm, help promote Stanford and the research that goes on here or promote a particular cause if, uhm, someone is very interested in environmental issues or energy issues or the arts.


[User] Please transcribe this text

[Assistant] They would want to, to promote that worldview or that cause. Uhm, show thought leadership. Kind of be someone who's worth following. Uhm, engage with people.


[User] Please transcribe this text

[Assistant] So talk about these issues in a way that other people might want to comment on or engage with you on. You might want to be driving traffic to, uh, to a website or a blog.


[User] Please transcribe this text

[Assistant] Uhm, or promote a personal brand. Maybe you want to be the person who, who is the go-to on some particular issue. Uhm, or just connect with like-minded people.


[User] Please transcribe this text

[Assistant] You want to think through this before you engage in social media, in part because social media takes time and why spend that time if you're not pretty sure you know what you're going to achieve.


[User] Please transcribe this text

[Assistant] So I mentioned at the beginning that, uhm, if you have something to say, you want to be in the room where most of the people are.


[User] Please transcribe this text

[Assistant] Room of social media these days. Uhm, there was a, the Pew Research Center has done some really good polls on how people are using social media and so I just wanted to provide some numbers.


[User] Please transcribe this text

[Assistant] Uhm, one of the questions was what percentage of social media users get their news from those channels. And this shows for both Twitter and Facebook, between 2013 and 2015, numbers have gone up considerably for people who get news and information from social media.


[User] Please transcribe this text

[Assistant] And then the next, uh, the next graph shows what percentage of people who get their news from those channels try to keep up in a, in a pretty current way from those channels. And you'll see that Twitter, um, 59% of people who get news from Twitter are really trying to keep up in a, in a very current way. So, um, it's definitely an engaged.


[User] Please transcribe this text

[Assistant] You see, Twitter users are much more likely to follow news outlets than Facebook. Issue-based group, roughly similar. Political party, definitely Twitter. Twitter, people on Twitter are often


[User] Please transcribe this text

[Assistant] Uh, there to follow specific topics of information. Whereas Facebook can be used to follow your mother and your...


[User] Please transcribe this text

[Assistant] Aunts and uncles, and your friends from high school, and also maybe a news organization. So it's a little less focused.


[User] Please transcribe this text

[Assistant] Okay, since Twitter might be new for some people, I'm going to go over what a tweet looks like so that we are all talking the same language. So this is a tweet from the Woods Institute, which is an environmental institute here on campus.


[User] Please transcribe this text

[Assistant] Uh, that I happen to like, and uhm, this tweet has all the relevant elements, so I'm using this. Uhm, so here we see there's an image off to the side. That's the profile image. So whatever image you upload into your Twitter profile is gonna show right there, and if it's a


[User] Please transcribe this text

[Assistant] Busy picture of you and your entire lab. What it's going to look like in a Twitter feed is a tiny little blob.


[User] Please transcribe this text

[Assistant] Uhm, so you want an image there that people can actually see and will remind them of who that tweet is from.


[User] Please transcribe this text

[Assistant] Okay, your handle. So at the top, uhm, it'll be your name and your handle.


[User] Please transcribe this text

[Assistant] You want to pick a handle that is indicative of who you are so that when people see the handle, they know they are who they're seeing it from.


[User] Please transcribe this text

[Assistant] Alright, so within a tweet, you can use hashtags and these indicate topics people might want to follow. So you don't want to hashtag every word in a tweet because that's annoying.


[User] Please transcribe this text

[Assistant] Uh, and it wastes characters because you're using the hashtag all the time.


[User] Please transcribe this text

[Assistant] But the hashtag can indicate, so a topic person, a person might want to follow. So in this case, the Woods Institute is assuming that there might be people out there on Twitter who are following the hashtag climate.


[User] Please transcribe this text

[Assistant] Those people might not know about the Woods Institute, might not be following the Woods Institute, but by putting the hashtag climate in their tweet.


[User] Please transcribe this text

[Assistant] This tweet will show up for someone who's following a climate conversation. So hashtags are a great way of getting your tweet in front of people who are following a topic or a conversation but might not know about you.


[User] Please transcribe this text

[Assistant] So in this case, uhm, in their tweet, they refer to Stanford and they, to do that, they use the hashtag or that they handle for Stanford. So what that means is that


[User] Please transcribe this text

[Assistant] Uhm, Stanford will see that tweet and people following the handle Stanford will see that tweet. So it's another way of getting your tweet in front of people, uhm, who might be following a particular handle.


[User] Please transcribe this text

[Assistant] Here is a URL, uhm, so that people who think this tweet is interesting can click the URL and learn more. So this is a way to get more bang out of your tweet.


[User] Please transcribe this text

[Assistant] So you only have 140 characters, including spaces, in your tweet. Uhm, so you cannot say a lot.


[User] Please transcribe this text

[Assistant] But if you are sufficiently compelling in your tweet, you can drive people to a URL that has quite a bit more information.


[User] Please transcribe this text

[Assistant] Tweets with images, uhm, you're gonna get more engagement, more retweets, uhm, more people clicking the URLs. That's just the world we live in these days. People like good images.


[User] Please transcribe this text

[Assistant] Okay, and there's a bunch of different ways of following tweets.


[User] Please transcribe this text

[Assistant] So I would encourage people to go into Twitter. You don't have to have an account.


[User] Please transcribe this text

[Assistant] To, to look at tweets, uhm, so people who are maybe not ready to jump in can still get a sense of what's out there. But you can look at tweets in a bunch of different ways.


[User] Please transcribe this text

[Assistant] Find a whole bunch of people to follow. So I follow several different science journalists. I follow a bunch of scientists who are really interested on Twitter. Uhm, and I follow a couple of science organizations because I think those are the people who are tweeting about things I care about.


[User] Please transcribe this text

[Assistant] Uhm, so what you'll see is a mix of tweets from a number of different, uhm, kinds of people. Uhm, all about fascinating science. So that's one way of doing it. You can only look at tweets this way if you, uhm, if you have an account and you go in and specifically decide to follow people.


[User] Please transcribe this text

[Assistant] Okay, in this case we're looking at tweets by a specific, uhm, Twitter handle. So this is Stanford BioX, which is a great life sciences, uhm,


[User] Please transcribe this text

[Assistant] Uhm, institute on campus, on Stanford campus. Uhm, and so if I were to follow, uhm, BioX on Twitter, this is what I might see. Uhm, a bunch of interesting tweets about, uhm, the research that's going on in the institute. You'll see images, uhm, you'll see a quote from a news story.


[User] Please transcribe this text

[Assistant] Uhm, you see something, uhm, from Stanford Medicine that they re-tweeted. So you can look at tweets this way without an account. You can go to any, uhm, any handle within Twitter and you can look at everything they've posted, uhm, and it'll just update automatically as, as those people post. So this is one way of looking at tweets. And finally,


[User] Please transcribe this text

[Assistant] These are my tweets and retweets. So uhm, you can follow an individual person. Uhm, so you can see the kinds of things that I've tweeted myself or retweeted.


[User] Please transcribe this text

[Assistant] Alright, my favorite way of doing all these different things, uhm, is a tweet deck.


[User] Please transcribe this text

[Assistant] So uhm, you can go to twitter.com and you can look up tweets in many different ways. I guess I didn't show here you could, uh, one of the ways of looking at tweets could have been to look at a particular hashtag. I didn't show you guys that.


[User] Please transcribe this text

[Assistant] Uhm, within tweet deck you can have, uhm, several columns and each one could be looking at a different set of tweets. So at any one time, uhm, I show three columns here. I generally have four or five open.


[User] Please transcribe this text

[Assistant] Uhm, I'm following BioX, I'm following what appears to be the middle column, uhm, people who, who I follow.


[User] Please transcribe this text

[Assistant] And it looks like I'm following immunotherapy on that other side, the hashtag immunotherapy.


[User] Please transcribe this text

[Assistant] Uhm, so there's a great quote, I tried to look this up and I don't know who it's by, keep your friends close and your enemies in a tweet.column. Kind of the point here, keep your enemies in a tweet.column.


[User] Please transcribe this text

[Assistant] People don't need to know that you're following them. So if you specifically go in and click, I, Amy Adams, want to follow you, Stanford BioX, uhm, they can see that I'm following them. But if I'm just following a hashtag, uhm,


[User] Please transcribe this text

[Assistant] Uh, none of these people, uhm, know that I'm following them. Their, their tweet appeared in my tweet deck column just because I was following that hashtag. So you can follow a wide variety of, uhm, conversations and hashtags and different handles, uhm, and you're kind of invisible.


[User] Please transcribe this text

[Assistant] Which is, I point out, uhm, not to encourage spying, but to say that when you are tweeting, you don't necessarily know how many people are seeing your content. It can spread far and wide. Okay, so I want to show a case story of, uhm,


[User] Please transcribe this text

[Assistant] Of a series of tweets. So, uh, this is the headline of a story I wrote. Stanford chemists develop a new method of cancer immunotherapy. Nice straightforward headline. Stanford University


[User] Please transcribe this text

[Assistant] On their channel tweeted this, a team led by Carolyn Bertosi has developed a new approach to cancer immunotherapy. A nice straightforward tweet with a link and a nice image of Carolyn and, uh, her students.


[User] Please transcribe this text

[Assistant] Okay, StanfordBioX tweeted, Carolyn Bertosi created a chemical lawnmower as a novel method for cancer immunotherapy. Uhm, so this is a little spunkier, uhm, but still a pretty straightforward description of what they did.


