[User] Please transcribe this text

[Assistant] Let's see from a practical implementation some of the variations and discuss some of the variations. Some of these have some interesting properties that we discussed for the sake of their interesting properties and some of them are things that we are interested in because they


[User] Please transcribe this text

[Assistant] I have implications in the design. So one of the things that we said, so let's just start.


[User] Please transcribe this text

[Assistant] Maybe keeping this here, move to the next whiteboard.


[User] Please transcribe this text

[Assistant] And you're trying to get gain, but let's say we have another constraint for the sake of argument right now. This constraint may or may not exist in your design, uh, depending on the parameters, but let's say the other constraint is that we, like we said, the passive devices are big and expensive. And active devices transistors are cheap and small.


[User] Please transcribe this text

[Assistant] So they are expensive because they are large. So let's say you wanna make the load, but you don't wanna make a load in such a way that a resistor is an expensive load.


[User] Please transcribe this text

[Assistant] Let's say, if you wanted to make that. And this, there are scenarios where this is the case. I mean, the resistor can be large, so you want to make something small, and you don't want to use a resistor. So what kind of loads can we use? What can we use instead of, so we initially had this Rd, right, to Vdd. So this was the Vin, and this was the Vout, and this guy had the W over L, L1, and


[User] Please transcribe this text

[Assistant] So it becomes a two terminal device, right? Essentially make it into a two terminal device to replace my resistor. Now, what are the properties of this load?


[User] Please transcribe this text

[Assistant] This is a good load, bad load. So I'm connecting it, connecting it again off of the VDD. So this is hanging off of the VDD here, the rail. And you have, so this becomes your V out.


[User] Please transcribe this text

[Assistant] The question is that, what are the properties of this load? This is by the way sometimes called the, quote unquote, diet-connected MOSFET.


[User] Please transcribe this text

[Assistant] Because if you do this to a bipolar transistor, you will get a p-n junction diode, right? Because you will get the base emitter diode.


[User] Please transcribe this text

[Assistant] That's how you make a diode out of a transistor, right? When you just basically short circuit the collector and the base, you're left with the base emitter diode. So this is also called a diode-connected MOSFET, although technically it's not a diode.


[User] Please transcribe this text

[Assistant] Well, it's a diode, it's not a p-n junction diode. I should take it back. It is a one-directional load.


[User] Please transcribe this text

[Assistant] Okay, but let's see what it does. So how do we know, what can we say about this? How can we characterize this? Well, we can look at the large signal one, so let's call this ID.


[User] Please transcribe this text

[Assistant] We can write a relationship between vout and id. Now, first thing we need to know is that what region of operation the transistor is in. Because they have different equations, right? Different regions of operation. So what region of operation is this transistor in, if your vt is positive? If your threshold voltage is positive? What is the con- what- What are the conditions for? What are the-


[User] Please transcribe this text

[Assistant] The requirements for different regions of operation. What are the constraints to be?


[User] Please transcribe this text

[Assistant] What is the requi- what, what, what do we need to happen? What is the- yes.


[User] Please transcribe this text

[Assistant] I mean, for us, our definition for the transistors to be on, of course we know that below that there's soft threshold currents, etc., etc., but for the basic classical model is


[User] Please transcribe this text

[Assistant] So basically, this voltage needs to be at least


[User] Please transcribe this text

[Assistant] Characteristics, so this is ID versus V out.


[User] Please transcribe this text

[Assistant] So if you, if this voltage is higher than VDD minus VT, the VGS is not even above the VT. So,


[User] Please transcribe this text

[Assistant] Other than sub-threshold, which we are ignoring for this time being, for a second, uhm, it's nothing. There's no current.


[User] Please transcribe this text

[Assistant] Again, we have to look at the constraints for different regions. There are two regions, right, when it's on. There's pinch-off and there's tri-off. The requirement for pinch-off


[User] Please transcribe this text

[Assistant] Do you agree that when you connect it this way VDS and VGS are the same?


[User] Please transcribe this text

[Assistant] So then VDS, if VT is positive, VDS is always greater than VGS minus VT. Right? Because it's the same number divided


[User] Please transcribe this text

[Assistant] So it's going to go into a pinch-off and remains in the pinch-off all along. So if it's in pinch-off, we know its current voltage characteristic. So then what we know is that we know VDD minus VT minus Vout.


[User] Please transcribe this text

[Assistant] Or let's say minus vout minus vt. Because this is vgs. This is the vgs. Vdd minus vout is the vgs here.


[User] Please transcribe this text

[Assistant] We are ignoring the channel life modulation for the time being.


[User] Please transcribe this text

[Assistant] That's basically just the basic quadratic equation that you get. Uhm, now,


[User] Please transcribe this text

[Assistant] And remember, now, so this is different from a resistive load line, because a resistive load line looks like this.


[User] Please transcribe this text

[Assistant] So this is a new load line we've defined with this load. So we are making it, essentially what we are doing, we are making a stage that looks like this.


[User] Please transcribe this text

[Assistant] W2 over L2, and that's what we are looking at. Okay, so with this, uh, what is the input-output characteristic?


[User] Please transcribe this text

[Assistant] So as we are changing the v in, this operation, with the resistive load line you went from this point to this point to this point, with this load line you're going from this point to this point to that point.


[User] Please transcribe this text

[Assistant] Is it better or worse than the resistive load line, in your opinion? And if, if it's better, why? If it's worse, why?


[User] Please transcribe this text

[Assistant] Let's see, let's write the expression. So we know this is id, right? So this is the id of this thing, which is the same thing as id here. So let's write the large signal equation. And that id is mu n cox over 2


[User] Please transcribe this text

[Assistant] Square. And this is really VT1 and this is really VT2 because that, because of the body effect the threshold voltages may differ. But, we can take it into account or not, doesn't matter.


[User] Please transcribe this text

[Assistant] They're both n-types, so you can cancel that. And then what you get also is you can divide both sides by double, you can solve for vout as a function of vn. And when you do that, you get vout is vdd minus vt2 minus


[User] Please transcribe this text

[Assistant] You get square root of W1 over L1, W2 over L2, V in minus VT1. So that becomes a V in V out relationship. And as you can see, if I ask you what's the gain,


[User] Please transcribe this text

[Assistant] If we ignore the body effect, so let's assume that VT is constant for now. It's not a function of V in or V out. What is the gain?


[User] Please transcribe this text

[Assistant] What is the small signal gain? You should be able to tell me from this expression, right?


[User] Please transcribe this text

[Assistant] Yeaah, negative of the square root. The ratio of the w over l's. The square root of the ratio of the w over l's, right?


[User] Please transcribe this text

[Assistant] The assumption is that there is no dependence on vt1 and vt2 on vout. But we know that vt2, if it's connected, if this is connected to ground, will be a function of the


[User] Please transcribe this text

[Assistant] You see the gain is basically the derivative. It's dv out to dv in, which is basically this.


[User] Please transcribe this text

[Assistant] Because our gains before were some sort of a ratio of resist RD over RM, right, or GMRD, right, for this stage. For this stage it was GMRD, right? Negative GMRD. So now it has become square root of W1 over L1 divided by W2 over L2.


[User] Please transcribe this text

[Assistant] It has less distortion. Okay, so, yes, correct. Can you elaborate a little bit more? Why does this have less distortion?


[User] Please transcribe this text

[Assistant] It's not the small signal one, of course the small signal one is always linear. By definition.


[User] Please transcribe this text

[Assistant] But this is a large signal characteristic that has become linear. And furthermore,


[User] Please transcribe this text

[Assistant] So your gain is defined by ratios of physical dimensions.


[User] Please transcribe this text

[Assistant] And this is usually a good thing, because it makes it less dependent on environmental parameters, things of that sort. If the temperature changes, things of that sort, that would not affect the gain as much.


[User] Please transcribe this text

[Assistant] So in general, we would like our gain to be defined as ratios of similar things.


[User] Please transcribe this text

[Assistant] And this is a general design principle that you will see over and over.


[User] Please transcribe this text

[Assistant] That ideally you want your game to be, if you want to make something that's robust.


[User] Please transcribe this text

[Assistant] That with temperature change and changes in the parameter of the supply or the process changes less, you want to make sure that your parameters of design, including gain, the parameters of interest, are defined as ratios of similar things. Ratio of resistors.


[User] Please transcribe this text

[Assistant] So that could be a positive thing about this thing. Yes, you'd get less gain, that is true, but that's a positive thing. So let's see, if I wanted to make this gain large, what do I need to do from a design perspective? Because my design parameters are w1 and l1. When you make a MOSFET, you can choose its w and you can choose its l. Of course there's a minimum length that you cannot go below, but you can make the w and l within reason some whatever you want.


[User] Please transcribe this text

[Assistant] So, how would you choose them if you want to increase your game? Yes.


[User] Please transcribe this text

[Assistant] Make L, exactly, make W two, this, this, this one small, this one large, this one large, this one small, right?


[User] Please transcribe this text

[Assistant] And you make a very long transistor, a very narrow one. Let's say another, I don't know, one to ten.


[User] Please transcribe this text

[Assistant] You can make it more, let's say. So you get a factor of a thousand, so you get the square root of a thousand, you get thirty. Ish. Gay.


[User] Please transcribe this text

[Assistant] But if you're looking for something that's stable and doesn't change that much, this may be something to consider.


[User] Please transcribe this text

[Assistant] Okay, and we can see it also from a small signal perspective. So let's do the small signal perspective really quickly. So from a small signal perspective, when the transistor is connected this way, let's look at the model and see what it looks like.


[User] Please transcribe this text

[Assistant] So, for that top transistor, so you have this grounded.


[User] Please transcribe this text

[Assistant] Because this terminal is actually shorter, so it's shorter to cross, so whatever current it has will flow inside this loop and would not affect anything outside, right?


[User] Please transcribe this text

[Assistant] So, I don't need to really worry about that. Do I need to worry about this current source? No, because it's shorted across both sides, so again, whatever current it induces. So what do I have looking into here? I have this resistor connected to ground here, this resistor connected to ground, and this resistor connected to ground.


[User] Please transcribe this text

[Assistant] So what is my total resistance? See? It's a parallel combination of those three resistors, right? So what we see here,


[User] Please transcribe this text

[Assistant] So RM1, RM, sorry, RMB, RM2, RMB2, and RO2. It's a parallel combination of those three resistors. That's what the load is.


