[Music] [Music] [Music] [Music] [Music] [Music] greetings welcome to electronic circuits one I am basil Razavi and this is lecture number 32 today we will look at the mass device in more depth and understand the one of the interesting second-order effects that occurs in this device and how that translates to the device characteristics first let's look at what we did last time and remember the equations that we derived for the MOSFET so last time we went past this boundary here this boundary here and we allowed VDS to increase beyond the vgs minus vth and what we observed was that as VDS becomes larger and larger the depletion of the channel begins to pinch off near the drain end meaning that the channel the charge density begins to drop to zero and once that happens and the current that we calculate by integrating along the channel comes out to be constant assuming that that channel that pinch off point doesn't move much inside the channel so we saw that now we can identify to the regions of operation for the mass device depending on whether VDS is greater than this magical number which we call the overdrive or less so VDS is less than this value we are in the triode region and if is greater than this value we are in saturation we saw that in saturation the current doesn't change with VD s so this drain current is constant even though this videos changes so if you put this in a black box it looks like a two terminal device again not worrying about the substrate and body and all that and we see that it operates as an ideal current source which means it has a constant current so in this region so long as we guarantee that VDS is greater than that value the device can be approximated by a current source so that's good that's what you were hoping to achieve when we built a little amplifier and then said can we build a voltage dependent current source indeed we have a current source and it's what is dependent because this value depends on this guy so it's important not to confuse this voltage with this voltage this is the current source as far as these two terminals are concerned and because when this voltage changes this current doesn't but it's a voltage dependent current source meaning that if I change this battery from one volt to two volts I get a neo current source and that new current source is also constant okay and we saw that in saturation the relation between ID and VDS is fixed by this constant it doesn't have dependence on VDS but we still have a relation between ID and vgs obviously because as vgs increases we increase the density of electrons in the channel so we have more current capability and that is reflected in this equation so this equation tells us ID as a function of vgs it doesn't have dependence on VDS as we see over here so in our circuits in this course we are mostly interested in mass devices operating in saturation so in saturation we say the current is constant may be relatively constant as we will see shortly as VDS changes but it's a function of vgs and it's a quadratic function of vgs so if vgs is equal to VT h we need at one threshold the devices right at the edge of off or on and then as we just exceed the threshold we fall in proportion to the square of the overdrive the current increases alright today we want to look at this characteristic a little more carefully and see if it is really as ideal as I portrayed it to be so we will talk about channel length modulation which is one departure from that ideal behavior that we saw and then we will approach the concept of biasing a very important concept that is necessary in many types of circuits so let's begin with channel length modulation and see what this is about now if you are familiar with bipolar devices this is similar to early effect in bipolar transistors but I will assume that you're not so we'll just start from scratch and talk about it all right so when people build MOSFETs for the first time and measured their ID versus VDS the first surprise was that ID did not follow a parabolic behavior all the way as we saw before so I started as a parabola reached here the the maximum but then I didn't fall we said that it went and stayed constant okay but when they looked at this very carefully they saw that this is actually not quite constant so in reality if I change the color to something else what we see is that this keeps going up by some small amount as we increase VDS and it's almost a straight line so it goes up as a parabola and then from here on almost like a straight line it keeps increasing so the question is why does this happen well again we have to go inside the transistors see what's going on and maybe we understand this phenomenon so let's draw the transistor again apply our voltages around it and see what we get so the source is grounded the gate is connected to a battery so this is the setup that we have here this drain is connected to a battery and we have some channel and because we are a saturation there's a pinch off point right there so somewhere around here the channels pinched off so that's the channel we have lots of electrons around here and fewer and fewer as we go this way a density of electrons okay alright so as VDS increases but now let's assume that the source is zero so we just increase the drain voltage right the drain voltage increases what happens to this pinch off point well if you remember last time we said that because this voltage is higher at this point has to move back a little bit because this point happens where the voltage difference between