[User] Please transcribe this text

[Assistant] Uhm, in this case, we inserted the hashtag cancer and hashtag immunotherapy with the idea that there might be people out there following the words cancer and immunotherapy because those are, uhm, hot topics and Stanford University did not do that.


[User] Please transcribe this text

[Assistant] I always enjoy working with Carolyn Bertozzi. This time she created a novel method for cancer immunotherapy. So this is an important transition. Uhm, when tweeting as myself, I try to


[User] Please transcribe this text

[Assistant] Scientists who tweet very impersonally, uhm, who tweet more like a Stanford University or Stanford BioX, it is going to be harder to reach a large number of people because you sound


[User] Please transcribe this text

[Assistant] Aloof and remote, uh, and not real. I think Twitter is a place where people can get an insight into what it's like to be a scientist and be doing scientists and sort of show


[User] Please transcribe this text

[Assistant] What's real and interesting and exciting in science? So when you start thinking about tweeting, uhm, think about how you can be


[User] Please transcribe this text

[Assistant] This is the headline for a blog that then also got written. Another nice straightforward.


[User] Please transcribe this text

[Assistant] Uhm, headline and then Stanford Medicine tweeted that blog to say, Stanford research develop a new target for immunotherapy, sugars, with a link to the blog entry. So back to a more Stanford University style.


[User] Please transcribe this text

[Assistant] So before I move onto Facebook, I want to give, uhm, a summary of Twitter. And that is, it's incredibly powerful. You can reach an incredible number of people.


[User] Please transcribe this text

[Assistant] Uhm, you can also spend an extraordinary amount of time following people and seeing what people are saying and uhm, and trying to get a sense of what's out there. It's both powerful and time consuming. Uhm, so it is worth thinking about what you are going to achieve before you dive in.


[User] Please transcribe this text

[Assistant] It's also worth following a number of other scientists for a while, maybe a good couple of weeks, to see what other people are saying, what kinds of success they're having, what kind of Twitter style.


[User] Please transcribe this text

[Assistant] Uhm, seems natural, uhm, and people are having success with before you jump in, uh, before you make that time commitment. So I want to talk, uhm, more quickly about Facebook because I think it's a less effective way of reaching a general audience and that's because


[User] Please transcribe this text

[Assistant] People will follow organizations to learn what's going on within organizations.


[User] Please transcribe this text

[Assistant] Facebook is incredibly image and video driven. Uhm, if you don't have good images or videos, your Facebook post will not go far.


[User] Please transcribe this text

[Assistant] Uhm, it is a hard way to really spread much information because you're restricted by who you are friends with. Uhm, so if you are


[User] Please transcribe this text

[Assistant] Something like Science Friday, Cal Academy, uhm, some other people


[User] Please transcribe this text

[Assistant] They can get pretty, pretty widespread, um, viewership, but even they have to work very hard to have great images and great video in order for their posts to spread far. As an individual, you are better off


[User] Please transcribe this text

[Assistant] Thinking about how you can use Facebook to educate existing friends. So if you have a friend base within Facebook, think about this. Some of those are people who don't really know science. Uh, they don't know what it's like to do science and you can


[User] Please transcribe this text

[Assistant] In how you post about your work, talk about things that make science more accessible and interesting.


[User] Please transcribe this text

[Assistant] So it is a way of existing, uh, uh, educating existing friends.


[User] Please transcribe this text

[Assistant] Okay, and so another quick comment on blogs, uhm, and medium.


[User] Please transcribe this text

[Assistant] Uh, is wrapped up in this. So, who's gonna blog about science and where is that gonna get you? So, there are individual scientists who write about their, their work on blogs. There's, uhm, lab pages that, uh, or lab...


[User] Please transcribe this text

[Assistant] Specifically if you then use social media to try to drive traffic to your blog. Uhm, so if you have a blog, you would need a good Twitter account to help drive traffic to it.


[User] Please transcribe this text

[Assistant] Uhm, science organizations will often blog about the work they fund. News outlets will sometimes have a blog to, uhm, to cover other topics, like Nature Jobs is a blog that's a good example.


[User] Please transcribe this text

[Assistant] Uhm, but blogs also take a lot of time and they are very hard to get a following for. So on the one hand, it's a great way to convey a large chunk of information to people. On the other hand, you really need to think hard about how you're going to drive people to that blog because just putting it up and writing great stuff isn't going to get you far. Uhm, so here's your roundup.


[User] Please transcribe this text

[Assistant] Think about what you're trying to achieve before you dive in. When you do dive in, it's your chance to be real. Show the public


[User] Please transcribe this text

[Assistant] What real scientists are thinking about and doing and what their lives are like.


[User] Please transcribe this text

[Assistant] And to pull that off, you need to know who you're talking to and why those people would follow you.


[User] Please transcribe this text

[Assistant] And then the biggie is engage and don't teach. If you're doing nothing but


[User] Please transcribe this text

[Assistant] Whereas you could use social media to engage people and get them thinking about something that is not what they're already doing. And that is a great use for social media.


[User] Please transcribe this text

[Assistant] I hope you don't. You'd probably miss and hit me in the eye.


[User] Please transcribe this text

[Assistant] The two-tang monk should recite the tight headband spell.


[User] Please transcribe this text

[Assistant] Only the Tang Monk, Guan Yin, and Buddha know the spell. We'll know right away which Tang Monk is fake.


[User] Please transcribe this text

[Assistant] Welcome to this last video of the class.


[User] Please transcribe this text

[Assistant] Congratulations, you have sat through a pretty long course. Anyways, I hope that you've learned some new tricks inside Firmware Pro. Basically you've seen almost everything that Firmware Pro has to offer. So what else is there for you now? Just go out and create stuff. Film stuff, edit stuff.


[User] Please transcribe this text

[Assistant] And just keep creating and keep using the different tools inside Premiere and you'll become a better editor. But as far as the technical explanation goes there's nothing more, I mean there are a few things of course, but there aren't any major features anymore that should be covered.


[User] Please transcribe this text

[Assistant] Anyways, I've been talking about Lumetri throughout this class as well. Now those things are more specific. We have a dedicated class on color grading inside Premiere Pro. So there is no use to go through every single tool and every single detail. The same thing goes with audio. There is still a lot that can be covered there. Mostly practical information and still a lot of technical information.


[User] Please transcribe this text

[Assistant] Definitely stuff in Adobe Audition, but that again could be a whole different class, so definitely make sure to follow me here on Skillshare to stay updated for those.


[User] Please transcribe this text

[Assistant] Anyways, you can download all of the project files, all of the media,


[User] Please transcribe this text

[Assistant] So that I can have a look at it and give you guys some feedback. Anyways, more information there on that project page on Skillshare.


[User] Please transcribe this text

[Assistant] Thank you so much for watching, it was a real pleasure to show you guys around in Adobe Premiere Pro, and I wish you the best of luck with your future edits. Thank you so much again, and as always, stay creative.


[User] Please transcribe this text

[Assistant] For these next two modules, I'm thrilled to have a guest speaker. Amy Adams is the Director of Science Communications at Stanford, and she's going to be talking to you about interviewing scientists and using social media.


[User] Please transcribe this text

[Assistant] Check out this macaroni necklace that Travis made me when he was one.


[User] Please transcribe this text

[Assistant] The majority of the galaxies are spiral, similar to our own Milky Way.


[User] Please transcribe this text

[Assistant] How rare galaxies form and evolve.


[User] Please transcribe this text

[Assistant] It's a mystery how the stars in the outer ring are just floating there in such an orderly manner. That's interesting, right? Hold on. Things are about to get more mysterious.


[User] Please transcribe this text

[Assistant] You know, sometimes you search and search for these objects, and you find nothing.


[User] Please transcribe this text

[Assistant] Well, actually, this is one of the nearby galaxies. I studied this object in different light.


[User] Please transcribe this text

[Assistant] Small details on our body, like a scar or wrinkles, tell the story of our lives.


[User] Please transcribe this text

[Assistant] I modeled the bright central body and removed my model from the image to check for any hidden features, because a bright structure in a galaxy may blind our views of faint features, just like


[User] Please transcribe this text

[Assistant] Using sunglasses when you are blinded by the intense light. The result was a big surprise. This galaxy doesn't just have an outer ring, it has an additional diffuse inner ring.


[User] Please transcribe this text

[Assistant] There is currently no known mechanism that can explain the existence of an inner ring in such a peculiar galaxy. So, the discovery of Burçin's galaxy clearly highlights the gap in our knowledge of galaxy evolution.


[User] Please transcribe this text

[Assistant] Can provide us with new clues on how the universe works.


[User] Please transcribe this text

[Assistant] Uh, there's a researcher at Penn State named Barbara Rolls and I'll discuss some of her work later on in a different context.


[User] Please transcribe this text

[Assistant] What we did last time in the last lecture was to look into the basic semiconductor properties. We started with the energy band diagrams, we saw that


[User] Please transcribe this text

[Assistant] Based on the basic assumptions of quantum mechanics, which is the basically wave particle duality, you arrive at situations that for bounded electrons or bounded particles, you will have discrete energy levels. And we saw that when you bring multiple atoms close to each other, due to exclusion, Pauli exclusion principle,


[User] Please transcribe this text

[Assistant] These atoms interact with each other and then you have degeneracy. And what that does is that you split the energy band diagrams. And if you form a crystal by bringing a large number of atoms together, what happens is that these energy levels become energy bands because they become so numerous and then they become degeneracy so they basically split and open.