[User] Please transcribe this text

[Assistant] And what do you see looking down, so again, going forward, we'll look at this and write RM2, parallel RMB2, and going forward we will not write RO2, but let's for now write it. Because that's much larger than those two. And looking down, what do you see?


[User] Please transcribe this text

[Assistant] What is that resistance you will see looking into the drain of this transistor?


[User] Please transcribe this text

[Assistant] R, R0 to R0. R01, right? Because the source is grounded here. Right? So you see basically R01.


[User] Please transcribe this text

[Assistant] And what is the drive? There's a current source driving this. It's the GM. So it's GM1, Vin. If this is Vin. So that is the AC drive current.


[User] Please transcribe this text

[Assistant] Now that AC drive current is being pulled out of that node that has this total resistance to ground.


[User] Please transcribe this text

[Assistant] RMB2, RM, RM2 parallel RMB2. So what is the total gain? It's negative GM1 from a small signal perspective. It's negative GM1 RM2 parallel RMB2. Now, this combination


[User] Please transcribe this text

[Assistant] So you can write, what is RM2 parallel RMB2? I think we went, we had that, uhm, here.


[User] Please transcribe this text

[Assistant] So in the previous whiteboard. And what we had here was basically, it was Rm over 1 plus chi. So this ratio is basically negative gm1 Rm2 over 1 plus chi.


[User] Please transcribe this text

[Assistant] Or you can write it as GM1 over GM2 divided by 1, 1 plus chi.


[User] Please transcribe this text

[Assistant] And if you want to see if you get the same result or not, let's express gm1 and gm2 for the specific transistor which we have, which is the


[User] Please transcribe this text

[Assistant] And if you just look at it the way it is, obviously these guys cancel.


[User] Please transcribe this text

[Assistant] And then you'll, well, if currents are the same, these guys cancel, and you get this ratio, but there's a correction factor of 1 over 1 plus chi. Because we had ignored body effect here,


[User] Please transcribe this text

[Assistant] Which now, if you take it into account, it scales the gain down by factor 1 over 1 plus chi.


[User] Please transcribe this text

[Assistant] So it reduces it by 10, 20% from that. But you can see the same result there.


[User] Please transcribe this text

[Assistant] And why is it linear? You look, you can see, kind of what you see really, I mean, it's easiest to see than the equations, but you can,


[User] Please transcribe this text

[Assistant] When you look at this nonlinearity, it's compensated by this nonlinearity, because they are the same nonlinearity.


[User] Please transcribe this text

[Assistant] That for an arbitrary device, where this relationship is some sort of a nonlinear relationship, since this relationship would be the same nonlinear relationship, if you make it for any arbitrary three terminal device, this kind of load, you can linearize it.


[User] Please transcribe this text

[Assistant] Because then you will have the inverse function of this and inverse function of that coming into play. The challenge would be that


[User] Please transcribe this text

[Assistant] And if this function is a very strong one, let's say for example an exponential, this inverse function is gonna almost kill your game.


[User] Please transcribe this text

[Assistant] The weaker the functionality of the drain current with respect to the input is, the more effective this technique would be.


[User] Please transcribe this text

[Assistant] For an arbitrary trans- three terminal device.


[User] Please transcribe this text

[Assistant] Right, so we can basically say if you, if you make this out of a arbitrary three terminal device, it will look like this, right?


[User] Please transcribe this text

[Assistant] So, these techniques are quite general if you think about them.


[User] Please transcribe this text

[Assistant] So you can always, if you need to linearize a device, you can always, this is one of the things that you can do. Have it fight itself.


[User] Please transcribe this text

[Assistant] Okay, so, so that's for that. Fine. We got the gain and we got the correction factor for the gain due to the, uh, uh, let's say,


[User] Please transcribe this text

[Assistant] Body effect, but let's talk about this a little bit more. Let's say, let me ask you this question. Let's look at this equation again. This was our gain.


[User] Please transcribe this text

[Assistant] And see if there are some modifications and changes we can make to increase this gain. So now let's say we are not worried too much about linearity. We are going back on our quest for large gain. Let's see what we can learn from this equation. Because equations, like I said, are useless.


[User] Please transcribe this text

[Assistant] If you don't understand what they mean, and if you don't understand the implications, then it's like that's coming out of them.


[User] Please transcribe this text

[Assistant] This or any other class or any other subject should not be exercised in writing equations.


[User] Please transcribe this text

[Assistant] We just write the sufficient minimum amount of equations that you need to get to the maximum amount of results. So anyway, here, let's go back to this equation.


[User] Please transcribe this text

[Assistant] And see, what else could we do to these two devices?


[User] Please transcribe this text

[Assistant] To make sure that we get more gain. If I wanted to increase my gain.


[User] Please transcribe this text

[Assistant] Well, you can say, I can get rid of your body effect, fine. That would give me 10% more.


[User] Please transcribe this text

[Assistant] What else can we do? Look at the equations and the parameters and see what we can do.


[User] Please transcribe this text

[Assistant] So can you think of, make suggestions based on those parameters? There are a couple of parameters that cancelled out, right? Yes.


[User] Please transcribe this text

[Assistant] So if you make it, okay, so that's an interesting suggestion. If you make the top transistor a PMOS, somehow, I mean we have to think about how we bias it, etc. If you made it out of a PMOS, would it help or hurt?


[User] Please transcribe this text

[Assistant] Which one would it change? It would change this. Why would PMOS help? You're right, but why would it help?


[User] Please transcribe this text

[Assistant] It has a smaller mobility, it has a smaller mu, right? So that's one way to do it. It will have some implications in the design, but yea.


[User] Please transcribe this text

[Assistant] Yeah, so in PMOS you can also get rid of the, connect the source to the bulk and you can get rid of that. So if you, but if you, well the PMOS would be upside down. It would be basically connected like this.


[User] Please transcribe this text

[Assistant] But yeaah, you're right. And then you also connect that source to the bulk. So.


[User] Please transcribe this text

[Assistant] Body effect goes away, so you get that 10%, and as you pointed out, and as you pointed out, you can also get the advantage of ratio of that.


[User] Please transcribe this text

[Assistant] So you get a square root of 3 improvement. Square root of 2, square root of 3. You get another factor. You get a factor of 1.5. Okay, so these are good suggestions.


[User] Please transcribe this text

[Assistant] But I want you to think about other suggestions you can make. And that goes back to design. Electronic design.


[User] Please transcribe this text

[Assistant] I mean, you can change the topology, you can add stuff, you can move things around. Yes.


[User] Please transcribe this text

[Assistant] The Seahawks, so that's a, that's a, that's a, true, I mean, that's a, you could try to change the Seahawks, but again, that's not a,


[User] Please transcribe this text

[Assistant] Electronic design perspective, that's really a device physics thing. And it's probably, out of all these parameters, the toughest one to change. In the sense that it is possible, but it means that you have to make a thick transistor. But now, what you're saying is not, uh,


[User] Please transcribe this text

[Assistant] Completely meritless. In modern processes, you sometimes have


[User] Please transcribe this text

[Assistant] And this is a very common option in many of the processes. Because if you want to deal with large voltages, the drivers for the outside circuitry, they have the thick gate oxide that can handle a couple of volts and then you have the


[User] Please transcribe this text

[Assistant] You have a cancellation of currents. So we have to change the current.


[User] Please transcribe this text

[Assistant] Well, so let's say you have a total fixed current. How can we change the current? We want to change the current in the sense that the current through NMOS and TMOS should not be the same.


[User] Please transcribe this text

[Assistant] You can add a parallel current source, right? You can steal some of this current away. You can reroute it. So not all the current that's going through here ends up here. So if you had an ideal current source,


[User] Please transcribe this text

[Assistant] Is not ID. This is ID over 10. Right? You have to think about it as design, right? Now if it's ID over 10,


[User] Please transcribe this text

[Assistant] That increases your gain by a factor of square root of 10. Which basically is another factor of 3. But if it's such a great thing, how about


[User] Please transcribe this text

[Assistant] And if it is such a great thing, why stop there?


[User] Please transcribe this text

[Assistant] 0.999. And make this a factor of a thousand. And we can keep going forever, but we are not.


[User] Please transcribe this text

[Assistant] There's something wrong in there, right? What is wrong? Two things are wrong, but go ahead.


[User] Please transcribe this text

[Assistant] Yeah, but even, but our equation is suggesting something that doesn't appear to be physical, right?


[User] Please transcribe this text

[Assistant] What is, what is the assumption, so we've, I've started violating some of the simplifying assumptions that I've made when I did this, when we did this, right? What are those simplifying assumptions that we violated?


[User] Please transcribe this text

[Assistant] ROs, right? Because ROs are not part of this expression.


[User] Please transcribe this text

[Assistant] Because we've ignored them, we assume that they are much larger than R. So now our O's start kicking in and say, well look, you know, you make, you're getting rid of that, but the R0 is sufficiently small compared to these other things that R0 matters. So your output resistance is not infinity.


[User] Please transcribe this text

[Assistant] And the idea again is similar to what we discussed about bipolar transistors, is that you can use a PFET, and keep it at some V bias, and you have a VDD, and you have a VIN, and you have a VOUT. So this is W over L,


[User] Please transcribe this text

[Assistant] When you start designing even in this class, when you start working with simulators and all, say, the first thing you need to verify is that if your transistor is really in the region you wanted it to operate, if, are your transistors in pinch-off region, or are they in some other region? Because if they're in some other region, all this small signal doesn't apply, really, the one that you're using.


[User] Please transcribe this text

[Assistant] So first things first. So what would happen as I increase? What is the V in V out characteristic of this stage? So this is V in V out DC characteristic.


[User] Please transcribe this text

[Assistant] How does that look like? Well, you start from VIN of zero, this transistor is off. Because it's input, VGS is smaller than VT. So you need to get to a VT.


[User] Please transcribe this text

[Assistant] ETN, really. And then you go there, and then it just starts turning on. As soon as it turns on, what region of operation is this going to be? This transistor. NFET is in.


[User] Please transcribe this text

[Assistant] Its VDS is relatively large, and VGS is small, so that's in pinch-off.


