 Let us start with a brief recap of last class. Last class we looked at extrinsic semiconductors. Another name for these are doped semiconductors. Now the reason why we wanted to doped was because we found out that if you had an intrinsic or a pure semiconductor, the carrier concentration at room temperature was very small. Correspondingly the conductivity was also small. For example in the case of silicon, we found that the room temperature carrier concentration in an intrinsic semiconductor was only 10 to the 10 per centimeter cube and the corresponding conductivity very low was around 3 times 10 to the minus 6 ohm inverse centimeters. So we wanted to increase the conductivity. So we selectively added impurities or dopants to the silicon in order to make it an extrinsic semiconductor. We also saw that in any semiconductor the law of mass action must always be satisfied. So that Np is equal to Ni square which is a constant at a given temperature. This meant that we can either increase the concentration of electrons or increase the concentrations of holes. We cannot do both. So we also saw that there were two types of dopants. One was your N type, the other was the P type. We saw that an N type dopant was found in the caves of silicon by adding group 5 elements. The typical elements that we add are phosphorus, arsenic, antimony. All of these have one extra electron compared to the silicon atom which means you have one extra electron in the conduction band. So these are N type. Similarly we saw that we could form P type by doping with group 3, boron, aluminum, gallium and indium. These form energy states that are close to the conduction band in the case of N type or close to the valence band in the case of P type. So that these are ionized at room temperature. So let us look some more today on the properties of extrinsic semiconductors. So far you only talked about silicon. So what about other semiconductor materials and doping in them? If you look at germanium, germanium lies in the same group as silicon which means all the elements that we used as dopants for silicon could also be used for germanium. So we can use your group 5 elements like phosphorus or arsenic as an N type dopant could also use your group 3 elements like boron as a P type dopant. Once again in the case of germanium we can calculate the ionization energies for these dopants. We could use the hydrogenic model. The only difference is the effective mass Me star for germanium will be different from that of silicon and similarly your relative permittivity for germanium the value is around 16 while for silicon epsilon R is around 11.9. But we can use the same hydrogenic model in order to calculate the ionization energies. So let us look at some actual values and just to compare silicon and germanium. Let us look at N type, phosphorus, arsenic and antimony IO3 dopants silicon and germanium. If you fill in the numbers in the case of silicon we found out that the ionization energy in milli electron volts. So energy is in milli electron volts was around 45, 54, 39. Similarly in the case of germanium we will find that the ionization energies are very small. So this is around 12, 12.7, 9.6. If you remember the band gap of germanium is also smaller. So at room temperature E g for silicon is around 1.10 electron volts. E g for germanium is around 0.67 electron volts. You can do the same thing with P type impurities or P type dopants boron, aluminum, gallium and indium. So once again let me write silicon and germanium 45, 57, 65, 157. Again we are writing E g in milli, we are writing E energy in milli electron volts. Germanium is around 10.4, 0.2, 10.8 and 11.2. So what this means is whether you have silicon or you have germanium both lie in the same group. So you can dope both of them in a similar fashion. If you want an n type dopant you are going to add group 5 elements. If you want to add P type dopants you are going to add group 3 elements. Usually the dopant concentration is very small. We saw earlier the case of silicon that your typical dopant concentration was around parts per million or parts per billion. So parts per million is 10 to the minus 6, parts per billion was 10 to the minus 9 and even the small amount was enough to increase the conductivity by orders of magnitude. So we saw for silicon we had an n type dopant with 10 to the 15 atoms per centimeter cube which was approximately 20 parts per billion. Your conductivity sigma went up to around 0.2 ohm inverse centimeter inverse. One thing we have not looked much is how we actually do the doping. So later when we look at the fabrication part we will spend some more time on how we actually dope. But if you think of a parallel to metallurgy one way we can think of is in the case of stainless steel or in the case of steel we do something called carburizing where we increase the carbon content. So typically what we do is we have your sample at high temperature you have a carbon source so that the carbon then just diffuses into the surface and into the bulk of your steel. So we can do something similar in the case of doping in semiconductors as well. So if you want to dope n type you have a source of your n type material. This could be in the gas phase. These could be ion implanted onto the surface. You will see what ion implantation is later so that these can then diffuse into the surface and from the surface into the bulk. If you want to do compensation doping we saw compensation doping last class where you start with an n type semiconductor and then you make it p type by adding excess dopants of the