 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 21 and we'll talk about pn junction diode and its IV characteristics if you remember from the last lecture we talked about the potential solve the Poisson equation in the presence equilibrium where there is no current to begin it so no there is no approximation but also in the presence of a bias forward and reverse bias and in that case of course there was a current but I assume that there is no current I solve the Poisson equation in isolation and I told you the rules now what will happen of course the most interesting thing is IV characteristics so let's get started so first I like to draw derive the forward bias formula then we'll talk about the nonlinear regime and then ambipolar regime and finally conclude now this term may not mean much to you right now but hopefully by the end of the lecture you will see what they mean so we'll be talking about DC characteristics of a diode today and the reason I mentioned this before also the reason I put it in this format in this matrix is because I want you to realize that the way the techniques I'll be using to solve the DC of a diode DC characteristics of a diode exactly the same essentially for short key barrier for BJT is the MOSFET the tricks are the same so if you understand what why am I doing a certain class of things you'll see you will see the whole process repeated over and over again so it will be very easy for you if you understand the basics so again I today I am going to focus on this part of solving equations assuming the Poisson equation I have already solved by drawing the band diagram I have mentioned in the last class that that in equilibrium you draw the Fermi level by following the rules flat quasi-Fermi level and associated continuity of the vacuum levels and then you go back and do this this is a simplified case where both sides of the junction is the same material homo junctions and therefore I have this simple rule when I apply a bias on this device the first rule is that of course I ground it on the right hand side let's say that's an assumption and then apply a corresponding bias negative or positive depending on the configuration on the other side the inside and then the barrier height gets reduced by that applied bias that was the forward bias forward bias case we know how to draw this I'm just putting it here for reference and you also know that how to calculate the width of the depletion region by replacing QVBI minus V with QVBI minus V and correspondingly calculate the depletion width for the forward and the further reverse bias now these are all from last class so therefore you should be able to understand this easily but let me go on to the main topic of interest in today's lecture so this is what I wanted to first mention physically that what happens during in forward bias let's say you see in equilibrium there is lots of electrons on the n side and a very few very few electrons on the p side in equilibrium that is ni square divided by na so there is a huge density gradient I mean you cannot believe the amount of density gradient 10 to the power 18 1 1 side let's say humongous number and 100 electrons trying to move around on the other side right huge density gradient and the electrons if you had any chance it will immediately spread out but as it begins to spread out as you realize that it depletes the region and as a result of electric field comes up and the electric field pushes that huge amount of electron from diffusing into it and that's what the equilibrium situation looks like a very intricate and detailed balance between diffusion and drift that keeps the current to zero current is zero but the component that is forcing the current are humongous probably in the order of let's say even the net is zero but individual components are significant so as soon as you disturb the balance by applying a bias then what happens the diffusion remains the same do you realize why because one side is ND doped right same 10 to the power 18 the other side is na doped accepted dope same hundred that I have not changed so the diffusion force will remain exactly the same but the drift component of it previously I had Q V B I dropped across the junction now I have a reduced bias dropped across the junction also so it's much smaller field and immediately current begins to flow as has been shown on the right side you can see now the green circles are gradually moving to the right now do you realize one thing that the current flow J sub M electron current flow should be going the other direction because anytime electron flows from left to right the current flow associated with the electrons must be going from right to left similarly for the holes the barrier has been reduced for the holes as well and therefore the red circles which are supposed to be the holes now can flow in to the other side and as a result there'll be a hole current but hole are positively charged therefore the hole current also goes the same way same direction from right to left now therefore you realize there's a very interesting thing here but just by looking at the direction of the current in your emitter you cannot say whether it's electron or hole because both goes in the same direction how would you know if something is electrons or holes in a semiconductor what experiment hall experiment precisely right that's why we have to do hall experiment because hall gave me a correct sign to tell me whether it's a p-dope material or an n-dope material from here in a two-terminal device you can never say which side is what whether it's the