 In today's assignment, we will be looking at extrinsic semiconductors. Assignment 3, in assignment 2, we focus exclusively on intrinsic or pure semiconductors. And today we will be looking purely on extrinsic semiconductors. So, before we look at the numerical problems, let me do a brief recap. Intrinsic semiconductors are also called doped semiconductors. So, the whole process of doping is to selectively increase your carrier concentration. So, either N or P. The law of mass action is something that has to be obeyed. So, N i square, where N i is your intrinsic carrier concentration, must be equal to N P. So, if we dope with donor type impurities to increase the concentration of electrons. So, if N goes up, P has to go down. And similarly, if P goes up by doping with acceptors, then N has to go down. We also saw the conductivity equation last time, sigma is N e mu e plus P e mu h. In the case of an extrinsic semiconductor, we usually dope, such that either N is much greater than P or P is much greater than N. Either way, only one of these terms will usually dominate. So, the conductivity will be either due to the motion of electrons or due to the motion of holes. The case of an extrinsic semiconductor, once again because we do not have equal number of electrons and holes, the Fermi level will shift from the center of the gap. It will shift closer to the conduction band for an N type semiconductor and closer to the valence band, if it is a P type semiconductor. This will also depend upon the temperature and whether all the donors or acceptors are ionized. So, these are some of the concepts that we will touch in today's assignment. So, let me look at question one. So, we have a group for semiconductor is doped with donor atoms N d. The donor atom concentration is 10 to the 18 per centimeter cube. The intrinsic carrier concentration is given. So, N i is given 2.3 times 10 to the 13 per centimeter cube. The values of N c and N v, the effective density of states is also given. So, N c equal to N v equal to 7 times 10 to the minus 10 to the 19 per centimeter cube. So, the sample is essentially at room temperature. So, temperature T is 300 Kelvin. So, the first question says what is the whole concentration at 300 Kelvin? So, it is doped with donor ions. So, it is an N type semiconductor. So, N is equal to N d at room temperature, the impurities are usually fully ionized. So, N is equal to N d equal to 10 to the 18. To calculate the whole concentration, we can use the law of mass action. So, N p is equal to N i square, the value of N i is also given. So, p works out to be 5.3 times 10 to the 8 per centimeter cube. So, the concentration of holes is much smaller. So, nearly 10 orders of magnitude smaller than that of the electrons. What is the band gap of the semiconductor? So, to calculate the band gap, we can actually use the intrinsic equation, which we saw in assignment 2. So, N i is N c N v exponential minus E g over 2 k T. So, N c and N v values are given. N i is known. So, the only thing that is remaining is E g and E g is 0.772 electron volts. We can sort of make a guess that the material is germanium, but germanium usually has a band gap of around 0.67, because in this particular problem we have taken N c is equal to N v and that is not true for germanium, but this is the band gap. The band gap is a low value, you can see that because N i is around 10 to the 13. For pure silicon, the value of N i at room temperature is 10 to the 10. So, you have calculated the whole concentration and also the band gap. Then we want to know the position of the Fermi level in the doped semiconductor with respect to the intrinsic Fermi level. So, E f N minus E f i. So, E f N is the position of the Fermi level in the N type semiconductor. E f i is the position in the intrinsic semiconductor is nothing, but k t ln of N over N i. In this particular case N is nothing, but N d. So, we can substitute all the values. We can use k in the SI units, but then we need to divide by 1.6 times 10 to the minus 19. So, that we converted back to electron volts and if you do this E f N minus E f i is 0.276 electron volts. In this particular case N c is equal to N v. So, we can also calculate an absolute value for E f i which is nothing, but E g over 2. E g is 0.772. So, this is nothing, but 0.386. So, we can substitute for the value of E f i in here and get the position of the Fermi level with respect to the valence band as well. So, let us now move to question number 2. So, we have a semiconductor with the value of N i. So, N i is 10 to the 16 per meter cube. So, this we can keep as meter cube or we can convert it to centimeter cube. So, this will be 10 to the 10 per centimeter cube, but for now we will just work with per