the gate and the local potential is one threshold so if I increase this voltage which means in this vicinity I'm increasing the voltage this point has to go back the pinch off has to happen or they have a shorter distance so that the voltage difference between these two is one threshold so in other words as VDS increases the channel length actually decreases the channel like defined as the length of the channel from here to here right the channel length decreases by a small amount so before we said let's let's assume it doesn't change and that's how we got a straight line but now we are considering that second-order effect so what happens that the channel length decreases well we go back to our integration now we have to integrate from zero to L but this L is not the original L right for some drain voltage we had L over here for a new drain voltage we have L over here and so on so the L that appears eventually in our equation as you remember from here in this equation slightly decreases as a drain voltage increases so we lied when we said ID does not depend on VDS right this is straight no what really happens that as VDS increases L slightly decreases which means IDs like the increases and that's what the phenomenon that we see so that is the situation and the question is how do you model this so that it is a manageable in our calculations to really want to put an L in the denominator and make it a voltage dependent quantity that's a little messy so instead this is what we will do will make an approximation so we say ID is equal to 1/2 of MU n SI ox w / L you know is the original L the effective L from here to here okay that's the L that we use here now if the Hydra at the the higher drain voltage pushes the pinch off point to the left I will take care of that not in this L term but somewhere else so I will write everything else like before and I will have a correction factor to 2 into reflect reflect this behavior here so what should the correction in fact you look like okay so we need VDS that's for sure right as VDS goes up you want the current to go up all right but it cannot be just multiplied by VDS because that means that this creates a very strong relationship like this we just want to have a mild relationship like this but we have to add some constant to VDS and then we have some sort of coefficient so that the units are right right this is one has no dimension this has to have no dimension so lambda is 1 over volt so this correction factor tells us that yes as VDS increases this begins to increase right lambda is some number so lambda which has a unit of 1 over volt is called the channel length modulation coefficient and this phenomenon is called channel length modulation the length of the channel is modulated as the drain voltage goes up and down so now we have a very simple model with this extra coefficient here to represent this behavior right so that's pretty nice it's not that complicated and it allows us to see what happens now this tells us that the in saturation even though we are in saturation here the current of the device this device here is not exactly constant as this voltage goes up and down right so we see that the current is slightly changes so it's not an ideal current source it's an approximation of a current source and that's something we have to remember now when do we take this into account and when do we not existent into account it depends on the situation so as engineers our first reaction should be to use the simplest model possible so when I see a mass device and are you know by some calculations that is in saturation I the only thing I consider is first this right and then I go along and see what I end up if the results are not very sensible then I have to try to trace back and say ah well maybe channelized modulation is important as I should include that okay all right so with channel length modulation we have this behavior for the device and let me see if there's anything else I want to mention about this so I can still redraw the model of the transistor that we did last time it drew B last time we said the device has a gate a drain any source between gate and source we don't have really any current the gate is insulated from the rest of the device is some modern devices we have a bit of leakage between the gate and the channel but we don't worry about that so it's an open circuit there is no current flowing through it at low frequencies so we just say what we have here is the vgs and this current source is the current that flows from drain to source and its value is given by some amount if we know we are in saturation so if we know the saturation this is the equation that we have so you should have 1/2 of mu n SI ox WL vgs minus vth squared 1 plus lambda VDS this is a more complete model of the most transistor that we can use if you are given a mass device usually you are given mu + c ox then W and L are things that you can choose if you want to design the mass device you pick some W some L and lambda is also a known quantity for the technology at hand so lambda is also known value okay this is what we call channel if modulation and in our calculations from now on sometimes we included sometimes we don't depending on the accuracies that we are looking for so if I write lambda equals zero that means channel length modulation is neglected and those of you who are familiar with bipolar devices see the resemblance between this behavior here channel if modulation and the early effect in bipolar transistors okay all right in the next step