 In today's assignment, we are going to look at metal semiconductor junctions. This is assignment 4. So in class, we looked at two kinds of metal semiconductor junctions. One was the short key junction or the short key contact and the other was the omic junction or an omic contact. So we saw that a short key junction forms when the work function of the metal, so phi m, is greater than the work function of the semiconductor. The case of an omic junction, it is the reverse. The work function of the metal is less than the work function of the semiconductor. We also saw that a short key junction essentially behaves like a rectifier. So it conducts the current in the forward bias and does not have any conduction in the reverse bias. So in this way, a short key junction is similar to a p-n junction which is also a rectifier. An omic junction on the other hand from the name is just a pure resistor. It conducts both in the forward and the reverse bias and the conductivity is defined by the conductivity or the resistivity of the semiconductor material. So in today's assignment, we will be looking mostly at short key junctions. We will do some calculations in the short key barrier, the contact potential and also the current in the forward and reverse bias. So let us go to problem number 1. So we want to show how a short key junction is formed between a metal and a p-type semiconductor. So we can do this by sketching the band diagram under equilibrium, forward and reverse bias. So in class, when we looked at the example of a short key junction, we looked at a metal and an n-type semiconductor. So let me draw that first under equilibrium and then from there, we will look at a metal and a p-type semiconductor. So we said a short key junction is formed when phi m is more than phi semi. So we will start with the metal here. We just draw this slightly up. So this represents the vacuum level. This is the Fermi level of the metal and this space is the work function of the metal. So all the energy levels below this are completely full. So we will start off with an n-type material. So this is your n-type material. You have EC and EV. So EV is the valence band and EC is the conduction band. It is an n-type. So the Fermi level is close to the conduction band. Once again here, the distance between the Fermi level and the vacuum level is the work function of the semiconductor. So when these two are bought together in contact, we know that in equilibrium, the Fermi levels must line up. So we have excess electrons that are there in the conduction band of the semiconductor. These will go and occupy all the empty states in the metal. So there is a net positive charge on the semiconductor side, a net negative charge on the metal side. The electric field goes from positive to negative and the bands bend up in the direction of the field. So if you were to draw this under equilibrium, so I will just mark my junction. This is EF. So this is my metal side and this is the n-type semiconductor. This is the EF of the semiconductor. So far away from the junction, the semiconductor will still be n-type. So let me draw the bands slightly closer. So this is n-type EC and EV. There is a net positive charge on the semiconductor, a net negative charge on the metal and the bands will bend up. So this in turn forms the depletion region and this is the band diagram at equilibrium. So this is a case of a metal and an n-type. This is similar to what we saw in class. So let us now draw one for a metal and a p-type. So once again we have the vacuum level. You have the Fermi level of the metal and this is phi m. So this is the metal. We now have a p-type semiconductor. So this is EC, this is EF and this is EV. So it is a p-type semiconductor so that the Fermi level is close to the valence band. So we can use the same argument that we used for a metal and an n-type. But now that the argument is reversed. So once again when the junction forms the Fermi levels must line up but instead of excess electrons going from the semiconductor to metal we now have the electrons moving from the metal to the semiconductor or the holes moving from the semiconductor to the metal. So that there is a net positive charge on the metal side, a net negative charge on the semiconductor side and bands will bend down as we go from the semiconductor to the metal. So this if we draw in equilibrium the Fermi levels must line up. So I will put an interface so EF metal and p-type semiconductor far away from the junction your material is still p-type and then the bands bend down so that there is a net negative charge and a net positive charge and this is the depletion region. So the band bending here is similar to that of a metal and an n-type but it goes the other way. So we now want to draw the energy band diagram in forward and reverse bias. So in the case of a forward bias so this is my metal, this is my p-type. The metal is connected to a negative charge, the p-type is connected to a positive charge. So in this case the Fermi levels shift and the barrier essentially is lowered so we can once again draw this that is my interface that is my metal. Now the Fermi levels no longer align and for the p-type Fermi level is shifted down and this shifting down is given by your external potential that is V0. So here the barrier for the motion of the electrons and holes is reduced so that there is an increasing current when you apply an increasing voltage. The case of a reverse bias so m and p the metal is connected to positive, the semiconductor is connected to negative. Once again the Fermi levels do not line up but instead of shifting down the Fermi level is now shifted up. So that is my interface EC and EF. This is the metal, this is the p-type semiconductor. So this is a situation where we have a metal and a p-type in equilibrium is the energy band diagram, forward bias and reverse bias. So let us now go to problem 2. In problem 2 we have an n-type silicon with 10 to the 16 donors per centimeter cube. Now n silicon with nB is 10 to the 16 per centimeter cube. The 2 ends of the sample are labeled B and C. So we have 2 ends. So they are essentially 2 metals at the either end. So if I were to draw a schematic this is my n silicon and I have B and C on both sides. So let me just shade them to show you that they are essentially metal contacts. The electron affinity of silicon is