other type. Similarly in the case of compensation doping could take a small portion of your sample that is typically n type and add excess of p type. So you could add excess of boron and then make it a p type semiconductor. These dopants are usually very stable at room temperature just to give you some numbers. Let us take the case of silicon. I have two dopants one is boron, one is phosphorus. Boron is a p type, phosphorus is n type. We will write down the values for the activation energy and the diffusion coefficients for these dopants. So if I write D0, EA is the activation energy for diffusion and D0 is a diffusion constant. So in the case of boron D0 has a value for on 0.76 activation energy is around 3.5 electron volts, phosphorus 3.85 EA is around 3.6 electron volts. What this means is in order to have boron or phosphorus diffusing into your material you essentially need a high temperature process. The corollary of this is that at room temperature your doping concentrations are inherently very stable in the case of silicon. So far we have seen two semiconductors silicon and germanium. We have looked at some examples of dopants calculated the ionization energies. Since both silicon and germanium belong to the same group we essentially use the same elements. Things become a little more complicated when we look at other semiconductors. So let us look first at gallium arsenide. We saw earlier that gallium arsenide is an example of a 3 phi compound semiconductor. So gallium is from group 3, arsenic is from group 5. The bonding in gallium arsenide is mainly covalent but you also have some ionic character to the bond. Instead of gallium arsenide if you have something like zinc oxide which is a 2 6 semiconductor there will be higher ionic character as compared to 3 phi. So it is also possible to dope gallium arsenide but there are now more possibilities in this case. To understand this let us go back to the periodic table especially the portion around gallium arsenide. So let me write the periodic table. We have done this earlier when we looked at elemental and compound semiconductors. So group 2, 3, 4, 5 and 6 boron, carbon, nitrogen, oxygen, aluminum, silicon, phosphorus, sulphur, zinc, gallium, germanium, arsenic, selenium, we have cadmium, indium, tin, antimony, tellurium, mercury, thallium, lead, vismuth and polonium. So this is just the portion of the periodic table around groups 3 and 5. So the semiconductor we have is gallium arsenide and we want to know what sort of elements could be added as dopants to gallium arsenide to make it P or N type. So when we think of dopants these are usually substitutional which means they are going to replace either the gallium or the arsenic atom. So let us say I want to make an N type gallium arsenide. So I would typically choose elements from group 6 because these elements have one more electron compared to arsenic. So if an element from group 6 replaces arsenic we will have one extra electron and if this extra electron is in a shallow state if it is easily ionizable it can get ionized to the conduction band and be available for conduction. So we are going to go to form N type, we are going to replace arsenic by group 6. We can use the same argument if you want to make a P type semiconductor. So we replace gallium by any of the group 2's which has one less electron or you have one hole. So that will make it a P type. If you want to make something P type you can replace gallium by group 2. Question is what happens if you have a group 4 impurity? Let us say I have silicon. Now silicon is an impurity in gallium arsenide. So there are essentially two possibilities the silicon can replace the gallium the silicon can replace the arsenic and depending on that you can either get an N type or a P type. So if the silicon replaces the gallium then silicon has one more electron because this is group 4, this is group 3. So the extra electron is available for conduction it will make it N type. On the other hand if silicon replaces arsenic it has one less electron and it will make it P type. So silicon added to gallium arsenide can be either N or P type and this type of doping is called amperage. Amphoteric doping. This term ampho essentially means both means you have the same element can act both as an N type or as an P type doping. The only issue here it is really hard to control the position of silicon. It is very hard to get silicon selectively doped into gallium or to selectively doped into arsenic. So if you have silicon as a doping it is very hard to control the type of doping and the concentration of doping. So we saw that these are all the possible elements that can be used to dope gallium arsenide. The only other question is where are the energy levels located. So let me just do that by drawing the band gap of gallium arsenide. So this is my conduction band that is my valence band. The difference between the conduction band and the valence band is your band gap. The case of gallium arsenide the band gap is 1.42 electron volts. Also mark the center of the gap. So if you look at the various elements the group 2 and the group 6 elements these will essentially form impurity states in the band gap of gallium arsenide and depending upon whether they are N type or P type they will be located either close to the conduction band edge or the valence band edge. So if you look at the group 6 elements these are N type. So this one is sulphur selenium is also very similar, tellurium is here again all the dopant energies are in milli electron volts. Which means the diagram is not really to scale but