electron dope or it's a whole dope transport material so let me get started on the diffusion limited regime the diffusion limited regime for the diode will be one piece on the forward bias side on the green region you can see I marked it one but I have also marked a piece of it in the reverse bias side in the ash colored region with a flat dotted line so both actually this one on the forward side and one in the reverse side they actually belong to the same curve and in an undergraduate class you just focus on one and then you go home and in that region the key characteristics is that the current increases exponentially with voltage and the proportionality factor is q over kt q over kt is a proportionality factor kt over q in room temperature is 1 over 40 right so therefore you have that corresponding scale factor okay so why does this exponential thing comes in let's try to solve this problem and we should be able to do that you can see I have drawn a forward biased junction characteristics qvbi has been replaced by qvbi minus v you see that I have applied a minus v voltage over there now how will I calculate current in such a structure so this is how it works it's very beautiful actually that people can solve such a complicated problem you see if I don't assume any recombination assume no recombination then any electron that comes in from the left side must get out from the right side and in steady state now steady state I'm talking about current must be continuous right if you have 10 to the power 18 number of electrons the current must go very slow if you on the other hand on the right side you have hundred electrons current must go very fast but current must be continuous right and in that case if the current is continuous it really doesn't matter where I calculate the current in fact I can calculate the current wherever I wish and as a result I will solve it somewhere well it went to be the simplest and the simplest when it's a minority carrier because there's just a few electrons to think about I could have solved it on the left hand side also in the drift region but that would have been more complicated but the continuity allows me to calculate anywhere and then translate that current to anywhere else so let's solve it we already know this solution right we have already done this in a in an example problem in the class have we not steady state which term do I get rid of right there no generation no recombination no photon coming in I said no traps so therefore Rn and gn g7 those go also now do you remember in the minority carrier when you have a minority carrier in that case the electric field is approximately equal to zero I gave a detailed explanation right why the electric field is zero it is not zero it is approximately equal to zero on the order of a microvolt or so per centimeter and so I drop that first term also and I have dndx now this should be now very simple for you so my solution of that region shown here in the in that sort of the material color and that's that will be equal to the secondary derivative of n multiplied by dn equals zero now this even I can solve this is very easy now before you take that second derivative and set it to zero which is very easy you have to understand about boundary conditions which is not as easy to understand it is easy once I explained assume that I want to know about this region in the minority carrier side where I have shown here in the ray triangle the profile of the solution of the second derivative of the equation what I want to know is the boundary condition that I have to put on the left side and on the right-hand side what would be the boundary condition on the left side well first of all in general you know that n of x I have already explained to you that how is it related to ni of equilibrium right and since it applies to any x it must apply to x equals zero plus which is the just to the right of the ray triangle on the bottom figure now this you have also seen the product of n and p is equal to ni squared and the difference of the quasi-farm levels fn and fp now do you realize that what this trick of keeping the Fermi level quasi Fermi level flat to the other side of the junction help helps me with because now if you look at this yellow region the line going vertically across the yellow region what is the difference of the quasi Fermi level this must be equal to the applied voltage because I have kept the red line flat I have kept the blue line flat out to the other side of the junction and since there is no drop whatever voltage I have applied that voltage that difference has stayed flat all the way to the line associated with that yellow circle as a result I can immediately put the difference of the quasi Fermi level equal to uvba beta is 1 over kt so I just remind you this okay now if this is the case of equilibrium of course you see when va is 0 np is ni squared equilibrium is equal to ni squared that's fine now at the yellow point I know it's a majority carriers are the holes how many holes do I have is equal to na because do you see that the blue line and the green region for the holes that difference has remained flat throughout this region so at the yellow point how many holes do I have it must be equal to the number of acceptors so I have that but that immediately tells me how many minority carriers I have at 0 plus that in equilibrium I simply have ni squared over na but as soon as I apply a bias then it increases exponentially right which started let's say with hundred immediately becomes a thousand ten to the power four ten to the power five as I keep