[User] Please transcribe this text

[Assistant] Conduct, valence band, lower energy band and completely empty conduction band. So in that situation,


[User] Please transcribe this text

[Assistant] The thermal energy would only be able, even at room temperature, to release or ionize very few electrons. And what that does is that you will have very few carriers in the conduction band, which are electrons, and you will have very few opening, open states in the valence band, which are the holes.


[User] Please transcribe this text

[Assistant] And the movements of these things are dictated by electric field, but you don't have that many carriers, so it would not be a conductor. Now if you, if your band gap becomes smaller, then your conductivity can increase.


[User] Please transcribe this text

[Assistant] But the other thing that we saw last time was that we saw that if you basically take a semiconductor material and in which it, let's say a pure semiconductor material type, and it doesn't have to be pure, there are like compound semiconductors, but like silicon or germanium or some, or what have you, silicon most of the times nowadays,


[User] Please transcribe this text

[Assistant] What happens is that you can introduce column 3 or column 5 elements. And when you introduce column 5 elements,


[User] Please transcribe this text

[Assistant] They will form four covalent bonds with the remaining atoms, let's say of silicon, but the fifth electron will be very loosely bound. We saw that the energy of attachment of the bounding energy of that is about one hundred times smaller than the bounding energy of the rest of the electron.


[User] Please transcribe this text

[Assistant] And therefore, as soon as the temperature raises, goes up, they will be released.


[User] Please transcribe this text

[Assistant] So we took that, and we said, this way we can introduce a situation where you have a lot more electrons than holes. And likewise, if you introduce a column three element, as an impurity, we call it doping,


[User] Please transcribe this text

[Assistant] What happens is that you have one missing spot that can be easily filled by breaking an adjacent covalent bond which takes a lot less energy to break a bond and form a new bond as opposed to just break the bond. And then you will have a hole. So by introducing this impurity, you can control the number of electrons or holes independently.


[User] Please transcribe this text

[Assistant] And that's an element that's very important to design a device.


[User] Please transcribe this text

[Assistant] So that was the base, the starting point. We also looked at energy distributions, thermal energy distribution. We saw that electrons basically being fermions follow the Fermi-Dirac distribution, and that dictates the way they behave. The number of electrons was given by, is related to the difference between the Fermi energy in thermal equilibrium,


[User] Please transcribe this text

[Assistant] And the intrinsic energy level between exponential and that's related to Ni. So these are the things that we covered in the last lecture. Now, so,


[User] Please transcribe this text

[Assistant] The objective of this series of discussions has been to arrive at conductivity or resistivity of the material.


[User] Please transcribe this text

[Assistant] From microscopic viewpoint. We started in a macroscopic view. We said, okay, well the resistivity rho times A divided by L gives you the resistance. But we are trying to find that rho, or that sigma, that conductivity.


[User] Please transcribe this text

[Assistant] So we are close, but we are not quite there yet. So we'll finish that conversation today. And we'll arrive at that first.


[User] Please transcribe this text

[Assistant] So, the next thing we'll do is really we need to understand the nature of the carrier movement inside a set piece of semiconductor. Now, so far we've looked at the distribution of carriers, but we want to see how it moves in response to electric field.


[User] Please transcribe this text

[Assistant] So let me ask you a question first. Let's start saying that in vacuum. If I had, if you had vacuum, and if I had an electron in vacuum, let's forget about gravity, although the mass of the electron is not that high anyway, uhm, but shouldn't matter anyhow, uhm,


[User] Please transcribe this text

[Assistant] So, and you apply electric fields. So let's say this is an electron here in space, and you're applying an electric field. Let's say you have two parallel plates, and then you apply an electric field. What would happen to this electron? What is the relationship between its velocity and the electric field?


[User] Please transcribe this text

[Assistant] Let's say the only force it experiences is a constant electric field. What would happen?


[User] Please transcribe this text

[Assistant] It will accelerate, exactly, right? And the amount of acceleration is proportional to the force. The force is q times, in this case, negative q times electric field. So if the electric field is pointing this way, the electron will accelerate that way because it's negatively charged, right? But it will accelerate. Now,


[User] Please transcribe this text

[Assistant] A different kind of situation can arise in a piece of semiconductor, and we will see why.


[User] Please transcribe this text

[Assistant] But let me ask you another question that I typically ask, and some of you have already been asked this question. Uhm, is...


[User] Please transcribe this text

[Assistant] Do the electrons, let's say inside a piece of conductor or inside a piece of semiconductor, you have a lot of electrons, right? They're moving around. Are they moving around fast or slow? Really fast, really slow, right? So, how many people say they're moving very fast?


[User] Please transcribe this text

[Assistant] And how many people say they are moving very slow?


[User] Please transcribe this text

[Assistant] Okay, so, and some of you raised your hand for both.


[User] Please transcribe this text

[Assistant] And that's probably, that is the correct answer, right? Depends on what movement we are talking about. So, now the thermal movement, the energy, the thermal movement of electrons, do we think it's fast or slow? Just purely due to the thermal energy. We think


[User] Please transcribe this text

[Assistant] Why, why would you say, if some, if you wanted to explain it to somebody, uhm, who doesn't know all the details of semiconductor physics, what would you say?


[User] Please transcribe this text

[Assistant] Correct. So they have a very small mass, they're very light.


[User] Please transcribe this text

[Assistant] Now why would that matter? That really goes back to the statistical mechanics in the sense that, you know, so the movement of carriers, if you think about it, in steady, in thermal equilibrium, each degree of freedom, there's an equipartition theorem called, that basically says that each degree of freedom in thermal equilibrium takes one half of kT of energy on average. So the average energy for each state


[User] Please transcribe this text

[Assistant] Now, let's say an electron is a classical particle.


[User] Please transcribe this text

[Assistant] 3, x, y, and z, right? So, the total energy is 3 times that, so the total average energy would be 3 half kT.


[User] Please transcribe this text

[Assistant] Now this is, it's kinetic energy, right? So if you assume that it's one half m,


[User] Please transcribe this text

[Assistant] V squared, and then what I put, I just put a little m star here, and star here basically implies the effective mass. Effective mass is a measure of, and we'll talk about it more explicitly later, but effective mass is a measure of the


[User] Please transcribe this text

[Assistant] M-m-mass that the electron appears to have when it's inside the crystal. And the same thing is true about a hole.


[User] Please transcribe this text

[Assistant] And you can imagine, you remember our water-filled analogy, if you remember our water-filled tube, two tubes filled with water, one filled with water, one empty and taking a droplet from one and putting it in the other one, is that if you shift it, the bubble, which was the kind of like a representative or analogy of a hole, moves more slowly generally than the, than the, than the,


[User] Please transcribe this text

[Assistant] So this equation, there's an assumption here for the velocity, so this is a root mean average. Really, we are calculating.


[User] Please transcribe this text

[Assistant] Uh, or mean average, not weak mean average, mean average velocity that you're calculating, uhm,


[User] Please transcribe this text

[Assistant] And then basic, sorry, root square, uh, uh, so this is square average. Now, what happens here is that


[User] Please transcribe this text

[Assistant] I'm just trying to calculate this velocity. So let's estimate this velocity. So v squared.


[User] Please transcribe this text

[Assistant] And I'm gonna just show it here, as v squared. It's gonna be 3kT over m.


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[Assistant] It's 1.38, 10 to the minus, uhm, 23 joules per Kelvin. Let's say at room temperature, 300 Kelvin.


[User] Please transcribe this text

[Assistant] Approximately 10 to the minus 30, but kilograms.


[User] Please transcribe this text

[Assistant] And we said it's quarter of that, so it's gonna be times, uh, or you can divide this, you can multiply this by four.


[User] Please transcribe this text

[Assistant] So what do we get? Roughly we have twelve, fifteen, nine. So let's say it's just like twelve-ish or


[User] Please transcribe this text

[Assistant] So, 12 here, times, so you have 10 to the minus 30, uh, oh no, there's another 3.


[User] Please transcribe this text

[Assistant] So it's 10 to the minus 21 here, there's a 10 to the minus 31 here, so it's a 10 to the 10, so that becomes 5, 10 to the 11 roughly.


[User] Please transcribe this text

[Assistant] If I'm not mistaken. And so that's the V squared. So V you're looking at.


[User] Please transcribe this text

[Assistant] Right? So the electrons jumping around like crazy. Going around. Electrons are going around. Crazy.


[User] Please transcribe this text

[Assistant] Now, what happens when you apply electric field? So the electron, let's say an average electron would be just going around like that. It's bouncing around. And this is happening over very short time periods, right? It's getting scattered by what? It's usually mostly getting scattered by the ions in the lattice. So they just bounce around. Now what happens is that now you apply an electric field.


[User] Please transcribe this text

[Assistant] And as you apply an electric field, let's say you apply an electric field this way, what happens? When does it accelerate?


[User] Please transcribe this text

[Assistant] Well, it accelerates all the time, which is basically these all the times that it's between the collisions. So it starts accelerating, so it goes a little bit off this way, so it's moving a little bit this way, so it's adding


[User] Please transcribe this text

[Assistant] It's displacing here, and of course the same scattering would not happen because now it has moved, so, but let's assume that if I wanted to kind of show it the same way, it would be moving a little bit this way. But now, once this collision happens,


[User] Please transcribe this text

[Assistant] This additional velocity is completely randomised, right? It just like, there's a slight difference, it just gets randomised every time. And this collision has happened on the order of sub picoseconds. At room temperature. Because of the very high speed that you have. The electrons have.


[User] Please transcribe this text

[Assistant] So this is constantly just accelerating, but then getting randomized. So as a result, it appears that this entire electronic gas is slightly moving, slowly moving, with respect to this.


