 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub.org and download the PDF file containing the slides for this presentation print them out and turn each page when you hear the following sound enjoy the show all right so this is lecture 23 on AC response and let me explain why AC response is interesting DC is because of course the photo diodes and normal diode all the time operates in this region but it's not really great I mean you have to really in order to use it you have to do something more and the thing that people did almost like a hundred years ago a little bit more hundred years ago was to use this diode for a radio receiver you know this transatlantic Marconi and others so this transatlantic transmission that was essentially a diode that was diode with a big antenna and that caught the signal the antenna called the signal and the diode demodulated it it extracted back the signal of the voice signal that you wanted to have beyond the carrier and I'll show you so this is a for my daughter is planning to buy this this is a actually a simple radio I'm not a radio but you can see that that's a diode do you see that long one and I will ask you to convince yourself that why this works as a radio because it looks like it's not even connected you see it's one side is open the other side has a switch but this is actually try to convince yourself that this is actually a simple diode diode you can see that transparent small cylindrical region and that coil is that inductor that wire acts like an antenna and then that one that small bulb like thing that's a little speaker and so you can show that essentially it looks like a small radio and it will work like a radio you can buy it here of course the idea is that you catch you have you bias your diode at a given voltage right with a battery with your triple-a battery you bias it at a given voltage and then the small signal that is coming in that's coming from the air that's the antenna catching a small signal on the order of maybe a microvolt or so and as a result on top of the DC bias there is this AC signal moving around left and right or the top and bottom and as a result there is a current flow in this and we want to know how that works so essentially instead of drawing that complicated figure we'll draw a pn junction we'll have a applied voltage given as VA and the small VA we'll use that as the AC voltage showing that that's coming from the antenna of your of your circuit now what I want to show you that when you have a want to analyze a circuit like this you can replace the diode with simply a register and two capacitors one is called a junction capacitor and another diffusion capacitor and I will explain that in a minute but by the way and this only depends on the small signal where is the applied voltage VA hiding in all this well it will hide in the magnitude of the resistance and capacitance so you will see the applied voltage VA how it gets in this magnitude of the current so for a given bias condition you will have a set of resistance capacitances if you change the DC bias you will have a different set of resistance and capacitances but once you have a DC bias then the AC circuit the AC voltage will just see a combination of capacitance and resistance now why isn't there any inductor in it that's a question I'm going to ask you later so how do I know the resistance or conductance the G in blue how do I do that or the series resistance RS series resistance comes from because when you have a high bias then remember that P and N side has a quasi-formal level drop and as a result that was the series resistance part so you will write the equation and you now know right that this equation we have derived this M can be one or two if it's in recombination generation two if it is diffusion only one it's ambipolar two again so depending on where you have biased your circuit that will be one or two take a log on both sides and this I don't have to tell you right how to how to rearrange this take a log on both sides and differentiate with respect to V differentiate with respect to V but differentiate with respect to I sorry that's because I want to get the conductance if I differentiate with respect to I do you understand the left hand side it's a log so log of X differentiation one over X so you can see that one plus I plus I not has flipped over and da divided by di that's the first term and you can see the Q and beta has gone to the left hand side just a little bit of rearrangement as a result you can say that delta VA that's the applied voltage small signal applied voltage and the current that's my G delta VA over delta I has a dimension of a resistance so one over resistance that I'll call G and this is only applies for the forward bias right that curve applies for the forward bias so therefore I have written a F of B you can see the series resistance so by just measuring it you can get the series resistance you know M right you know M and from DC characteristics you know I not beta is what is beta one over KT right so that's what the one so I actually know all these quantities so I could calculate G reverse bias side well if VA is negative then the first term is gone so what is the reverse bias resistance that's the constant that does not depend on voltage right so is it generally infinity yes if you do not have any recombination then indeed it is infinity recombination or generation it's indeed infinity but if you generally have traps in the junction region so you always have that extra term do you remember ni divided by 2 tau and then square root of V do you remember that now if you take a differentiation of that you will pick up no derivative from the first term a derivative from the second term so from that you will get a reverse bias conductance and that's the G on the reverse bias side no rocket science here right this is simple so we can calculate G no problem forward and reverse bias side remember depending on where you have biased it now what about junction capacitance the capacitance I have to draw now how do I calculate capacitance the way to calculate capacitance would be that I have biased it do you see the formula will split I have biased it in the forward bias side right forward