 Let us start with a brief review of last class. Last class we started looking at p-n junctions. A p-n junction is formed by combining a p-type semiconductor and an n-type semiconductor. So we said that these junctions are ideal so that there are no defects and both the p and the n-type are from the same material. So you could have a junction between p-type silicon and n-type silicon. When we form a p-n junction, we found that electrons from the n-type move to the p-type, holes from the p move to the n and they can recombine so that we have a depletion region. We also found that because these electrons and holes recombine they leave behind positively charged donors on the n-side, negatively charged acceptors on the p-side. So that there is an electric field and contact potential between the junction. So we also calculated this value of the contact potential from last class v0 where na and nd are the concentrations of acceptors and the donors on the p and n-side. We also found that the width of the depletion region was inversely proportional to the concentration of your dopants. So that wp over wn equal to nd over na. So today we are going to continue to look further at p-n junctions but we are going to consider cases when we bias the junction. So we will look at both forward bias and a reverse bias. Before we do that I want to link the p-n junction to the energy band diagram in the case of a semiconductor. So let me go back to the example that we looked at last class. So we had a p-n junction with the material being silicon. So ni for silicon is 10 to the power 10 per centimeter cube. We said that we had a p-side with an acceptor concentration of 10 to the 17 and we have an n-side with a donor concentration of 10 to the 16. So last class we calculated the contact potential in this case and when we substitute the values we get v0 to be 0.78 volts. So what I want to do is to link the contact potential to the location of the Fermi levels in the p and the n-side. So let us consider the p-side separately and the n-side separately. So let me draw them here. So this is my p-type, it is my n-type. They are both silicon so the band gap is the same. So what I want to do is to put the Fermi level on the p and the n-side. To calculate the position of the Fermi level we go back to the formula that we used in the case of extrinsic semiconductors. So Efp-Efi is minus kT ln of nA over ni. So if you do the numbers this is equal to minus 0.42 electron volts. So in the case of a p-type the Fermi level is located minus 0.42 electron volts below the intrinsic Fermi level. So just for simplicity I will take the intrinsic Fermi level to be exactly at the band gap. In the case of silicon this is not entirely true. It will be slightly shifted because the effective masses are not the same but for all practical purposes it is very close to the center of the band gap. So Efi is 0.55. So Efp we can get by substituting the value of Efi. So if I mark it here Efi is the center and Efp which is the Fermi level on the p-type is 0.42 electron volts below Efi. We can do the same calculation for the n-side. If we do the numbers Efn minus Efi is kT ln instead of nA it will be nD over ni. Once again we can substitute in the values and we get this to be 0.36 electron volts above the intrinsic Fermi level. So in this case Efn is 0.36 volts. So when I bring my p and n-type semiconductor together I know that the Fermi levels must line up. So you can either say that the p has shifted up or the n has shifted down and this shift is equal to the distance between the two Fermi energies. So this overall distance is nothing but my contact potential Ev0 and if you look at it Ev0 is Efn minus Efi. So that this entire distance which is equal to 0.78 electron volts which gives you V0 to be 0.78 volts which is the same value that we calculate from the formula. So the contact potential is nothing but the difference between the Fermi levels on the p and the n-side. Now this is all when we have a p-n junction in equilibrium. So what happens when I bias my junction? So the first thing we are going to look at is forward biasing the p-n junction. The case of forward bias this is my schematic p and n. I am going to connect the p to the positive side and n to the negative. So let this be V external. There is a contact potential at the junction that is your V0. So if you look at forward bias you are connecting p to the positive and to the negative which means you are injecting electrons onto the n-side on your injecting holes onto the p-side. Another way if you look at it you have an external potential V that opposes the contact potential V0. So this is something you have seen earlier in the case of a short key junction in forward bias where we saw that the external potential will oppose the potential between the metal and the semiconductor. So in the case of forward bias because this opposes the contact potential the barrier is lowered and we can show this again in the energy band diagram. So let me draw the energy band diagram first in equilibrium. So in equilibrium the Fermi levels line up. So this is in