[User] Please transcribe this text

[Assistant] Well, what's the voltage here? What's the voltage drop across the drain source of the PFET?


[User] Please transcribe this text

[Assistant] Very small though, right? So it's very small, which basically means that its VGS is reasonably large, probably biased it properly so that it's going to be in trial.


[User] Please transcribe this text

[Assistant] So it starts off with PFAT in triode and NFAT in pin chop. Then...


[User] Please transcribe this text

[Assistant] If your voltages are right, there will be a region where both of them are in pinch off.


[User] Please transcribe this text

[Assistant] And then what happens once this voltage starts going too low,


[User] Please transcribe this text

[Assistant] Continuing going forward. So it's really this region.


[User] Please transcribe this text

[Assistant] You need to make sure that you design it in such a way that it is. Now, if you use it with feedback properly, the feedback would guarantee that it remains in this region. And that's the beauty of the system.


[User] Please transcribe this text

[Assistant] We are right now assuming that there's some mechanism that keeps it in that region. Which we haven't discussed yet.


[User] Please transcribe this text

[Assistant] This current is going to be gmvin, gmnvin. And what is the total resistance here, if they're both in pin job? This is rop, and this is ron. And therefore, the total gain of the stage


[User] Please transcribe this text

[Assistant] It's gonna be negative GMN, RON, parallel ROP.


[User] Please transcribe this text

[Assistant] Divided by, so let's just write it in more generalities, L divided by ID, some DXD, DVDS.


[User] Please transcribe this text

[Assistant] And you can plug in some numbers and get some values, but what, one of the things that you see that there's a proportionality with w and l and inversely with id. Which we saw before.


[User] Please transcribe this text

[Assistant] And we know that this fails for small ids, so it's not going to go to infinity for zero current, because you're now going to enter sub threshold.


[User] Please transcribe this text

[Assistant] And when you intercept threshold, this is gonna behave more like a bipolar transistor, it's gonna be exponential, so its gm is not gonna be this expression, its gm is going to be id over vt, which is gonna cancel this id, so you're gonna get to some fixed gain. So the gain versus current,


[User] Please transcribe this text

[Assistant] Is not gonna explode, as this equation suggests. Rather, it's gonna become a flat value once you get to subthreshold.


[User] Please transcribe this text

[Assistant] So yeaah, so that's what that is, and in terms of load lines, if you think about this, the way it looks is pretty much a straightforward thing. So if you look at the IV,


[User] Please transcribe this text

[Assistant] Okay, so, so this is the load line, id versus vds, or vout, let's say.


[User] Please transcribe this text

[Assistant] Tire Town School. Episode 29. The Plan.


[User] Please transcribe this text

[Assistant] Roar! Eddie's engine started.


[User] Please transcribe this text

[Assistant] Eddie looks fast today, said Jeb. Yes, said Dipper.


[User] Please transcribe this text

[Assistant] Mr. Grill came outside. Time for class, he said. Eddie drove over.


[User] Please transcribe this text

[Assistant] Mr. Grill smiled. We will stop, he said. Stop! cried Zippy. That's right, said Mr. Grill.


[User] Please transcribe this text

[Assistant] And they came in the wake of two terrible racial riots or massacres in the south, which really got northern attention.


[User] Please transcribe this text

[Assistant] Now, and then we'll use those to develop further stages and go to more complex stages. So these stages that you're talking about are essentially, if you think about them, are the Lego building blocks.


[User] Please transcribe this text

[Assistant] So we'll talk about that and we'll see how it works, but there are two more


[User] Please transcribe this text

[Assistant] The next block to think about is the counterpart of what we call the emitter follower or commons collector. So this would be basically common source or, I'm sorry, common drain or source follower.


[User] Please transcribe this text

[Assistant] And the output now is taken from the source. So this is vout and this is vin.


[User] Please transcribe this text

[Assistant] Now from an, and then this device has a W over L, and this is some sort of an RS. Now, for a device like this,


[User] Please transcribe this text

[Assistant] You can always think about writing the expressions, and you know what the ID is. ID is basically related to Vin minus Vout, because that's Vgs. So Vgs of this transistor is Vin minus Vout. So ID of this transistor is mu n Cox over 2 W over L Vin minus Vout minus Vt squared.


[User] Please transcribe this text

[Assistant] If you're really keen on solving this equation, you can go solve it, you get a quadratic solution of a quadratic equation that will have square roots in it and all those things, and you will get something.


[User] Please transcribe this text

[Assistant] The key part is that, that if this current is not changing a lot, the VGS minus VT,


[User] Please transcribe this text

[Assistant] Or VGS in general is not going to change that much. You can keep it relatively constant.


[User] Please transcribe this text

[Assistant] And if that's the case, this node will follow that. Now, the other interesting thing in the MOSFET, which is not as much of a parameter of design in bipolar, because the exponential is so much stronger than the quadratic, is that you can actually, by changing these parameters of the W and L, you can control what this voltage drop is.


[User] Please transcribe this text

[Assistant] So you can not only provide a follower where this output kind of like try to produce the input.


[User] Please transcribe this text

[Assistant] More or less faithfully. The other thing is that you can actually make this DC drop controllable. So you can actually use it as a level shifter, what we call. So you can actually use this to adjust the level of the DC level. So for example, if this is too high, you can lower it to this level.


[User] Please transcribe this text

[Assistant] This level shift, right? Do I need a larger W over L or a smaller W over L?


[User] Please transcribe this text

[Assistant] A smaller one, right? Because a smaller one would require a larger gate overdrive to produce the same current.


[User] Please transcribe this text

[Assistant] That was going through here, then you will see the same thing. But anyway. So now, from a small signal perspective,


[User] Please transcribe this text

[Assistant] What does this look like? Again, your source is not grounded, so T model would produce more natural results. So this is what you're looking at.


[User] Please transcribe this text

[Assistant] And this is going there. This is grounded. So this, this current source plays no role, really. Neither does this one, if you think about them. Right?


[User] Please transcribe this text

[Assistant] Because, first of all, this resistor is in parallel with the RS.


[User] Please transcribe this text

[Assistant] So I can bring it down here and put it as rmb here, if I'm really interested in iOS 2, which I'm not, I would write it down here. And then this current source is shorted on both sides.


[User] Please transcribe this text

[Assistant] So the current circulates inside through the ground, doesn't matter, doesn't really play a role. So this is my equivalent circuit.


[User] Please transcribe this text

[Assistant] Even this guy is not doing anything really because it's a voltage source on this side. So it's a voltage source parallel to the current source.


[User] Please transcribe this text

[Assistant] Now what's a voltage source in parallel, an ideal voltage source in parallel with an ideal current source? It's just a voltage source.


[User] Please transcribe this text

[Assistant] What's an ideal voltage source in series with an ideal current source?


[User] Please transcribe this text

[Assistant] Alright, if you want to get rid of RMB right here and there, now this is chi RM. No, I'm sorry, RM over chi.


[User] Please transcribe this text

[Assistant] So you can do the calculations this way. This is the parallel combination of RS and RM, uh, RMB. So you get RS parallel RMB.


[User] Please transcribe this text

[Assistant] And it's just some simplifications here of basic algebra. Let's quickly do that. So it becomes rsrm over rs plus rm divided by rsrm plus rsrmb plus uhm,


[User] Please transcribe this text

[Assistant] It's a little bit different because it's really, in here, you have the RS scale by OnePlus Kai.


[User] Please transcribe this text

[Assistant] And if body effect is not there, chi is gonna be 0, so you're gonna get back to that original calculation that we didn't write what we talked about. So, the gain is always less than 1. But, well, the input impedance is high. The input impedance of the previous stage was high too.


[User] Please transcribe this text

[Assistant] So in terms of impedance adjustment this stage is not particularly


[User] Please transcribe this text

[Assistant] Very useful because in mass transistors the input impedance is high anyway, so it's not as commonly used for impedance transformation for the input side.


[User] Please transcribe this text

[Assistant] And you will see this will have an impact on the topologies of op-amps when you do them in MOSFET and bipolar.


[User] Please transcribe this text

[Assistant] When you make a bipolar op-amp and a MOSFET op-amp, you will see that in bipolar op-amps, there's several stages, several of these stages used between the gain stages because you want to make sure that you present a large enough input resistance to the nodes that have a high impedance, your gain nodes.


[User] Please transcribe this text

[Assistant] The output resistance is smaller because the output resistance when you null the input is the parallel combination of these three resistors, right?


[User] Please transcribe this text

[Assistant] What is that? But the output resistance is RS parallel RM parallel RMB. So it's smaller than RM.


[User] Please transcribe this text

[Assistant] And this parallel combination, you can also write it as basically rs parallel rm over one plus chi.


[User] Please transcribe this text

[Assistant] And it's not a very large resistance. You need to have something with a relatively low source resistance, which basically means that you need a small output resistance. So as a driver, it can be quite useful.


[User] Please transcribe this text

[Assistant] And we will see how it's used again later. Uhm, so that's for what we call the common gate.


[User] Please transcribe this text

[Assistant] Stage. Sorry, common source. Common drain. I said everything. Common drain. Or source follower.


[User] Please transcribe this text

[Assistant] Now, the last stage we will be talking about is what we call the common gate, which is the counter, bipolar counterpart is common base stitch. So,


[User] Please transcribe this text

[Assistant] You can write a V in V out characteristic as well, right? I'm sorry, this is not V, this is V bias. This is a fixed voltage, and this is V.


[User] Please transcribe this text

[Assistant] So the input is applied to the source and the output is taken out of the drain.


[User] Please transcribe this text

[Assistant] You can do it large signal characteristic for this, which is simple. You can say, well, the drain current of this thing, ID, is mu n Cox over 2, W over L,


[User] Please transcribe this text

[Assistant] And that current is what's drawn out of the resistor to produce Vout. So Vout is going to be Vdd minus Rd.


[User] Please transcribe this text

[Assistant] ID. And you can plug this into here, and you will get a relationship between the out and the in.


[User] Please transcribe this text

[Assistant] And if you look at that equation carefully, what you will see is that it's a non-inverting stage. Meaning that as you increase your VIN,


[User] Please transcribe this text

[Assistant] Uhm, I am going to ask you this question. So think about the fourth trans- the second transistor, the fourth terminal. It's connected this way. What do you think is going to happen intuitively? Now you have two transistors in parallel that are actually all three terminals of these two transistors are in parallel from an AC perspective, right?


