 you can follow along with this presentation using printed slides from the nano hub visit www.nano hub dot 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 okay so this is lecture 22 and non-ideal effects and in this by this non-ideal effect I mean there are some additional complexity that we want to look into for the IV characteristics the first one is junction recombination because we haven't been sort of putting the recombination away but you know you cannot do it forever and we'll talk a little bit of impact ionization before concluding we are still talking about the DC characteristics of a diode and at various bias conditions and today we are going to talk about this six region six this is trap assisted recombination generation so it's almost it happens at a only at a very low bias that is where you can see it happens at high bias also but I'll show you in a second why you will not see that effect in a high bias case I'll show you in a second and then we'll talk about this other other aspect of it now it's a little bit mathematically complex I cannot explain every step I can explain every step but I shouldn't go through them all you should be able to follow the math once you go home and trace through the argument one argument we had been making that in if you have the device in equilibrium the Fermi level is flat there is a built-in potential of q vbi and the carrier concentration has a certain profile as shown here on the right-hand side you can see the blue remains a majority carrier on the left-hand side becomes a minority carrier on the right-hand side and falls off in between similarly the red large on the right-hand side very small on the left-hand side and falls in between but the idea is that in equilibrium if you take any two points at a given location multiply them what is that value so always equal to ni square so therefore there's no recombination generation because the numerator of the recombination generation is NP minus ns square in equilibrium no net generation or recombination however as soon as you apply a forward bias let's say by the amount of v then the carrier concentration will not remain equal to equal to ni squared anymore right so that is what I have shown on the bottom right figure that the carrier concentration is no longer the same and in the junction region especially it has been modified quite a bit you can see the linearly drooping region on the blue car that's the diffusion remember the triangular region so that's I'm trying to show it there but now I want to just focus on the junction because junction is where recombination generation plays a important role now why in the junction it is because when you diffuse a dopant in in a let's say in a p-type material you diffuse a dopant in in the junction region many times they do not find their chair meaning they did not find the spot they should sit in and therefore junction is always prone to a lot of defects and right processing is very important so you primarily focus on the junction region where recombination generation might play an important role so let's let's try to work it out I just want to calculate the current due to recombination alone assuming I have calculated all other currents previously now what happens that the electron and hole comes in and you can see that the red and the blue they will come and they can come halfway in the junction and they can recombine right why don't they recombine on the right and the left hand side because on the right and left hand side carrier concentrations are close to equilibrium I'll show here that this is a huge amount of carrier concentration in the junction and that causes a significant recombination this expression you know and fear and this is a complicated expression but this is what expression is this this is Shockley Reed Hall expression right but this time I have written it as a function of nx and px because carrier concentrations are not constant everywhere you have solved this problem also in your homework now let's try to calculate this nx and px and we'll insert it in and see see what happens now if I assume tau n is equal to tau p for a particular case it simply means that their velocities are approximately the same okay e i is equal to e t that all the traps are mid-gaps sort of so that n1 and p1 you know this right becomes equal to n i okay it's my simplifies my life doesn't change the physics so I have a very simple expression where np is given by n i squared and exponential of the voltage and then in the denominator tau have been taken out n and p together and n1 and p1 has been written as 2n i okay this is no rocket science yet hopefully you remember the homework now this is I have mentioned this already that in non-equilibrium case the splitting of the quasi Fermi level at low current levels low current levels right otherwise they'll be drooping the Fermi level and in the junction the separation will have a delta fn and delta fp that I am not writing low current level so the carrier concentration goes like this but first let me convince you that the carrier concentration will be given by this red line do you agree n i exponential of fn minus e i that you agree right that's the traditional expression now fn at low current is flat up to the other side of the junction so I shouldn't really have to touch fn so I keep it keep it the same but e i well as the junction as the potential is varying e i you should follow along so e i will have a reference e i sub l which is whatever the i is on the left hand contact and the potential whatever potential does e i does the same thing so you put it in there and that's the second expression so only x dependence is coming through the potential vx okay so what about px do you agree because n and p is equal to ni squared the exponential that's on the numerator on the denominator I have simply written an expression borrowed the expression for n sub x and just put it in the right hand side denominator and if you divide you can see that ni squared and ni will take you take care of one ni here this is a constant that comes here and the whole thing gets flipped so you have a minus sign but other than that I haven't done anything here see okay now let's see so we'll redefine this term I'll go through the next two things because difficult to explain in a board