you want to talk about the concept of biasing again this is something that we talked about extensively in bipolar circuits but I'm assuming that you haven't washed those necessarily so I'll start from the beginning and talk about why we need biasing all right so let's take a step back and remember what we did before introducing the mass device we said we want to build an amplifier so let's do that again so let's go ahead and try to build an amplifier because I have a device I have a device that I can buy that can serve as a voltage dependent current source right now this is nonlinear so that might worry you but for now don't worry about it okay so voltage dependent current source so if you remember this is what we did before we said I buy a voltage dependent current source now previously I didn't connect these doesn't matter ready we said this particular source says if you give me v1 I give you k v1 right that was the voltage dependent current source that we used and we said well if you pass this current through a resistor to obtain a voltage so is the voltage and we give it a voltage an input voltage some sort of maybe a sinusoid right for example my voice go into this microphone uh my voice is not a sinusoid but let's say a sinusoid then what we expect is that V out becomes proportional to V in in fact if you remember V out was equal to minus K times RL times V in and if KRL this constant times this RL is large enough we have amplification I say Karos and so if we have a one-minute ball swing here we get a 10 millivolt swing at the output and that's amplification now we have a voltage dependent current source so let's go and build an amplifier out of this device all right here it is we have our voltage dependent current source like this I'll just copy everything down here so I have RL and then I have a an input for example from a microphone right so I will draw a microphone signal a microphone symbol here Mike and again this microphone signal looks like this it has a signal that goes up and down like this and it has some amplitude some peak amplitude let's say 5 millivolts just as an example okay now the microphone signal starts from 0 goes up by 5 millivolts and goes down by 5 millivolts around zero okay so do we have an amplifier here well let's look at this transistor here and in particular let's look at vgs now what we do know from this type of characteristic is that if a device is in saturation ID versus vgs has this behavior you don't have any current up to one threshold and then beyond that we have a square law behavior according to this equation now in all these calculations we'll assume lambda equals zero so that we don't have to worry about that too much alright so with this set up I'm hoping to build an amplifier but if I look carefully something is not quite right the microphone signal perturbs this vgs from 0 to 5 millivolts plus 5 millivolts or from 0 to minus 5 millivolts right so on this characteristic I'm right here a vgs goes up by 5 millivolts or goes down by 5 millivolts how much ID do I have zero the device is off the device is dead so this circuit cannot possibly be an amplifier because the voltage coming from the microphone is not capable of creating any significant current or current change in the mass device so I hope that's clear alright so if the signal is 5 millivolts and the threshold is let say 0.5 volts or 0.3 volts obviously the device is never turned on ok so we'll call this first attempt in our amplifier construction endeavor and that was not very successful so what should I do well the problem is that this signal is just way too small the device is not turned on at all so maybe what I should do is add something else to the signal so that even when there's no signal this device this transistor has some life in it right and by that I mean this so let's go to our second attempt and why don't I do this let's try we have the microphone still producing that type of signal I come along and place a battery here and I picked this battery voltage to be one threshold and I still have the rest of this stuff resistor and all that and now I would like to see what happens okay now is the situation better well maybe but is it good enough so let's see well if you look at this carefully we see that now if the microphone is quiet so if I'm not talking to the microphone then there's no signal the signal is zero so the voltage difference between these two is zero that means that vgs of the device in the absence of a microphone signal is equal to one threshold so we are right here okay now the signal comes in the microphone signal comes in and raises this voltage by 5 millivolts so we have this much plus 5 millivolts right so we go over here so this is V th + 5 millivolts okay and we get some current so is that a good scenario can be used this and build an amplifier so I'll just give you some example numbers and some representative numbers to see what we end up with right if the numbers are okay sure we have an amplifier if not then we have to go and go to a third attempt alright so as an example like last time let's assume that the UN Co ox is 100 micro amps per volt squared W / L is 10 and the threshold is for example 25 volts it does too much and you would like to see how much current we get when vgs is VT h plus this much plus 5 millivolts right the microphone raises its voltage by 5 millivolts okay so then we can write I D is equal to 1/2 off so I'm writing this equation without the connection due to channel length modulation so 1/2 of MU n see