[User] Please transcribe this text

[Assistant] Cause from an AC perspective, this terminal is what? Is ground. Cause it's now fixed bias. The gate is ground. So is the bulk.


[User] Please transcribe this text

[Assistant] So now you have two transistors whose all three terminals are in parallel.


[User] Please transcribe this text

[Assistant] Now if you have all three terminals of a transist- two transistors in parallel, what do you expect their behavior to do? Kinda like, in aggregate. They're gonna help each other presumably, right? They're gonna do the same thing.


[User] Please transcribe this text

[Assistant] So let's see what body effect does in this case. And you can see it from the small signal model. Again, if you look at this. So you have an RD.


[User] Please transcribe this text

[Assistant] You have an IS1, you have an IS1RM, and then you have the back gate transistor, IS2,


[User] Please transcribe this text

[Assistant] RM2. Ground. And this is ground. And this is your V in. And this is your V out.


[User] Please transcribe this text

[Assistant] If you look at this very carefully, or just a little bit carefully, not even very carefully,


[User] Please transcribe this text

[Assistant] And this being, I call this iS total.


[User] Please transcribe this text

[Assistant] And then I have a current source, or I could make it change the direction, really. It's called this IS prime. And this is gonna be IS prime.


[User] Please transcribe this text

[Assistant] Going straight into the RD, which is basically the sum of those two currents flip direction. Do you agree that this is the same circuit? Because you can see that these two are in parallel.


[User] Please transcribe this text

[Assistant] One end grounded and one end shared. And these two are in parallel.


[User] Please transcribe this text

[Assistant] So what is the total gain of this thing? V in converts to this through RM and RMB and then multiplies by this. So you can easily see that AV is negative GM, oh, I'm sorry, not negative, GM, uhm, RD times 1 plus chi.


[User] Please transcribe this text

[Assistant] And if it's not obvious, just you need to expand this and divide this to calculate IS1 and plug it in here. So, 1 plus chi appears there too.


[User] Please transcribe this text

[Assistant] The ratio, the correction factor due to the back gate transistor. But this time it's appearing somewhere good.


[User] Please transcribe this text

[Assistant] Are the key ingredients of the design. But now, there are all sorts of other things. So let's go back to our common source.


[User] Please transcribe this text

[Assistant] And talk about the gain. So we said the last thing that we said before the break was that if you wanted to get gain, let's say from our common gate stage,


[User] Please transcribe this text

[Assistant] We had something like this. So you had the VDD, you had the ground, and then you had a V in, and we had a V out.


[User] Please transcribe this text

[Assistant] And then let's say we kept this as some v bias, v b1. This is w, let's say, 2 over L2, w1 over L1.


[User] Please transcribe this text

[Assistant] We calculated the gain of this stage, right? What was the gain of this stage? We said if there's a fluctuation V in here, this fluctuation,


[User] Please transcribe this text

[Assistant] And then it's gonna be pulled out of this node, which will have its own resistances. What are the small signal resistances you see? Looking up here, what do you see? Let's call this ROP, or RO2, let's say, or ROP, and then let's call this RO1.


[User] Please transcribe this text

[Assistant] So the gain, as we calculated before, of this thing is going to be negative gm1, ro1 parallel ro2.


[User] Please transcribe this text

[Assistant] Some meaningful numbers. So let's say mu and Cox is 50 microamps per volt squared. And let's say mu p Cox is 25 microamps per volt squared. This may be more characteristic of


[User] Please transcribe this text

[Assistant] Uh, not super th- short channel devices, but yeaah, I mean that's a starting point.


[User] Please transcribe this text

[Assistant] Of N of 0.1 micron per volts and let's say you have a DXD DVDS of P of 0.05 microns per volt. Uhm, yeaah. Let's just keep those. Again, these are not particularly short channel devices, but just to get a sense.


[User] Please transcribe this text

[Assistant] And then let's also say our W over L for these things, we pick them some, some, so let's say pick the W over L to be, I don't know, let's say 100 to 1 mic.


[User] Please transcribe this text

[Assistant] Let's say 100 microns 1. Again, this is a long channel device. But for those numbers, let's see what we get. So, gm, n, is gonna be square root of 2, uh, mu n, Cox,


[User] Please transcribe this text

[Assistant] We get 004 and 3, so it's like 10 to the 7 microamps. Ah, I should have picked something.


[User] Please transcribe this text

[Assistant] Let's make this 100 milliamp, uh, 100 microamps to make things, actually actually, that's probably even better. Microamps, make our lives a little bit easier. It's a design problem, we can design it the way we want.


[User] Please transcribe this text

[Assistant] So yeaah, so that basically gives me 1 milliamps per volt. So that gives me 1 millisiemens.


[User] Please transcribe this text

[Assistant] Let's say the W over L is the same, so I get square root of 2 of that, so I get 0.7 millisiemens.


[User] Please transcribe this text

[Assistant] So let's say both of them have a W over L of 100.


[User] Please transcribe this text

[Assistant] So I get 0.7 because mu Pc ox I said is like half of that mu Npc ox. Now also we have to calculate the Ron and Rop. Let's calculate those. So Ron is


[User] Please transcribe this text

[Assistant] Uhm, or divided by, let's say, 1 over 0.1 microns per volts for the N.


[User] Please transcribe this text

[Assistant] Right, so the micron cancels micron, you end up with, so ten, so you end up with, uhm,


[User] Please transcribe this text

[Assistant] Because 1 volt by 1 microamp would be 1 mega ohm. So this is like 100 kilo ohms. And for ROP,


[User] Please transcribe this text

[Assistant] Well, this ratio is half, so this becomes like 0.05, and the


[User] Please transcribe this text

[Assistant] Dimensions and the currents are the same, so you get 200 kilo ohms based on this.


[User] Please transcribe this text

[Assistant] So, so those are the parameters, those are the numbers I have, 100 kilo ohms, 200 kilo ohms, 1 millisiemens and 0.7 millisiemens, right, if I need it. So, here let's look at this. So what is the gain, at this, for this stage, that you calculate from this, so we have 1 millisiemens times


[User] Please transcribe this text

[Assistant] One hundred kilo ohm parallel, two hundred kilo ohm. What is that? It's, uh, sixty-seven kilo ohms, right? It's two thirds.


[User] Please transcribe this text

[Assistant] If we want to increase our gain for the same amount of current, let's say, what are we, what are our choices? So,


[User] Please transcribe this text

[Assistant] Always when we are trying to solve a problem, we have to first identify the biggest source of the challenge. The biggest source of the problem. What's limiting? So,


[User] Please transcribe this text

[Assistant] Let's see, what is the most limiting parameter of these three? What is the one that has the most impact on us?


[User] Please transcribe this text

[Assistant] Well, yes, well, GM has an impact, but GM would cost, generally speaking, right? It can cost either in current,


[User] Please transcribe this text

[Assistant] Or it can cost in W over L, so how does it cost in W over L? Well, we haven't talked about speed yet.


[User] Please transcribe this text

[Assistant] But we know that the capacitances are W times L. So if you start increasing the size of the device, then you get more capacitance.


[User] Please transcribe this text

[Assistant] And let's say we want to, what else is left? R01 and R02, right? If I could change one of them, which one would you change first?


[User] Please transcribe this text

[Assistant] R01, because that's the smaller one, that's the more limiting one. How would I do that? Well...


[User] Please transcribe this text

[Assistant] Would it help if I just put a resistor here and increase R1? R1? Yes? No? No?


[User] Please transcribe this text

[Assistant] Alright, again of the stage, because there's a ratio of this resistance to that resistance. So if you make this larger,


[User] Please transcribe this text

[Assistant] It's not going to help by itself. So we have to do it in a different way. But we can actually do it by cascode. If we introduce a second transistor here, what that does, so this is V bias 2.


[User] Please transcribe this text

[Assistant] So the output resistance of this thing will be RO3, so let's say this guy is transistor 3, in general W3 over L3, times what? 1 plus GM RS, which is now RO,


[User] Please transcribe this text

[Assistant] 1, so this is GM3, RO1, 1 plus chi.


[User] Please transcribe this text

[Assistant] Now, you can write all of these terms, or you can say roughly, this is RO3 times, well, it's GM3, RO3,


[User] Please transcribe this text

[Assistant] I'm essentially keeping this thing. If you want to keep the OnePlus car, you can keep it, but it's okay.


[User] Please transcribe this text

[Assistant] So it's roughly GM3R03R01. So if you're designing it, I'm really focused on this term.


[User] Please transcribe this text

[Assistant] And the good news is that there are a little bit of extra few terms that are increasing my output resistance, so I'm keeping some margin for myself. It's not a lot. So if my design works for this number, it's gonna work for this number too.


[User] Please transcribe this text

[Assistant] times R03, so it's basically, I'm saying that it's 100 times R03, whatever, R03. So my output resistance,


[User] Please transcribe this text

[Assistant] Is going up by the intrinsic gain of this stage times the output resistance of what's in the source.


[User] Please transcribe this text

[Assistant] Okay, so now, what is this number? So let's just calculate this number, right? So what did I have before? So it's gm3.


[User] Please transcribe this text

[Assistant] R02, which is the PFET. Now this we said is 10 mega ohm. What was this thing before? It was 200 kilo ohms.


[User] Please transcribe this text

[Assistant] Uh, parallel with 200 kilobes, it's 200 kilobes.


[User] Please transcribe this text

[Assistant] I mean, if you come and say, no, it's 197, 198, say well, you remember how many approximations we've been making, right? Don't bother.


[User] Please transcribe this text

[Assistant] I mean it's too waste of time. I mean honestly, if you start doing that too much, it shows that you don't know what you're doing.


[User] Please transcribe this text

[Assistant] And people do this all the time, but they don't know what they're doing. I mean, these numbers, this is what matters right now. Okay? So, it's gonna be 1 millisiemens times 200 kilo ohms. What is the gain? 200. So we went from 67 to 200. Okay. That was not much of an improvement.


[User] Please transcribe this text

[Assistant] So we went to 200. Okay, we got, well, we can say, oh, we got a factor of 3. But, we can do a lot more.


