 Last two classes, we looked at intrinsic semiconductors. So, intrinsic semiconductors or pure semiconductors are those, where electrons and holes are generated in pairs. We saw that the electron concentration is the same as the hole concentration and we called it n i, where n i is the intrinsic carrier concentration. We also looked at some calculations for n i. In the case of silicon at room temperature, we found that n i has a value of around 1 times 10 to the 10 per centimeter cube. So, this number is really low when we looked at the concentration of atoms. Correspondingly, the conductivity is also low. We got a value of 3 times 10 to the minus 6 ohm inverse and centimeter inverse. We also saw that in the case of an intrinsic semiconductor, the carrier concentration n i is a function of temperature. So, the only way of increasing the conductivity, if you are not allowed to change the material, is by increasing temperature. Now, that is not practical. So, what we need is a way to increase the concentration of electrons or holes, while keeping the material at room temperature. To do that, we do doping and that comes to extrinsic semiconductors. We also saw this law of mass action, which states that the product of electron and hole concentration at equilibrium is equal to n i square. So, this is constant at a given temperature. What this means is, we can either increase electron n or we can increase hole. We cannot increase n and p at the same time, increase n and p together. Today, we will start looking at extrinsic semiconductors and I am going to start with the definition of extrinsic semiconductors. In an extrinsic semiconductor, we add small amount of impurities in your intrinsic material. Since we are dealing mostly with silicon, I will say a small amount of impurities in pure silicon to preferentially increase carriers of one polarity. That is either electrons or holes. So, the preferential part comes from the law of mass action, which says n into p is a constant at a given temperature. So, if we manage to increase n, we will decrease p and vice versa. There are two other operative terms here. One is small. We will see how small is small. The other word here is impurities. So, when we think of impurities in the context of metallurgy, we usually think of impurities as small amounts of unwanted materials that is there because of any processing problem. But in the case of semiconductors, impurities are carefully controlled materials that are added to your pure semiconductor to change the conductivity. So, another word for impurities or a more appropriate word if you think about it is dopants. So, we can say that an extrinsic semiconductor is one where we add a small amount of dopant in order to preferentially generate electrons or holes. First, we will start with n-type doping. As the name implies, in the case of n-type doping, we preferentially increase the number of electrons. So, we have more electrons than holes. Another way of stating is n will be greater than p. So, let us see how we can achieve this n-type doping. So, let us go back to our model of silicon. We said silicon has 4 electrons in the outer shell, 2 in the s and then 2 in the p which forms 4 sp3 hybrid orbitals which gives you 4 electrons. Now, these orbitals are arranged in 3 dimension in a diamond lattice. But just for the sake of illustration, I will show them on a 2D lattice. So, these are your silicon atoms. Each silicon atom has 4 bonds. Remember this is 3 dimensional, but just for the sake of explanation, I am giving it as 2D which forms bonds with other silicon atoms and then the picture repeats. Now, let us say we add a small amount of a pentavalent material. That is, we add of material from group 5. Examples of such material could be phosphorus, arsenic, antimony. Again, we only add a small amount so that the original silicon lattice is not disturbed. But what instead it does is the group 5 material goes and sits in the place of silicon. We do not disturb the silicon lattice. So, let us go back to this picture where now instead of a silicon atom, I have a phosphorus atom. I will just erase this, replace the silicon with phosphorus. Now, once again you see that phosphorus has 5 electrons. 4 of these electrons can easily form a bond with silicon and that leaves behind 1 extra electron. Now, this extra electron that is available can be delocalized and can be made available for conduction. Thus, by adding phosphorus atoms for each phosphorus atom, you have created 1 extra electron in the system. The question of course is how much energy does this electron have and is it easy to make it delocalized at room temperature. Let us look at that calculation next. So, what you are interested to find is the energy required in order to take the extra electron from phosphorus could be any other pentavalent impurity like arsenic or antimony to take that extra electron and delocalize it. So, we need to find the ionization energy for the extra electron in the silicon lattice. This extra electron is coming from the pentavalent impurity. So, one we can do a very precise energy calculation to find this energy. We can also do a more easier back of the envelope calculation. To do that, we simply say that we have phosphorus ion with 1 extra electron and we will treat this as a hydrogen atom with 1 electron. So, we get rid of the complexity of the problem and then treated as a hydrogen atom. This will give us a simple calculation for the ionization energy of the electron. If you have a hydrogen