 You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.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. So this is lecture 24 on large signal response of PN junction diode. Now if you remember that we are using the PN junction diode as a prototype of a very simple device. You can only simpler thing would be a homogenous chunk of semiconductor and which is basically a register and we know how to calculate resistance and other things. And this is the simplest one because you have a n type doping and a p type doping just two pieces a junction and we are going through this exercise of think the finding out how this device responds. So in equilibrium we draw the band diagrams of the Poisson equation and how saw how the built-in potential and how the depletion weight and all those things how they come about. We have talked about DC. The DC was a lot of discussion right because when you apply a bias we found that although the current is asymmetric positive bias it has a type of current which is exponentially rising with voltage and the negative bias it was saturating but there were lots of very interesting regimes. One was this trap assisted or defect assisted low voltage regime diffusion limited regime ambipolar and similarly on the reverse bias side we talked about Zener tunneling and Avalanche breakdown right. So all pieces are very interesting and if you understand them this forms almost a comprehensive list. In the last class we talked about small signal response that beyond the DC if you in addition catch a signal from air let's say you are listening to a radio station and then you have a diode then you have a DC voltage which is your battery to it you know your triple A or double A battery but in addition you pick up this microvolt amount of signal from the air and how would the circuit or the diode in equivalence respond to that small voltage that we found right we found that there are diffusion capacitance for minority carriers majority carriers there was junction capacitance and of course the conductance that depends on the bias point. Today we will be talking about this large signal response this is more like digital signals that you have let's say 1 volt and instantaneously you turn it to minus 2 volts let's say logic level 1 to 0 or go the other way around go from minus 2 volts let's say to plus 1 volt and if you make this transition too fast then what you have learned before wouldn't apply exactly and that is what would be the discussion topic of discussion today. So I will start with a charge control model and I will explain how that comes about we will start by talking about the turn of characteristics which is going from logic level 1 to logic level 0. Now remember logic level 0 doesn't mean voltage 0 logic level 0 might mean voltage of minus 2 volts for example this is just two separate largely separated voltages and we'll talk about a little bit turn on characteristics and then we'll conclude consider a diode in a circuit and the voltage source instead of being a triple A battery is actually transitioning between two voltage levels voltage level 1 maybe a volt and voltage level 0 which may be minus 2 volts let's say any voltage any specific voltage. Now the diode is in series you are measuring the current and if you just think about the diode itself you will get this IV characteristics right it is biased at a certain point and you can see the P side of the device is connected when it's logic 1 connected to the P side N to N side and so it's forward biased and so you can see the blue dot on the IV characteristics indicating that there is a finite amount of current flowing if it is at logic level 0 through the diode. Now if you gradually turned it off you know over a period of seconds then what would have happened that at every instant the diode would have adjusted to the DC potential almost and it would have gradually followed this curve. It is that the last point is let's say minus 2 volts every point is sort of inquires the equilibrium it knows the IV characteristics so beyond the DC calculation that we already did if the transition is slow then we don't really need anything more although the transition is not DC it is almost DC at every point however if you make the transition very fast let's say in your computer these things runs as tens of gigahertz in a nanosecond if you change it that fast then all in a sudden you will see that it cannot really follow this path what it is going to do is follow that red dotted path that instantaneously it will flip the current and flip the current you can see that immediately going keeping the voltage the same why the voltage remains the same I'll explain in a second but what is going to do from the positive current you see where y-axis is positive it immediately will flip the flip the car direction of the current as the voltage source wants the current to go the other direction it immediately responds to that dictate that current should go the other direction but then you can see for a while the current remains almost the same as the voltage is decaying off from a positive voltage gradually going to zero and then for a little bit it has the final final descent to the final final equilibrium point of course the final equilibrium point has to be the same because that's the final state what I'd like to explain to you today is the physics of this that why it takes this particular path under what conditions and how to understand that that's the purpose of this lecture now if you want to look it at it in a slightly different way you could plot the voltage and the current as a function of time you see this is voltage versus time on the right hand plot with time as a parameter so every point here