[User] Please transcribe this text

[Assistant] What is that speed roughly like? Right? So we can actually calculate. So we can find out, kind of like, what it is. So, the change in the, in the,


[User] Please transcribe this text

[Assistant] That the change in the velocity of this thing is dictated by Newton's second law, right? It says basically f is dp dt.


[User] Please transcribe this text

[Assistant] The derivative of the momentum, the change, the rate of change of momentum, this is Newton's second law, right, is f. And when you think about that, so what is f?


[User] Please transcribe this text

[Assistant] F is negative Q, electronic charge, times E, the electric field.


[User] Please transcribe this text

[Assistant] And then what is delta, so we can write this as delta P over delta T. Now delta T,


[User] Please transcribe this text

[Assistant] We call it the mean free path, the mean free time between collisions. We can show it as tau c. And I said it's like sub picoseconds.


[User] Please transcribe this text

[Assistant] And what is delta p? Not delta p, delta p. Delta p is the mass times delta v. Right? On the left.


[User] Please transcribe this text

[Assistant] So now, if you think about it, you can think about it like effective math.


[User] Please transcribe this text

[Assistant] So, I can calculate delta v from this. Delta v is the velocity, the change in the velocity at the end of a single


[User] Please transcribe this text

[Assistant] So that's what we calculate for the velocity at the end of the


[User] Please transcribe this text

[Assistant] So this velocity is just a change in velocity.


[User] Please transcribe this text

[Assistant] In this small segment. Now, so if I wanted to find the average velocity,


[User] Please transcribe this text

[Assistant] In this process, it would be fair to assume it's half of that, right? Because at the beginning, this, this component, this really small component of the velocity is zero, and at the end, it is this much.


[User] Please transcribe this text

[Assistant] So you can assume that basically this velocity, on the average,


[User] Please transcribe this text

[Assistant] Is going to be half of that. Negative q e tau c over m star, 2m star, uhm, times, and let me just put the e out.


[User] Please transcribe this text

[Assistant] So now something interesting is happening, right? The interesting thing is that


[User] Please transcribe this text

[Assistant] Then now, this component of the velocity, which we call the drift velocity, right, is now proportional to the electric field.


[User] Please transcribe this text

[Assistant] Now if you had a free electron, its acceleration would be proportional to the electric field. But this is different. Now this is really a situation where this, this component of velocity becomes proportional to the electric field.


[User] Please transcribe this text

[Assistant] So, in this case it's the mobility of an electron. So the drift velocity,


[User] Please transcribe this text

[Assistant] And the negative sign comes of course because of the direction that the electron takes. Because it goes in the opposite direction of the electric field because it's negatively charged.


[User] Please transcribe this text

[Assistant] So we don't make a negative mobility. We just call it mobility and we give it a negative sign for the electron. Okay, so that's the mobility.


[User] Please transcribe this text

[Assistant] And now, we are almost there as far as calculating the, uhm,


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[Assistant] Conductivity or resistivity is concerned. So let's, let's do that.


[User] Please transcribe this text

[Assistant] Now, why do they say it slowly? How do we know it's slow? We haven't, we don't know yet, right? Uhm, now let's think about it. Let's think about it.


[User] Please transcribe this text

[Assistant] Why this would be very slow, this component of the velocity? Any thoughts?


[User] Please transcribe this text

[Assistant] Right, so from an observational perspective, for a typical current, it's not a massive amount of charge, right? And let's say quantitative, let's make that a little bit quantitative, that's the correct answer. Let's make that a little bit more quantitative. So here's the thing, how many electrons, order of magnitude electrons do you expect to have per cubic centimeter of, let's say, copper? Ballpark.


[User] Please transcribe this text

[Assistant] Well, you first need to start knowing how many atoms you will have for cubic centimeter of copper. A cubic centimeter of copper is a few grams, right?


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[Assistant] And, so it would be an order of a mole. I don't know, it would be half a mole, or, we can, we can calculate, but it doesn't matter. It's an order of a mole. And a mole, how many atoms are in a mole? That's probably one of the most important numbers in science, right? It's called Avogadro's number. 6.02, 10 to the, 10 to the 23. Yes, exactly. 10 to the 23. So,


[User] Please transcribe this text

[Assistant] It's about 10 to the minus 19, right? It's 1.6 and 10 to the minus 19, 10 to the 24, it's basically 10 to the 5 coulombs of charge. So if you're carrying 1 ampere of current, this thing, right?


[User] Please transcribe this text

[Assistant] Now, the short answer to the second why this velocity is very small, the same way that the short answer to the first why the thermal velocity is very high was that the electrons are very high, eh, very light. The short answer to this one is that the electrons are very numerous, or there are lots of electrons in there. And as a result, you don't need, they don't need to move very fast to carry a lot of charge.


[User] Please transcribe this text

[Assistant] So there are these two components of current carrying. One is completely random, the thermal random. And you can see it's incredibly more prominent if you think about it in terms of the velocity.


[User] Please transcribe this text

[Assistant] And you will see this will have an impact on the behavior of VJTs or PN junctions and all those things. This random thermal movement. Being so fast and being so dominant. But, it's random.


[User] Please transcribe this text

[Assistant] So, it's basically not purposeful in a way. It doesn't carry net current.


[User] Please transcribe this text

[Assistant] So the net current is really carried by that drift, this component, as we'll see. So, let's go back to the calculation that we promised, which is basically calculating what is the conductivity.


[User] Please transcribe this text

[Assistant] So if you think about it, if you go back to our slab that we drew in the last lecture.


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[Assistant] And then we say, okay, well, let's say there's a cross section of the slab.


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[Assistant] Within a unit of time, I want to see how much charge goes through this cross section. So that's basically how much volume you fill up


[User] Please transcribe this text

[Assistant] If you move at the velocity vd. So let's say the electronic particles are moving at the average velocity vd. How much charge are you carrying? Well, it depends on how many electrons you have. So the question is how many electrons are


[User] Please transcribe this text

[Assistant] Inside this volume. So, this length is Vd times dt. Let's say over a short period of time dt. That's the distance they travel. An average electron will travel.


[User] Please transcribe this text

[Assistant] Due to the drift, right? This is not forgetting the, the,


[User] Please transcribe this text

[Assistant] Thermal movement. It's just a slow movement, net movement.


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[Assistant] And what is the volume of this thing? The volume of this thing is, oh sorry, yes, V, DT.


[User] Please transcribe this text

[Assistant] So that's this dx times a is the volume of this thing, right? Now, how do I calculate how much charge that is?


[User] Please transcribe this text

[Assistant] Well, first I need to know how many electrons that is, so I multiply it by the density of electrons, which is the number of electrons per unit volume, per cubic centimeter or cubic meter, and then multiply it by the charge of electron.


[User] Please transcribe this text

[Assistant] So that's the total charge that passes through this. Now we know current i is dq dt. Right? This is the dq.


[User] Please transcribe this text

[Assistant] So if I divide it by DT, I get negative QNAVD.


[User] Please transcribe this text

[Assistant] And if I'm interested in J, which is the current density, it's I divided by A for uniform current distribution, which would basically be, so this is Jn, this is In, this is the current through the electrons. It's negative QnVd.


[User] Please transcribe this text

[Assistant] But now I have a relationship, we have a relationship for VD in terms of the electric field. Because we are really trying to relate the electric field to the junk, because Ohm's law is, in microscopic form, is written like this, right? This is Ohm's law. J is sigma E. This is the


[User] Please transcribe this text

[Assistant] Microscopic equivalent of saying I is G, V. Or V equals


[User] Please transcribe this text

[Assistant] One was microscopic, one was macroscopic. Macroscopic and one was microscopic. So, what is VD?


[User] Please transcribe this text

[Assistant] Negative q, n, and then vd, we know it, we can now define it in terms of mobility. So the two minus signs cancel, you get mu n, e.


[User] Please transcribe this text

[Assistant] Can we say something about this quantity? What is it? Seemingly.


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[Assistant] It looks like conductivity, right? It looks like sigma. Except for one thing. This is not the whole picture.


[User] Please transcribe this text

[Assistant] Because these are not the only charge carriers. There's another group of charge carriers that can actually carry net current for you. Right? What are they?


[User] Please transcribe this text

[Assistant] Also carrying current. Now how about, so that's why I wrote this as Jn, not Jj, right? So there's a Jp here.


[User] Please transcribe this text

[Assistant] Which is basically similar, now the charge of the hole, which is positive, times P, times the V drift of the holes, which would be Q, P, mu, P, E. So the total current, J,


[User] Please transcribe this text

[Assistant] Time zone. And this, now, is the conductivity.


[User] Please transcribe this text

[Assistant] If you wanted to know the resistivity, of course it would be one over the conductivity, one over that.


[User] Please transcribe this text

[Assistant] So now you can see that we went full circle, we came back and we were able to calculate what our, uhm,


[User] Please transcribe this text

[Assistant] Conductivity is, in a much more sophisticated fashion, in terms of the microscopic properties.


[User] Please transcribe this text

[Assistant] of the material. And of course, this is a useful thing to know, and it's a valuable thing. Now, this mobility business, this kind of like proportionality that we have here, we say, we wrote it as V drift equals, I don't know, negative mu N E or mu P E. Now,


[User] Please transcribe this text

[Assistant] Do we think that this is gonna hold forever, this kind of relationship? Is there a place where it can break down?


[User] Please transcribe this text

[Assistant] So what can happen for this to break down?