like this all the steps but I'm just renormalizing things and calling variables various names so that I can write it a little simply don't have to carry around all the kts and all other things and I can emphasize on the shape of the function now do you see on the denominator why it will become a cosine hyperbolic do you see that what is the expression for cosine hyperbolic e to the power x plus e to the power minus x divided by 2 right so you can see why there should be a cosine hyperbolic do you see why should there should be a sine hyperbolic because if you pull out e exponential of ua over 2 pull it out then the remaining piece will become a sine hyperbolic because it has a exponential 2 exponential subtractive now the applied voltage is a constant it doesn't depend on a position so I should be able to pull it out of the integral that's what I have done and only place where the x dependence remains is in u right u is the new v here normalized v so there's the only place it remains and then if I want to integrate this out you know I have just written that if I want to integrate it out I just provide the two steps that is behind that step I just made that why there's an exponential of cosine and there's a sine hyperbolic in here this is just go back do your in your room for a few minutes that should be simple so once you have that you can integrate this and that requires the integration table and I will not do that but only point I will point out is that u at x at a given point generally one can find out the from this triangular electric field what the potential is at various points but in general that's difficult so I will replace it with a constant electric field and call that e max and then e max multiplied by x is what that u is this is an approximation but now I can do the integration and that's it that's the integration and this one has two features that I want to emphasize one is the effective width do you agree kt over q what is that that's a dimension of a voltage if you divide with e max which is electric field what do you get you get a dimension of a space or the distance so that is of sort of the effective distance over which recombination is occurring the right hand side do you agree that that's essentially the extra carrier concentration that you have in the junction region exponential of q va over kt now why is that to because it's middle of the junction because the whole thing is separated by va in the middle of the junction is separated by va over 2 so therefore you have that and this feature that it goes half of the slope of the classical diffusion that is an indicator that you have a huge amount of traps in your system you better take care of it otherwise your diode is not going anywhere that's signature of that to in the low current part in the high current part what is the signature of that's the ambipolar part right ambipolar has lots of electrons and holes together that's where the 2 comes off that's high current but a low current if you see also you have half the slope then you know that that's because there's a lot of traps in your system and where is the trap hiding by the way do you see what the trap is in the tau right because tau is inversely proportional to the number of traps if you have a huge amount of traps then they recombine very quickly tau is small if you have no traps tau is infinity no recombination current so that makes us and this is a very important problem by the way because we had been thinking about just one-dimensional cut most of the time it doesn't look too bad but you see the problem is that many times most of the time diodes are two-dimensional you cover it up with the oxide so that dark doesn't go through from the top it doesn't go through the junction is protected but do you realize that any time you have a curvature on the corners the electric field will be concentrated on the curvatures right when do you know that that generally if you have a sphere then you have a certain electric field but they the lightening rod on many of the top of the homes has a very asymmetrical shape right it's very long and very sharp because sharp points have high electric field so what's going to happen that although you did your design on one-dimensional thing that I just showed you in the previous one in the real device you will have these corners where the electric field will be very high and in that case you can have a significant amount of recombination on those points so one has to be very careful about designing junctions for practical devices now that was all about forward bias side that's fine on the reverse bias side the answer is very simple fortunately because in the reverse bias side again you have done it in a homework have you in the reverse bias side you do not have any carriers n and p are both zero and so therefore the generation current is ni divided by tau you have done this in last homework so you can integrate it out now this time even I can do this integration because ni divided by tau that's a constant so I can pull it out I will have a w and what is that w said to a saying and telling us this is the width of the depletion do you remember that the depletion with as you apply reverse bias that becomes larger and larger and goes as square root of the voltage right applied voltage va has a negative sign here so that overall thing is positive and so therefore this current in region 4 increases with the square root of voltage and that's why this current keeps increasing otherwise ideally it should have been flat the diffusion kind should have been flat because the recombination generation generation in the depletion region you will have a square root dependency on the voltage again that's another signature to know that you have too much too much defects okay I will start doing this non-ideal effect associated with impact ionization a little bit let me just show you one slide and then we'll continue so the final point one is to discuss is this breakdown due to impact ionization and this is how it occurs now you are in reverse bias right hand side is grounded left hand side you have applied a huge positive bias right positive on an n side is a bad thing means it's in reverse bias what happens that think about an electron coming from the top from the top side it's like a one skier is sort