we will just show it semi qualitatively. So we have a rough idea. Silicon we said silicon can go either to gallium or to arsenic. So silicon as an N type is around 6 milli e V. Oxygen on the other hand is around 400. Similarly we can look at the group 2 elements. So we have beryllium is around 28, zinc is 31, cadmium is slightly higher 35. We can also have silicon as a P type that is around 35 again. Carbon is around 26. So if you have gallium arsenide depending upon the type of impurity we can either have both N type and P type. Similarly for other 3, 5 semiconductors can once again have doping by choosing the appropriate group 2 or group 6 material. If you have a 2, 6 semiconductor something like zinc oxide or cadmium sulphide or cadmium selenide. Similarly we can choose appropriate dopants to get both N type and P type. Most of what you have done here. So whether we are talking about silicon or germanium or gallium arsenide. We typically want single crystals of these materials. Whether we talk about an intrinsic semiconductor or whether we talk about an extrinsic semiconductor with dopants. We want an ideal single crystal with no defects. This is because whenever you have a polycrystalline material or if you have defects. So whenever you have polycrystals you always have grain boundaries. Grain boundaries are a source of defects. You could also have impurities. You can have vacancies, dislocations. All of these are defects in your crystal and whenever you have defects you always have defect states. There are 2 types of defect states. One is shallow. Shallow states are those that are located either close to the conduction band or close to the valence band. So they are close to the band edges. And because they are close to the band edges they can easily get ionized at room temperature. So shallow states mostly affect the conductivity. It can either increase or decrease the conductivity. The other type of defect states supposed to shallow states are deep states. So deep states are states that are located much closer to the middle of the band gap or they are located far away from the valence band and the conduction band. So in this particular case you have oxygen that is located approximately 400 milli electron volts or 0.4 electron volts below the conduction band. So deep states are located close to the center of the band gap. Deep states can essentially act as straps for electron and hole. So they could modify the conductivity especially in the case of direct band gap semiconductors where we are looking at optical properties. Deep states can also act as straps for these electrons and holes and then decrease the efficiency of any optical recombination. We will see defect states later when we also look at conductivity in an extrinsic semiconductor. The next thing we are going to do is look at how the Fermi level position changes in an extrinsic semiconductor. So we want to find out the position of E f when you have dopants whether you have P or n type dopant. In the case of an intrinsic semiconductor we have electron and hole concentration to be nearly the same. So we wrote an expression for the Fermi level position E g over 2 k T. So this is for the case for the case for the case of an intrinsic semiconductor where n equal to P equal to n i and we found out that E f i is very close to the center of the gap. So if n c and n v are exactly the same is exactly E f i will be just E g over 2 but if they are not the same will be slightly shifted from the center of the gap but for most cases it is very close to the center. In the case of an extrinsic semiconductor n is no longer equal to P. In fact if you have an n type semiconductor we see that n is much larger than P. You have a P type semiconductor P is much larger than n. So in order to compensate for this difference in n and P the Fermi level will also shift from the middle. One way to look at it or to calculate the shift is to use equation for the electron concentration. So let me start with an n type semiconductor with n d being the concentration of the donors. The donors are all completely ionized. So the electron concentration same as the donor concentration. This we can write as n c exponential minus E c minus E f n over k t. So E f n refers to the Fermi level position in the n type semiconductor. So we can take this expression and then rearrange the terms a bit and what we will get is E f n minus E f i is nothing but k t d over n i. How we get this expression is that we start with this. We can write the same equation for intrinsic as n i equal to n c exponential minus E c minus E c minus E f n minus E c minus E f i over k t. So we start with 1 and 2 and then we divide 1 and 2 and rearrange the terms to get the expression for the Fermi level position in an extrinsic n type semiconductor with respect to the Fermi level position in an intrinsic semiconductor. You can write a similar equation for a p type material as well. So you have E f p minus E f i which is minus k t ln of n a over n i. So the similar expression to this except that in an n type we have the concentration of the donors. In p type you have a concentration of acceptors and there is a negative sign here. So let us do some numbers to get a sense of where the Fermi level is located. So let us take n type silicon with n d is around 10 to the 15 per centimeter cube. We know the band gap of silicon E g at room temperature is 1.10 electron volts. For simplicity let us just say n c is equal to n v. So the intrinsic position of the Fermi level E f i will be just E g over 2. We know this to be 0.55 electron volts. Now