equilibrium where the barrier is nothing but EV0. I now apply an external potential that opposes V0. So the equivalent of saying this is that this barrier is lowered. So what we can say is that the n-side has shifted up and it is shifted up depending upon the value of the external potential. So if I redraw this in forward bias this is my p-side. Now my n-side the Fermi level has shifted up so that the Fermi levels no longer line up EFP, EFN this shift nothing but E times V external and the barrier is E times V0-V external. So let me just mark the conduction band and the valence band. So because your barrier is reduced you now have an injection of carriers. So you have electrons injecting from the n to the p and holds from the p to the n. So we have carrier injection because the barrier is lowered. So this carrier injection if you look at it you have electrons going from n to p, holds going from p to n. So these constitute the current in a p-n junction in the forward bias and electrons are the minority carriers in the p-side, holds are the minority carriers on the n-side. So what this means is that current is due to the diffusion of minority carriers. So let me now draw the concentration of the electrons in holds across the p-n junction in equilibrium and what happens when we apply a forward bias. So once again I will go back to my picture of the p-n junction just a schematic. This is the interface between p and n. We have a depletion region so that this side is all negative, this side it is all positive. So we looked at how the concentration of electrons in holds changes as a function of distance. We looked at this last class when you have a p-n junction in equilibrium. So this is log of n or log of p versus distance. So let me mark my interface and also the depletion region. So if I plotted log of n, this value is nothing but n, n naught which is equal to n d and then as you go across the value of n drops then we had n on the p-side. So n p naught. Similarly we can plot the concentration of holds, starts of high so which is p, p naught and then goes down p n naught. Last class we saw that n n naught was n d, p n naught is nothing but n i square over n d, p p naught which is the p-side n a and n p naught is n i square over n a. So this is my n-side, that is my p-side. So this is in the case of equilibrium. So now we apply a forward bias. So when we apply a forward bias we are injecting electrons into the n-side and we are injecting holds onto the p-side. So if I once again plot my concentration, this is log of n on log of p, can again mark the boundaries. So we are injecting some carriers which means there is an extra concentration of electrons on the p-side and ultimately it goes back to your base line. So this we just draw it straight, this is n n naught, this is n p naught, we have some extra concentration of electrons on the p-side. Same way we have some extra holds that are being injected from the p-side. So once again if you draw this, so that you have p p naught, p n naught and you have some extra holds. So these extra electrons and holds are what that are responsible for the current. So we have these extra electrons and holds because you have reduced the barrier for the electrons and holds to move across. So if I call the concentration of electrons at the junction as n p of 0, so n p of 0 is the extra electrons that are injected onto the p-side. Similarly you have p n of 0, can write an expression for that exponential minus e v naught is v external over k t. Similarly p n naught is nothing but p p naught exponential minus e v naught minus v external over k t. So these two terms represent the excess electrons and the excess holds that are injected into your p-n junction because of the forward bias. So these are the excess electrons and these are the excess holds. The value is directly proportional to your v external, so higher the external potential lower the barrier and then higher the number of electrons and holds. So these electrons and holds are still minority carriers the case of p-n junction which means they can diffuse for some distance but ultimately they will combine with the majority carriers and get destroyed. So we can define a diffusion length and the diffusion coefficient for these electrons and holds. So the diffusion length for the extra electrons and holds we are going to call them L H or L E. This is called the minority carrier diffusion length. We can relate L to a diffusion coefficient so that L H is nothing but d H times tau H square root L E similarly d E tau E. So d H and d E are diffusion coefficients and T H and tau H and tau E are the minority carrier lifetime. So these define the time the electrons and holds can travel in the material before they get recombined and destroyed. So d H and d E are related to the mobilities so that d H is k T mu H over E, d E is k T mu E over E where we have seen earlier that mu H and mu E are the mobilities. So let me just put down some numbers in the case of intrinsic silicon we had mu H to be 450 centimeter square volts per second and we had mu E to be 1350. So then we can calculate