arcs so 1/2 times 100 micro amps per volt squared times W over L times 10 times what goes in here vgs minus vth vgs is this much we change it this much so we are above we th by 5 millivolts our orb drive overdrive resulting from the microphone signal is 5 millivolts so 5 millivolts squared we calculate this what do we get we get 12 point 5 nano amperes this color looks awfully small and will see what we mean by small okay so if the current increased by 12 and 1/2 nano amps when the microphone signal ray rose by 5 millivolts then I can pass this current through a resistor to have some sort of voltage right that's how I'm trying to build an amplifier okay so how much are L should I pick to have some reasonable amplification so again let's look for an application factor of 10 so from 5 millivolts I'm hoping to build 50 millivolts right at this output so I'm hoping that at the output I will have a swing like this wall it's actually inverted but doesn't matter and I'm hoping that this peak will be 50 millivolts right okay so this current are twelve and a half nano X has to pass through RL and give me 50 millivolts so I can find or L so RL is equal to the voltage that I need 50 millivolts divided by the current that the transistor has produced to organ half nano amps and that comes out to be four mega ohms four mega ohms and that's a very very large resistor so we could buy a four McGann resistor in if you in discrete form but we don't want to build an amplifier like that an amplifier would not be very good alright so something is not quite right here even though we added a voltage we brought the operating point from here where the transfer was totally dead to here where Chancellor is partially dead right so can i improve further so that I don't end up with such a small current here well maybe what I should do is instead of operating at this point and when the signal comes you know I just go up here maybe I should start somewhere out here so that when the signal comes in I go further in the current domain so let me elaborate on that I need to add a page and we'll see how that comes about all right so that's our third attempt and here's what we're thinking we take the transistor we have our good old microphone here and then on top of that we have a voltage and this voltage is actually more than a threshold so we'll call it V zero and then of course we have our resistor or L and so on and what we are thinking is this if ID versus vgs we have nothing up to one threshold and then we have something like this what if V zero is then over here so now the microphone signal comes in it comes in and this vgs is now modulated so to speak right so in the absence of a microphone signal and this difference is zero we just is equal to V zero when the signal comes in vgs goes up a little and goes down let's say by 5 millivolts this way and 5 millivolts this way okay so we can do a quick calculation and see if the situation has improved now that this battery is chosen to be somewhat above the threshold voltage so as an example I chose V 0 to be point nine volts remember that the threshold was point 4 volts right point 5 volts so Vth is point 5 volts ok so what I will do is this I will calculate two quantities first no signal if there's no signal we have this much battery and we have a certain current how do you find the current that's like this we have an equation for that current we call this I be 0 in the second step I allow the microphone signal to come in and we added the 2 V 0 so V 0 VG s goes from V 0 up by 5 millivolts so we go over here we go up up by 5 millivolts let me erase that for now there we go up by so this is V 0 we go up by 5 millivolts and then we get a new current ib1 and then we want to see as we went from here to here how much did we how much increase did we see no idea maybe this is better than the previous situation and maybe the value of this resistor will be more reasonable ok so we calculate our ie D 0 by D 0 is again 1/2 of MU n c ox so 1/2 of MU n c ox w 1l vgs how much is vgs v 0 no signal so V 0 is a point nine volts point nine volts minus the threshold 25 volts squared and that gets us approximately 80 micro amps so we have 80 my clamps of current so there's no signal we have a current of 80 micro amps flowing from drain to source and we have a gate source voltage of 0.9 volts and the signal comes in it perturbs things so it perturbs vgs by 5 millivolts so vgs now goes to 900 5 millivolts alright so then i D 1 is calculated as all of this here is 0.9 Oh 5 volts minus 0.5 volts squared right and this gives us 82 micro amps so we have 82 micro amps ok so the current through the resistor increased by 2 micro amps when the microphone signal came in and raised the gate-source voltage by 5 millivolts so what kind of resistor value should I choose here so that I get a swing of 50 millivolts at the output like before all right so we say change in ID is equal to 2 micro amps you realize why we look at the change because this 80 micrograms was with no signal this is with the signal so the difference between these two indicates how much signal has come in and been amplified and so on right okay so 2 micro amps and that means that if I want to obtain a 50 minimal swing ramaa these these two micro amps I have to provide that by 2 micro amps and that gives me 25 kilo ohms so that if somebody is much more reasonable than before right before we had how much 4 mega ohms now we have 2.5 - 25 kilos so that's much more respectable all right so this tells us something very important