by side and then I am doing showing you that little arrows up and down indicating my AC bias is sometimes making the total forward bias a little bit more and sometimes a little bit less so it's going jumping up and down smoothly moving up moving up and down so that's what I'm saying when it moves up a little a little bit more current flows when it moves down a little a little less current flows so you can see that there will be oscillation in the current in the DC case but similarly what's going to happen that the depletion region is also going to modulate little bit right different because you're forward bias a little bit more or a little bit less is going to modulate and therefore what's going to happen if you just look at this junction region you will see sometimes a little bit positive charge moving up and then a little bit negative charge moving up right and these are all majority carriers because majority carriers can respond very quickly holes will be coming from the right hand contact into the junction and similarly electron will be coming from the left hand contact into the junction and next to the junction you will see very rapid change in the charges and this charge this extra charge if you just focus on the extra charge you will see when the bias is a little bit more then you have charge in one one side this extra charge and when the bias is a little bit less then you see from the equilibrium position the charge will be a little bit less so you can see the charges will be flipping back and forth next day at the end of the junction but this is exactly what happens in a capacitor in a capacitor separated by insulator not semiconductor insulator when you apply AC bias charges dance around these two electrodes right go up and down in magnitude same thing is happening so therefore this is a majority carrier effect this very important to understand minority carriers couldn't do it and I will explain in a second but the majority carriers can do this and if it can do so you can easily calculate what the capacitances you have done this many times right epsilon A over D these are distance of the charge separation and that's WN over WP you put that formula back in and you are actually done you see why the capacitance junction capacitance should depend on where you are operating the device look at the VA on the numerator so the numerator contains the applied bias and that you can help you to define the junction capacitance at that bias point okay so this is not difficult if you measure it as a function of VA it goes as 1 over VA right this is in the denominator square root of 1 over VA so therefore the red line essentially will look like this as a function of VA now as you increase the voltage the capacitance goes down is that right because it's 1 over VA and so you can see the red line there are more negative bias you apply then red line is gradually getting smaller down the road now this is the first time remember then we'll be able to measure VA as one thing I have mentioned about this course that every time you see a theory you must make sure that you understand how this is measured if you didn't understand it then it's not good enough I told you about how to calculate VBI before built-in voltage I did not tell you how to measure it because how do you know your theory is correct you know so the only way is to measure it and you can easily measure it in this way the capacitance you can measure and then if you plot it in a particular way and the particular way is 1 over CJ squared square of the capacitance then you realize that this becomes a straight line and the point of intercept in the x-axis right that is VBI and from that you can calculate back what the VBI is you see it's a straight line and that's one then right hand side so you can calculate from here the right hand intercept point what the VBI is now this one also allows you to do variable doping if your junction doesn't have a fixed doping remember what this does is that it separates the charges plus and minus charges at a certain distance and it only proves the doping at that point the amount of plus and minus charge so if you deplete it a little bit more and it goes to a separate place where the charge is a little bit different then this capacitance formula will pick it up because ngx is the doping at the point of the junction end of the junction so again you can do the same formulation and from the local derivative and this I will ask you to verify from the local derivative of 1 over CJ squared and a VA curve this this one is always the intercept is always QVBI but from the local derivative you can pick out how the doping is actually changing across the substrate and so this is a very interesting one but take a look when you have a few moments now one thing that is sort of important to understand that when you modulate the charge on the left hand side why is it that it immediately comes to the junction you know why doesn't it take a huge amount of time when you turn on a switch for the light why is it that light immediately lights up the electrons don't sort of walk around and then say okay wake up everybody and then tries to you know light turn on the light that's a very important thing and that's why in these circuits or models in the majority carrier side there's no capacitor or inductor to represent the majority carrier side just the register RS and this you need to understand because that is called a dielectric relaxation time in the majority carrier side why there is no capacitance or inductance in the majority carrier side I only have drift current is that right because that's most of the electrons are over there and let's let me assume no recombination and generation but I'm looking at a transient that I'm oscillating on the left hand side and try to see how long will it take for the signal to propagate to the junction that's what I'm trying to answer okay so that equation I already know now do you agree with these statements because I'm doing dn dt only the excess delta n is what I need to focus for and for the current I only have the drift current so I have put it in and therefore I can get an expression for ddx so that delta n dt goes is proportional to ddx nothing