the values for the diffusion coefficient so that d H is k T mu H over E. So we can substitute the numbers that gives you a diffusion coefficient of around 11.64 centimeter square per second. Can do the same thing for the electron k T mu E over E this is nothing but 3493. So the diffusion coefficient for the electron is higher simply because the mobility is higher. So tau H and tau E I said these were the minority carrier recombination times typically tau H and tau E are of the order of nanoseconds 10 power minus 9 second. So these numbers are different from the scattering times that we saw earlier. So scattering times refers to the time between two scattering events. These times refer to the time it takes for the electron and hole to recombine. So these electrons can undergo multiple scatterings because we saw that a scattering time is around picoseconds. So the electron and hole can undergo multiple scattering before they recombine. If I take tau H equal to tau E to be 1 nanosecond we can calculate the diffusion length L H L is nothing but square root of D times T. So you get L H to be around 1.08 micrometers and L E can do a similar calculation to be around 1.9 micrometers. So L H and L E refer to the diffusion length of these extra electrons and holes before they ultimately recombine with the majority carriers and get eliminated. So let us go ahead and now calculate the current in a P n junction the case of forward bias. So I am going to redraw the plot of the change in concentration versus distance. So here is my P n junction. This is my interface. These are the depletion regions P that is n. We have a P n junction in forward bias. So P is connected to the positive and n is connected to negative. We saw in this case that we can plot the concentration of holes. So that there is some excess holes and do the same with electrons. So you have some excess electrons. So we can mark the different regions. This is P n naught, P P naught, P naught and the extra carriers. So it will be P of 0. So it will be n, P. So for the sake of derivation I am going to take my origin on the n side at the interface between the depletion region and the bulk of the n side. So I am going to fix that as the origin. N P naught which is your excess electrons exponential minus P n naught exponential minus E V naught minus V external over K. We saw this before. Another way of writing this is nothing but P n naught, P P naught exponential. So this is in the case of equilibrium and this is when we have a forward bias. These equations together are called law of junction. So they describe how the junction behaves when you apply a bias. So we saw that we have these excess carriers and as you move away from the depletion region, the excess carriers start to recombine and they ultimately are un-highlighted. So we can calculate or we can write down the expression for the excess holes on the n side as a function of distance x. X is measured from the boundary between the depletion region and the bulk of the n side. So this excess is related to the diffusion length. So it is P n naught or P n zero times exponential minus X over L H. So L H refers to the diffusion length and X refers to the distance from the origin. You can also write an expression for the excess carriers. So delta P n of X should be nothing but delta P n of zero times exponential minus X over L H. This delta is nothing but delta P n of X is P n X minus P n naught. So this refers to your extra holes on the n side as a function of distance. We can convert this to a hole current. So we can write a hole diffusion current going to call it J and the hole diffusion current is minus E the diffusion coefficient of the holes times D P n over DX. This is also the same as writing minus E D H D delta P of n as a function of X over DX. So this diffusion equation is very similar to your fixed first law. So fixed first law says that the flux is minus a diffusion coefficient divided by your concentration gradient. So now your flux is nothing but the hole diffusion current which is equal to minus of a diffusion coefficient and D P over DX or D delta P over DX is nothing but a concentration gradient. So we can substitute the expression for P n that we got and do the differential so that J D hole is nothing but E D H L H. I will only write the final expression delta P n naught exponential minus X over L H. You can write a similar expression for the current due to the electrons. So the current due to the holes is a function of distance. The current due to the electrons will also be a function of distance but the total current the sum of these two will be a constant. So if I were to plot that my distance X that is the total current. Let me just mark my two depletion regions W P W n in the case of holes. So you have a majority current due to holes. So let me just redraw that in the case of holes you have a majority current on the P side and starts to reduce as we reach the depletion region and then it becomes a minority carrier on the N side. Same way for electrons you have electrons are majority carriers on the N side then it starts to