[User] Please transcribe this text

[Assistant] This is a pretty useful relationship, right? Because you can say, you might drift velocity, the average net move velocity of the movement of this electronic gas or electronic cloud is Vd proportional to the electric charge. But what happens is that now as it moves,


[User] Please transcribe this text

[Assistant] So, is there some value of E or some conditions where this can fail?


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[Assistant] Inside a semiconductor or inside a solid state situation. Well,


[User] Please transcribe this text

[Assistant] You remember what the assumption behind this was? The assumption was that this randomization is happening, but this is a really, this, this change in the velocity, this drift velocity is a very small fraction of the total velocity. Right?


[User] Please transcribe this text

[Assistant] The cat. So as you apply higher and higher electric, uh, electric, uhm, field,


[User] Please transcribe this text

[Assistant] This is because of the very high acceleration. And then what you have, we call it the velocity saturation. So you double your electric field.


[User] Please transcribe this text

[Assistant] This velocity doesn't double because you have so many impacts that it basically does not allow it to happen. So the mobility itself starts saturating.


[User] Please transcribe this text

[Assistant] And at some point, basically instead of having a velocity that's proportional to the electric field, you have a velocity that we call the Vsat.


[User] Please transcribe this text

[Assistant] You could say that they are still accelerating right between the collisions. Yes?


[User] Please transcribe this text

[Assistant] But the time between, the average time between the collisions is becoming smaller.


[User] Please transcribe this text

[Assistant] So yes, you accelerate more, but these things happen, the collisions will happen more frequently, and as a result, you have an effective saturated velocity. Basically, there's a velocity you cannot exceed. And this VSAT is very important in today's devices. Because many of the transistors that we actually use of


[User] Please transcribe this text

[Assistant] Typically today, in integrated circuits, are actually, especially the MOSFETs, are operating in this VSAT region.


[User] Please transcribe this text

[Assistant] So we arrived at what we promised we arrived, which is basically the conductivity from a microscopic viewpoint. Any questions on that?


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[Assistant] That's that, huh coach? That awful, huh?


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[Assistant] When you're in nothing, you have a hard time understanding anything.


[User] Please transcribe this text

[Assistant] Nothing. Probably because she isn't here today.


[User] Please transcribe this text

[Assistant] The teacher thinks I have a problem with my eye, and says I should go see the nurse. What am I going to tell the nurse?


[User] Please transcribe this text

[Assistant] Yes ma'am, our teacher told me to go see the nurse. She thinks there's something wrong with my eye, and well, I, I was sort of winking. You know, like this.


[User] Please transcribe this text

[Assistant] The way things have been going, I may never wink again. I can't believe this is happening.


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[Assistant] I wink at that little red-haired girl who wasn't even there.


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[Assistant] The teacher sends me to the nurse. The nurse sends me to the eye doctor. I'm lucky I didn't try to give her a hug. They'd have sent me to an orthopedist to fix my arms.


[User] Please transcribe this text

[Assistant] Journey to the West, Chapter 70, The Star Lord.


[User] Please transcribe this text

[Assistant] You cannot go back to that cave, said Guanyin. But we must go back, said Wukong. That woman kidnapped the Tang monk. Guanyin sh-


[User] Please transcribe this text

[Assistant] Wukong shuddered, thinking about the pain he'd felt. I won't go near her, said Guan Yin.


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[Assistant] Go to heaven, and ask the Star Lord for help, said Quan Yin.


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[Assistant] The bodhisattva then disappeared in a flash of light.


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[Assistant] A little while later, Wukong stood at the southern gate of heaven.


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[Assistant] The guard told Wukong the way to the Star Lord's palace.


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[Assistant] A few minutes later, the monkey knocked on the palace door.


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[Assistant] Wukong led the Star Lord out of heaven and down to earth.


[User] Please transcribe this text

[Assistant] They landed on the mountain, where Bajie was still rubbing his nose.


[User] Please transcribe this text

[Assistant] Said Wukong. The Star Lord looked at Bajie's nose.


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[Assistant] Bajay walked over to the Star Lord, who then blew on the pig's nose. The injury healed instantly.


[User] Please transcribe this text

[Assistant] Wukong stood in front of the cave entrance and shouted.


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[Assistant] Then he ran as fast as he could away from the cave.


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[Assistant] The woman rushed out of the cave, spinning wildly.


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[Assistant] Suddenly the Star Lord turned into an enormous rooster.


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[Assistant] So glad you're breathing. So Dina, this is amazing. The positive applications are incredible. What is the framework, though, like the ethical framework around this? What are you doing to prevent this technology from being used for other, perhaps less positive, types of applications?


[User] Please transcribe this text

[Assistant] How did you get here so fast? What are you talking about?


[User] Please transcribe this text

[Assistant] I'm going there right now. I'm going to destroy him. I'll go with you.


[User] Please transcribe this text

[Assistant] I am Sun Wukong, King of the Fruit and Flower Mountain.


[User] Please transcribe this text

[Assistant] Or I'll recite the tight headband spell until your head explodes.


[User] Please transcribe this text

[Assistant] Uh, but what we will do first today is just, we are going to spend a little bit of time looking at some of the bipolar stages, just reviewing quickly what the core stages are and how they behave and what, how do we model them. So, one of the things that we looked at last time, the first thing we looked at was, really the only amplifying stage we looked at was the common emitter, for example, or common source.


[User] Please transcribe this text

[Assistant] Without source degeneration. So this is the way we show it, again, just to refresh your memory. So this is VCC, this is RC, and this is VIN, and this is VOUT.


[User] Please transcribe this text

[Assistant] Now of course one of the things we saw when we looked at that was that even from a large signal perspective, if you just think about this as a V in, V out characteristic,


[User] Please transcribe this text

[Assistant] You can draw the V in V out characteristic and we did derive it. Basically you can write the output voltage as Vcc minus the voltage drop across here. So V out was Vcc minus Rc times the current, the collector current.


[User] Please transcribe this text

[Assistant] Which is related to the base emitter voltage, which is now Vin. So def, and we know how that dependence works, we know that it's primarily exponential. So it's Is.


[User] Please transcribe this text

[Assistant] E to V in minus V out, uh, V in I'm saying, over VT. The thermal voltage, KT over Q.


[User] Please transcribe this text

[Assistant] And if you plot that, of course, that's for forward active region. So you start from here, you start from VCC when the V in is zero. So this is V in, this is V out.


[User] Please transcribe this text

[Assistant] And you initially started turning the transistor on, it's...


[User] Please transcribe this text

[Assistant] The current is small first, it's exponential, but then it grows exponentially.


[User] Please transcribe this text

[Assistant] And of course it's not gonna go negative because once this voltage gets below a certain level, which is like 0.1, 0.2, this transistor is gonna enter saturation.


[User] Please transcribe this text

[Assistant] You can see this is a very non-linear behavior, of course, right? And then what the other thing that we did, we looked at it from the Pi model, uh, perspective. We said that generally, uhm, Pi model is more useful when the emitter or source is grounded or connected to a fixed potential.


[User] Please transcribe this text

[Assistant] Now, if generally when collector or source or terminal one in the ATD, arbitrary transistor, uh, three terminal device, uh, is not grounded, you will see that the T model is useful.


[User] Please transcribe this text

[Assistant] Which is a lot of times. Uh, a team model works better. Again, you are welcome to use whichever model you feel more comfortable, you feel more comfortable with. This is just a suggestion, not a rule.


[User] Please transcribe this text

[Assistant] So you can do anyway you want. If you do it properly, you will get the correct result.


[User] Please transcribe this text

[Assistant] So, but when we did this, so this is from last time, uhm, we, we had r pi here, and here of course the DC part of this thing, the bias part will disappear, you only are left with the V in, the lower case V in, which is the AC part of it, the perturbation.


[User] Please transcribe this text

[Assistant] Because in the large signal model it's connected to VCC. It's connected to a constant potential. Which basically from an AC perspective is zero. So there's no difference between this and that from an AC perspective. They're both constant. The derivative of both is zero.


[User] Please transcribe this text

[Assistant] So, okay, so we had this, and then we calculated the gain to be, so AV, we demonstrated the small signal gain to be GM, uhm,


[User] Please transcribe this text

[Assistant] And we saw that the maximum that you can get from this kind of stage was essentially GMRO, assuming that you present a very large resistance. We said that you can actually, if you could load it, if you had an ideal current source here, it would become an open loop.


[User] Please transcribe this text

[Assistant] It will allow you to bias the device, but it will still remain open loop. So what happens is that we were basically GMRO. So AB max was GMRO.


[User] Please transcribe this text

[Assistant] Which we basically saw gm being ic over vt and ro being va over ic was essentially ratio of va over vt.


[User] Please transcribe this text

[Assistant] So that was the thing. Now also there's an input and output resistance associated with this. So there's, if you look into the input port from an AC, from a perturbation perspective, you see what we call an R in. And if you look into the output terminal,


[User] Please transcribe this text

[Assistant] And you null the independent drives, which is namely this guy.


[User] Please transcribe this text

[Assistant] You will see a resistance. And it's clear what R in is. I mean this is obvious. By looking at it you can see what it is.


[User] Please transcribe this text

[Assistant] So Rn is really R pi. But if it was not as obvious, how do we generally determine that? We apply a test voltage, let's call it Vx, or Ix, and measure the other, calculate the other one in terms of it. So if you're applying a Vx, you calculate Ix in terms of Vx, and you calculate the ratio of Vx to Ix. And we'll see examples of that.