atom, the binding energy of an electron and hydrogen, we can do the calculation by solving Schrodinger's equation. I will just write down the expression. This is nothing but M e e to the 4 divided by 8 epsilon naught square h square. M e is the rest mass of the electron, e is the electric charge, epsilon naught is the permittivity of free space and that is also a constant. It is equal to 8.854 times 10 to the minus 12 farads per meter. So, M e mass of the electron e to the 4 divided by 8 epsilon naught square h square. This is the binding energy of an electron in the hydrogen atom. If you solve this, you get a value of minus 13.6 electron volts. The negative energy tells you that it is a binding energy slower than 0. We will take this expression and modify this for the phosphorus atom within the silicon lattice. So, we will replace M e by M e star which is the effective mass of the electron in silicon and we will replace epsilon naught by epsilon naught epsilon r where epsilon r is the relative permittivity of silicon. So, let me rewrite this expression again. So, we want to calculate the binding energy of an electron of an extra electron in the silicon atom because we added a group phi impurity to do the binding energy calculation. We started with the binding energy for hydrogen and we replace it with the effective mass of the electron. So, it is M e star and we replace epsilon naught by epsilon naught square epsilon r square h square where epsilon r is the relative permittivity of silicon has a value of 11.9. So, once again we can substitute the values M e star for silicon approximately 0.3 M e and epsilon r is 11.9. So, we substitute the values this gives us an answer of 0.032 electron volts or if you write this in milli electron volts it is 32 milli electron volts. So, this energy 32 milli electron volts is the energy that is required in order to ionize that extra electron to the conduction band. The energy ionize the extra electron and we did this calculation by simplifying the system and taking it to be a hydrogen atom. If you do the actual calculation in the case of phosphorus the actual ionization energy is around 45 milli electron volts which is very close to the value that we got by assuming a simple hydrogen atom. Just for comparison if you looked at the value of K T which is a thermal energy at room temperature K T has a value of 25 milli electron volts. So, the ionization energy is very close to the thermal energy which means that at room temperature it is very easy in order to ionize these electrons and take them to the conduction band and make available for conduction. We will also show later on explicitly that all of these impurity atoms are ionized at room temperature when we do the calculation for electron and hole concentration as a function of temperature. But for now just by looking at the ionization energies and the thermal energies we will say that all of these impurity atoms are ionized. So, the actual value for phosphorus is around 45 milli E V. We said there are other group 5 dopants like arsenic or antimony. If you look at ionization energies for these these are also very similar. If you have arsenic the ionization energy is around 54 milli E V and if you have antimony the ionization energy E V is around 39 milli E V. All of these numbers are very close to the thermal energy at room temperature. So, in all of these cases if you add them as dopants to the silicon the extra electron will be ionized. We can show these ionizations energies can show these energy levels on a silicon band diagram. If you have a silicon band diagram you have a valence band and a conduction band. The distance between the valence band and the conduction band is the band gap. In case of silicon the band gap is 1.10 electron volts at room temperature. To this silicon we add some pentavalent impurities. So, let us say we are adding arsenic. We already saw that arsenic has one extra electron and this one extra electron can easily ionize to the conduction band and the ionization energy for arsenic is around 54 milli electron volts. We add a small amount of arsenic. So, that these are localized states they do not come into contact with each other and form a band and these states are located very close to the conduction band. So, where to depict this I will show the arsenic in terms of dotted lines. The dotted line just tells you it is a localized state. You have individual arsenic atoms within the solid and the difference between the arsenic level and the conduction band is nothing but E b which is 54 milli E b. If the concentration of these impurities the pentavalent impurities is written as N d each atom will give one electron and at room temperature all of these electrons are ionized which means the concentration of electrons N is the same as N d. This is true when N d is much higher than the intrinsic carrier concentration. So, as long as N d is much higher than N i and all the atoms are ionized which is true in the case of silicon will find that N is same as N d. To calculate P we use the law of mass action and say N P is N i square which implies P is N i square over N this is N i square. Favour to write these two together the case of an N type material with N d being the concentration of the impurities N is the same as N d and P is N i square over N d. Another name for a pentavalent impurity or an N type impurity is called donors and these are called donors because they have one extra electron which they donate. So, they donate the one extra electron