you could stamp a time but in addition you could split it up also you could plot the voltage as a function of time and current as a function of time do you see that the the right hand side and the top drafts essentially contains the same information let me explain look at the the yellow points the yellow points are the voltages before this transition occurred I had a forward bias you can see on the left hand side the voltage versus time curve I have also a forward bias positive current and that's called I sub f you know f for forward now if you switch it very quickly again then you can see you will have this green circle the green circle is the current that has dropped down and you can immediately see also in the current versus time plot that the current has switched from I f to IR and it has done so instantaneously there's no time in there right so you can see almost instantaneously but for the voltage look at it the voltage the green circle is sitting on top of the yellow square so voltage has not changed now that's strange voltage is the same current going the opposite direction I mean how can that happen so we'll see that and then finally you see that great triangle great triangle is when the voltage has gradually decayed to its final value essentially going to voltage equals zero and on the left hand top side you can see the triangle and similarly but for the current look at this going from the green circle to the red arrow or a triangle the current has more or less remained the same now that's very strange and we like to explain that now let me explain some terminology so that I can help you guide through this concepts by the way when you do 0 to 1 minus 2 volts to plus 1 volt again it is going to take a non-trivial path is not going to follow the steady state DC characteristics and that's shown here in blue dotted line now a few quick definitions the definition is the we'll call this time T sub s storage time is the time when the current essentially remains the same during its transition current remains essentially the same we'll call that T sub s and for many circuits many minority carrier circuits mean sorry minority carrier devices like APN junction diode this time is a dominant time this is the storage time and I'll explain why they call it storage but it is only typically true for that minority carrier type devices that this time would belong and dominant then there is this TR the TR time is when the voltage has gone to zero and it's gradually swinging in back to the equilibrium reverse bias point you know gradually going down that's the T sub R now T sub R for this device will not be as big but for many devices as we'll show short key barrier diode majority carrier diode in that case that's the next class TR might become the dominant one and the total time is TRR which is called the reverse recovery time this is the time it takes for the signal to go voltage to go from positive side to the negative side now why do I what why am I interested in this time because if you try to flip the current or flip the voltage faster than this time your circuit isn't going to work so you calculate this time and design your circuit slower than this so that the voltage can reach the final point without any trouble okay now I have to solve this equation now that's going to be a challenge you can see Poisson equation I have to solve it from minus plus 1 to minus 2 volts now that's not something that's very good this time I cannot say n is n not e to the power j omega t it is not the signal coming from your 9 20 a.m radio signal with a micro volt from the air this is going from 1 to minus 1 volt or minus 1 to plus 1 to minus 2 volts I cannot drop the entity well I cannot really do much with this so I will have to stay with this and solve this it turns out that indeed on the computer you can do that people do that often but I can simplify this and that will in fact give me a lot of information a simplification of this where I get rid of the space coordinate of this equation and I explain to you how the way if we can get rid of that then I will have instead of those two complicated equations I will have this very simple first order differential equation and similarly instead of this whole transport equation drift and diffusion and all those things I will have this very simple equation you know it's approximate but for the transient what are you going to do if I had to solve this partial differential equation back of the envelope your real your envelope has to be pretty pretty large in order to contain all this calculation so these approximations are called charge control equations it is an approximation when you have large signal time transient you do so I will explain to you how this comes about but let's first talk about the what is happening in the device and how does this current flip without voltage flipping how does it do that so you see let's me assume that I have forward bias this diode do you see why it is forward bias you can see the Fermi levels quasi-Fermi levels has been split by that amount and quasi-Fermi level on the n side assuming the right hand side is grounded the left hand side has been pushed up forward bias minority carriers have been injected now in this particular case I have assumed that there is recombination on the minority carrier side how do you know that because this curve in xt is actually exponentially dying off had there been no traps then it should have been a straight line going across okay but then what happens now if I all on a sudden go from vdc from the dc voltage to a reverse bias voltage right so the Fermi level on the left hand side will go down significantly immediately in that case this is what's going to happen the