 In today's assignment class, we will be looking fully at intrinsic semiconductors. This is assignment 2 and we will be focusing on intrinsic semiconductors. So before we start looking at the problems, we just do a brief review. So intrinsic semiconductors or pure semiconductors are essentially single crystals. We say that there are no defects in the semiconductor because these defects can again create electrons and holes of their own. In the case of an intrinsic semiconductor, we say that the electron concentration in the conduction band that is n is equal to the hole concentration in the valence band that is p and it is equal to something which we denote as ni and ni we call the intrinsic carrier concentration. We also say that ni is a function of the band gap of the material e g and also a function of temperature. So typically ni is written as n c n v exponential minus e g over 2 k t. So the intrinsic carrier concentration depends exponentially on the band gap. The temperature term enters in this exponential factor but n c and n v which are the effective density of states at the valence band edge and the conduction band edge are also a function of temperature. So n c and n v are also a function of temperature. Typically they are proportional to temperature to the 3 over 2 but the exponential term is the one that dominates. We also saw the general equation for conductivity sigma is nothing but n e mu e and p e mu h. In the case of an intrinsic semiconductor, this just becomes ni e mu e plus mu h. So these are just a few points about intrinsic semiconductor. We will be using them today during the course of the assignment. So let us first look at problem 1. So what fraction of current in intrinsic silicon is carried by holes? So we have silicon and it is intrinsic which means n equal to p equal to ni and the question asks what fraction of current or what fraction of conductivity is defined by the holes. So if you just say n is equal to p is equal to ni that means there is a 50% contribution that is a very simplistic answer. The reason is the conductivity not only depends on n it also depends upon mu e and mu h which is the mobility of the electrons and holes. So we can write the conductivity equation n e mu e plus p e mu h. This represent the fraction carried by the electrons, fraction due to electrons. This is the fraction due to holes. So we can include the numbers for mu e and mu h. So fraction carried by the holes we can write this in the form of a ratio it is nothing but p e mu h divided by the total that is n e mu e plus p e mu h. So for an intrinsic semiconductor n is equal to p is equal to ni so these terms cancel e will also cancel. So this is nothing but mu h over mu e plus mu h. So the fraction of current carried by holes is directly proportional to the whole mobility. We can plug in the numbers for silicon here. So for silicon mu h is 450 centimeter square per volt per second mu e is 1350. So the mobility of the electrons is higher. So we can plug in these numbers and the fraction is 0.25. So even though we have equal concentration of electrons and holes they do not have the same mobility and this is because your electrons are moving in the conduction band and the holes are moving in the balance band and this ultimately determines what fraction dominates whether the electron conductivity dominates or the hole conductivity dominates. Whether when we see an extrinsic semiconductor we will find that n and p are not the same one is much higher than the other and then there one of the term dominates because of the difference in concentration. So let us now move to question 2. So we have a pure semiconductor or an intrinsic semiconductor the band gap is 1.25 e v. So the effective masses of the electrons and holes are also given. The effective mass of the electron me star is 0.1 times me where me is the mass of the electron and m h star is 0.5 me. This is the effective mass for the hole. Once again we have seen the concept of effective mass before. So effective mass does not mean a change in the actual mass of the electron or the hole. It just represents the cumulative of all the forces of the atoms in the lattice that basically acts on the electrons and holes. Once again these numbers are different because you have electrons that are moving in the conduction band and holes that are moving in the valence band. So the band gap is given. The effective mass values are given. The carrier scattering time is temperature dependent and that is given of the form. So tau which is your scattering time is a function of temperature and this is 1-10 to the minus t. So 1 times 10 to the minus 10 divided by temperature and the units are seconds. So the effective masses are given. The band gap is given and the temperature dependence of the carrier scattering time is also given. This we will use to calculate the mobilities. So we want to find the following at two temperatures. One is 77 kelvin and the other is 300 kelvin. So 300 kelvin is room temperature, 77 kelvin is typically your liquid nitrogen boiling point. So that is a low temperature. The first one we want to find is the concentration of electrons or holes. Because this is a pure semiconductor, what