of standing on the top of a mountain now that electron comes in if it is not scattered by the phonons which is lattice vibration then it can have a certain amount of energy right if it has enough energy larger than the band gap then and in fact it turns out three halves band gap yeah it's a 50% more than the band gap if it can somehow gain that amount of energy then it can kick a way it can scatter with an electron from the valence band and kick that electron up in the conduction band right now it has two electrons one hole this two electrons keep going it's like a second skier has joined two electrons starts going let's say phonons done doesn't steal its energy it gets three halves kt a three halves the band gap approximately then this two scatters with two more electrons from the valence band gets his friends up in the conduction band now I have four electron going and how many holes I had one from before this time I have generated two I have three electrons three holes going and you can very quickly see that 1 to 2 2 to 4 4 to 8 8 to 16 what is this progression this is an exponential progression geometric progression so therefore eventually in exponential progression and therefore huge amount of current will start flowing through such a system and this is called an impact ionization and impact ionization is also maybe called as a inverse oj process because oj force them to combine and go down here electrons are being generated why doesn't it happen and his bias is low well if the bias is low no matter how much bias you put in or how much what you do it never has enough energy larger than the band gap and if it doesn't have any better larger than the band gap no hope of impact ionization so therefore in a forward bias case right or in the small reverse bias case impact ionization doesn't occur and so this is this is it keeps going down like this and somebody helped me to do this animation let's see how it works not too bad what you say so but you get the idea right how many electrons and how it will rapidly multiply now the math part of it I will take up in the next slide okay alright so what you just saw in the previous slide was this idea that if you have an electron from the top of the conduction band shown here in single red circle then when it gains enough energy from the electric field this is a reverse bias p-n junction you remember then it is going to gain enough energy larger than the band gap and most of the time almost 50% larger than the band gap in that case if it can survive that far without scattering then it is going to excite an electron from the valence band which is shown here in blue it will leave behind a hole and correspondingly an electron will go up and then there are two electrons and in this process one electron generates two electrons two electrons generates four four to eight and this process continues like an avalanche and the same thing is going to happen for the holes because if you consider the holes that have been generated in this process when they the world is upside down for the holes and so holes would like to go from left to right and this electric field assist them and when in a high electric field it has enough energy the holes can scatter and it can generate among themselves and it can also pump an electron up to the conduction band so both this avalanche goes out in goes in parallel and as a result there's a huge flux of electrons flowing through so let's see how if you take a small chunk between 0 and w a small grid delta x let's say 100 you divide it and you take one small chunk in that case if you look at the electron number then I will say that the electron number at x plus delta x out of that little box is whatever electron came in in x and then the electron multiplied by its impact ionization probability because not all electrons had energy or were able to gain energy equal to the band gap or more than the band gap in only a fraction did and that fraction is let's say alpha n that were able to ionize the electrons and therefore put them in the conduction band so this impact ionization would be alpha n multiplied by i n that's the contribution from the electron side but the holes are also doing their own multiplication and every time a new hole is generated a new electron is generated as well as a result the term in blue that in is the whole contribution to this impact ionization process remember all this top line is all electrons actually this is just contribution from various places now one thing I would like you to notice is the x axis I have drawn here is in a typical in the form right to left so that the signs will remain consistent okay so this is a very simple equation that we know how to solve because if I just bring the right hand i n x to the left hand side divide throughout by delta x then you realize that this becomes a simple first order differential equation of how the electron current grows as a function of distance right because of the impact ionization and you can see this equation you may have seen before if there were no impact ionization in that case alpha n and alpha p is 0 and you see what this equation is this is simply divergence of j is equal to 0 so this right hand side is like a generation term you see previously have seen divergence of j is equal to generation minus recombination so this whole thing that I'm seeing showing you here is actually nothing new it is just the generation term but this time we can solve it so for example the electron current remember this is steady state there is no derivative term for time in steady state the current must be continuous that whether you look at in the right hand side in the middle or in the left hand side the total current some of the electron and hole current that must be continuous and that total current if I call that total current i sub t whatever that current is I don't know yet if I call that current i sub t I shouldn't carry around a x around it because it is independent of position and the hole current since the current must be conserved must be written as it minus i sub n now this is only in steady state of course if you are doing a transient problem you cannot write it this way because then it will be depending on both time and position okay so if you have this then you realize that immediately this becomes a