meter cube. A semiconductor with N i equal to 10 to the 16 per meter cube is doped with acceptor impurities. The concentration of acceptor impurities is also given. So, acceptor impurities the value of N a and that is 10 to the 23 per meter cube. So, donor impurities basically produce electrons. If you go back to the class donors are typically group 5 elements. So, if you think about silicon donor impurities are phosphorus, arsenic, antimony which donate the extra electron. Acceptor impurities are elements like boron, aluminum and gallium which accept the extra electron from silicon and create a hole. So, acceptor impurities produce excess holes donor impurities produce excess electrons. So, we have acceptor impurities. So, once again we want to calculate the electron and hole concentration. So, the temperature is given it is 300 Kelvin. So, N a is fully ionized. So, all the acceptors are ionized. So, P which is the hole concentration is equal to N a is equal to 10 to the 23 per meter cube N. We can again use the law of mass action N is equal to N i square over P. So, we can do the numbers N is 10 to the 9 per meter cube. If you want to convert these to centimeter cube divide by 10 to the 6. So, this is 10 to the 3 per centimeter cube and this is 10 to the 17 per centimeter cube. Assuming that the effective masses so, in part B. So, assuming that the effective masses of electrons and holes are equal to the free electron mass calculate E g the band gap and E f the position of the Fermi level. So, this is part A in part B. We are given that M e star which is the effective mass of the electron this M h star is nothing but M e M e is the rest mass of the electron. So, once again we have to calculate E g we know the value of N i. So, N i is square root of N c N v exponential minus E g over 2 k T. The thing here is we are not given the values of N c and N v, but these can be calculated from the mass of the electrons and holes. So, N c is nothing but 2 times 2 pi M e star k T over h square whole power 3 over 2 N v is 2 times 2 pi M h star k T over h square whole power 3 over 2. So, M e star and M h star values are given they are equal to the mass of the electron. So, we can evaluate N c and N v. So, N c is equal to N v and it is equal to 2.5 into 5 times 10 to the 25 per meter cube. So, you can convert this to centimeter cube as well you will just have to divide by 10 to the 6. So, now, we know the values of N c and N v N i is known N i is given to be 10 to the 16. So, we can go ahead and calculate E g the value of E g is 1.12 electron volts. Once again the material that we are talking about here is silicon. You can clearly see that N i is 10 to the 10 per centimeter cube which is what the room temperature intrinsic carrier concentration of silicon is. So, we now have to calculate the position of the Fermi level. This is a p type material because we have acceptors rather than donors. So, in the case of donors the Fermi level moves closer to the conduction band. In the case of acceptors the Fermi level moves closer to the valence band. So, E f p minus E f i is nothing but minus k t ln of p over N i. So, p we know N i is known we can substitute and this is minus 0.42 E v. So, this is below the intrinsic Fermi level and it is minus 0.42 E v below the Fermi level. We can calculate the value of E f i. E f i is nothing but just E g over 2. So, E g is calculated to be 1.12. So, E f i is 0.56. So, that we can substitute here and that gives you E f n is nothing but E f i minus 0.42 which is 0.14 E v above the valence band. So, in a way problem 2 the part b is very similar to what we did in problem 1 except that now we all have acceptor impurities instead of donor impurities. So, in part c question says donor impurities are now added to a concentration of phi times 10 to the 22. What are the new values for the 4 quantities calculated above? So, we now add donors and the concentration N d is 5 times 10 to the 22 per meter cube. The sample already has acceptors acceptor concentration N a is 10 to the 23 per meter cube. So, this is an example of compensation doping where we have both donors and acceptors. In this particular case N a is greater than N d. So, ultimately the material becomes p type and the concentration of holes is just N a minus N d. So, p is 5 times 10 to the 22 per meter cube and we can calculate as before again using the law of mass action n i square over p which is 2 times 10 to the 9 per meter cube. We can go ahead and calculate the position of the Fermi level. So, E f p we can repeat the calculations that we did before except using the new values of p. So, E f p if you calculate becomes 0.16 electron volts above E v. The band gap will not change because the band gap is calculated based