we want to find is the value of the intrinsic carrier concentration. So we can go back to the equation Ni is nothing but square root of Nc and Nv exponential minus Eg over 2 kT. So the problem is we do not have the values of Nc and Nv. These are the effective densities of states at the band edges but these we can calculate once we know the effective mass. So Nc which is the density of states at the conduction band edge is nothing but 2 pi Nv star kT over h square whole power 3 over 2 Nv we can do the same for the valence band edge 2 pi mh star over kT by h square whole power 3 over 2. So we have the values for Nc and Nv. Then we see both are temperature dependent they are proportional to T to the power 3 half. So once we calculate Nc and Nv for both the temperatures we can plug in here and calculate the value for Ni the band gap is also known. So typically you have to keep all of this in SI units. So you have to convert Eg from electron volts to joules and that we can do by just multiplying by 1.6 times 10 to the minus 19 kB is also in joules. So it is your Boltzmann's constant that has a standard value. So once we plug in the numbers I am just going to write the final answers but you can just go through and check. So Nc at 77 Kelvin is 1.03 times 10 to the 23 per meter cube. So if you remember the definition of the effective density of states is the total number of states per unit volume that is available for the electron to occupy or the hole to occupy. Nc Nv is 1.151 times 10 to the 24 per meter cube. We can do the same calculations for 300 Kelvin again just write down the answers. So Nc is higher 7.92 times 10 to the 23 Nv is 8.85 times 10 to the 24 per meter cube. So compared to 300 Kelvin Nc and Nv are higher this is because we have more density of states available at higher temperature simply because they are directly proportional to T to the 3 over 2. So we can substitute these values of Nc and Nv in this expression and calculate the value for Ni. So let me just write that down Ni at 77 Kelvin is 4.63 times 10 to the minus 18 meter cube. So that is a really small number Ni at 300 Kelvin is 8.56 times 10 to the 13 per meter cube. So I can also write this in centimeter cube or 8.56 times 10 to the 7 centimeter cube. So your Nc and Nv values if you look or off by one order of magnitude simply because you have a rise in temperature but because your Ni depends exponentially on the band gap there is a huge variation between 77 Kelvin which is your liquid nitrogen temperature and 300 Kelvin which is room temperature. So here you have a value that is 10 to the minus 18 and at room temperature you have a value for Ni that is close to 10 to the 13. So that overall there is a 31 orders of magnitude change as we go from liquid nitrogen to room temperature. This is why we say that out of these two terms Nc and Nv and the exponential term the exponential term is the one that dominates in determining the value for Ni. So this is part A where we want to calculate the concentration of electrons and holes. Part B we want to calculate the Fermi energy or the location of the Fermi level. So part B we want the location of Efi. This is again a standard expression Efi is Eg over 2 minus three fourths kT ln of Me star over Mh star. So if Me and Mh were equal then Efi will be exactly at the center of the band. If you remember Efi is nothing but a representation of the chemical potential or the amount of work that needs to be done in order to remove an electron from a semiconductor. So even though your electron and hole concentrations are the same so N is equal to P because you have different effective masses your Efi is slightly shifted from the center of the gap. We can once again plug in the numbers so 77 kelvin 300 kelvin Me star and Mh star values are given temperature is also known. So Efi here if you do the substitution is 0.633 electron volts at room temperature Efi is 0.656. Eg over 2 if you look at it is just 1.25 over 2 which is 0.625 electron volts. So the values are very close to the center of the band gap but they are slightly shifted and the shift becomes higher the higher the temperature. So this is 0.633 and this is 0.656 so slightly deviated away from the center of the band gap. This is part B in part C we want to calculate the electron and hole mobilities. So we want to calculate the values of Mu E and Mu H. So Mu E and Mu H are related to the effective mass of the electrons and holes and they are also related to the scattering time. So Mu E is nothing but E tau E over Me star and Mu H is E tau H over Mh star. So Me star and Mh star are given tau E and tau H are your scattering times and you said that tau is nothing but 1 times 10 to the minus 10 over temperature and the unit is seconds. So once again we can calculate the values of tau. In this particular question tau E and tau H are both the same because we do not distinguish between electrons and holes. We only say it depends upon temperature. Once we calculate tau we can go ahead and calculate Mu E and Mu H and get it for the two different temperatures. So let me again write down this side is 77 Kelvin this side is 300 Kelvin. So tau if you calculate is 1.3 times 10 to the