current will go down a little bit and then the charge will gradually its charges gradually disappearing where is the charge going the charge is going to the other side getting back where it came from and part of it is recombining with the traps and with the majority carriers so you had a big build up of charge and gradually discharge is going away and eventually this charge will all be gone and you will get to the reverse side but one thing I want you to see do you see what is happening in here although initially the total amount of charge was exactly the same right when I turned it off look at the slope on the left hand side the slope I have changed the slope although I have not changed the charge more or less but since I have changed the slope at that point going now the current will be going in the negative direction so it follows the dictate that the current must be going in the negative direction simply by just nodding its head a little bit in that direction in the opposite direction because dndx diffusion coefficient multiplied by dndx is the current so all I care about is not the n n could be anything so long I allow the dndx in the right direction current will be flowing in that direction so you can see that by keeping this profile my current will all be going it will be a constant and going in the negative direction although the charge may be huge and actually be corresponding to a forward bias charge okay so this is a plot I will be showing you often the idea is you should see that this is a time sequence of the charge that is sitting inside the p n junction inside the minority carrier as you are going from the forward bias to the reverse bias now this is the equation transient equation I want to derive that charge control equation I will explain how first of all it's minority carrier right so I should always be able to take out the electric field as I said taking that out is a little subtle and I have explained to you you should learn it once so that later on when you drop it off you sort of know that you can go back and explain to a student but for the time being minority carrier side majority carrier holds the potential I drop the electric field so I have the diffusion term now on the diffusion term I have this n naught which is the equilibrium density well n naught does not depend on time I get that rid of that and I have delta n which is whatever the charge you see on the top right hand side the charge profile this delta n is a function of x and is a function of time you can see that the top side it has changes with space and also changing with time do you see why there should be a second derivative with respect to the diffusion coefficient of course because you insert the expression for j sub n and that's what you have and if this is essentially a small injection condition then is in the shock lid hall term becomes delta n over tau sub n right when you have put a small amount of charge in so that's what you have delta n divided by tau sub n that's the shock lid hall condition it's a partial differential equation still not probably easy to solve but let's see whether we can do some more this we could do let's say we take that delta n multiplied by q and a and delta x and then integrate from 0 to wp 0 to wp what is that that's the minority carrier region on the p side right and so that I'm just integrating the whole thing now if you integrate now that whole quantity q a delta n well that's changing with time right do you remember it was a big initial triangle and it was gradually going down so of course that's changing with time let me call that thing whatever is inside area under this curve q q is changing with time of course but I have integrated over wp and the wp is from 0 to wp is end of this region so that will be my area under the curve and so therefore I could write do you see a whether I could write this q is the whole thing and d dt for that will give me the first term dq dt that's no problem now what about the second term the diffusion term do you agree how I have written the second term in this particular way look at this the second term was a second derivative with respect to x so those the second derivative I have written as dx and dx and then I have cross multiplied with dx that will take care of one dx so I have a differential of q a delta n right and if I have a differential can I not integrate it out this is a differential I I can always take that differential and integrate it out and that you will see that says that I have the first term is the current that is going out from the right hand side and the second term with respect to diffusion coefficient which is current coming into the into this box that's the first and second time and the last term well that's the recombination term q n divided by 1000 now you have seen this type of integration before why have you seen this do you remember the delta function business the second derivative anytime in the Schrodinger equation when I gave you a delta function you had to integrate it across a delta function and the two sides you had this first derivative on one side and the first derivative on the other side that was separated by the discontinuity well the same trick here I haven't done anything else you just multiply dx and integrate it out okay there is a typo I can see that w n should be w p on the top side okay but you get you get the idea so the current so what does this equation say well this equation is telling me this equation is telling me that the charge that is building up in this box is equal to the current that's coming in from the left hand side you see x equals 0 that's the current that's coming in and the current that's getting out x equals w p and so the