[User] Please transcribe this text

[Assistant] But it would be obvious here to apply it. And again, for the R out, the same deal. So we can actually try doing that here just to make sure that you're on the same page. So if you wanted to calculate what R out is, you apply some sort of a Vx. Let's see. And then you're trying to calculate Ix.


[User] Please transcribe this text

[Assistant] And you have to know the sources. So when you know the sources, where are the independent sources, other independent sources? This guy. Right?


[User] Please transcribe this text

[Assistant] So what are you left with? You're left with these two, right?


[User] Please transcribe this text

[Assistant] And therefore if I wanted to go through the procedure formally, R out would be ratio of Vx


[User] Please transcribe this text

[Assistant] Over Ix, which from this calculation you can easily see it's gonna be RC parallel aura.


[User] Please transcribe this text

[Assistant] Now, we want to also, as we go forward, we want to develop some sort of an intuition about the devices, transistors in general. So for example, if you think about it, and again, this is some sort of a rough way of thinking about things, but it's useful to start thinking about it that way, because we'll refer to it. So if you have a transistor,


[User] Please transcribe this text

[Assistant] Really, what these resistors that you're talking about are, are really


[User] Please transcribe this text

[Assistant] The Thevenin resistance of that thing, how is this calculated? The Thevenin resistance of this entire thing. You null all the independent sources. We nulled this. And you determine what's the resistance that's seen.


[User] Please transcribe this text

[Assistant] Right? Or the Norton equivalent. That's how we calculate. So it's basically the Norton equivalent of that box.


[User] Please transcribe this text

[Assistant] From a network circuit theory perspective. But, now in general if you have a transistor, let's say ABJT,


[User] Please transcribe this text

[Assistant] It's good to develop a sense about what are the resistances seen looking into the three terminals of the transistor.


[User] Please transcribe this text

[Assistant] Not only what, what the values are, also kind of get a feel for how large or how small they are. So let's talk about that briefly. So,


[User] Please transcribe this text

[Assistant] What is the resistance, and when we say that, we kind of implicitly mean, when we talk about it, really it's the resistance that you see if the other two terminals were AC ground.


[User] Please transcribe this text

[Assistant] Meaning that the transistor still is biased properly, so it's in the forward active region.


[User] Please transcribe this text

[Assistant] You're not short-circuiting everything to ground, really. You're biasing it, but there's no perturbation on it. So from an


[User] Please transcribe this text

[Assistant] In this model, what is the resistance that you see looking into the base? To ground.


[User] Please transcribe this text

[Assistant] Let me keep my color code consistent going forward.


[User] Please transcribe this text

[Assistant] So what is this resistance that you will see looking into the base? What would you see if the other two terminals were AC ground?


[User] Please transcribe this text

[Assistant] R pi, right? Because if your AC ground is, this stuff disappears, so what you see is R pi.


[User] Please transcribe this text

[Assistant] Yes, but RC is not there, so we said just the transistor, right? Correct. RO in parallel with RC, but there's no RC because we said we have AC ground. It's like connecting it to some supply voltage directly, collector. And looking at that. Or some bias current. It's RO, it's just the RO, right? And what is the resistance you see looking into the emitter?


[User] Please transcribe this text

[Assistant] R0 parallel to rpi. Is that so? Because you know there's a current source now. You have to take it into account. And it's actually easier to see it in the T model. So let's go to the T model. And see how this thing looks in the T model. Looks like a T model. So you have alpha rm.


[User] Please transcribe this text

[Assistant] So for that, of course, to have meaning, then the emitter is not grounded, because, you know, you want to look at it. So, what do you see? If I ground this terminal, and ground this terminal, right? Would the value of that current source matter?


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[Assistant] I mean the current source is not zero, the current, if I apply a test voltage, let's say a test voltage of Vx.


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[Assistant] The x applied here. Well, bad drawing.


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[Assistant] But, which induces a current source here. A current through the current source. But this current source is shorted across.


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[Assistant] Because it's shorter on both ends. So that current will flow inside that loop, so it wouldn't affect anything else.


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[Assistant] Right, it's a shorted current source.


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[Assistant] The loop, the closed loop here. So fine. So that doesn't matter. So what else do you see? You see two resistors in parallel. Alpha RM in parallel with RO.


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[Assistant] Right? So what you see here is alpha RM parallel with RO.


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[Assistant] Right? What is RM? Let's say for 1 milliamp.


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[Assistant] Yes, which is basically, at 1 milliamp is IC over VT. So it's like 1 milliamp divided by 25 millivolts is 40 millisiemens. It's GM, so it's 25 ohms at 1 milliamp. And it scales. The smaller the current, the larger the RM.


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[Assistant] So this is like a, so r-pi, by the way, what's r-pi? r-pi was beta rm, right?


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[Assistant] So let's say weight of 100 for the sake of argument. So you look at 2.5 kiloohms. So this is in kiloohms.


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[Assistant] Right? 1 kilo ohm, 2 kilo ohms. If the current is smaller, like 10 kilo ohms, things like that. RM!


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[Assistant] Let's say for 1 milliamp, this is 25 ohms. This is 2.5 kiloohms. And RO. What is RO? Let's say a VA of


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[Assistant] So each one of them is about two orders of magnitude different from. So this is two orders of magnitude higher than that, this is two orders of magnitude higher than that. Roughly.


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[Assistant] So this is four orders of magnitude higher than that. So it's really ridiculous for me to say it's alpha Rm parallel with Ro.


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[Assistant] But these are good things to remember, kind of like that, for a transistor, looking into this emitter, you see roughly alpha RM, or RM. Looking into the base, you see R pi, and looking into the collector, you see RO. So you can see that we have three terminals, from very low, low impedance, to middle, mid-level impedance, to high impedance.


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[Assistant] And this is important when you design circuits because there are places where you need low impedance and there are places where you need high impedances. And that would affect the way you use the transistor. The topology of the design.


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[Assistant] Is really dictated by these impedance levels. Uh, primarily, really.


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[Assistant] When you apply Vx, so if this voltage is Vx, what is this current? Ie. That's the key current, right? What is it?


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[Assistant] You have a voltage VX across a resistor alpha RM. What is it's current?


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[Assistant] Now, if that's Vx over r- so what is Ix? What is this current, in terms of Ie?


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[Assistant] This current is IE, this current is alpha IE. So this current is gonna be 1 minus alpha IE. So Ix


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[Assistant] So, this is, this is really 1 over beta, so this becomes vx over beta rm. So if you look at rn,


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[Assistant] So obviously it should give you the same result because it's an equivalent model.


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[Assistant] Next, see how it's useful when the emitter is not grounded.


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[Assistant] So, this is kind of like what we call the basic emitter, common emitter.


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[Assistant] If you're trying to use it as an amplifier, let's say to amplify audio.


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[Assistant] Let's say you want to take this and you attach it to a microphone and amplify the voice. Would this be, what would it do to the audio the way it is?


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[Assistant] In exponentials are particularly not, not particularly linear. So,


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[Assistant] It's coming from this exponential, right? Which is coming from the transistor, which is coming from the Boltzmann distribution, which is coming from the Dirac distribution, from, well, so on and so forth, right? So, it's this exponential. Now, why do you see this exponential in its raw, fully exposed form?


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[Assistant] Because you're applying your input directly, all of it to the, between the base and the emitter, right?


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[Assistant] So all of it is applied directly there. So every fluctuation that you have goes directly through the exponential.


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[Assistant] So one of the things that you can do is to create a situation where this voltage is actually divided between that and something else. So one of the ways of doing that


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[Assistant] And a reasonably good way of doing that is to introduce a resistance in the emitter. Now once you have a resistance in the emitter,


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[Assistant] Or in the source later on, as we'll see, or in terminal one.


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[Assistant] Now you have a resistor here, so what happens is that when you apply a voltage V in here, it drops across the VBE here, which in a small signal appears like an alpha RM, and RE. Right?


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[Assistant] So there's a voltage divider, this is a linear resistor, and this is a non-linear resistor really, which in the small signal model becomes linearized.


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[Assistant] So, what do we get for gain and other properties of a stage like this? What is the gain of this stage?


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[Assistant] Our month, kinda like I guess our rule of thumb suggests that since the emitter is not connected to


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[Assistant] Ground. Let's use the T model. So you have alpha Rn and Re. And for now, let's ignore the Ro for a second.


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[Assistant] And then we'll come back to it. So we have the RC here that goes to where? To ground.


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[Assistant] So this is V out, lower case, V in, small signal, small signal. And this is IE, and this is alpha IE.


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[Assistant] It's that current source, right? That dependent current source. It's pulled out of the out, out of RC, sorry, to produce the out. So we know V out is gonna be what? This current source, negative alpha IE times RC. And why negative? Because it's being pulled out of it as opposed to being injected into it. It's just a polarity question. So, the question is what is IE?


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[Assistant] Well, it's a pretty straightforward thing. You know this voltage is VN. There are two resistors here. Right? So it's this voltage divided by these two resistors. So IE is simply


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[Assistant] And if you plug that back into that guy, you will get V out in terms of V in. So if you do that, you see V out over V in.


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[Assistant] So if you set RE to zero, the alphas cancel, you have RC over RM. Or GMRC. Negative GMRC.


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[Assistant] Right? Is there a better way to think about this? Yes, there is a better way to think about this.


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[Assistant] This is actually a pretty decent way to look at it. Better than a pie model, I think.


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[Assistant] Because what you see is basically you see the transistor.


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[Assistant] This is the transistor. This is where you have the potential you're blurred, and this is the current that gets injected to the collector.


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[Assistant] The electrons that get injected into the collector, or if it's PMP, the holes that get injected into the collector.