applying the bias very quickly it goes up so that will be my boundary condition on the left side you realize this I'm looking at delta n is amount that I have above the equilibrium value and so therefore I have a minus one which is simply whatever I have in the presence of a bias minus whatever I had in equilibrium so that's what I have the extra value that I have okay this is one side what's about the other side the other side you already know this other side is a method has been terminated is the metal with infinite surface recombination velocity excess carrier is zero and as a result if the excess carrier is zero delta n will be equal to zero W sub P is the bulk region width of the bulk region for the P side it may be several microns let's say this is not depletion right we are thinking about the bulk region I have the boundary condition so I don't need to do anything more right the solution of that equation is C plus DX at X equals WP I know what the delta n is that's equal to zero therefore C must be equal to minus D multiplied by WP you realize this one and as X equals zero I also know what the value is so that gives me the value of C and as a result I'm done that's my carrier concentration as a function of position does it make sense to you do you see at X equals WP if you put it in what is the value of delta n zero which is what I know I should be if VA is zero applied bias is zero then what should be the value of delta n zero again right do you see that you VBA multiplied by beta put VA is zero that gives me one minus one that's that's zero again so this looks about right this is excess carrier right not the original amount original amount is ni square divided by na okay so I have the carrier concentration but I don't care about carrier concentration what I really need is the amount of current that's what I'm after so you can do the current the current is simply diffusion term for the minority no drift and so therefore you take a derivative you can take a derivative one minus X divided by WP and that gives you a minus sign right minus WP and you see that's the current do you agree with this current why is there a minus sign because it's diffusing to the right right more on the left less on the right electrons diffusing to the right which way electrons are going electron current is going to go the opposite direction that's why you have a minus sign and if of course if it's easy to diffuse current is high if the depletion region is very large then the gradient is small so therefore your current will be smaller and as a result you can get the current for the electrons now can you get the current for the holes well why not the current for the holes I'm sorry I should current for the holes is again same procedure because barriers for the holes have also reduced the same time barriers for the electrons have reduced as a result the blues are blue triangle simply says the amount of extra holes you have again a diffusion you do the same process and you get the hole current again you see there is a negative sign the negative sign is because holes go the same direction as the current does so both electron and hole current when I add them up they will amplify each other and that makes sense the total amount of current now I want to calculate the total amount of current and this is the trick I have calculated the current for the electrons on the plus side of the junction to the right I've calculated the current as a minority carrier for the holes on the left side of the junction can I simply add this to to get the total current of course not because current must be added at a given point if you are looking at the traffic flow between Chicago and Lafayette one person cannot sit on the Lafayette side and look at the not bound lane and another person cannot sit in Chicago and look at the southbound lane and then simply add them together right that will not be appropriate only time you can do so is if there is no car transfer going on between between the north side and the south side only then and that was remember my assumption because the hole current although I calculated it to the left of the junction however in the absence of any recombination that current must be continuous to the majority carrier to the right side of the junction now at a given point I can add only in the absence of a recombination so I added and I get the total current you can see how it depends on NA and ND right and from that I can easily show you that in the forward bias case in strong forward bias case this would be the formula do you agree the first term within the bracket on the right hand side that depends on the diffusion coefficient and d sub p also so things in the bracket that's a constant I know the doping if I know the bandgap so I know ni squared silicon remember so silicon outside I know the whole thing is a constant as soon as I make a device those are given now if I apply a little bit higher few voltage higher than a few kt then q exponential of q v a multiplied by beta that will be much larger than 1 so I can drop that minus 1 if I drop that and take log on both sides do you agree with me that you essentially will get a curve that increases exponentially with voltage that's good forward side that increases exponentially what about the reverse side though the reverse side will essentially be just a constant why because you see as soon as you put va is a negative number then it's going to go to 0 and then you have a minus sign and as soon as you have a minus sign then as a result the current will become a constant independent of bias now where