total amount of charge that is building in is the equal to the net amount of electron that is flowing in that's why it's called charge control and of course a certain amount of charge is going away through recombination do you see that q divided by tau n term that is the amount of charge that's getting out so only thing I have done and this is a general trick by the way anytime you have a differential equation short engine equation diffusion equation any equation that you can have second order partial differential equation if you want a big signal transient you can do do this in any case in any problem it's nothing to do with this particular problem but it's a general physics problem or general mathematical trick okay so I have a beautiful simple equation now I may be able to solve something so let's look at the turn off characteristics I'm making an equivalence on the left hand side I'm going from 1 to 0 and to reflect that similar situation in a you know undergraduate circuit type context look at this right hand plot right hand top plot I initially have the p-n junction forward biased with the voltage v sub f and you look at the position of the switch the switch is connected to the forward bias diode right then once a time t equals 0 the switch goes from the forward bias the battery and goes to immediately to the right hand side to getting connected to the reverse bias there battery do you see why it's called reverse bias because look at the negative terminal of the battery that is connected to the positive of the positive of the diode so it's reverse bias so if I can understand this circuit right hand top circuit then actually I have understood the left hand circuit and that's what we are going to do so let's take a look I have to solve this equation in that in that top equation or top figure context at top right context and see try to calculate the storage time the red t sub s is what I am after t sub r will be a little complicated so we will not get there for this device t sub s is what we are after now t less than 0 of course the equation is given by this now do you see whether you agree dq dt that's fine if I sub f well there is a forward current flowing dc forward current flowing so my i diffusion is i f and that was not dependent on time it was a dc thing so that's what I have now what was this dq dt it was steady state was there any change in with respect to time of course now there was a constant profile sitting there so dq dt would be 0 before I have started switching there is no change in change in chart and so the dq dt is 0 now the question is that why would the current remain the same after the transition also that's that's what I'm going to explain to you a little bit later now one thing I'm going to let me before I go there one thing is this current is going to remain the same do you remember in the last class when we are talking about this small signal response we had a register a diffusion capacitor and a junction capacitor now can you change the voltage across a capacitor instantaneously you cannot right that will require because it's half cv square if you change v instantaneously then that will require infinite amount of energy so you cannot change the voltage instantaneously across the capacitor and so and thereby since you cannot change the capacitance therefore you not sorry cannot change the voltage therefore you cannot change the charge as well because q is c multiplied by b c is a constant v you cannot change between 0 minus and 0 class and therefore q you cannot change also so from here can I write this that for t less than 0 dq dt is 0 steady state i f is whatever current I was flowing in the forward bias case I know the initial bias 1 volt I know the current so I have that and so q at 0 minus it is going to be i f multiplied by tau sub n cross multiplied dq dt is 0 okay now the charge cannot change right because the voltage across the capacitor cannot change total integrated charge cannot change so q 0 minus is equal to q 0 plus so that's fine I'm trying to solve that equation during the transient period so I'm setting up the boundary condition you know boundary condition anytime in the circuit you have done this many times so I should really be level at this point okay so how do I solve this equation then now t greater than 0 t greater than 0 the current is the equation is now dq dt equals minus ir now why is minus ir because you see it's a reverse biased diode the current must be flowing in the other direction whatever the vj is whatever the junction voltage is what is the maximum that it can be in the forward bias side equal to the band gap right it cannot be larger than the band gap because the vbi the maximum voltage of vbi the whole barrier it can be if you dope it on both size degenerate then the maximum barrier it can be is equal to the band gap so whatever you have let's say you have a close to 0.6 0.5 0.6 volt and then if vr is minus 2 volts then you can just look at the circuit and immediately realize that given a r the current in the output circuit must be flowing in the negative direction it doesn't matter what the diode wants to do given that diode can only have let's say 0.6 volts in plus vr let's say is minus 2 volts the output current in R must be negative going in the negative direction so the diode has no option but to supply the negative current it it doesn't want to change the voltage fine but it must supply the negative current otherwise the circuit will not be satisfied and it can do so easily so by the way this current and I'll show that this current so long the vj is smaller than much smaller than vr that current through R will be simply vr divided by r right that's a small voltage diode voltage