minus 12 seconds. We can then calculate Mu E which is 2.283 unit is meter square per volts per second. Sometimes centimeter square per volts per second are also there. It depends upon which you want to use. Mu H is 0.456 meter square volts per second and Mu H the mobility of the holes is lower because the whole effective mass is higher than that of the electron. We can do the same for 300 Kelvin. In this case tau is 3.3 times 10 to the minus 13 seconds. So higher the temperature smaller is the scattering time. So at low temperature this is minus 12 seconds this is minus 13 seconds. So one way to think about this is higher the temperature faster the electrons and holes are moving because they have higher thermal velocities. So they can scatter of the atoms quicker. Mu E is 0.586 meter square per volts per second Mu H 0.117 meter square per volts per second. So the last part of the question we want to calculate the electrical conductivity sigma is nothing but N i E Mu E plus Mu H. So N i we got in the first part of this question Mu E and Mu H we just calculated. So we can just plug in the numbers. So sigma at 77 Kelvin is very small because if you remember N i is very small. So 10 to the minus 36 ohm inverse meter inverse you can also have ohm inverse and centimeter inverse depending upon what your values the units for N i and Mu E and Mu H are. So the same thing we can do at room temperature and sigma is 9.6 minus 10 to the minus 6. So by looking at an intrinsic semiconductor at two different temperatures one thing we find is that the carrier concentration increases exponentially with temperature. Similarly the conductivity will also increase because the carrier concentration increases. This again is determined by the value of the band gap. So higher the value of E g more steeper is this dependence. So if instead of 1.25 we had done the same problem with say 2 electron volts your answers will also be different but the difference between 77 and 300 Kelvin will also be more pronounced. So that is something you can always work out can take the same values but change the value of E g to 2 electron volts and do this question and you can see the difference between 77 and 300 Kelvin. So let us now move to question 3. So question 3 we have gallium arsenide which is a direct band gap semiconductor with E g of 1.42 electron volts at 300 Kelvin. So gallium arsenide it has a higher band gap than silicon it is also a direct band gap material but that is not relevant for this question. It is at room temperature so temperature is 300 Kelvin. Take N c equal to N v equal to 5 times 10 to the 18 per centimeter cube and also independent of temperature. So the N c and N v values are given and just for this question we are assuming that both are same and that they are also independent of temperature. Simply speaking this is not true but as you see for this particular question this is a very valid assumption. So first we want to calculate the intrinsic carrier concentration at room temperature. So that is pretty straight forward you have seen the formula before. So N c N v exponential minus E g over 2 k T. So we can plug in the numbers the value of E g is given. So N i is 6.05 times 10 to the 6 per centimeter cube. So this is the intrinsic carrier concentration at room temperature N i is a pretty small number for comparison silicon has a value of N i of 10 to the 10 so 4 orders of magnitude higher but this is because gallium arsenide has a higher band gap. So the next part of the question says explain numerically how the carrier concentration can be doubled without adding dopants. So we want to keep this semiconductor your pure semiconductor but at the same time we want to increase the value of N i. So the new value of N i that we want is double of the N i value at room temperature. If you are not allowed to add dopants and if you look at this equation the only way to increase N i is to increase temperature because N c and N v are both temperature dependent terms N i also depends on temperature through this exponential term minus E g over 2 k T. So increasing temperature will once again increase N i. So the only way to increase N i without adding dopants is to increase temperature. So we want to know what the new temperature is when your value of N i is 2 times the N i at room temperature. So we will once again use this expression N c and N v are constant so it is not a function of temperature if it were a function of temperature that will also have to be taken into account but N c and N v are constant we know the new value of N i so it is 2 N i at room temperature square root of N c N v exponential minus E g over 2 k let me call this temperature T prime. So T prime is the only thing we want to know is the only unknown this is known these are all known we can put this and recalculate and this gives you the value of T prime to be 307.7 Kelvin. So you increase the temperature by 7 degrees so delta T is 7.7 Kelvin you can double the concentration of N i. So temperature equal to 300 is approximately 27 degrees so 300 Kelvin is 27 degree Celsius so 307 is 34.7 degree Celsius. So we find that even for a small increase in the value of N i so you are only doubling the