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[Assistant] If I show you a stage and I ask you for the game, and you've done a little problem already on this, so you should have developed a little bit of a intuition on this. So, here's the thing.


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[Assistant] We need to read it for designing. We will be talking about this like 10 second design, uh, exercises.


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[Assistant] So what I wanted to be able to do, you to do and us to do, is that if we have a stage and we say we want to kind of come up with values that make it have, let's say, make it a stage that has a gain of five. Right?


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[Assistant] You should be within 5 seconds come up with something that's within 10-20%. 10 seconds you should come up with something that's within 10-20% of


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[Assistant] That's something that gives you a gain of 5. So if you give me a gain of 4.7, 4.8, I'm okay.


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[Assistant] If you give me something that has a gain of 4.5 or 5.5, it's okay. If you give me something that has a gain of 50, or a gain of .5, then we are not okay.


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[Assistant] So, we are gonna be building towards that. So for that we need to really understand what's happening. What's really happening?


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[Assistant] So, it is natural for it to have some sort of a relationship. If you have a resistor here and a resistor here, the voltage fluctuations here are inducing voltage fluctuations here. So they should be related to each other.


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[Assistant] So let's look at this. I claim that the total gain of a common emitter in general is negative alpha times the total resistance in the collector to the total resistance in the emitter.


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[Assistant] If you look at this, you will see it. This is right now based on this model of the total resistance in the collector. Right? It's Rc.


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[Assistant] And the total resistance in the emitter is RE, which is explicit, and alpha RM, which is part of the transistor.


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[Assistant] So if you want to make a stage that has a gain of close to 5, how would you choose these? I'd make this 5 kilo ohm and I'd make this 1 kilo ohm.


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[Assistant] If you want to make it exactly five, you can make, just like, play with that resistor or something like that. So,


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[Assistant] So the same current flows through this guy and that guy. So whatever fluctuation you see here, it will be reflected here.


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[Assistant] Going through that resistor, now if this resistor is larger than that, the same current fluctuation will produce a larger voltage. Why is there an alpha here?


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[Assistant] So instead of thinking about it, so now people sometimes write this in a different way. They write it as


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[Assistant] Negative GMRC divided by 1 plus GMRE. This is a more classic textbook way of writing. In my opinion, it's an inferior way of writing.


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[Assistant] Don't take me wrong. We do write it from time to time this way, even if it's necessary. But if you're designing, this is the way you should be thinking about it.


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[Assistant] So the voltage gain is really the ratio of the total resistance in the collector to the total resistance in the emitter, times alpha. And you will see that essentially, once you are at a point where this Rm part is small compared to this Rc2Re,


[User] Please transcribe this text

[Assistant] The behavior, the primary parameters of the transistor, the exact specific IV characteristic of the transistor becomes irrelevant.


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[Assistant] If you put a resistor in the col- it's in terminal 2, and the resistor is in term- is terminal 1, if you have an ATD,


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[Assistant] So, that's an important thing, I mean, it may appear simple, but it's important. Now, of course, what happens to the IV characteristic for the large signal? So as you're doing this, so V in, V out,


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[Assistant] Because you are, the absolute slope would go down, really. The absolute value of the slope. It becomes shallower, right?


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[Assistant] And you may think that, okay, well this is gonna go all the way down, and you know the answer already. Right?


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[Assistant] It doesn't, because at some point this voltage becomes high enough that this junction, basically these two get close to each other. This device goes into saturation and beyond that essentially this guy will track that because this junction is forward biased. So you have something like this and then it may come back up if you just use it like this and calculate the V in V out.


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[Assistant] So that's a great question. The question is that do we need to really think about it as what's happening physically or do we think about it as small signal model? See, we are, we want to create abstractions that enable us to deal with more complex circuits. We'll soon we'll be dealing with like circuits with like 30, 40, 50 transistors in them.


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[Assistant] And it's good to have the abstractions handy in a way that allows you to think about it quickly and have a first pass thing. Like these kind of approximations. Well, if you want to make something like that, it's the ratio of these things. And we'll see a lot of these as we go.


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[Assistant] It's also important to know where these things come from, what are the underlying assumptions behind them, and what's happening physically. Because when things don't work out a certain way, or if you're looking for something that's not normal, something new or novel, a lot of times you can take advantage of that view.


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[Assistant] So the short answer is that you need to be able to switch it on and off in your head. Sometimes you just say, okay, I'm thinking only about abstractions. I want to make it as simple as possible.


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[Assistant] I don't want to think about what's happening to every individual electron that's in there, because it becomes intractable.


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[Assistant] Sometimes you have to go say, okay now, something interesting is happening, or I need something unusual, what's happening really inside, how can I use that?


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[Assistant] So it's the ability to go up and down these levels of abstraction that enables you to, but to do that you need to know what these are, what's happening at each one of these levels of abstraction.


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[Assistant] So that's a good question, somewhat of a philosophical question about it, but it's an important one.


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[Assistant] Get into too much nitty-gritty of this thing, right? So let's talk about that. So as I increase VIN, when VIN is zero,


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[Assistant] Right? What is Vout? Vout is at VDD, right? Or VCC. As I increase V in, I start drawing current through here, which basically means I start drawing current through here, which would mean that this voltage gets pulled down. Because I'm drawing more current out of it and it's just pulling it down. Now, this voltage is going up, this voltage is coming down.


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[Assistant] So this, what you see here, this red plot, is what that collector voltage is doing. But if you look at the input voltage, which is the base voltage, that voltage is actually going up, right?


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[Assistant] So that is essentially this voltage. It's going up. It's V in.


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[Assistant] So at some point, right, so think about it. There is a diode here, right? This is an NPN transistor, so it basically means that, uh, it's like this. There's a PN junction diode.


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[Assistant] So when this voltage becomes about 0.5 volt higher than that voltage, when the purple becomes about 0.5 volts or 0.6 volts greater than the red, what happens? This junction now is forward biased.


[User] Please transcribe this text

[Assistant] Right? So this diode is forward biased, which basically means that the transistor is saturated. There are two ways to think about that.


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[Assistant] If you, let's think about it as a diode. Now, as I increase V in, now this diode is forward biased, so its voltage is going to remain 0.5, 0.6, whatever, right, 0.7. So this voltage is going to track this voltage, with a 0.7 volt or 0.6 volt difference.


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[Assistant] So this was going down at the exponential, once it gets to, once this voltage becomes like 0.5 or 0.6 volts, then this guy tracks it, because now the transistor is really not in forward active region anymore. It's in triode. I'm sorry, in saturation. Right? So it just basically keeps going up.


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[Assistant] Another way to see that is that essentially this voltage is going to be, I'm sorry, yes, this voltage is going to be about 0.1 volts above that voltage.


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[Assistant] About the emitter voltage, right? What is the emitter voltage with respect to the base voltage? If the transistor is on, this is going to be .7 below it. Roughly.


[User] Please transcribe this text

[Assistant] Right, maybe on. I mean, it varies, but not much, because it's a logarithm. So that voltage is gonna be here. This is where it's like 0.7 smaller.


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[Assistant] This makes a good interview question, by the way. Let's see if somebody knows what's going on in the transistor.


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[Assistant] Uh, but anyway, so you get that red behavior. So it's kind of like very bizarre, right, like initially. But that's what actually happens. And if you look, and this was actually one of the problems you did. So uhm, so this is what's happening with it. Okay?


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[Assistant] Right, right, so basically what happened is that initially we had like something like this, without the, with Re equals zero.


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[Assistant] Yeaah, well the diet argument doesn't apply because what happens is that your emitter


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[Assistant] See, this voltage, as it goes, it happens over a short period of time anyway. So it does apply, but when it applies, essentially you are here. You've already made it here.


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[Assistant] Cause for that to happen, it means that this voltage now, if it's ground, if it's like, if there is no emitter degeneration. Right?


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[Assistant] But at that point means that you have already reached, this voltage has already reached, because this is also 0.7 above this. Which means that this is 0.2. So it already has to have reached here.


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[Assistant] So it does happen. The argument does apply. It just happens by the time it gets down to the bottom.


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[Assistant] Regardless, right, so you can think about this voltage being around 0.7, 0.8, right? So if this voltage is 0.7, for this to turn on, you need 0.5, which means that this voltage around is going to be 0.2. If there is no resistor here, this is going to happen. Of course, this is a simplified model, because there are internal resistors.


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[Assistant] Inside the transistor, right? You know, you had, you remember the base had a resistance in series, the collector has a resistance in series, the emitter has a resistance in series. So, if the, if you pick some lopsided values,


[User] Please transcribe this text

[Assistant] When you're drawing a lot of current, then the voltage drop across those resistors will matter. Because these voltages we are talking about are the voltages across the junction.


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[Assistant] Partly because of that. And also remember, this is not a ca- hard and fast number, I mean, or rule. This is an approximation that comes from the fact that the VBE is VT natural log of IC over IS. So it has a weak dependence on the collector current.


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[Assistant] Still it does have a collector current dependence. So logarithmic dependence. So for a range of currents that you're talking about typically, it may not change that much. And we'll do these numbers more accurately. Let me talk about biasing and all those things. But uhm, yeaah. Any other questions?


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[Assistant] Alright, okay, so let's talk about this, let's continue our discussion. So, so this is the gain of this common emitter, and this of course subsumes that. So you say, how does it subsume this case? Well, this is still valid, right? The total resistance in the collector is what? Rc. What is the total resistance in the emitter?