so in that case the current is approximately a constant not fully a constant approximately so ir is a constant and so this is a first order differential equation everybody can solve it right so you just cross multiply divide and then what will be the equation what will be the answer from here do you think it's some sort of log that will come out and the time on the other side will have a certain amount of time on the other side okay and when you integrate you will get an expression like this you can see the log right and then you can see the two boundary conditions q at 0 plus that's on one end and q at p sub s end of the storage time that's the another end right now this point is very important to understand and this is a conceptual thing that why ir can go in the opposite direction and I have explained that in a few minutes ago while not changing the charge too much between 0 minus and 0 plus you see all it has to do all it still has to do is can keep the charge almost the same but change the slope on the one end a little bit and that immediately gives it the requisite current minus ir and as it is losing the charge gradually over a period of time it just keeps the slope exactly equal to the value that the circuit dictates the circuit through the resistance are once a fixed negative curve and so it will just keep that profile okay all right so how much time then I have I still haven't solved the problem completely except if I make this assumption you see in the beginning I had a lot of charge forward biased diode it was exponentially delta n exponentially went with voltage and it was gradually dying off millions of electrons sitting in the minority car historic at the end when the charges have all gone away then it's a reverse biased diode on one side I have ni square divided by n a and another side I have zero because in the reverse bias side the boundary condition requires that that's zero and so I essentially have no charge left you know ni square over n a in silicon let's say 10 to the power 18 dope I have 100 minority carriers and so one end I have 100 another end I have zero I have minuscule amount of charge in the beginning I had millions of charges as a result I can get rid of this q t sub s because that's the final amount of charge I'm left with and if I do that then I have a equation which is actually very simple you see ir plus if why does that if come from do you remember q at zero plus is equal to q of zero minus and q of zero minus I relate it to the forward bias current so ir plus if divided by ir now there is something very interesting here so do you realize that if you forward bias the device very hard if you forward biased it very hard then it will take a long time for the charge to go away because tau sub s is how long if the charge how long it required for the charge to go away and if ir if is very large you forward biased it very hard it will take more time that makes sense right you can see that that immediately makes sense tau sub n when you have a lot of traps is tau sub n larger small small right tau sub n small so when you have a lot of charges it makes sense that charge will decay very fast it makes sense right as soon as because this was a lot of area under this curve and if you have a lot of traps then it will go away very fast if tau sub n is not very large then it will require me a lot of time because all the electrons has to go out through the door the other door meaning the other side of the junction nothing is going out through the majority carrier side on the door acceptor side and so it will take a lot of time now this is a very important point tau sub n because generally traps are bad we don't want traps right because it is non-ideal characteristics and all those things but when we want ultrafast switching many times people intentionally put in traps in the minority carrier side because by putting that in you can see that you can turn on and off this diode very fast and it will i'll show you later that it behaves almost like a majority carrier device at that time if you put a lots of tau sub n and that's why this is a very important circuit trick that people put it by hand in order to analyze the devices okay all right what about the turn on characteristics or the turn on voltage transient well if you want to do the voltage you can again start looking at the total charge q sub n instead of integrating between 0 to t tau sub s you know that's what we did in the previous equation you just integrate between 0 to any given time t so not all the way to t sub s but any given time t you will get the charge q sub n and once you know the charge you can get the voltage how much voltage is how will you get that you will put this charge in in n p 0 and t so this is that charge so you will just put this charge back in n sub p is zero that's the minority carrier n i square divided by n a so you know how much the diode is forward biased and from that you can calculate what the forward bias voltage must have been so again you can calculate this and characterize this transition from the red triangle in the v a car to the green circle where it gradually turns off so when you go home make sure that maybe you can put it in a mat lab or in your calculator see how this works out put some value forward bias current you know a milliamp maybe a reverse bias current maybe a micro amp put it in and see whether you can actually calculate these things this is something I will not explain but what I want to point out that if you worked a little harder not too much hard then you can also calculate this storage reversed storage time at this t sub r and the t sub r has this complicated relationship but you can see you know them all right you