it says if we want to use a MOS device for example as an amplifying transistor then we can't just bring the signal in and give it to the mass device and hope that something will work first we have to set up some conditions around the mass device even before the signal comes in and those conditions are called biasing bias conditions for example we have a certain battery between the gate and source so as to create a certain vgs over here and a certain amount of current so even when there is no signal the device has certain vgs and a certain drain current those are called bias quantities and in fact this point is called the operating point so we see that need to bias the transistor by applying or maybe I should say creating by creating proper current and terminal voltages so caught up inside the device and voltage it around the device the gate and the source of the drain and so on so that so maybe I should say voltages in the absence of signals okay so when there's no signal there's no signal here we have a certain vgs this much we have certain current this much so that's there's a proper current and voltages and the reason for this is that so that the device can amplify the signal okay so we know that a signal will come in at some point but before the signal comes in we have to set up proper biasing and this point is called the operating point okay so as you can see biasing is a critical issue without biasing different stories dead doesn't have any ability to do anything so we do need to have some biases bias currents and voltages around the device before the signal comes here this is a question that we always ask students in different classes why do we bias transistors and sometimes oftentimes they don't really have a clear answer for it so to make them work but by what I'm able to make them work so what we mean is we have created the proper current the proper voltages so that the device is capable of amplification all right as if something if you can define as YB bias transistors okay so let me look at the following alright so let me make some observations before we wrap this up all right let's change the color so observations okay so number one what we have seen is that the mass device can operate as a voltage dependent current source so we give it a voltage it gives us a current such a device is also set to convert and voltage to a current okay so we say a MOS device converts a voltage to a current in fact we might call it a voltage to current converter so you say a mass device can be considered a voltage V to current I converters so that's how we sometimes look at it it's a voltage to current converter we give it a voltage you get it you get a current of course it's nonlinearities things doesn't matter at this point but that's that's what it is okay and a voltage to current converter has another name in the circuit design which you should also know and that is called a trans conductor so sometimes we say a MOS device is there transconductor and these are all fancy names for what it does right it just converts a voltage to current is able to depend on current source okay all right now based on this let's see I would like to make another observation so I will write it here so which operating point is preferred here are two scenarios and I want you to tell me which one of these is preferred ID versus VG s so we have a device in saturation so always a saturation unless otherwise specified so we have a device in saturation and what we see is this general behavior right okay so I choose one operating point to be here so let's call this v1 and this gives us some current i1 this is operating point number one or I pick another one over here v2 some operating point and some current i2 so that's operating point number one this is operating point number two which one of these better and why I'll give you 90 seconds to think about that okay so what did you decide well what we recognize is that this curve is not linear it's a second-order function is a quadratic function which means the slope of this curve increases with vgs right okay so if I sit here and I apply a signal let's say from the microphone on top of this amount of vgs this bias value then i have a certain perturbation the current if the if this V changes by Delta V this current changes by Delta I and we know that Delta I is equal to Delta V times the derivative similarly here if I change this by Delta V this current changes by Delta I and that Delta I is given by this Delta V times the derivative so we see that the derivative is higher here than here so for a given input signal swing from a microphone for example let's say 5 millivolts of swing here we get a larger current change around this point than around this point because if that fiber loss is considered small Phi below was a change here is multiplied by this derivative to give us this current change and here this 5 millivolts of change is multiplied by this derivative to give us this current change and because the derivative is higher here than here we have a more favorable situation here meaning that a 5 millivolts signal coming in will give us a greater current change at this point than at this point meaning that the transistor in a sense is stronger here than here it's a better voltage to current converter here than it is here when we perturb this voltage by some amount we get some current change by that perturbation and over here the transfer is stronger for a given voltage perturbation gives us larger current changes so the transistor is stronger here than