[User] Please transcribe this text

[Assistant] Now what if, how, what's the next thing to do? You have to take care of this.


[User] Please transcribe this text

[Assistant] Okay, let's go with the top one, right? So we just basically introduce another one here.


[User] Please transcribe this text

[Assistant] VB3 or whatever. Let's make this consistent. VB2, VB4, VB3.


[User] Please transcribe this text

[Assistant] And then you have a W4 over L4. Let's say you keep it the same as that one to make life easier on us. So what is this input resistance? What is the, well, the output resistance here?


[User] Please transcribe this text

[Assistant] It's this, whatever that was. What was that? We said 200 kilo ohms. So this was 200 kilo ohms times GMRO of this transistor.


[User] Please transcribe this text

[Assistant] Is 140, so you get 140 times that. Which becomes, what, 28 mega ohms? Right? Which is basically roughly, well, that, that is


[User] Please transcribe this text

[Assistant] So now, you replace this one with that parameter. So now you're going, we're going to GM R03 R01 GM.


[User] Please transcribe this text

[Assistant] Yeah, so that's a seven megaolms roughly, right?


[User] Please transcribe this text

[Assistant] 7 mega ohms times 1 millisiemens, what is the gain? So now, what's our gain? 1 millisiemens times 7 mega ohms. 7,000, right? 7,000. So now we are talking.


[User] Please transcribe this text

[Assistant] Right? So what's the next step if I want to get more gain? What should we do? Well in MOSFETs, unlike bipolar, there's no limit. If I introduce a third one,


[User] Please transcribe this text

[Assistant] So this is down, this is GM1, yeaah. Cause that's your drive, that's what converts your voltage to the current.


[User] Please transcribe this text

[Assistant] Uh, yeaah, this is because GM, no, this should be, sorry, this is incorrect. This is GM3. Thank you. Thank you.


[User] Please transcribe this text

[Assistant] Yes, thank you. That was it. That was an error. Uhm, yeaah. So...


[User] Please transcribe this text

[Assistant] That's, but now what's the next level? Next level is basically make it a triple cascode, right? Now, I mean, just for the sake of argument, let's do it and we'll discuss it.


[User] Please transcribe this text

[Assistant] So let's say that very quickly we do that. I'm not going to bother too much. What are we going to expect to see?


[User] Please transcribe this text

[Assistant] You're gonna expect these things to get multiplied by their respect and other GMRO.


[User] Please transcribe this text

[Assistant] Another factor of a hundred. So then it means that your resistances you're talking about are on the order of a gigaohm.


[User] Please transcribe this text

[Assistant] So if you could make a giga ohm resistor, you would get another factor of 100 increase or another.


[User] Please transcribe this text

[Assistant] Now you need to really think about the physics. You can't let your mathematical abstraction that you created fool you. It's kind of almost like idle worshiping, right? You make it and then you worship it. You have to realize you made this.


[User] Please transcribe this text

[Assistant] Whatever kind of idol you want to make, right? Math can be that idol too. So it's just basically...


[User] Please transcribe this text

[Assistant] You make some abstractions, yes, gigaohms. And then you get like, hundred thousands. What is, what is gonna break down?


[User] Please transcribe this text

[Assistant] Body effect. Body effect could come in. Yeah, body effect. But body effect is having some limited effect, right? It's a 10% effect.


[User] Please transcribe this text

[Assistant] There's something even, that's a good observation. We didn't take body effect into account. But there's something even more basic that's gonna happen. And make a gigaohm of resistance.


[User] Please transcribe this text

[Assistant] When you drop a voltage of one volt across a gigaohm resistor that's one gigaohm, what is the current that's flowing through it? A nanoamp.


[User] Please transcribe this text

[Assistant] So if there's anything, anywhere that's drawing a nanoamp of leakage current,


[User] Please transcribe this text

[Assistant] Would mean that the resistance on this node really is not a giggle. And you can't get it to giggles.


[User] Please transcribe this text

[Assistant] Even if you didn't have that, there's surface state, there's current that can pass through the imperfection of the surface.


[User] Please transcribe this text

[Assistant] If you have a discrete device, you touch it, there's, there's, there's some sort of a...


[User] Please transcribe this text

[Assistant] Grease or some sort of an impurity on the surface, some leakage current will pass through it. Even the plastic is not a perfect insulator.


[User] Please transcribe this text

[Assistant] And that happens with more modern transistors because that intrinsic gain is smaller. So it doesn't mean that triple cascode is a useless thing.


[User] Please transcribe this text

[Assistant] It just means that you have to be cognizant of the numbers and see if in some design you got to the gigaohm, then it means that double cascode was good. If you got to megaohm with triple cascode, then it means that you are still good with triple cascode. Now, one last thing before we go, is that this question of stacking, it seems that we are like doing it without any limit.


[User] Please transcribe this text

[Assistant] Supply voltage. Both of you said the same thing in different words. One said headroom, one said supply voltage. Basically, how much supply voltage, I mean, you have to, each one of them, you have to sustain a certain voltage on them, right?


[User] Please transcribe this text

[Assistant] So there are limitations to that. And that's why certain other stages, for example, a folded cascode, and we'll talk about folded cascode a little bit more later, but is designed. So you can actually take this and say if I had a current source here,


[User] Please transcribe this text

[Assistant] I can take my cascode and make one of the transistor out of an NMOS, but the second transistor, this one, I fold it back.


[User] Please transcribe this text

[Assistant] And bring it down here. So this, so this is v-bias and this is v-in.


[User] Please transcribe this text

[Assistant] Would be a folded cascode. Not to be confused with differential pair. Differential pair has similar transistors. These are two different kinds. Although they may look like that.


[User] Please transcribe this text

[Assistant] And the idea here is that the AC current that you produce here, so this is Vin, is producing some GM Vin, and then this GM Vin,


[User] Please transcribe this text

[Assistant] Is going through here and being pulled out of here. So this is GMVN. And this is whatever your load is.


[User] Please transcribe this text

[Assistant] You may say, eh, well, what about this current source? This current source is not ideal. It has some resistance, right? Sure, but let's look at the resistor values. What do you see looking into here? You see R0.


[User] Please transcribe this text

[Assistant] What do you see looking into here? If you make it out of a transistor, you see another RO.


[User] Please transcribe this text

[Assistant] Right? What do you see looking into here? You're looking into a source.


[User] Please transcribe this text

[Assistant] So this guy wouldn't do anything when it's in parallel with that guy.


[User] Please transcribe this text

[Assistant] Even if it's not, if it's not an ideal current source, which it never will be.


[User] Please transcribe this text

[Assistant] But it's okay. But by doing this, you'll break down the voltage. We'll talk about this breakdown of the voltage and all those things later as we go. But just wanted to give you a heads up that this is an artifact of the topologies that we choose.


[User] Please transcribe this text

[Assistant] Because of the way the output resistance scales. Because of that beta that was in one equation and wasn't in the other one.


[User] Please transcribe this text

[Assistant] Uh, differential amplifiers. We spent a fair amount of time thinking about differential signaling and differential amplification. And one of the keys to differential signaling and differential amplification was that you have a current source that basically you have two amplifiers, typically two common emitter or common source amplifier, although it's not limited to that. You can actually have different kinds of


[User] Please transcribe this text

[Assistant] Differential stage. You can take a common base and turn it into a differential stage. You can take even a


[User] Please transcribe this text

[Assistant] That looks like this. So you have a common source, basically you have a current source that's shared between the two stages, and then you have two resistive loads.


[User] Please transcribe this text

[Assistant] Rc, rc, you have an equivalent one, you can draw a mosfet version here if you want. Just leave it here for now.


[User] Please transcribe this text

[Assistant] And this is the essence, so in MOSFET this would be RD, RD this would be called ISS, there could be some resistance RSS here.


[User] Please transcribe this text

[Assistant] So the two transistors need to be matched, we talked about different ways of making them match well, we talked about common centroids for example, structures that allow you to make this differential


[User] Please transcribe this text

[Assistant] And there are different variations on that.


[User] Please transcribe this text

[Assistant] That allows you to get different levels of matching and accuracy. But anyways, so this was the S essential stage. And one of the things we did, we said, okay, well, you can, if you have an arbitrary input, you can break it into two segments. You can say there's a differential part, any two arbitrary inputs could be broken down, VID over 2 and negative VID over 2.


[User] Please transcribe this text

[Assistant] And then there was a common mode part, right, which you can show as


[User] Please transcribe this text

[Assistant] Vic and Vic, which would be the same on both sides.


[User] Please transcribe this text

[Assistant] And one of the things that we did is that, okay, if you have this anti-symmetric element, the red, which is the anti-symmetric differential signal,


[User] Please transcribe this text

[Assistant] What it does, it creates, we saw that from a small signal perspective, this node would be a, what we call differential ground or virtual ground. And as a result, because it basically one side goes up by the same amount that the other side goes down, again from a small signal perspective,


[User] Please transcribe this text

[Assistant] This becomes a virtual ground, and by substitution theorem, we said, okay, if it's not moving, you can replace it with a voltage source of the same value. From an AC perspective, it would be a voltage source of zero value. So this resistor becomes unimportant. Stuff here becomes unimportant. This reduces to an equivalent half circuit.


[User] Please transcribe this text

[Assistant] So this would be VOD over 2, negative VOD over 2. So you can take one of these half circuits and say well the differential equivalent half circuit looks like this.


[User] Please transcribe this text

[Assistant] And this is VOD over 2. And the gain of the state is the GM of this transistor.


[User] Please transcribe this text

[Assistant] Well, it's a matter of definition, right? I mean, so, so I said, if this is plus, apply plus minus, this is a taken plus minus, and that becomes an inverting stage.


[User] Please transcribe this text

[Assistant] This is a good point. I mean, it's more of a matter of convention, the way it's defined. But this convention allows this gain to be actually, makes it inverting. Now, if I take my output the other way, then it becomes non-inverting. And that's one of the strengths of the differential.


[User] Please transcribe this text

[Assistant] Amplifier. Depending on the way you connect it, you can turn it into an inverting or non-inverting amplifier.


[User] Please transcribe this text

[Assistant] Hmm. So that's a good question. So he says, well, should I view this as VCC or not? This is really a small signal analysis, right? Because what we are doing is a small signal. So this is really graph.