at room temperature you have all the donor atoms are ionized. So n is equal to n d equal to 10 to the 15 per centimeter cube. We will then use the formula that we just wrote down. So that E f n minus E f i is k t of n d over n i. In the case of silicon n i is around 10 to the 10. So substituting the numbers this gives you a value of around 0.30 electron volts or in other words E f n is E f i plus 0.30 electron volts which is 0.55 plus 0.3 is 0.85. Adding an n type impurity so or adding a donor shifts the Fermi level above the intrinsic level. If you want to show this in a band gap diagram this is your valence band. This is the conduction band. This is the band gap. So this is 1.10 electron volts. E f i is equal to 10 to the 15 per centimeter cube. So this is the conduction band. So this is the conduction band. This is the conduction band. This is the conduction phi is located in the middle. So this is 0.55. So you have added an n type impurity. So this creates your donor levels E d. So these donor levels are very close to your conduction band. At room temperature all of the donors are ionized. And what we have is your Fermi level E f n located above the intrinsic level but below your donor level. In the case of a p type semiconductor. So let us do a situation where we have n a. So it is a p type. We just block it off with 10 to the 15. Instead of electrons you have holes. So your hole concentration is 10 to the 15. Your equation becomes E f p minus E f i minus k t which is minus 0.30 electron volts or E f p 0.55 minus 0.3 0.25 E v. So in the case of a p type semiconductor you have the Fermi level located below the intrinsic level. So depending upon the type of extrinsic semiconductor you have can either have the Fermi level shift up. If you have more electrons it shifts up towards the conduction band. If you have more holes it shifts below towards the valence band. So all these calculations that you have done so far is for an extrinsic semiconductor at room temperature where all the dopants are ionized whether they be donors or acceptors. So the next thing we are going to do is to look at how the doping or how the carrier concentration changes with temperature. So we want to look at the temperature dependence on the majority carrier concentration. So it is a temperature dependence of n if it is an n type or p if it is a p type. For simplicity we are only going to look at the n type case but whatever arguments we use for that we can use the same for the p type semiconductor. How is this different if you have an intrinsic semiconductor? In an intrinsic semiconductor we have essentially no dopants. So a carrier concentration n equal to p equal to n type n i per square root of n c n v exponential minus e g over 2 k t. So if you want to look at the temperature dependence of n it is an exponential dependence on temperature. So if you did a plot of the log of a carrier concentration over 1 over t it is approximately a square straight line as long as you assume that n c and n v are independent of temperature straight line with slope e g over 2 k. This is as far as an intrinsic semiconductor goes. In the case of an extrinsic semiconductor again that is an n type you have essentially two sources for the electrons. You can get the electrons from the donor level. Remember the donor atoms each have one more electron than the silicon that is why it is an n type. So you can get the electrons from the donor level and you can also get the electrons from the valence band. So because you have these two sources you have a different behavior for how the value of n changes with temperature. We are going to look at different regimes for that. So the first case let us look at a semiconductor at exactly 0 kelvin. So in this case you have an n type semiconductor at 0 kelvin. There are no ionized carriers. You have a valence band that is full and you have a conduction band that is empty and you also have a donor level because it is n type which is also full. So this is my conduction band. This is my valence band. So that is full. That is empty and these are my donor levels. The donor levels are located very close to the conduction band. I am just exaggerating the difference to show the fact that they are close to the conduction band. We can expand this portion of the diagram. So that now I will draw the conduction band. I will draw my donor level. At temperature equal to 0 kelvin and at temperatures very close you are going to have negligible number of electrons coming from the valence band. So we can ignore the valence band totally and think of this as an intrinsic semiconductor. So I will put intrinsic within brackets with the donor level acting as your valence band and the conduction band being the conduction band of silicon. So all I have done is just expand this portion here and take it to be an intrinsic semiconductor with the donor level being the valence band and Cb being the conduction band of silicon. At 0 kelvin you have a donor level that is completely full. You have a conduction band that is empty and then you will have your Fermi level in between the conduction band and the donor level. Now if you start to increase your temperature from 0 kelvin electrons are going to get excited from the donor level to the conduction band. Once again your temperature is low enough that you can ignore any contribution from the valence band of silicon. So we do this till some temperature which we are going to call Ts or a saturation temperature. So temperature is above 0 kelvin but it is below a certain temperature called Ts. We are going to