know if what was the forward bias current in the beginning IR well that was a final final reverse bias current t sub r is unknown also p what is that that's the recombination time you had more traps the recombination will be more you can put this whole thing in and you can see how you can calculate the t sub r the error function well I am sure you have forgotten but you can look it up and the point is that once you express it in a very simple form you can calculate this the point is that you can do it so we will not do it for the time being and you can sum them up and look at the total current and you can see that in most of the time the t sub s the t sub r is typically very small you can see the dotted line and the solid line in general and so therefore the simple calculation we made is fine for the time being is fine no need to understand it in detail but if you really wanted to understand it you can look at these book at page 117 and then we can move on from there but for the for the time being we will not get into that that detail turn on characteristics again that's very simple turn on characteristics let's say you're initially you had a voltage and you now from a reverse bias this time you just want to turn it on and go to the forward bias and if you wanted to go in the forward bias what do you do well one thing you can say that at t equals infinity whatever it is that t equals infinity is going to be steady state again so if it is steady state and you also know at t equals infinity the current will get to the forward bias whatever the steady state forward bias current is i sub s and dq dt at t equals infinity is zero right steady state again because you have transistor turned it on steady state and so therefore at time t equals infinity your current is i f tau 7 right that makes sense because that will be the steady state current and then again the charge will exactly now go in the opposite direction it will start with essentially zero as soon as it's forward biased although it doesn't have the full amount of charge it will tilt the potential so that it can keep supplying the forward bias current and you can see why the slope of the profile is essentially the same dndx at that point gives you the same forward bias current i f and this will go on until it reaches steady state and so in the forward bias case do you realize why this i f must be the same because the output circuit wants the current to be the same and therefore i f is the same and again you can integrate it out no rocket science here this is the result of the integration because you started from time t equals zero therefore it looks a little bit simpler does this final answer look about right when t equals infinity when t equals infinity what will happen to the exponential do you think that will go away so of course at q at t equals infinity they must be equal to each other right and t equals infinity i already know this i sub i sub f multiplied by tau multiplied by tau tau sub n right that's i derived it in the last slide and so you can see that over time in the beginning when t equals zero current charges zero t equals zero exponential gives you exponential of zero gives you one one minus one is zero so in the beginning you don't have any charge makes sense right and then when you have t equals infinity forward bias current is gradually filling this tank up you know this storage tank up and gradually you will have the full final potential so you can calculate how the charge would behave as a function of time now on few other applications of how this charge control model that you have just seen used to do large signal transient how it can be effectively used to solve many other important problems so one of them which is very important is calculating this diffusion time by this charge control method and what I mean by that is that assume that a device is forward biased an electron has just been injected on the left hand side let's say of the device just at the age of the minority carrier and then this carrier will spend some time within this minority carrier region before eventually being collected by the right hand contact the question is how long does it take for the electron to go from the age of this region to directly to the other contact how long does it take and that's the diffusion time because the electrons are moving by diffusion now if this recombination was a dominant thing then of course you expect so you expect that the electrons would come and directly will recombine with the holes below and that will be whatever is the response time right because that is how long it takes for the electrons to go but if you do not have traps and as you can see this triangular profile is saying that we do not have traps because the second derivative of diffusion equation is zero ax plus b equals zero ax plus b equals the carrier concentration so no traps and in that case what the electron does is is sort of this this issue that it takes an electron takes many path and eventually gets out and everybody takes a different path why does it go back and forth phone on scattering ionized impurity scattering so it scatters back and forth many times before it goes out and the question is how long does it take now although the path looks complicated a very simple formulation of the charge control can actually help us answer this question so here I do not have any recombination so I will drop that term now the way I can answer this question about how long it takes is effectively to assume that in the beginning I had a certain amount of charge and then I dropped the voltage I turned the voltage off and if I look how long does it take for this charge to decay away from this minority carrier region