on the intrinsic values. But, because the values of p and N change the position of the Fermi level will change. So, let us now move to question 3. Question 3 using the hydrogenic model how much energy is required to ionize a donor atom in a semiconductor with a dielectric constant of 10 and an electron effective mass that is only 30 percent of the free electron mass. So, we looked at the ionization energy of a donor or an acceptor earlier. We found that these donor levels are close to the conduction band and the acceptor level is close to the valence band and typical ionization energies are of the order of 10s of milli electron volts. This is why these levels are usually fully ionized at room temperature. So, to look at the ionization energy E b we typically use a hydrogenic model. But, we modify the mass of the electron and also the dielectric constant. So, the equation for that is something similar to what we worked out in class is minus 13.6 Me star over Me times 1 over epsilon r square. This answer will be in electron volts. So, 13.6 is the ionization energy for the hydrogen atom. This value is in electron volts Me star over Me is nothing but the electron effective mass. In this particular problem it says the electron effective mass is 30 percent of the free electron mass. So, Me star over Me is 0.3 epsilon not r is the relative permittivity of the semiconductor. So, again for this particular problem the permittivity or the dielectric constant is given as 10. So, we have all the values that we need. We can substitute that and evaluate E b and E b if you calculate comes out to be minus 41 milli electron volts. So, this is the ionization energy of the donor atom. So, this is the energy that is required to remove the electron from the donor atom and take it to the conduction band. For comparison the thermal energy of an electron at room temperature is 25 milli electron volts. So, at room temperature it is possible to easily ionize the donors and take the electron to the conduction band. In the next problem in problem 4 we will look at those calculations explicitly. In part 2 of problem 3 we also need to calculate the bore radius of the donor atom. So, once again we are using the hydrogen model and we want to calculate the bore radius. In the hydrogen model the bore radius a naught is given by the expression epsilon naught head square over pi m e and e square. So, this is again based on calculations and the bore radius works out to be 0.53 angstroms. So, in this particular case we can use the same expression, but epsilon naught. So, for the donor atom the bore radius I am going to call it a naught prime is nothing but epsilon r epsilon naught head square. So, we replace epsilon naught by epsilon r times epsilon naught and m e is replaced by m e star. So, the expression is the same except that we are adding the dielectric constant and also the electron effective mass. So, we can plug in these numbers and this gives you a bore radius of around 18 angstroms. The way to think about this is that this represents the influence of the donor electron. So, it represents a size of the influence of the donor electron. So, in the case of an extrinsic semiconductor we usually treat these donor levels as individual atomic levels. So, the concentration is usually of the order of parts per million or parts per billion. So, that we treat them as individual atomic levels, but if you keep on increasing the concentration of the donors then these atomic levels will come together and when two donors see each other which means when the distance between them comes to be less than 18 angstroms then they will start to interact and they will form a band. At usually these high concentrations donors form instead of single atomic levels a donor energy band which can then overlap with the conduction band. So, these types of semiconductors are called degenerate semiconductors. So, to calculate the number of dopants or the dopant concentration when we have degenerate semiconductors. So, I will call it donor overlap it is nothing but 1 over the volume of the dopant atom. I will put the word atom here within parenthesis to actually talk about the sphere of influence of your donor. So, this is nothing but 1 over 4 by 3 pi a naught prime cube. So, this particular value works out to be 4.1 times 10 to the 19 per centimeter cube. So, at this concentration and at higher values of concentration your donor atoms are essentially too close. So, that the atomic levels mingle and we have a donor band. So, this determines the conditions for forming a degenerate semiconductor. So, when the concentration is greater than 4.1 then 4.1 times 10 to the 19 per centimeter cube we get a degenerate semiconductor. So, this is a simple back of