here so this point would be preferable but is it does it come for free or is there something some price we have to pay well yes we have to prep a price remember that when there is no signal right you do not talk in the microphone right anything we high bias current we have a drain current and that drain current is draining the battery out I have a battery in my transmitter here that batteries being drained out because I am drawing a certain amount of current so if I want to have a stronger device so to speak I have to go over here which means I have more bias current and more bias current will drain the battery more quickly so that is the price that we pay for a stronger transistor very well now with these we are ready to look at another interesting concept but before we do go there I need to show you a few other skills that we use often in our analysis and one of them is the following combining time response with i-v characteristics it is not that hard I just need to explain it to you so that say from now on anytime you use that you're familiar with what we mean ok so again let's go back to our simple amplifier where we have a microphone and we have a battery and so on ok and let's just focus on the drain current of this device ID we have some other stuff here but we don't worry about that too much just ID this current okay so again there's a certain bias to call it V zero so here's what we have we have IB as a function of V GS no current up to one threshold and then some behavior like that and v-0 the bias somewhere so let's say this is V zero and we have a certain amount of current ie okay now the signal comes in we're on talking to microphone the signal is coming in does he know let's say he's a sinusoid so let's draw that sinusoid here so this is V mic and just a simple sinusoid like that as a function of time all right now I can see things better if I try to combine this plot with this plot so let me explain what that means okay so what I will do is I will rotate this by 90 degrees clockwise so I get this so that's V mic and this is the signal and this is cheap so far so good okay now I realize that V Mike this voltage is added to this battery voltage right so let's go and add it to battery voltage then what do we get so I will plot V Mike + V 0 so that it look like well it's just shifted right so this is zero for both time and voltage and now we are over here so we have this and it starts at v-0 here that's okay right I'm just twisting your mind in a row but that's that's what we get you could have drawn this here and shifted this by v-0 and then turn by 90 degrees but that's what we have all right so this axis is this whole voltage here vgs because it's v-mic plus v0 now this axis is time and this one is going up and down around v-0 now let's take this and put it right on the bottom of this curve here okay so here's what we have the this axis is actually vgs so we just write vgs and this waveform is swinging around v-0 so there's a me zero right here so this this voltage is swinging around v-0 and this axis is time okay so the important point here is that this axis happens to be the same as this axis after all of these manipulations alright now what I would like to do is try to plot ie D as a function of time and this combining allows me to do that so if I change the color of my pen you can see how that is accomplished so I said well vgs has gone from v-0 to some amount v0 plus 5 minutes right okay so that comes up here and maps to this amount of current so when the gate source voltage starts from v-0 and goes up by 5 millivolts as a result of the microphone signal the current starts from its bias value so this should be i0 by the way and it goes to some new value okay and how about when the micro single goes down so we go below v-0 no problem so now the new vgs is this much from here to here right so that comes up here and mapped to another value so now I can go ahead here draw a ID as a function of T for this peak I got this much current so that's the current that I get here for this peak I got this much current that's the current I get here and in between I have something between right so that's the approximate behavior of the drain current as a function of time we know that if the gate source voltage is changing periodically the drain current also have to change periodically so that makes sense now what is the exact shape of this is it is it a sinusoid etc we don't know but at least qualitatively we have a picture of how ID changes as a function of time so this combining of the time response of the input with the IV characteristics to get the time response of the output is very useful and is handy in our calculations and analysis very well ok let's talk about one more interesting concept today which is critical in everything that we do from now on and for that I would like to go back to this example that I showed you before we said that the second operating point is preferable because this characteristic has a higher slope here and that higher slope signifies a stronger transistor because for a given signal change here you know like this kind of thing right we get a larger change here like kind of thing so this type of Swing is larger around this operating point that are on this operating point so perhaps we can define a quantity that tells us how strong it transistor is and what we are thinking is maybe just the slope of this curve tells us how strong the transistor is right here the transistor is very weak because when we increase the gate source voltage