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[Assistant] Right, think about tmodel, alpha rm. So it's alpha, it's rc over rm. So this is generally true. Sometimes write that as gmrc, but you can even write that as rc over rm.


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[Assistant] What is the R in and R out? Let's talk about the R in first. The input resistance. What is the input resistance of this thing? Do you have any feel for what it is? First of all, let me ask a question. Is it alpha Rm plus Re?


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[Assistant] It appears that there's an alpha or M plus or E there, but is that the input resistance?


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[Assistant] And why, why not? The answer is shaking your head is the correct answer. Uh, uh, why not? Because there's this guy here too. So the key question is what is this current? In terms of IE. What is that current in terms of IE?


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[Assistant] 1 minus alpha E, right? Because this is IE, this is alpha IE.


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[Assistant] Okay, so, so that's the current. So what that, what does that tell? So I have an interesting situation. You have this resistor, and the current through this resistor is clearly determined by this voltage. So if you apply a Vx here, let's say V in becomes Vx, right?


[User] Please transcribe this text

[Assistant] So if you apply a vx and you're trying to monitor what ix is, vn is vx, therefore ie,


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[Assistant] The voltage drop across it. So that's IE. But the current that you experience, IX, is gonna be 1 minus alpha IE.


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[Assistant] So what is that? So you just multiply by this, you get one minus alpha over alpha Rm plus Re, uhm,


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[Assistant] Times vx. And therefore, if you calculate r in,


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[Assistant] If you do it blindly with equations, with algebra you get alpha over one minus alpha Rm.


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[Assistant] Some equations, some algebra, so it's not particularly useful unless we understand what these terms mean. This is again a known quantity.


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[Assistant] That's beta, right? What is this? 1 over 1 minus alpha.


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[Assistant] Remember, beta was defined as alpha over one minus alpha.


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[Assistant] So unless your calculation is really effective, there are very rare cases where that plus one makes a difference in the calculation because something cancels something else and you have the difference of two things.


[User] Please transcribe this text

[Assistant] But if it isn't, beta and beta plus one are the same thing. As far as we are concerned from a design perspective. Because beta is a parameter that's relatively large, and it's very inaccurate. Because it has a sensitive dependence on alpha.


[User] Please transcribe this text

[Assistant] No, I wouldn't get too hung up on this beta plus one a lot of times. Shouldn't bother you.


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[Assistant] If it makes your life easier to attempt to make calculations, that's the assumption that you need to make.


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[Assistant] Because it's built upon layers of assumptions already anyway. So, it's, but what is beta rm? Let's write this thing. Beta rm is r pi.


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[Assistant] So it's rpi plus, so this is basically, I would say rpi roughly times beta re.


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[Assistant] So if I want to remember it one way, this is probably not a bad way to remember it.


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[Assistant] So remember, rpi was the resistance you had in the non-degenerate case. The input resistance, right? Was rpi.


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[Assistant] Why do you see some of the R, what R, so the RE, the emitter resistance is multiplied by a factor of beta. Why, what has happened?


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[Assistant] Why does it multiply by a factor of beta, well, beta plus one? Yes.


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[Assistant] Correct, because the current is beta plus one times smaller, because it's the emitter current versus the base current, right?


[User] Please transcribe this text

[Assistant] That's the ratio of the emitter current to the base current is beta plus one.


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[Assistant] So since you have the same voltage applied across here, but the current is beta times smaller, so the resistance appears beta times larger. Or beta plus time, one larger. This actually...


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[Assistant] So what is the reflection rule? Reflection rule is a relatively straightforward thing. Uhm, so what it says is that, and it's easy to see,


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[Assistant] So let's say you have a situation like this, so you have alpha RM, you have an RE, you have a resistance in the emitter, and then you


[User] Please transcribe this text

[Assistant] Wanna see, and let's say you have a resistance in the base too. Let's say RB.


[User] Please transcribe this text

[Assistant] So it says that the volt, but Vx minus Vx prime is the voltage drop across the resistor, right?


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[Assistant] This is the voltage across this resistor. So, if I wanna have the same voltage drop across this resistor, but if I wanted to put it in the emitter, could I do that?


[User] Please transcribe this text

[Assistant] The behavior of this circuit doesn't change if I take this RB and put it in the emitter and make it, as long as I scale it down,


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[Assistant] Why is beta plus one better than one plus beta? Because it puts the primary emphasis on the beta, not one.


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[Assistant] Alright, so this is alpha IE, and this is IE. So it says that from the perspective of the voltages that you see here, and the currents, these two are equivalent.


[User] Please transcribe this text

[Assistant] And this is called reflection rule, so it says that if you have a resistor in the base,


[User] Please transcribe this text

[Assistant] So what we did, we took alpha RM plus RE and multiplied them by beta plus one. And that's what you get. Exactly. That's what you get. If you do the algebra.


[User] Please transcribe this text

[Assistant] So, this is a useful thing. So if you're trying to determine resistances and impedances, it says, look, you don't really need to write all these equations. If you see stuff in the base,


[User] Please transcribe this text

[Assistant] Here's, here's, that's what it is. So now, you could actually do it this way. You could say, okay, what I have in the base is alpha rm plus re times beta plus one, which would give you exactly that. So regardless, the input impedance is larger.


[User] Please transcribe this text

[Assistant] Then what you had in the case of a non-degenerate, right, in the non-degenerate case of when there was no resistance, it's increased by what? So this was the non-degenerate resistance, if you set RE to zero, you will see that it reduces to that resistance, it reduces to r-pi, and whatever we put in the emitter gets multiplied by beta.


[User] Please transcribe this text

[Assistant] So it makes it what kind of a good kind of a what? Sensor for what? Is it a good current sensor or voltage sensor? That has a high impedance. A voltmeter, a good voltmeter or a good ammeter? A voltmeter. It's a good voltmeter.


[User] Please transcribe this text

[Assistant] So, it becomes better for measuring voltages because it loads it less.


[User] Please transcribe this text

[Assistant] Okay, how about this output resistance? Well, if I look at this like this, look at it like this, the way it is right now, the output resistance is pretty simple to calculate, right? What is the output resistance? You null all the independent sources, which means that you set this to zero, you set this to zero, the voltage across these two resistors is zero, IE will be zero, this current source becomes zero, what is the output resistance? Just RC.


[User] Please transcribe this text

[Assistant] But you can see that that's incomplete because we have ignored r0 in these calculations. So what if we take r0 into account? Let's do that. Let's take the r0 into account and see what that gives us.


[User] Please transcribe this text

[Assistant] Let's go to the model again. So now this time what we are trying to do, so this is again our T model, alpha RM.


[User] Please transcribe this text

[Assistant] And then you have the RC on the top. So let's not worry about that because whatever resistance you will see, and I'll show you what I mean by that, will be in parallel with RC.


[User] Please transcribe this text

[Assistant] So let's say this is RE. So we are trying to determine the output resistance here. R out.


[User] Please transcribe this text

[Assistant] Do you agree that this R out will be the parallel combination of this resistance, the resistance from this node to ground, looking up, and the resistance from this node to ground, looking down? And what do you see looking up? Rc. To ground, right? What do you see looking down? We don't know, we have to find out.


[User] Please transcribe this text

[Assistant] How do we set that, find that out? We null the independent sources, which basically means that we ground this node for that calculation.


[User] Please transcribe this text

[Assistant] We'll deal with that later. We are trying to find out what this intrinsic are out of the stages.


[User] Please transcribe this text

[Assistant] So what is it? Well, we have IE, so this current is IEE, and then this is alpha IE.


[User] Please transcribe this text

[Assistant] And we are applying a test voltage here. So we will apply a test voltage of Vx.


[User] Please transcribe this text

[Assistant] And you want to measure Ix to find out what R out is.


[User] Please transcribe this text

[Assistant] Okay, how do we do that? Well it's relatively straightforward if you look at it carefully. This resistor and this resistor are in parallel. Do you see that? These two resistors are in parallel. Because this end is also grounded.


[User] Please transcribe this text

[Assistant] So, do you mind if I take this resistor, detach it from here, put it down here?


[User] Please transcribe this text

[Assistant] So this is what I'm doing. I'm just putting it down here, connect it to ground, alpha Rm, and remember the current is now going this way. This is IE.


[User] Please transcribe this text

[Assistant] And then this is also ground, so I just keep it simple.


[User] Please transcribe this text

[Assistant] So what we are trying to find out is Ie. In terms of Vx, so from that we can calculate Ix.


[User] Please transcribe this text

[Assistant] So the first thing we need to do is we need to estimate, so this branch, what is the current through this branch? It's Vx divided by this resistor of this three combination of these three resistors, right? So this current. We're trying to find out. Now,


[User] Please transcribe this text

[Assistant] It will become a little bit of a messy equation, you can solve it, and then you can simplify it.


[User] Please transcribe this text

[Assistant] Or, you can make an observation about the relative sizes of these values.


[User] Please transcribe this text

[Assistant] So do you agree that this parallel combination cannot be greater than alpha RM, or RM?


[User] Please transcribe this text

[Assistant] What's the relative, typical relative ratio of these two?


[User] Please transcribe this text

[Assistant] Large, ten thousand, thousand, depending on what the parameters are. So this is a, this is a much larger thing compared to that. So if this is like a hundred kilo ohm, this is gonna be a hundred ohms or fifty ohms or something like that. So do I need to really worry about the exact value of these things when I'm thinking about finding this value? No. So what I can say is that this current is essentially vx over ro.


[User] Please transcribe this text

[Assistant] Cause that's what the dominant part of this total resistance is. It's roughly RO. Cause it's RO in series with some small resistance.