 Let us start with a brief review of last class. Last class, we started looking at junctions. These junctions can be formed between two semiconductors or between a metal and a semiconductor or between two metals. We weren't concerned about how we will form the junction. We just assumed that we have an ideal junction or an ideal interface between the two materials with no defects. Whenever we have a junction and a junction is at equilibrium, we said that the Fermi levels must line up. So the Fermi levels line up at equilibrium. An equilibrium is defined as when we have no external potential applied to the system. We first looked at a metal-metal junction. So we have two metals A and B with work functions phi A and phi B. And we also said that phi A is greater than phi B. In such a case, we found out that electrons will move from B to A and this will create a contact potential. We then looked at metal semiconductor junctions and we saw that there were two types. The first one was called short key junction. So this is a junction where the work function of the metal, we call it phi M is greater than the work function of a semiconductor. In such a case, we found that electrons will travel from the semiconductor to the metal and this will again create a depletion region in the semiconductor and a contact potential. We also looked at the short key junction under bias and found that it behaves as an rectifier. So that in the case of forward bias, the junction will conduct but when we apply a reverse bias, there is a small saturation current but the junction does not conduct. The other type of junction that we saw was the ohmic junction. In this case, the work function of the metal is smaller than that of the semiconductor so that you have electrons flowing the other way and the ohmic junction behaves as a resistor. This is where we left off from last class. Today, we are going to look at a junction between two semiconductors. So we are going to start with a PN junction. As the name implies, a PN junction is formed between a P type semiconductor and an N type semiconductor. Usually both are formed from the same material and such a kind of junction is called a homo junction. These are where both your P and N type are from the same material. We can also have a junction formed between different materials and such a type of junction is called hetero junction. So in this case, P and N are different materials. We will mostly look at homo junctions where they are the same material and we will develop concepts of the junction in equilibrium and under bias and also calculations for the current and towards the end we will also look at the effect of having different materials. So we will find that hetero junctions have some really interesting properties and later when we talk about optical devices as well like LEDs or lasers, we will find that hetero junctions have some advantage there. Homo junctions are easier to grow because you are essentially the same material. All you are doing is doping one side P type, the other side N type. So the interfaces are easier to form. The case of hetero junctions, the materials have to be chosen carefully so that we have a good interface with no defects. So that poses some restrictions on the kinds of junctions they are formed. But for the analysis we are going to do now, we are going to assume that we have an ideal junction that is there are no defects. Next to the time we will be talking about these PN junctions with respect to silicon. So silicon is our standard material of choice but we will also look at some examples of other materials and when we come to hetero junctions we will talk a lot about compound semiconductors. So let us consider a junction formed between P and N silicon. So let me start by drawing the band diagram for P type and N type silicon. So I will first put them far apart and then bring them together to form the junction. So here is the band diagram for P silicon. We have a conduction band EC, valence band EV. So EC is the bottom of the conduction band, EV is the top of the valence band. The material is P type. So we know that the Fermi level lies closer to the valence band. So let me put the Fermi level closer to the valence band and call it EFP. Now I have N type silicon, once again I have EC, EV and then EFN. So the position of EC and EV does not change because both of them correspond to silicon. So the band gap is maintained. The only difference is where we place the Fermi level. So we bring this P and N material together in order to form your PN junction and we said the first rule is that at equilibrium the Fermi levels must line up. Another way of thinking about it is that in the N side you have excess electrons in the conduction band. These electrons can move to the P side. On the P side you have excess holes. These are in the valence band and these holes can move to the N side. We have electrons moving here, holes moving the other way. Because we have electrons moving we will have a net positive charge on the N side because we have holes moving, we will have a net negative charge on the P side. So once again we will have a contact potential between the PN junction. So let me draw the junction in equilibrium. The first thing I will do is line up the Fermi level. Let me mark the interface region. So this I will call IF. So IF represents the interface between the P and N. Far away from the junction the material behaves as a P or an N type. So far away from the junction I have a P, I have an N. Now we said that electrons move from N to P so that we have a net positive charge on the N side and a net negative charge on the P side which means there