[User] Please transcribe this text

[Assistant] Sometimes I just draw a constant basically implying that from an AC perspective it's ground. But yea, that would be a better way of drawing it. Good questions. Any other questions?


[User] Please transcribe this text

[Assistant] Okay, so, so that's the differential equivalent half circuit, right? Now you can also have an equivalent half circuit for the common mode, right? So you can say this is VOC, this is VOC. Right? The blue stuff, the symmetric mode.


[User] Please transcribe this text

[Assistant] The symmetric mode, how would that look like? I mean, we did this breakdown again. Again, from an AC perspective, the DC peak, the constant current source is gonna disappear. What we did is that we saw that we could split this resistor into two


[User] Please transcribe this text

[Assistant] And two REE. And if this really bothers you, you can keep this here and then make it open circuit.


[User] Please transcribe this text

[Assistant] But if it's there, so this is IEE. So this from a large signal perspective is still the same circuit, right?


[User] Please transcribe this text

[Assistant] This is exactly the same circuit. Nothing has been happening. And from a small signal perspective, this disappears. And then when this disappears, what you see is that basically there would be a, the question you need to ask yourself is that which way is the current flowing from left to right to right to left? And the answer is that it can go, uh,


[User] Please transcribe this text

[Assistant] So this current has to be zero, then by substitute this current is zero, or small signal, rather.


[User] Please transcribe this text

[Assistant] And if it's small signal 0, then you can replace it by the current source of 0 value by substitution theorem again, which basically means that you have open circuited. So you can actually open circuit this.


[User] Please transcribe this text

[Assistant] Nothing will happen. Of course, from a small signal perspective, this will be gone anyway. So you can see that this two half circuits separate.


[User] Please transcribe this text

[Assistant] But the half circuit you end up with is a different half circuit, right?


[User] Please transcribe this text

[Assistant] It would be, and exactly the same thing happens for MOSFETs. So there's nothing fundamentally different.


[User] Please transcribe this text

[Assistant] And this would be vic and this would be voc, the common mode, the symmetric mode. The anti-symmetric mode is the differential one and the symmetric mode is the common mode.


[User] Please transcribe this text

[Assistant] And again, now we know how to calculate these gains, right? I mean we know, that's why we spend a fair amount of time understanding these individual building blocks, because that's what everything boils down to eventually. So it's the total resistance in the collector to the total resistance in the emitter.


[User] Please transcribe this text

[Assistant] Or total resistance in the drain to the total resistance in the source, or total resistance in terminal two to terminal one, et cetera, et cetera. So, it is going to the gain of this thing, obviously on the top, was...


[User] Please transcribe this text

[Assistant] RO parallel RC, if you wanted to think about it roughly, if RC is much smaller than RO, GM RC. And here, AV


[User] Please transcribe this text

[Assistant] Common mode is gonna be roughly the ratio of rc divided, there's an alpha there but we'll drop it, divided by rm plus 2ree. More accurately, it's alpha rc divided by alpha rm plus 2re. Uhm, yeaah. But we won't worry too much about that stuff.


[User] Please transcribe this text

[Assistant] Uhm, and then this you can also write it as negative GMRC divided by 1 plus 2 GMREE, the more textbook version of writing this expression.


[User] Please transcribe this text

[Assistant] And of course from these two we saw last time, and these are nothing new, CMRR,


[User] Please transcribe this text

[Assistant] The common mode rejection ratio, which is defined as the ratio of the differential gain to the common mode gain, which makes sense, which is the amount of suppression that you get. How much you amplify the differential and how much you attenuate the common mode is basically defined as this, and in this case you will get 2 plus gmRe.


[User] Please transcribe this text

[Assistant] 1 plus 2 gmRe, roughly, if Re is large, it's gonna be 2 gmRe.


[User] Please transcribe this text

[Assistant] So you can see the size of this, the output resistance here that you possess, that this output, the output resistance of this common, the current source is essential. And it makes sense because this current source's purpose was to keep this total current constant.


[User] Please transcribe this text

[Assistant] Right? So if there's a common, if something that changes in both of them, no, neither one of them change their operation point or current, and therefore the output voltage doesn't change.


[User] Please transcribe this text

[Assistant] And the less of an ideal current source you have, the smaller these resistors are, which is the one resistor really, the same thing, the lower this effect will be.


[User] Please transcribe this text

[Assistant] So we saw that, and that was the basis for the kind of basic operation. The next question is, how do we increase the gain? We did some gain calculations, right? So for example, I'm just picking some...


[User] Please transcribe this text

[Assistant] Nominal values, let's say, if this were, I don't know, 1 milliamp, right? What is your GM for each transistor? Well, each transistor, yes.


[User] Please transcribe this text

[Assistant] So, your question is that, what if the current source was really ideal? Meaning these resistors were infinity. So what would happen in that case?


[User] Please transcribe this text

[Assistant] Would you have any, so, so if you didn't have any channel length, any early voltage, or channel length modulation, right?


[User] Please transcribe this text

[Assistant] Your voltage cannot change here. I mean even if you did, if you have a current source that's fixed, there's a true current source, these two currents cannot change. Unless there's a differential voltage. So let's say there's no, the differential voltage is constant, right? So if I take this, move it up and down, as long as these transistors are in the forward active region,


[User] Please transcribe this text

[Assistant] There would not be a significant current drawn from the input terminal, the base or the gate, and therefore this current will, these two currents will remain the same. So your common mode rejection ratio is infinity in that case.


[User] Please transcribe this text

[Assistant] If any of these things bother you, put a resistor and take the limit to go to infinity. Yes.


[User] Please transcribe this text

[Assistant] Okay, so that's a great, that's a great question. So the question is that if I'm taking my differential, so, so half circuit is basically just capturing the,


[User] Please transcribe this text

[Assistant] The essence of the differential operation, right? So what really I should have written, which I didn't, uhm, is that I should have said AV differential is equal to VOD divided by VID, which is also equivalent to VOD over 2 divided by VID over 2.


[User] Please transcribe this text

[Assistant] But then the other question is that, why do you care? If you're taking the difference, output differentially anyway, you will not see any of the common mode, right?


[User] Please transcribe this text

[Assistant] If there's no mismatch, yes, exactly, so that's exactly right. So, there is another element which is the conversion of the common mode to differential. Which we haven't talked about yet, but we can talk about now.


[User] Please transcribe this text

[Assistant] Situations where this is not important. This becomes different and both of them are important. One is that if you're at some point, you're not taking your output fully differentially.


[User] Please transcribe this text

[Assistant] What if you're taking one of the outputs at some point? I will see examples of this. For example, a classic op-amp has a differential input, but has a single-ended output, right?


[User] Please transcribe this text

[Assistant] So then you have to really worry about how your common mode is changing. The second one, which is what you pointed out, is that if there's mismatch. So we can actually talk about that. Let's talk about it. Let's pick a new color, I don't know. Let's say this is plus delta r. Let's make that, right?


[User] Please transcribe this text

[Assistant] So if you have something like this now, so let's say one of the resistors is slightly different. Let's see what happens.


[User] Please transcribe this text

[Assistant] So if you have one of the resistors that's slightly higher, what happens is that on this side,


[User] Please transcribe this text

[Assistant] From the input perspective, if the early effect is not strong, still you are seeing the same behavior, right? But from the output, the same current would produce a different kind of output. So on this side, you produce what is basically gm vid over 2 times rc, right? Minus sign. This is the voltage you induce here.


[User] Please transcribe this text

[Assistant] But on this side, you will induce, so let me just get rid of these, negative gm vid over two rc plus delta r.


[User] Please transcribe this text

[Assistant] There is an other component here, right? Which is fine for now as far as differential is concerned, but now if you had a common mode,


[User] Please transcribe this text

[Assistant] Let's apply to these. What would you see on this side? The common mode induces on this side a negative, well, let's write it that way.


[User] Please transcribe this text

[Assistant] Now you can see that even in the presence of a purely common mode signal, you have induced a differential component, which is this.


[User] Please transcribe this text

[Assistant] Right? This is called common mode to differential conversion.


[User] Please transcribe this text

[Assistant] Because basically the mismatch has converted your common mode to differential mode.


[User] Please transcribe this text

[Assistant] And now once it's converted to differential, you can't get rid of it.


[User] Please transcribe this text

[Assistant] Because the circuit cannot differentiate between them, it's already become part of your signal.


[User] Please transcribe this text

[Assistant] Now, it is still attenuated by that factor, the common mode rejection ratio essentially, right?


[User] Please transcribe this text

[Assistant] But it is there. And that's why common mode rejection ratio matters because it even appears here. In that conversion. Did I answer your question?


[User] Please transcribe this text

[Assistant] No, no, no, but, but, so, so, remember, in a common mode case, the output voltages are the same. If you had ideal case, right? If this is going up and down, this is symmetric mode. So the two VOCs are the same.


[User] Please transcribe this text

[Assistant] No, but for common mode, what I'm trying to calculate, let's think about what physically we are trying to calculate. If I start wiggling both inputs together,


[User] Please transcribe this text

[Assistant] And if you're doing it together, if the circuit is ideal, if there's no mismatch like this, by symmetry the outputs need to move together too.


[User] Please transcribe this text

[Assistant] Right? So it's a question of the ratio of that fluctuation on the outputs that are moving together to the fluctuations on the input. And that's that definition. So this is really the definition of the common-mode gain. We define it that way for that physical. Now,


[User] Please transcribe this text

[Assistant] Granted, all of these issues pertaining to mismatch and all those things and conversion of the common mode to differential is there. But if you're thinking about purely common modes,


[User] Please transcribe this text

[Assistant] There is, it's a different representation. And this really goes back to, you remember we said any two arbitrary inputs to inputs could be broken down into an equivalent circuit like this.


[User] Please transcribe this text

[Assistant] So we are applying superposition essentially. We are applying these two once and then we are applying this separately.


[User] Please transcribe this text

[Assistant] And when we apply that, you can see that these two are shorted. When you are applying the IC, these are null.


[User] Please transcribe this text

[Assistant] They would be basically applying the same input to both and therefore by symmetry the same output will be present on both.