the envelope calculation where we use the effective bore radius of the donor atom to calculate the concentration where we get a degenerate semiconductor. So, let me now go to problem 4 with 10 to the 15 phosphorus atoms. So, phosphorus is a donor. So, N D is 10 to the 15 per centimeter cube. The donor energy level for phosphorus in silicon is 0.045 E V below the conduction band edge. So, delta E which is your donor ionization energy is 0.045 E V or 45 milli E V and this is below the conduction band. So, C B is your conduction band. So, where is the Fermi level located at 0 Kelvin? So, T equal to 0 Kelvin we want to know the position of the Fermi level. So, at low temperatures if you think about the model. So, let me draw a schematic of the band diagram. This is E C, this is E V, the material is silicon. So, your band gap is typically 1.10 electron volts. We have a donor level that is very close to the conduction band edge. So, this is your donor level and this energy is 0.045 E V. So, the diagram is not to scale, but this just shows you that the donor level is very close to the conduction band. So, at 0 Kelvin the donor level is not ionized. So, we basically have electrons here. The valence band is at a much lower energy level. So, we can sort of ignore the valence band and we can treat this as an intrinsic semiconductor with the donor level being the valence band and the conduction level being the conduction band of silicon. So, in this particular case E F will be located between the donor level E D and the conduction level E C. So, this is nothing but your donor level. So, let me mark that here as E D and the temperature is 0 Kelvin. So, if you use this expression E F I is E G over 2 minus k T ln. So, 3 4th k T ln is M E star over M H star. Temperature is 0 Kelvin. So, this term goes to 0. E G is nothing but delta E which is 0.04 y electron volts. So, that E F I is half of that. So, 0.225 electron volts. So, that E F I is half of that. So, 0.225 electron volts are right in the middle of the donor level and the conduction band. So, at what temperature? So, this is part A in part B. At what temperature is the donor 1 percent ionized and where is the Fermi level located at this temperature? So, we start at 0 Kelvin where we have a donor level that is completely that is not ionized at all and the conduction band that is completely empty. We then start to increase the temperature. So, that electrons from the donor level start to move and occupy the conduction band. So, we can again treat this as an intrinsic semiconductor. So, concentration of electrons is nothing but N C N D over 2 and the reason for the 2 is because these are individual atomic states. So, they can only take 1 electron minus E D or minus delta E over 2 k T. So, we are using the expression for an intrinsic semiconductor except that instead of writing N V we write N D over 2 and delta E is the ionization energy N the question says is 1 percent ionized. So, 0.01 N D. So, everything else is known except for the temperature. So, this we can substitute N. So, temperature we can calculate to be 29 Kelvin. So, all we have done is to take the expression for an intrinsic semiconductor and then modify it. We can calculate the position of the Fermi level. So, E F I once again we can use the expression for an intrinsic semiconductor. So, it is delta E over 2 minus 1 half k T and we will use the effective density of states. So, ln of N C over N D over 2. So, once again if you plug in the numbers this works out to be 10.1 milli electron volts and this is above E D. So, let us look at part C part C part C ask at what temperature does the Fermi level lie in the donor energy level. So, when the Fermi level is complete when the donor is completely ionized E F will be at E D. So, E F will be equal to E D when the donors are completely ionized. So, once again N is equal to N D which is square root of N C N D over 2 exponential minus delta E over 2 k T. So, everything else is known except for the temperature. So, the temperature gives calculation gives you a temperature of 61.3 Kelvin. So, at a relatively low temperature of around 60 Kelvin you get the donors to be completely ionized. So, estimate the temperature part D estimate the temperature when the sample behaves as if it is intrinsic. When the sample behaves like an intrinsic semiconductor N is around 1.1 N D. So, if you go back to the notes we say that the sample is intrinsic when the electron concentration is 10 percent more than the donor concentration. So, we have a regime between saturation and this is your saturation temperature and the intrinsic temperature where the concentration of electrons is within 10 percent of the donor concentration. So, N is 1.1 N D P is nothing but 0.1 N