is an electric field. So the electric field goes from positive to negative. In last class we also saw that whenever we have an electric field we have band bending and the bands bend in the direction of the field. So the bands bend up in the direction of E. So we have the bands bending on the N side up, we have bands bending on the P side down. So that is how the junction forms. You can do the same for the valence band. So this is my conduction band EC, this is my valence band EV and I have a contact potential at the junction. So V0 represents the contact potential. So we can draw a simplified picture of this. So let me just draw a schematic. So this is my P side, that is my N side, I form my PN junction. Electrons move from the N to the P. So when these electrons move you are left with donor ions. Since donor ions donate an electron and when these electrons move they have a net positive charge. So there is a net positive charge on the N side. So this net positive charge refers to the donor ions. Similarly there is a net negative charge on the P side because the electrons or the holes from P side move to the N side leaving behind your acceptor ions and the acceptor ions are negative which means you have an electric field E and a built-in potential V0. So we have excess electrons from N to P, we have excess holes moving from P to N and when these meet they recombine and get destroyed. So what you are left behind is a depletion region around the junction. So let me redraw this diagram and mark the depletion region. So we are forming a PN junction. So in the case of a PN junction we said that we have excess electrons on the N side move to P. We can think of this in terms of diffusion where diffusion usually occurs from a higher concentration gradient to a lower gradient. So the higher concentration of electrons is on the N side and this moves into P. Similarly we have holes which move from the P side to the N side and when these meet they recombine and form a depletion region. So let us consider a PN junction where the acceptor concentration NA this is on the P side is greater than the donor concentration ND and this is on the N side. So let me redraw my PN junction, this is the P that is the N. There is a depletion region that is formed because of the diffusion of carriers we call this W0. So W0 is the total width of the depletion region. So this forms at the interface of the PN junction and it extends to both the P and the N side. So you have sum width on the P side let me call it WP sum width on the N side WN. So that W0 is just the sum of the depletion region in the P side and the depletion region on the N side. So we said that in the case of a PN junction we have a net positive charge on the N side because the electrons are gone leaving behind positively charged donors and you have a net negative charge on the P side because the holes are gone leaving behind the acceptor ions. So if A is the cross section of the junction, so A is the cross sectional area of the junction we must have a balance of charge between the P and the N side. So the total charge on the depletion region in the P side is nothing but the concentration of acceptor ions times the volume, the volume is A times WP. So this is the concentration, this is the volume. We can similarly calculate the total charge on the N side which is ND which is the concentration of donor ions times the volume AWN. In order to maintain the neutrality of charge, this total positive charge must be equal to the total negative charge. So let me equate these two and write the expression. So if you want to maintain charge neutrality, you have AWP NA which is the total negative charge on the P side must be AWN ND. The cross sectional area is the same, so that can be removed. So what we are left with is WP NA is equal to WN ND. Another way of writing this is that WP over WN is equal to ND over NA. So the ratio of the depletion region which on the P and N side is inversely proportional to the concentration of the dopants whether they are donors or acceptors. So if NA is greater than ND, so NA is greater which means WP will be less than WN. So the depletion region is larger on the N side than on the P side. There are certain PN junctions that are formed between a heavily doped P plus region and an N region. So P plus refers to a heavily doped region. In this particular case, NA is usually much larger than ND. So using this above charge neutrality expression, you have WP is much smaller than WN. So that the depletion region is almost entirely on the N side. An extreme example of this is in the case of your metal semiconductor short key junction. We saw this last class where we formed the junction between the metal and the semiconductor and we had electrons moving from the semiconductor to the metal so that you had a depletion region and the depletion region was almost entirely on the semiconductor and this is again because if you try to look at an expression similar to here, the charge density in the case of a metal is much higher than that of a semiconductor. So in order to maintain charge neutrality, you have electrons coming not only from the surface of the semiconductor but also from the bulk creating a depletion region and this depletion region lies entirely in the semiconductor side. So the next thing to do is to calculate this built-in potential that comes between the P that forms when you have a PN junction. So we want to calculate the contact potential that forms in a PN junction. To do that, let me first draw how the carrier concentration that is the whole concentration changes and the electron concentration changes as we go from one end to