away from this region altogether and restore equilibrium that would be the amount of time the electron effectively on average everybody has different time on average how long it takes for the electrons to get out so you do something like this you will say that right after zero plus time c equals zero plus when the voltage was turned off I had a certain amount of charge and at t equals infinity well the charge will be very small right because that's the equilibrium charge and the time for this charge effectively requires to drain it away which is draining it away to the right hand side the contact at that I will call the tau diffusion so I can calculate the dq dt as a difference of initial and final charge divided by the time it takes for the charge to drain and the diffusion current now this I could drop because it's a small quantity and as a result I could write the tau diffusion is equal to this charge initial charge I had divided by the current but you realize the charge is simply the area under this triangle half the ordinate multiplied by wp wp is the base so that's the total amount of charge q in the beginning and the current is simply q dn dndx the derivative but derivative is same everywhere here this is straight line so that will simply be delta np divided by wp and the amazing thing is that it tells you that on the average how long does it take for on the average how long does it take for the electrons the red electrons to bump around and get out of this region and you can see it goes as wp squared so the length of the minority carrier squared divided by 2 dn now why does it have this particular formulation you can easily answer that question actually because you can see that I can write it approximately as wp which is a distance and dn divided by wp is a diffusion velocity right and so essentially you have this wp squared divided by dn that's what you have and why there is a half well half because some of the particles are sort of in the end in the beginning close to the right hand side some close to the left hand side and on the average you have a transit time which is sort of a mean of this time because not everybody is starting on the left and going out right for this formulation and so that gives you a formulation of the diffusion time simply by using the charge control method there are more complicated ways but this is the simplest now you can also do another thing very conveniently with the charge control method which is to calculate the ac dc diode current without having to go through solving all the minority carrier equation you know matching boundary condition we do all those things in the about two or three class ago about remember region one two three four all those complicated things you don't have to do any of this charge control will almost give you an answer for that very quickly for example this way so we are looking at steady state so dq dt is zero and i diffusion is the current that's coming in which I don't know and want to calculate but by the way this tau sub n if I don't have any recombination I just told you that tau 7 is equal to tau diffusion because that is how little it takes for the electrons to get out not through recombination but by diffusion so I put that in and so I can write that the diffusion current whatever that is is equal to q divided by tau diffusion and if I put it in then you will see the q is again half the ordinate and multiplied by the wp which is the base but this time the q is actually proportional to the voltage exponentially proportional to the voltage remember that was the boundary condition that we did for as a function of voltage fn minus fpe that gave me the quasi-farm level separation and excess charge so I can put that in tau diffusion I just calculated and do you remember this equation the final one dn divided by wp and i squared divided by na and the exponential of the whole thing that was the that was the diode current right but I don't did it just in two lines instead of having to go through all those complicated derivation this is really easy but of course you have to really understand the problem in depth before can use this type of two line derivation that requires practice and this is an exact expression that we actually got from the previous case okay and very quickly then we talked about large signal response is of great importance in digital switching of course I gave you a very simple example but in reality when you have a CMOS inverter a MOSFET turning on and off the resistance and output resistance capacitance those are more complicated so we'll have to reapply many of the things that we learned in actual situation but for the time being it gave you some idea this way of solving differential equation by charge control which is to get rid of the space derivative by integrating over a certain region is very powerful technique it's a general mathematical technique has got nothing to do with electrical engineering per se anytime you see a second order differential equation with time derivative built in you can always use this and impress your professor that you can you can do this but this is a general way and it's really very very powerful and finally the boundary conditions for these problems are required to be carefully done t equals zero minus and how it couples to t equals zero plus what happens to charge at infinity sometimes you will see charge goes away so you can drop the infinity term sometimes the charge builds up to a steady state then you shouldn't drop the the boundary condition right so this has to be it has to be carefully done and if it is done carefully this simple differential equation is very powerful you can do a huge number of things while using this okay