by 5 millivolts we got only 12 and a half nano amperes of current so the voltage to current conversion of the device was very weak 5 millivolts of change gave us only 12 and a half nano amperes of current later on we went to point 9 balls and things were better 5 maneuvers of change gave us 2 micro amps of change in the current so voltage to current conversion was better the transistor was stronger and of course here the transfer is dead because the slope is zero so nothing there okay so what we are thinking is maybe we can't call the slope of ID versus vgs something and use that to think to feel how the how strong the transistor is as far as voltage to current conversion is concerned so we do that we exactly define that so let me go to the next page and that's what we call the concept of trans conductance conductance so a transconductor like the transistor has certain amount of transconductance just the way a resistor has certain amount of resistance okay here is what we are thinking so we have ID versus vgs and we go up to one threshold beyond that we go up quadratically and we define officially the transconductance which we denote by g sub m to be the derivative of IB vs. VG s with respect to vgs okay the slope of this characteristic is called the transconductance you can see that the unit is either one over all because this is one of our resistance or there's also an official name for this which we say Zeman's and we denoted by uppercase s lowercase s is used for second uppercase s is used for Siemens so either way which you really like so GM is a lower here I are here higher here and zero here right so that's the transconductance of device and it tells us how strong the devices we saw that we can have a higher transconductance you can have a stronger device if you go farther and further over here to have a larger amount of current but then the penalty be pay is a hard larger amount of current drawing from the battery so a shorter battery lifetime all right now because we have an equation for ID as a function of vgs in the saturation region we should be able to calculate this GM so let's go ahead and do that ID is equal to one half of the u-n-c Arc's w / l vgs minus vth squared again lambda is zero and we just need to take the derivative of both sides with respect to vgs so that's easy right there's two kinds of this and we end up with IgM is equal to MU and C arcs W over L times vgs minus vth that's very simple this just tells us what how the slope varies right so it says if you were to plot the slope of this characteristic as a function of EGS what would you get y says this says uh if you were to plot the slope GM as a function of vgs well over here GM is zero no slope right up to V th and then beyond that we have this simple straight line so boom right this slope it keeps increasing so that's the slope of the characteristic so if we device if you bias the device here we have a higher vgs and we are still saturation that's the key and then we have a higher transconductance right that's that's what this equation tells us okay now it turns out that there are two more equations for transconductance that i should mention here the other one the the second one is this to ID over vgs minus vth this arises because I can take this to ID and divide it by vgs minus vth and what i left over off with is this right so if you combine these two you get this and there's yet a third equation so GM is equal to 2i d mu n SI ox WOL this is obtained if I go back here I find vgs from here and plug it in here so vgs is equal to ID divided by this whole thing square root of that and then I use that vgs minus vth over here and i find a new GM and that's an image here so we have three equations for the transconductance of a MOS device and it's very interesting because each of us tell each of these equations tells us something about how GM varies with various parameters GM varies with we just minus vth yes GM varies with ID yeah we saw that right we saw if we have biasing here we have a lower GM than if we are biasing here and then GM also varies with WOL as indicated here and here and also some interesting things so these are things that you have to study in the future but the transconductance of a mass device which tells us how strong the device is in converting a voltage to a current has these three expressions for it okay our time is up I just want to mention one more point before we go home and that's in relation to these amplifier designs that we looked at for example this one right now with our third attempt and we seem sort of happy with it we said yeah we have two microns of change in response to five levels of change the lister was twenty five kilo ohms everything looked good but one should ask is this configuration such that the device is in saturation we took care of the gate source voltage right we apply the battery here chose it to be this value and so on and I just drew ID as a function of vgs but remember for the device to be saturation the drain source voltage also has to be some value right it cannot be too low it is too low then we have a problem how much is drain source voltage here when there is no signal if there's no signal here we have a battery between gate and source but you don't have any battery connected to drain so then what's going on here looks like the device has is not capable of carrying any current here in this loop so maybe the device is actually not biased properly yet so for that you have to wait until the next lecture I will see you next time [Music]