the other. So let me just redraw my PN junction here. This is the interface. So the P side and the N side, we have NA and ND as the concentration of acceptors and the concentration of donors and we have NA greater than ND. We just saw that this means that the depletion width is larger on the N side. So there is a depletion region on the P side and there is a larger depletion width on the N side. So let me call this WP and then WN. So if you plot a log of how the concentration of electrons or holes change as a function of distance. So I will plot log of N or log of P as a function of distance. So let me again mark my interface. This is WP. This is WN. So this represents the depletion region. Let me also mark NI. NI is the intrinsic carrier concentration. In the case of electrons and you are on the N side, the concentration of electrons N in the N side, let me call it N, N0 is nothing but ND. The concentration of holes in the N side, we can use the law of mass action, it is nothing but NI square over ND. And usually this is much less than NNO. Similarly the concentration of holes in the P side is just NA and the concentration of electrons in the P side is just NI square over NA. So these are all just notations but we have done these calculations before when we looked at extrinsic semiconductors. If we go ahead and plot this, this is NNO. It is equal to ND. At the depletion region, the concentration begins to drop because we said we have a depletion region because electrons move from the N to the P side. So the concentration drops and then finally in the P side, the concentration becomes equal to NPO. We can do the same for the holes and we said that the hole concentration is higher than the electron concentration. We said NA is more than ND. So let me just draw these axes a bit up so that this is PPO. Once again, when we reach the depletion region, this number is going to drop because you have holes moving from the P side to the N side and then far away from the junction, we have the concentration PNO. So this graph shows you how the electron and the hole concentration change as we move from the N to the P or from P to N. So this difference is related to the built-in potential because what the potential does is that it prevents further motion of electrons from N to P or holes from P to N. So in this way, it is similar to what happens when we have a short key junction. So there also we had a built-in potential that prevents further motion of electrons. So V0 is the built-in potential. It is related to the concentration of electrons on the N and the P side. So NP0 which is the concentration of electrons in the P side by NN0 which is the concentration of electrons on the N side is equal to exponential minus EV0 over kT. So once again V0 represents the barrier that the electrons have to overcome in order to go from the N to the P side. We can substitute these values for N0, NP0 and NN0 and rearrange this expression to give you your contact potential. And V0 is nothing but kT over E ln of Na Nd over Ni square. We get this expression by taking natural ln on both sides and then substituting for these values and rearranging. So the contact potential in the case of a PN junction depends upon the concentration of the acceptors and the concentration of the donors and also the intrinsic carrier concentration. We can also calculate the width of the depletion region that forms when we have a PN junction. Let me just redraw the junction again. So we have a PN junction with Na greater than Nd so that we have some Wp which is smaller than Wn. So we said that in the depletion region the excess electrons and holes recombine and get annihilated so that you have a net negative charge on the P side and a net positive charge on the N side. I am putting two positive charges to indicate that we have a wider depletion region on the N side. If you plot the charge density as a function of distance so rho represents the charge density as a function of distance. We can usually say that the depletion region is devoid of carriers so that the charge density is just a delta function. Let me plot the interface, this is the N side. So on the N side the charge density is just given by the concentration of the donor ions. So this is minus E Nd or sorry this is plus Nd because you have a positive charge. On the P side the charge density is given by the acceptor ions which have a negative charge so that this is minus E Na and the total charge has to be 0 which means the positive charge on the N side has to balance the negative charge on the P side. So the area under these two graphs are the same. We can relate the charge density to the electric field. The equation is electric field E as a function of x is just 1 over epsilon where epsilon is the permittivity of the material integral rho net as a function of x dx. So we just said that the charge density is a delta function and it is a constant. So we saw that rho net is equal to minus E Na. This is the P side and it is equal to E Nd, it is the N side. We can substitute for this here and then integrate over the entire width of the depletion region to get the electric field E. So the expression for the electric field E is equal to E Nd over epsilon x or minus E Na epsilon x. We can define an E naught which is equal to minus E Nd Wn epsilon which is equal to minus E Na Wp and these two are the same because we know that to maintain charge neutrality Nd times Wn is equal to Na times Wp. So we can plot the electric field E as a function of distance. So I will use the same plot here, trick field E, the function of distance. The electric field is essentially negative. So let me just