is this current coming from this current there's no generation and recombination remember so therefore there is no current like that what's going to happen and I'll just give you a hint here maybe you will try to understand yourself what's going to happen the great triangle that you see the great triangle that you see the great triangle on the right hand side the concentration is ni squared divided by na that's the equilibrium value and on the left hand side you have the exponentially large value when you have it in a reverse bias the left side the left side of that great triangle will go to 0 will go to 0 and there will be a diffusion current going not from right to left sorry left to right but from right to left because one side is ni squared divided by na and the other side is 0 so this diffusion flux will go in the opposite direction and do you see therefore you have a plus sign there because that when you get rid of qbba then the minus one there and a minus q there gives you a plus sign so current will start flowing in the opposite direction no regeneration recombination it's just the diffusion in the bulk direction so we'll talk about now about this solution of the IV characteristics of a p-n junction diode in the nonlinear regime nonlinear in the exponential sense as I'll explain in the next slide one thing we have discussed so far is the forward junction forward bias p-n junction diode and we have seen that the current in region one marked by region one it increases exponentially with voltage and that's because as you applied a forward bias then the barrier effective barrier is reduced the diffusion current wins over the back forcing drift current and as a result I have an exponentially increasing carrier concentration near the junction and that forces in exponentially increasing current in region one and that is shown in the blue blue dotted line that the slope of that is q over kt now if we continued applying larger and larger bias you obviously realize that the current cannot go on increasing exponentially forever and what happens that is you pump more current through this diode that the maximum current you can put is almost equal maximum voltage you can put is almost like a band gap band gap of the device you cannot put anymore but even at that point the current begins to get nonlinear with respect to the voltage that's region 2 and 3 I will get started with region 3 in region 3 what you will see that one of the approximations we made of assuming that the quasi Fermi level is flat starting from the contact to the other side of the junction that is no longer correct and that approximation will now break down and this delta f sub n and delta f sub p is how much it deviates from is the ideal characteristics at a low current case so that's where we'll get started by the way this is an exact formula the other one was a case approximation where this delta fn and delta fp was very small and negligible and therefore we didn't have to write it at the end of the day what we'll see that this applied voltage will will be reduced effectively reduced by the current that is flowing in the device itself so you can see now it's a nonlinear function of that current itself no longer as simple right because Jt is a total current is the sum of electron current and the whole current and you can see on the right hand side on the top of the exponential also the electron and whole current sitting so it's a nonlinear dependence and that nonlinear dependence is reflected in region 3 so let's me explain the physics things will become clear at that point now this question about whether a quasi Fermi level is really flat up to the junction a undergraduate student wouldn't really ask this question because you're not expected to get into this part but for all modern junctions high injection that is what is called when you have a lot of current flowing is an integral part of the device operation and the student must know about it ahead of time now we have already said that a current let's think about electron current for the time being has a drift and a diffusion component and this we can rewrite this in a slightly different form something that you have already seen before we can say that the carrier concentration n at any point in non equilibrium condition is given by this exponential of the quasi Fermi level minus e i how far away is it from the quasi Fermi level in equilibrium the fn would have been e sub f that's something we know the beta is 1 over kt just remember that so if I wanted to write the diffusion component of the current and with that expression of n then I can immediately take a derivative of dn dx and you realize that there can be in principle in general a gradient of the quasi Fermi level taking a taken out from the exponential and the gradient of the intrinsic energy the intrinsic level e i because both are sitting on the top of the exponential so I can have a gradient there as well and that gradient of the e i is written as the electric field e in magenta so the second term on the numerate or on the exponential becomes an electric field and the whole exponential term shown here in red comes out as is you know this is an exponential function now one approximation or one assumption I have already made do you see what I have done here I've assumed that both side of the pn junction are at the same temperature right if the temperature is varying from point to point right in addition of quasi Fermi level and e i then of course my I would have taken out another derivative with respect to temperature but for the time being we'll