define Ts as saturation temperature. In this particular regime you are going to have electrons from the donor level getting ionized into the conduction band and this process is going to happen till all the donor atoms are ionized. So the saturation temperature is defined as the temperature in which the donors are 90% ionized. Some books will give you a definition of 99% ionized but we will just use 90% here. Here at which donors are in order to calculate the value of Ts we look at the expression for the electron concentration. We will treat this material as an intrinsic material with the donor level being your valence band. This case N is 1 half Nc instead of Nv you will have Nd which is your donor concentration exponential minus Ed over 2kt. The 1 half term is here because your donor levels are localized levels. So they can only take one electron of a normal band can have 2 electrons of opposite spin because your donors are localized levels. They can take only one electron which is why we have the term 1 half. We are treating this as an intrinsic with EC being your conduction band and the donor being the valence band and at the saturation temperature N is 90% of Nd. We can put this value here and try and solve this expression for Ts and if you do this we get a value of Ts to be approximately 32 kelvin. So in this assumption we will get the value of Nc to be independent of temperature. If you take in the temperature dependence of Nc you will get a Ts of around 60 kelvin but either way the number is much below room temperature which is around 300 kelvin. What this means is at a relatively low temperature in the case of silicon you have a situation where the donors are completely ionized. So we started at low temperature where we said that we will ignore all the electrons coming from the valence band. We found out that as we started going above 0 kelvin electrons are going to get ionized from the donor levels till we reach a saturation temperature when almost all the donors are ionized. So above the saturation temperature all the donor atoms are ionized so that N is nothing but Nd and P would be Ni square over Nd. Ni square over N this is nothing but Ni square over Nd. Now if we keep increasing temperature further there is going to come a point when electrons are starting to get ionized from the valence band. In this particular case the semiconductor now behaves as an intrinsic semiconductor and we define this temperature as Ti. So we define Ti as a temperature when N is equal to 1.1 Nd. So you have extra carriers that are now coming from the valence band of the silicon and this makes it an intrinsic semiconductor. So we had two temperatures Ts and Ti. So Ts was 0.9 Nd, Ti is 1.1 Nd. So we have a regime where the carrier concentration varies within 10%. So we have a regime where the carrier concentration varies within 10%. If you want to calculate the value of Ti, the intrinsic temperature we can look at the numbers. So N is 1.1 Nd and we are going to treat this as intrinsic. In any semiconductor the total charge should be balanced. So N is equal to P plus Nd. So that P is 0.1. We also know that Np is Ni square and then doing the math we get Ni to be 0.33 Nd. So N is 10% more than your donor concentration and at that temperature Ni is approximately 0.33 Nd. We can use this expression Ni equal the square root of NcNv exponential minus Eg over 2kt. But now the temperature will be Ti, which is your intrinsic temperature. You can substitute in the values. Once again we will assume that Nc and Nv are independent of temperature. That gives you a Ti of approximately 584 Kelvin. If you try to convert that into degrees, this is approximately 584 Kelvin. 310 degrees. Above this temperature the semiconductor essentially behaves as an intrinsic temperature. So in the case of silicon, we have a regime starting from the saturation temperature which is around 50 Kelvin or 60 Kelvin to an intrinsic temperature that is around 584 Kelvin, where the carrier concentration is essentially a constant and it is equal to the donor concentration. The difference is only within 10%. This means the conductivity will also be very stable because the conductivity depends upon the carrier concentration. If we put these numbers together and did a plot of the log of the carrier concentration versus 1 over T, we can compare that to your intrinsic semiconductor. In the case of an intrinsic, you have already done this before, log of N versus 1 over T is just a straight line with one slope. So this is intrinsic. You can do the same for extrinsic. So now instead of one slope, we will have two slopes and we have three regimes. In this particular case, since this is 1 over T, this is low temperature and this is high temperature. So at low temperature, you have electrons coming from your donor level. So you have a straight line with a slope that is given by your donor ionization. Then we have a situation where the donors are all ionized and your carrier concentration is almost a constant. At some high temperature, we are going to get new carriers from the valence band. The material behaves like intrinsic. So this is minus e g over 2 k. So we have three regimes, an ionization regime, a saturation regime where the carrier concentration is almost a constant and then finally, an intrinsic regime where the material starts to behave like an intrinsic semiconductor. So we will stop here for today. In the next class, we will look at how the conductivity changes. It is a function of temperature and we also look at the effect of doping on the mobilities of electrons and holes.