[User] Please transcribe this text

[Assistant] So now let's see what kind of gain are we talking about, right? Let's say we say IEE a 1 milliamp. So what kind of GM are we gonna get? What is the GM of each transistor?


[User] Please transcribe this text

[Assistant] So if I have 1 milliamp down here, how much do each one of my transistors get?


[User] Please transcribe this text

[Assistant] Half a milliamp. So my GM is gonna be 20 millisiemens, right? This is one of the things to keep in mind. Although the expressions look the same.


[User] Please transcribe this text

[Assistant] You're, for the same amount of current, you're getting half of the gm in a differential pair.


[User] Please transcribe this text

[Assistant] So you have to keep that in mind, I mean you're, you're using, you're breaking your current into two, I mean dividing it into two equal halves. So for each transistor, so IC is gonna be half a milliamp.


[User] Please transcribe this text

[Assistant] And therefore, gm for a bipolar would be 20, which is easier to calculate, 20 milliamps, 20 millisiemens, sorry. Millisiemens.


[User] Please transcribe this text

[Assistant] So that's your, that's, that allows you to calculate again. So let's see.


[User] Please transcribe this text

[Assistant] Let's say, what kind of supply are we talking about? Let's say you have a supply of, I don't know, three volts. And let's say this is ground. So, how should I pick this resistor if I'm running at one milliamp?


[User] Please transcribe this text

[Assistant] Well, I want to make sure that those transistors don't go into, uh, saturation, right, the bipolars. So, and let's say roughly this, if I had made this out of a current, this current source out of a transistor, the lowest this can go is some sort of Vsat, 0.1, 0.2.


[User] Please transcribe this text

[Assistant] So if this is at 0.1, 0.2, this means that this can go, this voltage can be as low as another 0.1, 0.2 above the ground. So let's say it can be 0.3, 0.4 above the ground. So I want to make sure that if, when all the current flows in one, what's the worst case scenario? The worst case scenario is that the state is completely steered to one side. Meaning all, almost all of the current is going through one branch.


[User] Please transcribe this text

[Assistant] From a DC perspective. And in that case, you're carrying 1 milliamp through RC and this voltage should not go below .3, let's say .4 volts.


[User] Please transcribe this text

[Assistant] Below the ground. So that gives you a very simple calculation, right? It said 3 volts minus .4, 2.6 volts,


[User] Please transcribe this text

[Assistant] Divided by this resistor is 1 milliamp. In other words, this resistor needs to be 2.0 greater than 2.6 kilo ohms. Is that clear? Let me just write this down while I just...


[User] Please transcribe this text

[Assistant] So, let's say you don't want this to exceed 0.2 volts. Let's say your piece has 0.2 volts, right? Ah, let me use a different color.


[User] Please transcribe this text

[Assistant] 0.2 volts. So, and you don't want to have another 0.2 volts here. So that basically means that the lowest of this voltage is 0.4 volts. This was at 3 volts. Right?


[User] Please transcribe this text

[Assistant] So when all the current flows through one side, meaning that this is 1 milliamp, this voltage should be 2.6 volts. Or no greater than 2.6 volts. Which basically for 2.6 volts that would be


[User] Please transcribe this text

[Assistant] To get 1 milliamp through something, produce 1, 2.6 volts, that thing has to be 2.6 kilo ohms, right? So let's say 2.5. To make things round and nice, right?


[User] Please transcribe this text

[Assistant] Well, we calculated the GM. What's the differential gain? Let's talk about that. The differential gain. GM is 20 millisiemens, 2.5 kilo ohms. What's the gain? 50. Right? So we have a gain of 50 in this.


[User] Please transcribe this text

[Assistant] Increase the resistance to increase the gain just by, instead of 2.5 kilo, what if I put a 10 kilo on there? What would happen?


[User] Please transcribe this text

[Assistant] So what if I put 10 kilo ohms here, if I could pass 1 milliamp, let's say if I had an ideal current source, what would this voltage be? In the limit, 10 volts, right?


[User] Please transcribe this text

[Assistant] When it's there to one side. So it would be 10 volts below 3 volts, so it would be negative 7 volts.


[User] Please transcribe this text

[Assistant] Now I don't have an ideal current source, so that I can't basically just go below that, so basically what it means is that if I put 10K ohms here, this transistor will go into saturation.


[User] Please transcribe this text

[Assistant] So it's not going to be operating properly, you put it in spice, you expect to gain a 50 or 100, you get to gain a .1.


[User] Please transcribe this text

[Assistant] So the transistor's not really operating in the proper range of operation. So, it seems like a dilemma, but you may say, okay, wait, how about you make this four times larger, make it ten kiloohms, but make this current four times smaller? That way the DC will work out, right?


[User] Please transcribe this text

[Assistant] GM, how, how different would it be in that case? What would it be if I reduced this by a factor of four? The GM would be what?


[User] Please transcribe this text

[Assistant] GM is proportional to IC, right? It's IC over VT.


[User] Please transcribe this text

[Assistant] If I make this factor of 4 small, it will become 5, right? It will become a factor of 4 small. So it becomes 5 millisiemens. Your resistor is 10 kilo ohms. What is the gain again?


[User] Please transcribe this text

[Assistant] Back 50, right? So we haven't gained much. In terms of gain. So, what happened here?


[User] Please transcribe this text

[Assistant] Well, you lost some GM, you get, you increase the resistance. Which one of these stages is better?


[User] Please transcribe this text

[Assistant] You have two choices, right, it seems to be presented, and you can of course take it anywhere on that spectrum of changing one, turning that knob, right? So, you had one with 1 milliamp at the tail, 2.5 kiloohms here, and you had one with, I don't know, a quarter of an amp here, and then 10 kiloohms here. And they have the same gain, differential gain-wise they're the same.


[User] Please transcribe this text

[Assistant] Is there any benefit, benefit, disadvantage or disadvantage to either one? Tradeoffs.


[User] Please transcribe this text

[Assistant] More robust. The larger current is more robust. The smaller current will happen.


[User] Please transcribe this text

[Assistant] So, both statements are good statements. Can you elaborate a little bit on how it would be robust? And then we have Alex.


[User] Please transcribe this text

[Assistant] That is correct. No, no, no, that's good, good, that's good conversation. So GM is going down. Right? That's right. But this REE, where is it coming from?


[User] Please transcribe this text

[Assistant] This is the output resistance of the transistor you use to make this. This represents that output resistance, right?


[User] Please transcribe this text

[Assistant] What is output resistance proportional to? It's basically, or given for bipolar, it's VA over IC. So, this, also this product remains constant, because this goes up and this goes down, so come on. There's something else we haven't really extensively talked about, but


[User] Please transcribe this text

[Assistant] In the light of our model, to think about these things, what determines the speed of how fast I can move something? How fast I can shake something, right?


[User] Please transcribe this text

[Assistant] There are two things really, right, that determine to first order how fast I can shake this. I can shake this a little bit faster than I can shake this in general, right? Well if I take this chair, it would be harder to shake it as fast. Why? This is heavier, right? What is that mass? What is that?


[User] Please transcribe this text

[Assistant] State of being heavy in electrical equivalence. It's the reactances, right, or capacitances, for example. Right? So you have larger capacitance, it slows you down.


[User] Please transcribe this text

[Assistant] The time constant is a product of capacitors and resistors or the ratio of inductors and resistors.


[User] Please transcribe this text

[Assistant] And we'll see that. But here's what's happening. So, now the other thing that controls how fast I can move this is how strong I am. Right? For a given weight. Now, what is that strength in electrical equivalence?


[User] Please transcribe this text

[Assistant] One of those stages, the one with more current versus the one with less current.


[User] Please transcribe this text

[Assistant] The higher current one, right? There are two ways to see that. One is that, okay, I have more drive current, I have a higher GM, so I can move this faster.


[User] Please transcribe this text

[Assistant] Which means that the time constants here are four times larger. These two are one and the same.


[User] Please transcribe this text

[Assistant] And we'll see this in ext- extensively when we talk about high frequency behavior.


[User] Please transcribe this text

[Assistant] Yet I cannot re- increase my resistance and increase my gain by increasing the resistance. Because that's what we need to do, right? I mean, from this expression you can see that you need to increase Rc and at some point it becomes comparable to Ro and that's where basically you get the higher gain. Where is this coming from fundamentally, if you think about it? When you use a resistive load. Yes?


[User] Please transcribe this text

[Assistant] Headroom, correct. And where does headroom limitations come from? It's basically there's a direct coupling between the DC and AC behavior of these two things.


[User] Please transcribe this text

[Assistant] Alright, the DC drop across here is determined by exactly the same parameter as the AC performance gain, right, which is both RC. RC determines both the gain as well as the DC drop you have here.


[User] Please transcribe this text

[Assistant] These two are defined by the same parameters so they are strongly coupled to each other so you can't, you want to ideally decouple these two if you could. How do you decouple them? You want something that gives you high


[User] Please transcribe this text

[Assistant] AC output resistance, but does not limit your current. And that's a well defined stage, really, if you think about it. It's a recurring theme, essentially, really, if you think about it.


[User] Please transcribe this text

[Assistant] What is it? What, what ideally, it's, you can think about it also in terms of load lines. We talked about this, right? If you look at this IV characteristic for MOSFET or bipolar, it doesn't matter. You, basically what you have here is that this kind of load represents this kind of load line.


[User] Please transcribe this text

[Assistant] Where you get more variations on the output, but you're getting less and less resist- eh, eh, current. Because you're increasing the resistance, right? This is i versus v. So what kind of load lines do you like? You want something that's more horizontal.


[User] Please transcribe this text

[Assistant] Something that can maintain current, but does not represent a lot of fluctuations with the current, which is basically a current source.


[User] Please transcribe this text

[Assistant] And how do we present a current source? How do we, so we basically want to have current sources here. The same way that we did it in single-ended stages, right? Because this is not fundamentally different, this is the same thing.


[User] Please transcribe this text

[Assistant] And we can do it in MOSFET and bipolar and we'll do the MOSFET shortly. I want current sources here as load. Right?


[User] Please transcribe this text

[Assistant] But you want, what kind of current source do you want? Do you want something that can source current? And that would be a PMP.


[User] Please transcribe this text

[Assistant] So if you keep this as some fee bias, fixed fee bias, then we can talk about how this is generated.