complicated I haven't done anything now from Poisson equation I also know ddx but do you realize why I have dropped all these terms na and p because it's the inside I better not have any na so I also dropped p and as a result nd is almost equal to n0 also right it's a majority carrier side and so what I will have is q delta n divided by epsilon I put it in the second equation back in ddx expression and I get an expression for delta n delta t now this one you can solve can you this is a first order equation and you can solve and this is the solution this says that if you start a oscillation on the left hand side how long does it take for that oscillation to essentially get out of this material right to the outside of the junction because this is how fast the whole perturbation decays as a function of time if you calculate how long it takes that's one picosecond within a picosecond or a fraction of a picosecond from one side of the signal it comes to the other side that's why when you turn on the light this light immediately turns your Bob's on you don't really have to wait because here the electrons are not going themselves one electron is saying to the next it's neighbor that move the next electron is saying to the next that move and so this cascading process goes on without a electron physically have to move all the way and that's why this process is so fast and there is no capacitance or inductors in the in this series resistance part of it it's very important that you understand this and you will see it in many many different context in in other years now the next one is the minority carrier diffusion equation junction capacitances only for the majority carriers minority carriers cannot respond this fast because minority carriers go by diffusion diffusion cannot goes propagate the signal as far as in diffusion electrons really have to move on their own no help from no group effort here now diffusion capacitance how will you do that let's say I have biased signal a biased diode so I have that triangular profile no recombination now I am forward biasing and a little bit with this with this small voltage from my antenna as a result what's going to happen that there will be a modulation of the corresponding carrier concentration why is it like this because it's like a rope or a string on which one side you oscillate and you can see that as you oscillate depending on the frequency there will be a wave going to the other side at low frequency there'll be a single wave as the frequency goes up you're pushing the electrons go this way before they have chance to get out from the other side you have reverse your bias and say come back and so they cannot decide to go or come back and therefore there is this periodic oscillation going down going down the road if the frequency is high then the oscillation will also go up right so that's what I'm saying if it's close to DC it will almost be flat okay now the math of it well that requires you to wake up now this requires that I solve this problem correctly this requires that this is a minority carrier side and I shouldn't have a drift term right previously I had a drift term because I was looking at majority carriers minority carrier on the right hand side the electrons so I drop the first term now I put that thing in here and this becomes this horrendous thing but let's think about it the left hand side is n dn dt n has three piece one is n not the equilibrium piece before I have applied any bias does not depend on position doesn't depend on time so that's n not I have applied a DC bias then of course it has changed the carrier concentration a little bit that's my triangle this magenta triangle and that's DC depends on position but doesn't depend on time is that right because of course nothing is moving and then I have an AC AC signal there extra thing is coming in that one depends on both time and you can see from this oscillation of the blue curve on the top on the top on right side it depends on time because it's a wave moving forward and it depends on position because you say slope you can see so therefore now you realize what I have to do in order to simplify this equation the first term n not an NDC will go because they are not changing with time in the second term only n not will go because the DC term and AC they depend on position so I shouldn't drop those and then you have the recombination term that's why of course the same will happen for the holes the holes are trying to go this way but we'll focus on the electrons and that will probably do it so you see this that there is this DC and AC term of course now I can I should be able to separate because I should be if this two equation two sides left and right hand side must be equal to each other then this might be equal to each other or all omega any frequency that you come in it should be equal to each other and if that is the case my DC and AC component must separately satisfy this equation moreover so let's take the first derivative I have for drop the n not an NDC term right you see that and do you see why I picked up a j omega because I've taken a derivative with respect to time I have a picked up a j omega do you see on the right hand side that I have dropped the second derivative with respect to position for n not I can do that that's the DC of course other things I cannot drop now I can separate this thing out assume that for my for me omega is equal to 0 so in that case only the DC components remain so you have seen this equation before right the DC second derivative and delta n over tau well that was my minority carrier equation I haven't done anything because I haven't applied any AC bias any antenna signal so therefore I should be and therefore as a result you know also the solution in terms of exponential because now you have a recombination term what about the AC the AC is whatever term you had on the top you just collect them all up you can see j omega tau n that I have picked up from the right hand side a left hand side and pulled it up in here now do you see and this is the crucial part that these two equations are essentially the same if I redefine tau n divided by this one j omega tau n plus one if I redefine the whole quantity as if