differential equation with the only unknown is i n and it whatever it is that one can calculate in magenta that will become that's a constant now this is a first order differential equation that you have solved many times in your life you may not recognize it here so people use it something called an integrating factor do you remember first order differential equation of this sort that has both the first derivative and a linear term so if you do not remember just open up a book or search it in google called integrating factor and you will immediately see that the answer you get is look something horrendous like this but it's not too bad because you know as electrical engineers will actually begin to throw away most of the terms and only keep the one that are most interesting but thing is that you can sort of see how why the solution has to be like this right you can see there are lots of exponentials floating around and when you have exponentials floating around that's sort of saying that if you put a small amount of current on the one side that is going to multiply geometrically you know one two four eight and eventually when it goes out from w by the time it goes out to the w a huge amount of current will go out so therefore there are lots of exponentials floating around and you can also see i n of zero i n of zero is the amount of injected current on the right hand side of the junction what is that coming current coming from do you remember that in reverse bias in the minority carrier side one side of the carrier concentration was suppressed to zero right in the reverse side as a result there was a diffusion flux is constant diffusion flux coming out from the right hand side so that minuscule amount of current that will be the seed of this avalanche and then it will proceed so that's the right hand side the one at red i n zero is the diffusion reverse diffusion current flowing through the diode right and this is something we have already calculated in our previous class and and then of course i n w is the total amount that is getting out from the left hand side after all this avalanche multiplication immediately you realize i n is very small i w is huge right because of this all this impact ionization process going on okay and alpha and alpha p you remember these are what are these these are impact ionization coefficients how many electrons does one electron generate as a result of impact ionization per unit length so this number can be 1000 electrons new electrons generated per injected electron per centimeter to a million that means so you can see it's a very efficient process at high voltages this i have already told you that the whole current and the electron current is constant at any point and if it's constant at any point obviously it's constant at w so i am writing it i sub t now look at what i did next what i said that i can drop i pe of w right the whole current and then at beginning of w is this correct this may be correct because you see current is constant and the electron current has drawn huge amount by the time it has reached w after all this impact ionization if the current has to be constant and this one component has grown exponentially the other piece must be negligible right so therefore i drop that term which is ip at w of course i couldn't drop ip at zero because ip at zero is a huge number so therefore you have to be careful how you drop it and also let me write this quantity that i enough zero as a fraction of it is some number i sub p it could be a million so that is the total current that is getting out at the point w relative to the current that was injected on the initial seed point that's really i will be calling a global multiplication factor m sub p so this number by the time it is done it could be you may inject one electron and you can get 10 000 out from the w point so that's what this number is a very large number okay so if i have these two relationships then do you see i can simplify it in this way do you see this on the right and look at the first equation on the top side i n zero i'm dropping it that's small do you see the magenta the show one shown in magenta that i'm cross multiplying and i n at w and it is that equal to one almost equal to one do you see that why because most of the current which is getting out at w is the electron current i do not have any whole current so therefore i n at w divided by it is very close to one so i have one i cross multiply the whole thing and just exchange sides and that will give me this very simple relationship of course it's not simple enough for me i need to simplify further so let's see what we do but you do you see what term next term you should drop one over mp this m sub p is 10 000 let's say so compared to one well that can go also so if that can go one over mp then i'm carrying around just one and if i'm carrying around just one then it's already simple enough but not really now i'm going to further assume that the number of electrons extra electron generated by an electron and number of extra electron generated by a hole they are the same now of course they cannot be the same right because electrons can gain energy they have a different effective mass so they can gain energy a little bit faster let's say then the holes holes are heavy and sluggish they don't get energy from the field as easily and phonons can steal its energy very easily so in general if you look at your book reference books and other things you will see that alpha n and alpha p are not the same but you know for the time being let me assume they are the same if you want you can put the right values in and you can do the integral but if i set alpha n and alpha p equal to each other then what should be the fate of the exponential well that will be gone because that's e to the power zero now that's getting simple enough further further i'm going to assume that alpha doesn't really depend on position because if the electric field is constant in this starting from zero to w if it's so constant then i can say alpha is a constant but of course it's not a constant you can see the electric field shown here in the bottom is like a triangle right it's not a constant but what is going to happen that most i'm going to show you a little bit later that most of the impact ionization occurs at the maximum