assume temperature on both sides are the same now if I have that you see you can see that the quantity in the bracket in the red is again n i one more time right so therefore I should be able to write that quantity in the bracket as n so if I insert that diffusion term diffusion term back in the original equation by assuming of course we know that Einstein relationship d over mu is kt over q a single line of algebra then you realize that the electric field term in magenta in the diffusion equation term comes as a with a negative sign and the first drift term in the first equation comes with a positive sign for the electric field and as a result when I sum them the current will only have dependence on the quasi Fermi level so the gradient of the quasi Fermi level is proportional to the current now this you know already you know that but you can immediately see therefore my assumption of having a flat quasi Fermi level implies that I do not have any current right but of course I have a current that is the whole part of the whole the whole derivation I did in the other day with this red arrows so I do have current but the point was that the current was small so when I have a large n in the majority carrier side I'm not talking about the right hand side arrow now I'm talking about the arrow on the left hand side in the n region in the n region I have a huge number of electrons right so therefore the gradient of the quasi Fermi level that I need in order to drive a certain amount of current is miniscule as a result you can see that n is equal to nd take a look at that expression if n is equal to nd jn it's a constant right because its current is flowing without any recombination from one side to another cross multiply with dx and divide by nd and mu n the whole thing is a constant integrate from the left side of the contact up to the junction that's wn and that gives me that the change in the quasi Fermi level the drop in the quasi Fermi level is proportional to the current and what is that factor by the way w n nd divided by or sorry mu n nd divided by w n what is that that's actually conductivity if you look go back and see that that's actually conductivity so therefore this quasi Fermi level drop is essentially proportional to the amount of resistance that you have multiplied by the current amount of resistance you have in the bulk of the material multiplied by the current that sort of makes sense right because as the current is flowing in the majority carrier side there will be this resistive drop or the series resistance before it comes to the junction and that is reflected in this delta fn okay so let me pictorially show you what delta fn really implies and what it means so as I mentioned that this is the part of the left hand arrow that I have focused on in getting the delta fn and making making sure that I understand how the red dotted line falls off from the left contact to the towards the junction side that's how much drop you'll have okay so now let's take a look a little bit carefully first let's look at the diagram below this is a high injection case lot of current flowing in you can see from the black dotted line that the quasi Fermi level on the two contacts have been pushed up by the applied voltage va that's fine no problem but you also realize that by the time the quasi Fermi level have come to the junction then it has dropped by delta fn from the n side because of this resistive drop and correspondingly delta fp as holes are also flowing through the majority carrier side towards the junction and so in the junction itself the applied splitting of the quasi Fermi level is no longer va va it was equal to the applied voltage when they are both parallel to each other right without any drop now if you have a little bit drop on both sides then that absolute splitting in the junction region will be how much it will be va of course multiply minus delta fn and minus delta fp because this is the drop because of the series resistance the junction does not see right as a result my original equation for the minority carriers on the right hand side of the junction which I used to previously write at ni square divided by na and the exponential of the difference of the quasi Fermi level red and blue you can see on the top equation that one I cannot simply assign it equal to va anymore applied voltage anymore but rather I must put that real difference at the junction and that's why I have taken out this delta fn and delta fp in the exponential now do you realize that delta fn has a magnitude or dimension of energy quasi Fermi level and so therefore I don't multiply with q the va of course is a voltage I have to multiply with q in order to get the dimension of energy and beta is one over kt once again so from here you can immediately see that if you wanted the excess carrier not the total one the excess carrier the equilibrium density is how much ni square divided by na right minority carrier equilibrium so I have subtracted a minus one to tell you how much excess I have okay now I'm done because now I can quickly calculate the current again you follow the same rule you have that relationship boundary condition you solve for the diffusion equation without recombination that gives you a straight line you take a derivative of that straight line and that gives you the current so you see everything is the same except I have delta fn and delta fp now this is very important if you realize I just explain the delta fn and delta fp both are proportional to the current right now if you're trying to pump more and more current what's going to