to silicon. Next we will look at a P type impurity or a P type dopant. We will look at P type doping the silicon atom is located in group 4 of the periodic table we said that donors which are N type dopants come from group 5. So, for P type doping we look at group 3 elements examples of such elements include boron, aluminum, gallium and so on. Can I again use that ball and stick model for silicon to explain P type doping. Start with silicon has four bonds and I am only using a 2D picture to illustrate actual molecule or actual solid is 3 dimensional. Silicon each of the silicon has further bonds. So, they just propagate through your solid. So, now you add a small amount of group 3 element. So, let us say add a small amount of boron. Once again the boron atoms just replace the silicon atoms in the lattice. The concentration is so small that it does not affect the lattice arrangement of silicon. I will take the silicon atom and I replace it with boron. Now, a boron has three electrons. So, I can easily form a bond or three bonds. So, a boron can form three bonds but this silicon atom has one extra electron. In order to form the bond with this silicon atom, an electron from the valence band is excited to the boron level so that it can form the bond with silicon. When an electron moves from the valence band you are left with the absence of an electron this forms a hole. We can do a similar calculation that we did for the n type impurities to look at the ionization energy. There is the energy required to excite the electron from the valence band to the boron level. If you do that once again we will find that the energies are the order of milli electron volts. If you do the actual calculation in the case of boron the ionization energy E b is around 45 milli electron volts. If you have aluminum it is around 57. If you have gallium it is around 72. Once again these numbers are very close to the thermal energy at room temperature. So, once again we can say that all of the boron or all of the atoms are ionized which means if you add Na boron atoms each boron atom will give you one hole. So, the concentration of holes at room temperature would be Na. We can draw a similar band picture in the case of acceptors. So, we will draw the band gap of silicon. This is E c this is E v. So, this is your valence band this is your conduction band. In the case of n type we said that they are localized levels very close to the conduction band. The case of p type they are localized levels very close to the valence band. So, that an electron from the valence band can be excited to the boron or the acceptor level. So, this is boron and the energy gap between them is around 45 milli electron volts. A p type dopant is called an acceptor n type is called a donor because it donates one electron this accepts one electron. So, it is called an acceptor and just the same way with n type if Na is your concentration of acceptor atoms and Na is much greater than Ni. You will find that p is Na, n is Ni square. These are the same arguments that we used when we describe your n type dopants. Let us actually put in some values now and try to calculate the electron and hole concentration and the conductivity. One of the reasons for doping is to increase the carrier concentration and to also increase the conductivity. We saw earlier that if you have intrinsic silicon your carrier concentration at room temperature is 10 to the 10 and your conductivity sigma is around 3 times 10 to the minus 6 ohm inverse and centimeter inverse. Let us take intrinsic silicon and add arsenic with a concentration of 10 to the 15 arsenic atoms per centimeter cube. So, N D is the concentration of arsenic. Arsenic is a group 5 material. So, it has one extra electron. So, it is an n type dopant. We also see that N D is much greater than Ni and at room temperature all the arsenic atoms are ionized. So, the concentration of electrons N same as N D 10 to the 15 centimeter cube. The concentration of holes is Ni square over N D. If you do the math Ni is 10 to the 10, N D is 10 to the 15. So, Ni square is 10 to the 20 divided by N D which is 10 to the 5. So, we have a concentration of electrons which is 10 to the 15 concentration of holes which is just 10 to the 5. So, P is 10 orders of magnitude less than N. So, it is much less than N. So, by doping your silicon with arsenic which is an n type dopant you have preferentially increased the concentration of electrons to more than 10 orders of magnitude compared to holes. The equation for conductivity you have seen this earlier N e mu e plus P e mu h. Now, N is so much larger than P we can ignore this term and then just write it as N e mu e. Mu e is the mobility for silicon. Later we will look at the effect of dopant concentration on mobility. You will see that dopants actually reduce the mobility, but for now we will just take the mobility of silicon. So, mu e is 1350 centimeter square holes per second. So, we substitute N which is the same as N D e and mu e and if we evaluate the expression you get sigma to be equal to 0.216 home inverse centimeter inverse. So, we started out with an intrinsic silicon which has a conductivity of around 10 to the minus 6. We added 10 to the 15 arsenic atoms and we found that your conductivity has increased by more than 5 orders of magnitude. So, this is the reason why we use extrinsic conductors. It is possible to increase the conductors by orders of magnitude by adding these dopants whether they are N type dopants or P type dopants. We say