electric field point so actually we are talking about a very small region close to the peak of that triangle and so i will be taking alpha p out of the integral as well now that's simple enough for me because if i take alpha p out if i set the exponential top of that zero then i have nothing left i have alpha p multiplied by w equals one and this is when you know this happens when significant avalanching is going on because without significant avalanching one over mp that i couldn't neglect compared to one right so this is this condition only happens when i have a significant avalanche current flowing then i will have this now this alpha p what is this alpha p is number of extra electron generated per injected electron and it depends on the electric field because the more electric field you have it is easier for the electrons to gain energy right as a result this has typically if you look at the experimental data people have done a lot of beautiful theory many famous professors have spent probably a significant part of their life looking into this type of impact ionization process especially in the 60s and 70s and one simple form that sort of looks correct in terms of electric field is this now do you realize why this equation might be right you see if the electric field is very small then the exponent is very large and e to the power a very large quantity is zero so you don't have much impact ionization when the field is small but if the field is very large then you can see the gradually the magnitude of this quantity will keep rising and as a result you will have a huge amount of impact ionization if the electric field is high okay now if that is the case what electric field should i put in well if i solve the diode equation correctly then you can see the blue and the red the triangles depending on various bias reverse biases i can always calculate that do you remember depletion approximation the charges and the triangles that we created so i can always create that and i realize that the maximum electric field will be essentially i can put the e max in place of e the electric field e and calculate alpha now if i are on the other way if i know w the depletion width then the point where the avalanche will begin is one is given by one over alpha but that electric field from that i can also calculate the electric field a not is known a not and b unknown material parameters so only unknown in that case is the electric field e and from that i can calculate the voltage right if i know the electric field i also know the voltage voltage at which this diode will break down by avalanche right you by just using this relationship do you see this electric field as i have already told you that this is a formula that we have derived already right do you remember that vbi minus va and it's reverse bias so this should be a plus sign vbi plus magnitude of va and then from this if we insert back in the equation for alpha p and require that condition alpha p multiplied by w that be satisfied i only thing unknown in that equation is the applied voltage and as a result i can immediately tell you at what point is this diode going to break down by avalanche multiplication now this is a very important problem of course in technology there's a very important problem there are places where this effect is actually used in great advantage and sometimes it is bad you know there's always good and bad on the top side where i say good these are could be imagers where you reverse bias a diode just before impact ionization point so it hasn't really impact ionizing a lot now a single photon comes in that's the seed the single photon comes in generate an electron hole pair and now you have an electron which is falling down this huge potential and similarly a hole falling down the potential potential well or not potential well that's a from the mountain top of a mountain going down sort of down the hill and in that case you can have a single photon detectors based on this avalanche process but sometimes it's bad for example there are places where you do not want impact ionization and the corners i have already mentioned the corners are very high electric field like a lightning rod and in that case let's say you have done a one-dimensional analysis everything is fine you put your diode in the product all in a sudden you see it's breaking down all the time and your diodes are burning right feel there is customers are returning your product that is because you haven't designed the junction properly junction electric field was too much so when you're applying a reverse bias this is breaking down from the corner an avalanche is occurring from the corner and then the diode cannot function so depending on the application this is different things might happen how would you avoid that it's a very nice trick how you can avoid this type of problem and the trick is this instead of having an abrupt junction between n and p all you need to do is to insert a little i region little i region in between so put a little bit lighter dope region now i'll see whether you understood the previous concepts now do you agree that the electric field on the left hand side for the abrupt junction is a triangle but that of a region which has a i in between is really like a trapezoid why is it because by the time you come to the intrinsic region whatever electric field you came with that electric field must continue right because there is no charges to capture the field so that is the base of the trapezoid and of course once you come to the other side the electric field is going to go down and go to the zero value because it must start at zero and ended zero now do you remember i have applied a given bias across this if i have applied a given external bias then the area under these two triangles at two cars must be the same because voltage is simply area under the integral of the electric field and electric field means the area under the two cars must be the same now if the area under the two cars is the same which one would have higher electric field do you think obviously the abrupt junction because this one will have significantly lower electric field so therefore just by inserting a thin insulating or intrinsic region in between the abrupt junction you can lower the electric field significantly and in