[User] Please transcribe this text

[Assistant] Something that can provide DC current, maintain the DC current, while presenting a high output resistance.


[User] Please transcribe this text

[Assistant] And, so therefore each one of the transistors, gm's would be, gmn, which would be gm, would be ic over vt, half a milliamp each, so that's 20 millisiemens, it's 1 over 50 ohms, for rm.


[User] Please transcribe this text

[Assistant] So this is gonna be gmVid over 2. So that's the current that's being pulled out of that node. That's the fluctuation, that's the small signal current. And the resistance on that node is of course the parallel combination of these two resistances, Rop and Ron. Right?


[User] Please transcribe this text

[Assistant] And what is ROP? What are ROP and ROR? Well, the ROs...


[User] Please transcribe this text

[Assistant] Over IC. So, let's say, for the sake of argument, this is 100 volts, this is 50 volts. So let's make this 100 volts divided by half a milliamp. And this is gonna be 50 volts divided by half a milliamp. So that's gonna be what? 200 kiloohms. And this is gonna be 100 kiloohms. I mean, the exact numbers really don't matter, it depends on the parameters.


[User] Please transcribe this text

[Assistant] I want to see. So you have a parallel combination. So what is your gain? If I ask you what's the gain, it's gm times the parallel combination of these two resistors, right?


[User] Please transcribe this text

[Assistant] The differential gain, which would be 20 millisiemens times whatever, 17 kilo ohms. Just a parallel combination of these two. So, gain.


[User] Please transcribe this text

[Assistant] Right, AB differential would be, uhm, negative 20 millisiemens.


[User] Please transcribe this text

[Assistant] Negative 20 millisiemens times, uh, 70 kilo ohm roughly, which is a parallel combination of those two. Uhm, so that's what, what is that?


[User] Please transcribe this text

[Assistant] You're at 1400 now. There's a substantially higher gain you can achieve with something like this.


[User] Please transcribe this text

[Assistant] Uhm, and if you were in MOSFETs, it would be less, because then you get less GM for the same current, and we'll do the numbers with MOSFETs too, you will see, but you can, you will still get higher gain than that. So that's basically what you get, uh, for this system. Now,


[User] Please transcribe this text

[Assistant] What's the next, and we are not going to be doing it because we've already done it, but let's say you wanted to go even beyond that from the same stage. What would you do? You need to increase your upward resistances, right? We know how to do that, right? But we cascode these two, right? So we make this like double,


[User] Please transcribe this text

[Assistant] NPNs and double TMPs and then you get a mock factor of beta, roughly, depending on how you do it.


[User] Please transcribe this text

[Assistant] Take a single-ended output. You don't need two outputs, you just have one. Let's say something that goes to the next stage that has a single-ended input.


[User] Please transcribe this text

[Assistant] It's to do what? It's just to take it from one of these notes, right? And just discard the other one. What happens if I do that?


[User] Please transcribe this text

[Assistant] Yes? My gain is half, right? Why is my gain half?


[User] Please transcribe this text

[Assistant] This is not VOD anymore. This is not VOD over 2 anymore. This is just VOD. If this is my output, let's say. One of these is my output, right? We can define it as a VOD. So it becomes VOD over VID over 2. Right?


[User] Please transcribe this text

[Assistant] So what you see here is twice the gain, because this is not here. So the gain would be half of that, as a result. Right?


[User] Please transcribe this text

[Assistant] So you get half as much gain. So 1400 becomes 700. You say, okay, I'm okay with that. I can, I mean that's plenty of gain, sure. If it's enough for you, you can take one output and then discard the other one. Yes.


[User] Please transcribe this text

[Assistant] Exactly. It also degrades your common mode rejection ratio. That's a very good point. Because, right now, what happens is that whatever, well, okay, so let me just rephrase. Your common mode rejection ratio is still the same.


[User] Please transcribe this text

[Assistant] Because if you, if you keep the resistor there, if you keep both resistors, right, the stage is symmetrical.


[User] Please transcribe this text

[Assistant] Or the next stages. Yeaah. So I'm assuming right now that that load is not presenting a major problem. And it's a big assumption.


[User] Please transcribe this text

[Assistant] We'll see, we'll see that. That this is actually an assumption we have to deal with.


[User] Please transcribe this text

[Assistant] But assuming that you have a very high impedance load that's not going to load this stage, which is not easy to do, but you will see how to do it, uhm, it is, yeaah, fine.


[User] Please transcribe this text

[Assistant] But yeah, you're right. If you're loading one side differently from the other side, you're absolutely right. You're basically introducing a change, delta r, that would also induce a


[User] Please transcribe this text

[Assistant] An additional conversion from common mode to differential, but you're not dealing with differential anyway.


[User] Please transcribe this text

[Assistant] Once you take your apple single-endedly, you have lost all differentiation between single-ended and common mode. You get what you get. So you're at the mercy of this factor already. At that point. You don't get anything beyond that. Whatever you got already, you got it.


[User] Please transcribe this text

[Assistant] Or repurposing this other half of the game. Can we kind of reclaim it, if you will?


[User] Please transcribe this text

[Assistant] Is there a way to not throw it away, but use it in a meaningful fashion? Because remember, if I'm taking the output only from one side, I'm throwing away the other half.


[User] Please transcribe this text

[Assistant] Is there a way to repurpose this or reclaim this game that I already have, just in the wrong place? Can we think of a solution?


[User] Please transcribe this text

[Assistant] Inverted and added. Okay, so, so, so that's one suggestion. You had a suggestion?


[User] Please transcribe this text

[Assistant] So inverted and added somehow to that. Okay, how do we do that though? Right?


[User] Please transcribe this text

[Assistant] What's the easiest way to add two things in circuits?


[User] Please transcribe this text

[Assistant] Two current sources. Put them in parallel. Two current sources in parallel.


[User] Please transcribe this text

[Assistant] And the reason that's easier than putting two voltage sources in series


[User] Please transcribe this text

[Assistant] The other thing, the other question is, okay, how do we do that inversion?


[User] Please transcribe this text

[Assistant] Because again, these blocks need to be implemented, right? With some, with transistors.


[User] Please transcribe this text

[Assistant] At the end, we want to have something that has only transistors or resistors and things of that sort in it. We don't want things with boxes and voltage sources and current sources because those are not the things that you can go buy or put on a chip.


[User] Please transcribe this text

[Assistant] What if we took this current fluctuation and reflect it down here? What would happen?


[User] Please transcribe this text

[Assistant] See if you can actually do this. So what do I mean by that? So let's take the same stage at the input. So you have the same differential pair.


[User] Please transcribe this text

[Assistant] With the tail current source, you may have this, you may not have it, doesn't matter. Uhm, and now you have a current.


[User] Please transcribe this text

[Assistant] So we are not using this side, so I'm saying, well, this side I was not using, what if I mirror it back here? Reflect it back here. I'm using an optics analogy here, right? Of course.


[User] Please transcribe this text

[Assistant] Well, one way, if it's BJTs for example, you can use a PFET current meter. So instead of using two, what if I use something like this? Now remember, if you ignore the base current,


[User] Please transcribe this text

[Assistant] Whatever this current is, is going to be reflected here. As long as this transistor is running the forward active region. Because they share the same base emitter voltage.


[User] Please transcribe this text

[Assistant] So if they share the same base emitter voltage, they will, if they are in forward active region, they are bound to have pretty much the same current. Ignoring the early effect for now.


[User] Please transcribe this text

[Assistant] So, if I pull a certain current out of here, it would be, or if it's sourcing a certain amount of current, it would be sourcing the same amount of current here.


[User] Please transcribe this text

[Assistant] Again, I'm ignoring the base current because there will be some base current here that I'm just assuming is very small. First order analysis. And we can do a more detailed analysis.


[User] Please transcribe this text

[Assistant] If necessary. But so if going to that, so if this is VID over 2 and this is negative VID over 2.


[User] Please transcribe this text

[Assistant] And this becomes gmvid over 2. And this would be what? Negative gmvid over 2, right? That's the small signal fluctuation. Those are the small fluctuation.


[User] Please transcribe this text

[Assistant] Now look at the polarities of these things. This is negative GMV ID. So I want everyone to look here. Look at this for a second.


[User] Please transcribe this text

[Assistant] So, do you agree that I can get, make this a plus and change the polarity of this thing?


[User] Please transcribe this text

[Assistant] Both of these two currents are being injected into that node. Look at the polarities, right? And they are the same.


[User] Please transcribe this text

[Assistant] So this guy is forcing GMVID over 2 into this node, and this one through the mirror is also injecting GMVID through this node. So if I ask you what is the gain,


[User] Please transcribe this text

[Assistant] What would you tell me? What is the total current? The two current sources both injecting current into the snow. That's your summation, right?


[User] Please transcribe this text

[Assistant] So, what's the total current being injected into this node? GMVID, right? So it's basically some of these two. And what is the total resistance in this node? That's how you calculate the time in the gate, right? The total impedance on this node, or resistance now, is this resistance in parallel with that resistance, right? What is this resistance looking up? Let's call it ROP. This is RON. Right? So you have a total current of


[User] Please transcribe this text

[Assistant] But the collector emitter voltages are not necessarily the same.


[User] Please transcribe this text

[Assistant] Right? Because the base emitter is the same, they're just connected, they're shorter to each other. Now, so you're concerned about the base current, I guess, or, or not, or the voltage.


[User] Please transcribe this text

[Assistant] How you went from that to this? Okay, here, this is truly symmetrical.


[User] Please transcribe this text

[Assistant] Right? Because these two are biased with same voltage, and they're fixed.


[User] Please transcribe this text

[Assistant] What we are doing here is that we are saying that well, this base emitter is not fixed, but it's the same for both of them.


[User] Please transcribe this text

[Assistant] Now, yeaah, it's kind of floating in the sense, floating in the sense that it's not connected to anything, but these base currents are being, for PNP, are coming out, right? You have two things that need to provide DC current.


[User] Please transcribe this text

[Assistant] Yes, small, but still DC current that needs to go somewhere. If you just leave it open, this will accumulate, it will just go to a voltage and it will saturate and nothing will happen. So there needs to be a pathway for this current to be drained out. This will shunt it out, right?