happen as you pump more and more current you're applying a more bias let's say applied voltage let's say you apply two volts and your band gap is one uv what's going to happen is my barrier completely disappear and it will go on the other side because the Fermi level separation is now two uv impossible that cannot happen what's going to happen that as you pump more current there will be more delta fn will rise it proportionally because it's proportional to the current and it will always keep making this thing smaller and smaller smaller than the band gap eventually and as a result their junction no matter what you do will not ever disappear and one thing of course and therefore and this I will come back in the next one next problem as well but this is how I the current starts deviating from a pure exponential rise with respect to voltage so that's something we should know now one approximation of course which is a little difficult now is because the quasi Fermi level is dropping the electric field is no longer zero as it was the case for the minority carrier right in the minority carrier case we assume that the majority carrier is holding the potential fixed and the minority carrier was sort of diffusing through that potential barrier no longer because you can see that there will be electric field electric field and potential drop even in the majority carrier side so this approximation well it's good but not great because but in this case you need a numerical simulation that you have been doing and no longer is possible to do analytical work any further but you can give get the physical feel that what happens at high bias why it's non-linear but the exact value well at this point numerical simulation is better now what happens if you have a slightly more complicated situation now this is called an ambipolar regime here the carrier concentrations are so high so high that the minority carrier assumption is completely gone it's no longer true so I want to case that very extreme case and we'll talk about ambipolar transport why ambipolar the word means two polars right so here my previously there was minority carrier and majority carrier when there is no minority and majority both are playing a role then it will be called an ambipolar transport regime again only numerical solutions are truly effective but we can also make some progress based on analytical expressions the main point I want to make in this expression I'll be talking about region 2 but many times region 2 and region 3 maybe interchange depends on the diode diode you have the main thing that I want you to notice of course we have that delta fn and delta fp that we just derived but there is a 2 sitting there and the whole purpose of this derivation to explain beta over 2 now in previous expressions I just had beta 1 over kt but now I had that factor of 2 sitting there and I want to explain that why that factor of 2 comes in and why in the ambipolar regime the slope in the exponential side becomes half of the diffusion limited regime that's what I want to explain now remember this is why I'm how I'm deriving it when you become a big name engineer then of course nobody will tell you what is region 1 and region 2 a technician will make a measurement or maybe your graduate student will make a measurement and they will bring a piece of paper that I don't understand what to do with this device not behaving correctly what you immediately do what professors do what managers do is that immediately they know the physics in their head so they immediately try to identify the slope of this region and by looking at where how it's breaking at what point it's breaking from one slope to another they immediately realize that what's going on in this device maybe the region has not been doped properly therefore the region 2 has start appearing much before then it should ideally and then they will tell the technician that go you must not have doped it properly and he'll be surprised but the reason is of course behind it is he's thinking about he or she is thinking about all the physics that defines this various pieces so this is very important to go from the other side I'm going I'm going we are doing analysis but how to analyze the problem from a given data is very important so this is the physics we all know that n multiplied by p is n s squared in equilibrium and if they are not in equilibrium fn minus fp in non-equilibrium case right now and we have already seen that fn minus fp is delta va qva I have made a mistake here you can see the q should be inside the bracket on the right hand side multiplying the exponential but in this case it's you get the idea that is applied voltage and the you take out the quasi-formulable drop because it's drooping on either side and you take them out so the right hand side is something you have seen already but look at the left hand side left hand side is a little troubling because n is fine n is ni squared divided by na because it's a minority carrier right in equilibrium ni squared divided by na and the little excess carriers that you have is delta n now in general the delta n used to be much larger than ni squared that's why you had the triangular profile much higher than the carrier background carrier concentration no problem think about the right hand right right hand side factor na plus delta p now delta p is to be so small right it's a majority and minority carrier delta p is to be so small that we used to drop delta p and therefore we used to write delta n is equal to ni squared divided by na and the whole thing that's what you used to write but