fact this is what you you will hear the term as we go on the term called lightly doped drain this is a this was a huge problem in 1980s for MOSFET design because MOSFETs were failing by hot electron injection and one person essentially came up with idea that you just put a lightly doped region in between the heavily doped region i will come back to this point again but the point is that actually allowed the scaling MOSFET scaling in your computer that to continue for many years in 1980 just based on the concept that i just showed you so none of them are just you know just to torture you these have actual and practical relevance so this is the point i wanted to make that if you insert an intrinsic region that can be excellent now one thing i must explain to you that this analysis that i showed you this date back to 1960s when the diodes were really big the junctions were huge and doping was one side 10 to the power 16 another side 10 to the power 17 this centimeter cube this type of doping low doping these days however in modern MOSFETs in many junctions the doping on one side could be 10 to the power 20 and another side it could be 10 to the power 18 huge doping if you have huge doping the junction is going to be very tiny right very small as a result when the electron comes in from one side it may be able to go from one side to the other side without ever scattering that's ballistic transport if you don't scatter of course you cannot really do an impact ionization right you need to scatter with other electrons so in case of ballistic transport in a highly doped junction everything that i told you might not be correct in this case an electron and that's called a dead space the space you need before an electron can impact ionize you know the first one it has to start now if it is many a huge junction you can impact ionize many times you can start with one get out of 10,000 if on the other hand it's extremely small let's say on the tens of nanometer in that case electron can get out without impact ionizing so you shouldn't any time you see this form situation you shouldn't immediately try to apply the classical formula you should see whether it is relevant or not now one final thing is how do you differentiate between zener tunneling and impact ionization do you remember both in the reverse side gives huge current one by tunneling another by impact ionization now generally one rule of thumb which is not a very good way of differentiating is that zener tunneling always occurs at a much lower voltage compared to impact ionization i'll ask you to verify this statement this is an interesting statement that zener tunneling would occur about a volt generally impact ionization occurs much later you should convince yourself why this should be the case but more importantly you see the interesting thing is zener tunneling is a tunneling process electrons slowly going to the other side and it doesn't have too much noise if you see it's a regular pattern of electrons sort of swimming through the band gap region and getting to the other side very calm current very not very very quiet current on the other hand impact ionization is a horrendous thing going on you know electrons holes are being generated so on the output side even if you have the same amount of current you will see impact ionization has significantly more noise and as a result that tells you that this is impact ionization versus zener tunneling and also impact ionization is not is a highly temperature sensitive function remember because phonons are taking away its energy band gap is changing because of this temperature on the other hand zener tunneling is much less temperature sensitive so by looking at this you know you have to find out why it's breaking down before you prescribe a fix fix to this problem and this is what you should be looking for the signatures a doctor like a doctor you should be looking for before you prescribe a medicine for the patient so that's it i explained the region five and two physics of region five in terms of zener tunneling or it could be impact ionization whichever occurs first so we have actually gone through all these dc characteristics region one purely diffusion region two what is that half the slope ambipolar transport because electron and hole where so much they essentially become equal to each other so the ni square divided by na it picked up a square root similarly region three where what is a huge injection so that there was a huge resistance drop in the bulk region itself delta p n and delta p region four what is this this is generation square root of the voltage as the depletion width is getting up then the thermal generation process increasing with a square root of delta v applied voltage region five either zener tunneling or impact ionization and you know the physics how to calculate that six and seven well six is trap assisted recombination in the forward side and again just like number two six has half the slope of the diffusion regime because the electron and hole must be equal to each other before they recombine so if you see a half slope region in our low voltage then you know your junction is horrible lots of trap sitting there you'll have to fix it right and seven well seven was again a tunneling process and that's the isaki tunneling you remember that how under heavy bias heavy doping condition electrons can flow from the conduction band to the valence band of the other side the extra current component wouldn't happen in low dope samples okay so i just try to describe the physics of various pieces and junction recombination is very important that's a process maturity tool that one can use it as a process maturity tool to see how many defects you have if you have very low this region six that you have sorry very high region six in low voltage regime you better go back and fix your process otherwise this product isn't going out of the door now as i said impact ionization is very important and wide variety of devices use this and these are commercial products that you can buy in various places go to radio shack you may be able to find some of this and now let's move on to ac response we are done with dc this is gone let's talk about ac now ac is i'll apply a small bias and small signal and see how it proceeds how the same thing proceeds okay