drop and then finally it becomes a minority carrier on the P side. So these are your holes these are the electrons but if you add both of them to get the total current the total current is a constant. So we can evaluate the value of the current at X equal to 0 so that we can have a simpler expression. So let me do that next. So I can write down the value of the hole current at X equal to 0 P dH over LH delta P n of 0. We saw that this is nothing but P dH over LH P n of 0 minus P n0. We can substitute the values for P n of 0 and P n0 and we take out the common terms we are left with E dH P n0 over LH times exponential E v external over kT minus 1. P n0 is nothing but ni square over N d. This expression becomes E dH ni square LH N d exponential E v external over kT minus 1. We can write a similar expression for the current due to the electrons and we said that the total current should be a constant. So if we add these two let me just write the current due to the electrons. So it will be E instead of dH you will have dE ni square instead of LH you will have Le and instead of Ne you will have Na. So remember again the electrons are moving on the P side and the holes are moving on the N side but the rest of the expression is the same external over kT minus 1. So if we take these two and add them we get an expression for the total current. So total current J is constant J S0 times exponential. I am going to drop the subscript v external and just write it as E v kT minus 1. Where J S0 is ni square E dH LH N d plus dE over Le. So this equation is called the Shockley equation and this constant term J S0 is called the reverse saturation current density. So this represents the total current in a P n junction in the case of a forward bias. The current is because of diffusion of the minority carriers. So electrons on the P side holes on the N side the current depends upon the external potential. So higher the value of v higher the current. So what happens if you now apply a reverse bias to the P n junction. So if you apply a reverse bias we connect the P to the negative and N to the positive. So this is your v external. v external now is in the same direction of v0 which is your contact potential. So just like we did for a Schottky junction we will find that the barrier has gone up. If you were to draw the band diagram for this this is E of P this is E of N. So that now the total potential since the P side is the N side plus v external. So in the case of a reverse bias there is a small reverse saturation current and the reverse saturation current happens because we have minority carriers that are holes on the N side get attracted to the negative charge on the P side and then diffuse similarly electrons on the P side will diffuse on to the N side. So there is a small current there is due to the diffusion of minority carriers and the value of the reverse saturation current is nothing but the JSO. So this is the reverse saturation current that we just derived. So let me put these two together to draw the IV characteristics for a P n junction. So let me draw the IV characteristics. So the first quadrant is your forward bias and the fourth quadrant is the reverse bias just extend this axis. So we found out that in the case of forward bias we have a current that increases exponentially with the applied voltage. So if you were to draw this have a current that increases exponentially typical values of the current or of the order of milliamps. So this is in the case of forward bias. In the case of a reverse bias we found that we have a small reverse saturation current that is due to the diffusion of the minority carriers. So the J is just J s naught is in reverse bias. If you were to draw this you would just draw a small reverse saturation current. The forward bias current is of the order of milliamps. The reverse bias current is either microamps or nanoamps. It is much smaller than the forward bias. So based on this can say that a P n junction is just a rectifier. So that it will conduct in one way that is it will conduct during forward bias and during reverse bias it will still conduct but the current is so small that you can say that the resistance is really large. So it is equivalent to saying it will not conduct. Similarly when we looked at a short key junction we found that a short key junction is also a rectifier. It will conduct in the forward direction but not in the reverse. The difference between a P n junction and a short key junction is that the reverse saturation current for a P n junction is much smaller. So later we will also look at some examples where we will compare a short key junction and a P n junction. So today we have looked at a P n junction in bias. So we found that a P n junction is essentially a rectifier. So it conducts in one way and not the other. So far we have only looked at junctions which are between the same material. So P and n are the same material. Next class we will first do an example to get some idea of some of the numbers that are involved and later we will look at what happens if we have a junction between different materials.