if you have bias the junction so much that you are pumping a huge amount of electrons over the barrier and in order to compensate that huge amount of minority carrier influx a huge amount of majority carrier is also flowing in the device right in that case what will happen and that's the plot that is shown here on the right hand side you can see that n sub n is equal to nd and n sub p sub p is equal to na and the carrier have been injected at a such a high level that both of them are now majority carriers in fact higher than the doping density in that case which term should I drop this time I have to drop both i ni squared over na which is negligible anyway had been dropping all along but this time I'll also have to drop na because the carriers that have come in is actually larger than the doping density so if I take it out do you see what's going to happen because n p is approximately equal to na in this high injection regime and therefore I will have to if I just wanted to know what they are individually then I will set them equal to each other and they will individually be equal to the square root of the whole thing now in the exponential part the apply voltage is huge first exponential term is large minus one drops out if I take out the minus one do you see where the two is coming from the two is essentially just the square root of the exponential sitting inside now that's very important because this factor of two is saying the physics of the diode has fundamentally changed because previously it was minority carrier transport a little bit of diffusion going on nothing funny but in this case as soon as you see a factor of two appearing you realize that you have flooded the pn junction with carriers so there is no majority carrier per se both have become comparable and as a result there is a change in the slope so in a log log plot you'd expect the slope to be half right half as before once again once you know the delta n at x equals 0 then you can calculate the corresponding electron current and the whole current from this expression one thing is again I have just assumed diffusion obviously you can see with this amount of delta fn in the majority and a minority carrier side obviously it is not equal to just the diffusion current look at the right hand side I mean the band EC and EV are drooping significantly significant electric field and so my neglect of electric field well that helps me to explain some of these things but not really 100% correct okay but the point I want to make and this is something people often don't understand 100% that no matter how much bias you apply across a junction the junction itself the potential barrier in the junction is never going to disappear there is always going to be a barrier yes with large current there'll be a lot of drop in the majority carrier side but no matter what you do unless you burn the diode up the junction remains okay so we are proceeding through with various nonlinear part now let me talk about something that earned a young scientist or not by the time he won the Nobel Prize he was no longer young but he is a Japanese person Isaki and I want to explain that how he not only addressed the PN junction problem a very important issue but he finally made quantum mechanics relevant for solid state devices because before 1960s the solid state devices the way the quantum mechanics actually applies or not there was a lot of doubt and this finally ended it all very important and interesting interesting aspect of the IV characteristics so let's talk about it this is called an Isaki diode and I'm looking at region 7 of the IV characteristics and this region 7 has a little bit different characteristics than 1 2 and 3 I talked to you about first of all it's a very occurs at a very low current regime that's first so there's no business of majority minority and ambi polar nothing like that second this only occurs but that they didn't know in the beginning it only occurs if both sides are heavily doped you do you see that in this device in the inside of the diode the family level is actually above the conduction band so it is degenerate on the family level side and you also see the F sub P in equilibrium of course everything is flat so both sides are actually EF but this is below the valence band that means it's again on degenerate on the right hand side now the question is that that how would current flow in such a structure now first of all the types of problems we had been looking into is where when you apply a bias the minority carriers go over the hill on the other side and diffuse out right that's what you have been saying but now you will see a new current component will come through so first of all that if you just played it linearly on the top plot was logarithmic and the cell in the green region and if I plot it linearly then the current will have this exponential right I'm just replotting it just in order to emphasize the low current part of it now when you bias it a little bit then or that device in the state of following the standard dotted line on the right hand side picture it will now follow a strange blue car current of course in the beginning will start with zero in equilibrium you cannot have any current but apply a little bit of bias then this strange thing happens do you see what's happening here as you have applied a little bit of forward bias the electron formula level has moved up a little with respect to the grounded side on the right hand side when the electron moves up a little then the electrons above the Fermi level above the conduction band on the left hand