we added around 10 to the 15 arsenic atoms. So, what is that in terms of concentration? To know the concentration of arsenic we need to find out the number of silicon atoms per unit volume. So, this let me say is concentration of silicon. Concentration of silicon again we have done this before is nothing but the density of silicon divided by the atomic weight times the Avogadro number which is 5 times 10 to the 22 centimeter cube. So, if you looked at the concentration of arsenic it is 10 to the 15 by 5 times 10 to the 22 which if you do is around 20 parts per billion. Parts per million is 10 to the minus 6 parts per billion is 10 to the minus 9. So, by adding a really small amount of arsenic so much small that it is of the order of parts per billion have increased the conductivity by more than 5 orders of magnitude. So, also the reason why doping has to be such a precise process because we have to make sure that we have the right concentration of dopant atoms. So, that we have the right concentration of carriers and the conductivity that we desire. We can do a similar analysis in the case of p type impurity. So, let me just write that down briefly. So, once again we start with intrinsic silicon n i is 10 to the 10 centimeter cube but now instead of adding phosphorus let me add 10 to the 15 atoms of boron. Boron is a group 3 element. So, when you add boron to silicon you basically increase the concentration of holes. Once again at room temperature all of these holes are ionized or all of these atoms are ionized. So, the whole concentration is the same as n a and this is true again because n a is much greater than n i. We can once again calculate the electron concentration is n i square over n a which is 10 to the 5. So, in the case of a p type impurity we find that p is much greater than n. So, your conductivity equation which is n e mu e plus p e mu h will be just p e mu h. And once again do the numbers the case of silicon mu h is around 450 centimeter square volts per second. So, the conductivity will be slightly lower than that of an n type dopant because of the difference in mobilities but it will still be 5 orders of magnitude higher than the conductivity of intrinsic silicon. So, we have looked at an n type dopant we have looked at a p type dopant. Now, what will happen if we add both an n and a p type dopant together. So, we look at a situation where we have both the n type and the p type dopant. This kind of doping is called compensation doping. The first question is why do you actually have to dope both materials. Let me just give you a brief answer. So, let us say we are trying to form a junction between p and n type materials. Usually we need such junctions when we form devices we will talk about devices later. One way to form a junction is to start with a material that is completely n type and then dope a certain region of the material p type. So, we will see later that there are certain ways of doing this, but in this particular region you have both n type impurities and both p type impurities and we do it in such a way that the overall material is p type. So, this type of doping with both kinds of dopants is called compensation doping. So, consider a material where n d is the concentration of donors. Donors are your n type dopants and n a is the concentration of acceptors. You can have two scenarios. You can have a situation where n d is greater than n a. If you have a situation where n d is greater than n a the material is said to be overall n type with n the concentration of electrons nothing but n d minus n a. This will still be much greater than the intrinsic carrier concentration. Concentration of holes is an n i square. So, we have both donors and acceptors, but if the concentration of donors is more than acceptors the material is said to be overall n type material. On the other hand if you have n a greater than n d it is the reverse case overall it is a p type material. Concentration of holes is nothing but n a minus n d which is greater than n i and n the n i square n a minus n d. Let us just work out an example for this. So, consider an n type silicon with n d equal to 10 to the 15 centimeter cube. So, this is an n type material. So, n is 10 to the 15, n i for silicon is 10 to the 10. So, p is 10 to the 5 centimeter cube. Now, you take this and we add arsenic we add boron atoms. So, we add acceptors and the concentration of acceptors n a is 10 to the 16. So, we originally started out with an n type material. Now, you have a situation where the concentration of acceptors is more than that of donors. So, the material overall is p type. The concentration of holes is nothing but n a minus n d which is 10 to the 16 minus 10 to the 15. So, n a minus n d which is 9 times 10 to the 15 per centimeter cube and this is still greater than n i since n i is only 10 to the 10. Concentration of electrons is now n i square over p 1 times 10 to the 4 centimeter cube. So, thus we started out with an n type material, but by preferentially adding acceptors whose concentration is more than the concentration of donors. You have made this material into a p type material. We will stop the class here for today. The next class we will look at the position of the Fermi energy in the case of an extrinsic semiconductor and we will also look at the behavior of the carrier concentration with temperature. We will try and explain how we can say that all the donors or all the acceptors are ionized at room temperature. But we will look at that in the next class.