redraw it only on the negative side, that is my interface. This is the P side, this is the N side. Inside the depletion region, the electric field is 0 but within the depletion region E is a linear function of distance and the maximum value is E naught. So the maximum value of E is E naught. E is also related to the potential by the expression E dv over dx which means the potential v is integral of E dx. We can substitute this expression and do the integration. The total potential which is the contact potential which we are interested in goes from the P side to the N side and v naught is nothing but minus 1 half E naught and W naught where W naught is the entire width of the depletion region that is equal to E Na Nd W naught square over 2 epsilon Na plus Nd. We can plot v naught as a function of x. If you look at this expression, v is the integral of E dx, E is a linear function in x so the integral of a linear function is a parabolic function. So if you plot v naught over x that is my interface, that is the N side, that is the P side. Potential goes from 0 up to the contact potential v naught and the expression for v naught is given here. Let me rearrange this to get the total width of the depletion region in terms of the contact potential. If you rewrite this expression, we get W naught is nothing but square root of 2 epsilon Na plus Nd times v naught over E Na Nd. Epsilon which is the permittivity of the material is nothing but epsilon naught r, nothing but epsilon naught which is the permittivity of free space times epsilon r which is the relative permittivity of the material. In the case of silicon epsilon r, we saw this earlier as a value of 11.9. So we have an expression for the contact potential kT over E ln of Na over Ni square. We all have an expression for the total width of the depletion region and we also saw that the individual widths are inversely proportional to the concentrations. So let us plug in some numbers to get a sense of what these values are. So we will take the example of silicon with acceptor concentration Na equal to 10 to the 17 per centimeter cube and donor concentration is 10 to the 16 per centimeter cube. So Wp over Wn is nothing but Nd over Na which is 1 over 10. So the width of the depletion region in the P side is 10 times smaller than that of the N side and that is because the concentration of acceptors is 10 times more than the concentration of donors. The case of silicon, we know that the intrinsic carrier concentration is 10 to the 10 and we are doing these calculations at room temperature, so T is 300 Kelvin. So we can calculate the potential, we will just use this expression where we will plug in the values. If you do that, V0 is 0.78 electron volts and also calculate the width of the depletion region W0. Once again it is a straight case of using this equation and plugging in the numbers. If we do that, we get a value of W0 to be around 3.3 times 10 to the minus 7 meters or approximately 330 nanometers. So the total width of the depletion region is slightly less than 1 micron, it is around 0.3 micrometers. You can also calculate the individual widths by using the ratio. If you do that, we get Wn is around 100, sorry, if you do that we get Wn to be around 300 nanometers and Wp to be 30, so that this ratio of 1 over 10 is maintained. The width of the depletion region is inversely proportional to the concentration of the carriers. So if you have a higher value of Na or Nd, then the total width will be lower. If you want to rework this with Na, increased 10 times to 10 to the 18 per centimeter cube, similarly Nd is 10 to the 17. So we have increased both Na and Nd 10 times but the ratio is still the same. We can redo these calculations. Let me just write it down, V0 is also higher because V0 is equal to Na times Nd. The potential V0 is around 0.89 volts. The total width is lower, W0, it is around 113 nanometers and once again the width on the N side is 102.7 and the width on the P side is 10.3. So we can reduce the width of the depletion region by increasing the carrier concentration on the P and the N side. So this is the PN junction as far as silicon is concerned. Now let us just look at what happens if we change the material. What we want to know is how the contact potential changes when we change the material. Now V0 is related to kT over E ln of Na Nd over Ni square. For sake of comparison I am going to keep Na equal to 10 to the 17 and Nd equal to 10 to the 16 but V0 also depends upon Ni which is the intrinsic carrier concentration and that depends upon the band gap. So higher the value of Ni which happens if you have a lower band gap then smaller is the contact potential. So if you look at three materials, germanium, silicon, gallium arsenide. So we are forming PN junctions between two germanium P and N, same with silicon, same with gallium arsenide. The band gap values at room temperature 0.7, 1.1 and 1.4, Ni is different. We have done these calculations before for these intrinsic materials, 2.4 times 10 to the 13, 1 times 10 to the 10, 2.1 times 10 to the 6. So these are the values of Ni and if you plug in this equation V0 in volts is 0.37, 0.78 and 1.21. So the contact potential increases as the value of Ni goes down and this happens when you have a larger band gap. So today we have looked at a PN junction that is in equilibrium that is there are no external potentials applied. We saw that we have electrons moving from the N side to the P side, holes moving the other way and this creates a depletion region. In the next class we are going to look at the IV characteristics of a PN junction and what happens when we apply a bias to this junction.