side will now find empty space I'm sorry so it will it will find empty space on the right hand side why empty this is full of holes right remember anything above the conduction band above the Fermi level are full of holes holes is the absence of electrons so electron from the left hand side will now be able to come through and contribute to the current now this current component didn't exist before and it doesn't happen when they're doping is not strong this current if you forward bias it a little bit more you can see that this will try to come to this junction region and as a result there's no state there and so electrons will not be able to come from the left side to the junction at to the current and as a result the current will drop as a result the current will drop and therefore there is this non-linear characteristics so this was for the first time actually it was shown that band gap really exists because you know previously we had optical experiments showing that electron jumps from one side to another nothing very precise but this is for the first time by just by doping on the two sides you can control the current through tunneling and this was very unique and as a result this had a he was honored later on with a Nobel Prize for just this effect it turns out noise there's another giant in the semiconductor field recently there was a paper that about several years before is sake found it out about two years ago he in fact sketched it out in his notebook but only difference is a key actually experimentally demonstrated the effect noise didn't but he understood that if the PN junction theory is correct that under heavy doping conditions this must apply so I will post it an updated slide on this one or maybe I didn't delete the slide okay so this was the physics I'm trying to talk about where because of the current flow at a higher bias essentially the tunneling is suppressed and as a result you can see the magenta point the current essentially goes down eventually joins the original current why because the final current is over the barrier that diffusion current always stays it is this extra component that was unusual now there is another aspect of the of the tunneling that is also important now however in the reverse junction reverse bias side you see I'm going in the other side why I have said a plus VA because in a PN junction on my right side is grounded that's what I have assumed and so when I apply a positive bias then my quasi Fermi level always goes down electrons like to stay in the positive potential and so you can see the FN that has essentially been suppressed down as if a proportional to VA now again there's another tunneling but this time is not as mysterious because you can see for the yellow if the Fermi level on the right hand side is at FP below the Fermi level it's full of electrons isn't it right was a Fermi level so it's full of electrons below the Fermi level and above that lots of holes however now with the reverse bias you can see for the yellow region that electrons will essentially tunnel from the right hand side to the left hand side okay now if you apply a little bit more bias if you apply a little bit more bias then the current will actually increase and as you can see in the blue region the blue dot the current will increase exponentially now do you realize that you have actually solved this problem before you didn't have you might not have realized but you have already solved this problem before you remember that when we were doing chapter one or chapter two in quantum mechanics I told you about a rectangular barrier and how to calculate current through a rectangular barrier do you remember that this was that exercise was in preparation for this problem because you see for an electron to go from the valence band to the conduction band it essentially has that triangular barrier triangular barrier that it has to cross it could directly jump up and then go or it could tunnel through the barrier now if you apply a higher bias then what's going to happen the electric field will increase right the electric field in the junction region will increase and as a result if the electric field increases that means the gradient of the potential will increase that makes the triangular region a little thinner and as a result more current flows through the structure so in fact only modification you have to do is that in that example you assume free electron mass in here you will have to assume the effective mass because it's electrons are tunneling and the electrons on both sides are not free these are effective masses for the valence band on one side effective mass for the conduction band on the other side other than that you have already solved this problem before okay so the IV characteristics of the p-n junction has many phenomenon and we'll continue in the next lecture but I have explained some of them diffusion ambipolar transport and tunneling tunneling has two types one is this zener tunneling which is a reverse bias side and isaki tunneling which in the forward bias side isaki tunneling only occurs in heavy doping zener tunneling occurs all the time if you apply large enough bias it will occur and as I said that you have to understand the characteristics of the specific features of the side V characteristics so that you can identify them and troubleshoot your process so people may depend on you solve a particular problem in the process and you'll have to these are diagnostics that allows you to explore the physics inside of a dive and we'll talk about a few more non ideal effects in the next class