D this is something we can get by just doing a charge balance. So, that the net positive charge must be equal to the net negative charge N P is N I square which means N I is 0.1 N D. So, N P is 0.1 N 3 3 N D this is your intrinsic carrier concentration. So, this is now N C N V and these are the conduction and the valence band of the silicon minus E G over 2 k T. So, we can substitute the numbers temperature T is around 618 Kelvin. So, the question also gives you the number to the density of states. So, you know the value of N C and N V E G is known. So, the only thing that is unknown is temperature. So, this temperature T is called your intrinsic temperature. This is called your saturation temperature. So, that within the between the saturation and intrinsic your concentration of electrons which is your majority concentration is within 10 percent of the donor concentration. Part E you are asked to sketch a schematic. So, in part E sketch schematically the change in Fermi level with temperature. So, let me just draw it here. So, this is your conduction band E C this is the valence band E V these are my donor levels E D. So, just for schematic let me take this to be a temperature axis and this is the center of the band gap E G over 2. So, at 0 Kelvin your Fermi level starts to be here as a temperature rises electrons start to go from the donor level to the conduction band the Fermi level drops at some particular temperature which is T i or T s the saturation temperature the donors are completely ionized. And then the Fermi level starts to fall down and it really high temperatures it becomes equal to E G over 2. So, this temperature is T i and within this particular regime your electron concentration is almost a constant. So, let us now look at the fifth problem. So, we have an n type silicon which is doped with phosphorus atoms. So, N D is 10 to the 17 per centimeter cube the drift mobilities of electrons and holes in silicon depend upon the total concentration of the dopant and the expression is also given. So, usually we take the drift mobility to be a constant, but actually with increase in doping especially in extrinsic semiconductors the mobility actually decreases. You have seen this in class this is because we have the electron or the hole being scattered by the ionized donor or the acceptor. So, in this particular case mu E is given to be 88 plus 1252 divided by 1 plus 6.984 times 10 to the minus 18 N dopant the units are centimeter square per volts per second. So, in the first part we want to calculate the room temperature conductivity. So, we know N is equal to N D it is equal to N type semiconductor. So, P is N i square over N D is usually much smaller than N. So, if it is silicon N i is 10 to the 10. So, we can actually work it out P comes to be 10 to the 3. So, in this particular case conductivity sigma is N E mu E plus P E mu H P is much smaller than N. So, this term goes off. So, this is nothing but N E mu E. So, we can plug in the numbers to calculate mu E you plug in the dopant concentration which is 10 to the 17 and you get the value of mu E. So, if you do that the value of sigma comes to be 13.2 ohm inverse and centimeter inverse. In part B we want to do compensation doping in the system. So, we are going to add acceptors to make the sample P type with having the same conductivity value. So, now, we have a P type semiconductor and we do this by adding acceptors N A and this must be greater than the donor concentration N D. So, P in this case is nothing but N A minus N D and conductivity sigma is P E mu H. So, there is a similar expression for mu for the P type or similar expression for the mobility for the holes mu is 54.3 plus 407 1 plus 3.745 18 N dopant. So, the important thing to remember here is that this is the case where we have acceptor and donor impurities. So, the total dopant concentration in part B is equal to N A plus N D. So, we use the same equation sigma it is now P type. So, it is P E mu H P is nothing but N A minus N D and N D we know mu H we can use this expression except that N dopant will be N A plus N D. So, what we have is an equation where everything is known except for N A. You can simplify this expression I would not go through the math, but this is essentially a quadratic equation which you can solve and when you solve you get the final value of N A and the value of N A is 3.04 times 10 to the 17th. So, this is greater than the value of N D which is 10 to the 17th. So, what you have is essentially a P type semiconductor where you have done compensation doping by adding excess amount of acceptors. So, the main point here is that when you have whether an N type or P type the conductivity is essentially determined by the majority charge carriers.