you say AC lifetime then these two equations are mathematically exactly the same you see that as a result their solution will exact be exactly the same also except this ln which is square root of dn multiplied by tau n that changes but other than that is exactly the same equation now if I assume that this minority carrier region is long then of course you realize that d will go to 0 because it cannot go exponentially up forever only the c term will remain so that's only a redefinition here nothing more than that now the boundary condition is something we have to worry about why do you see why I cannot simply equate so I want to know what is the value of the excess AC concentration right here excess AC concentration there for the time dependent AC case now do you see why I cannot simply equate the two sides the rape components directly because on the right hand side the j omega t is sitting on two-story up it's not only sitting on top of a exponential it is sitting on another top of another exponential so therefore I will need to simplify this a little bit before I can equate so that's the tricky part in this whole business now do you agree that I can write it this way the exponential of two sums two terms can be written as this multiplication of the two exponential that's fine now omeg this AC voltage is minuscule right microvolt and you have applied a bias of a volt so it is e to the power x when x is very small in that case you can expand it e to the power x is equal to 1 plus x and that is what I have done the expansion for and once you expand it this time you can equate the j omega t term on both sides because they are on the same level and then you pick up the AC voltage AC extra AC component which is given by the magnitude of this oscillation that's given by this and it depends on the DC voltage because the more DC voltage you have then the correspondingly more AC voltage you have you can see that the VDC term over there so I'm done actually I can then this AC current will be the derivative sorry derivative of the AC extra signal extra carrier concentration and I'm sorry so this one you already know this solution you know the value for C I just derived the value for C you insert it in that expression take a derivative and that gives you the AC current and the AC current is here and there you can see the V AC the AC voltage so if I divide the current by the voltage what do I get I get a that's something that has a impedance or 1 over the impedance right admittance and so because it's a AC thing and that's what after I simplify this is what I will get now why did I get this j omega tau n why is this hiding ln exactly so ln was where remember the tau n was normalized by this factor so when I wrote it out the ln properly this got me flipped over and got me this extra term j omega term and you can see why they say admittance not simply a resistance because there is a j omega sitting here okay so one more line and we are done if I plot this capacitances and conductances that I have done so that's my admittance I can separate the real and imaginary part and if I separate the real and imaginary part this is the conductance and that's the capacitance now you can see when omega is small do you see what's going to happen when omega is small in that case this gd will essentially become a constant that's fine what happens when omega is large do you agree with this statement that when omega is large then it should go as a straight line on a square root of plot or in log log plot with a slope of half why is that take a look at that so if omega is huge then I can drop the one in comparison I can also drop the other plus one in comparison now I have a square root inside that gives me omega tau n then I have a square root outside that gives me another square root that's why it goes as a square root of omega what about the capacitance you say this I can drop again once again omega is so large I can drop the one and a minus one I can drop right and as a result what should I get on the top in the numerator side again a square root of omega but this capacitance is going the other way why is that because look at the left hand side I have a omega sitting on the left hand side so when I look at the capacitance it goes as one over a mega and do you see this amazing thing that the conductance multiplied by the capacitance is essentially frequency independent regardless of how quickly you modulate 100 kilohertz or a megahertz let's say you want to hear this am radio or FM radio everybody has a different frequency of frequency band right regardless of what you do at all those frequencies essentially the conductance multiplied by capacitance is independent of frequency although individually they are dependent now physically I'll end with this physically why is it omega dependent because of course you can see that when you have low frequency the carriers can really go from one side to another you have a resistance you don't have as much capacitors when it's oscillating very fast the carriers cannot go to the other side you're saying go and then say it's pulls it back and so there's a phase shift between the voltage you apply and the current which is trying to keep up and there is that phase shift frequency dependent phase shift that is getting reflected in the capacitance you see as a result this has this capacitance dependence so to conclude then this small signal response is very important for many analog circuits small signal and that was a center central to this argument then this parameters this capacitance inductors and sorry the capacitances and the resistances they always depend on this DC operating condition you saw that right the conductances has VDC inserted into it and it's very important that you distinguish between majority and minority carrier effects if you don't have any minority carrier no diffusion no diffusion term majority capacitance is always there but if you don't have a minority carrier transport there are many devices like your MOSFET may not be a minority carrier device in that case you will not have a diffusion capacitance so very important to get this various pieces in the right place so that you can apply it into a novel context okay thank you