value of N i you need to increase your temperature by 7 degrees. If you want the really high conductivities that we see in the extrinsic semiconductors can actually calculate that the temperature change must be much higher. This is one of the reason why intrinsic semiconductors are almost never used in the case of devices usually doped semiconductors are used because it is much more easier to control the dopant concentration and then control the carrier concentration and also the conductivity. So let us now go to problem 4. So problem 4 we want to calculate the intrinsic carrier concentration of germanium. So we have germanium with a band gap E g of 0.66 electron volts. This is lower than that of silicon in fact germanium was the first semiconductor that was used ME star is 0.56 ME and MH star is 0.40 ME. So we want to calculate N i so the equation is the same N i is N c N v exponential minus E g over 2 k T. So N c and N v are again related to ME star and MH star. So N c is 2 pi ME star k T over h square whole power 3 over 2 we can write a similar equation for MH star we saw that earlier during question 2. So once again we can calculate N c and N v plug it back in and get the value of N i. So we will do the numbers can write down the final answers N c is 1.05 times 10 to the 25 per meter cube N v which is the same equation except ME star is replaced by MH star N v is 6.33 times 10 to the 24 per meter cube. So N c and N v are known we can calculate N i. N i if you do is 2.36 times 10 to the 19 per meter cube I am just writing down the final answer the math can always be worked out or 2.36 times 10 to the 13 per centimeter cube. So we saw that gallium arsenide has a value of N i that is 4 times or 4 orders lower than that of silicon. Germanium on the other hand has a value of N i that is nearly 3 orders of magnitude higher than silicon. Once again the differences are all related to the band gap values. We can then calculate sigma is N i E mu E and mu H the value of mu E and mu H are given mu E is 3900 and mu H is 1900 centimeter square per volt per second. So that sigma is nothing but 0.022 ohm inverse and centimeter inverse if you also want to calculate the resistivity rho is 1 over sigma which is equal to 45.66 ohm centimeter. The question also asked to calculate the position of the Fermi level at room temperature. So that is again an application of the formula E f i is E g over 2 minus 3 fourths k t ln of M e star over M h star. So this again is very close but it is not exactly at the center of the band gap. So let us now look at problem 5. There is a particular semiconductor and it says that the effective density of states is a constant N c naught times temperature to the power 3 over 2 and same way N v is N v naught times temperature to the power 3 over 2. So the experimental values of N i at different temperatures are given. So we have temperature values of N i in centimeter cube so 200, 300, 400 and 500 and the values of N i 10 to the 7. So we can see that with increasing in temperature the value of N i also increases. So the question ask us to determine this product N c naught times N v naught and also the band gap. So both E g is not known and these two numbers are not known. You can go back to the original equation N i is N c N v exponential minus E g over 2 k T. N c and N v we can substitute these x terms so that this simplifies to T to the 3 over 2 square root of N c naught and N v naught which is a temperature independent term times exponential minus E g over 2 k T. So we can choose any two temperatures so we have four we can take any two temperatures and take the ratio of N i at these two temperatures. So N i some temperature T 1, N i at another temperature T 2 it is nothing but taking this ratio which is here this is a temperature independent term. So this becomes T 1 over T 2 whole to the power 3 over 2 exponential minus E g over 2 k 1 over T 1 minus 1 over T 2. So T 1 and T 2 values are known N i values are known for example your T 1 could be 200 Kelvin T 2 could be 300 Kelvin in which case the N i values are tabulated the only unknown here is E g. So I did these calculations taking 200 and 300 you can take it with any of the other temperatures we have also done and checked you substitute the values E g works out to be 1.25 electron volts. So this is the value of E g which is the band gap of the material once you know E g you can substitute E g for any of the temperatures and evaluate N c naught and N v naught this is a temperature independent term if you do that N c naught times N v naught is nothing but 1.188 times 10 to the 29 the units here are crucial N c and N v the square root of that should have the units of centimeter per centimeter cube or per meter cube in this particular question this also depends on T to the 3 over 2. So the units of this product is centimeter to the power minus 6 Kelvin to the power minus 3 that way when we substitute for square root units work out in the right way. So today we have looked at various problems related to intrinsic semiconductors the important thing to remember is that the intrinsic carrier concentration is a function of temperature and depends upon the band gap of the material.