given. So chi is 4.01 EV and there are 4 potential metals which can be used for these contacts and their work functions are given. So we have cesium, lithium, aluminum and gold and their work functions are essentially given. So the first thing we need to do is to calculate the work function for the semiconductor. So we can draw an energy band diagram. So this is vacuum. We are only drawing the semiconductor side, EC and EV. It is an n-type semiconductor. So the Fermi level will be closer to the conduction band. Now the electron affinity is the energy difference between the conduction level and the vacuum level. So this is essentially chi. We need the work function. So we need psi of the silicon. To know that we need to know the position of the Fermi level and that can be calculated from the concentration of donors in the material. So we can just say EFN-EFI is kT ln of n over ni. So n is nothing but nd. It is fully ionized. Ni for silicon is given and we also saw this during the previous assignment. It is nothing but 10 to the 10. From this we can calculate the value of EFI-EFN. So EFN-EFI we can calculate. We also know the position of the intrinsic Fermi level. So we can calculate it from the values of NC or NV. In this particular case, EFI is given to be 0.55 electron volts and the band gap of silicon is 1.1. So this whole thing is 1.1 EV, EFI is given to be the center of the band gap and this is 0.55. This distance is also known and the value for this is 0.192, 0.357, sorry, this is 0.357. So everything else is known except for this. So EC-EFN is nothing but 0.192. So from this we can calculate the work function of the silicon. So phi of silicon, since it is n type I will just write phi of n is nothing but chi plus EC-EFN which works out to be 4.20 EV. So if you look at the various parts of the question, part A asks which metals will result in a short key contact. So we have a short key contact when the work function of the metal is greater than the work function of the semiconductor. So those are essentially go aluminum and gold. So short key contact would be aluminum and gold which metals will result in an ohmic contact. So an ohmic contact is 1 where it is reverse. The work function of the semiconductor is higher. So it is just cesium and lithium. So part C sketch the IV characteristics when both B and C are ohmic contacts. So let me draw that. So B and C are both ohmic contacts. So an ohmic contact is nothing but a resistor. So when you have both B and C to be ohmic then the whole thing just acts as a resistor. So B and C are both ohmic and the whole thing is just a resistor. So the IV characteristics will just be a straight line and the slope of this line will be just 1 over R. Part D sketch the IV characteristics when B is ohmic. So B is ohmic and then C is short key. So we have one ohmic and one short key junction. So in the case of a forward bias you will have a short key junction will be essentially a high conductor. So it will start to conduct. In this particular case the resistance will be determined by the highest resistance point which is your ohmic contact. So here in the case of forward bias so if you were to draw I versus V you have a highly conducting junction which is junction C and you have a resistor which is junction B. So this will essentially behave like a resistor. The case of a reverse bias, C is essentially reverse biased so that there is very low conductivity through that. So that will essentially determine the conductivity of the entire circuit so that you have a very low conductivity in reverse bias. So the same is true when B is short key and C is ohmic the curve will be similar. In part E sketch the IV characteristics when both B and C are short key. So B and C are both short key. So in this particular case if one of the junctions is forward biased the other junction will be reverse biased and so on. So whether you are in the forward or the reverse there always be one junction that is reverse biased which will have very low conductivity. So the IV characteristics in this particular case will be a very low current in both forward and reverse bias. So this kind of a situation is very important when you are trying to make electrical contacts to a semiconductor usually we have to make two contacts. Ideally we want these contacts to be ohmic because we do not want the contact itself playing a role in determining the IV characteristics but there could be diodes based on the short key effect. These are short key diodes. In this particular case you would want one junction to be essentially a short key junction and the other to be an ohmic junction. So let us now go to problem 3. So problem 3 you have a short key junction diode between tungsten and n type silicon. The silicon is doped with 10 to the 16 donors per centimeter cube. The cross sectional area is given so A is 0.1 millimeter square. The electron affinity of silicon is given same as the last problem 4.01 EV and the work function of the metal is given to be 4.55 electron volts. So once again we need to calculate the work function of the semiconductor. So we can do the same thing that we did in the last problem. In this particular case the effective density of states in the conduction band is given. So that is 2.8 times 10 to the 19. So we could use this directly to calculate the position of the Fermi level. So N is nothing but NC exponential minus EC minus EFN over KT from which you could calculate EC minus EFN. So N is nothing but ND which is the concentration of donors. NC is given everything else is known. So EC minus EFN is essentially 0.205 electron volts. So from this you can calculate the work function of silicon to be 4.215 electron volts. So we can draw this in a band diagram. So I will just draw it schematically. This is my tungsten. Work function of tungsten is given. So 4.55 and this is my N type semiconductor. So EC, EV, this is 4.01, this whole thing is 4.215 and this is 4.55. So again we have a case of a short key junction between tungsten and N type silicon. So part A, we want to calculate the theoretical short key barrier. So we want to calculate phi B. Phi B is the short key barrier and that is essentially the work function of the metal minus the electron affinity of the silicon. The short key barrier represents the barrier for the electron to move from the metal to the semiconductor side. So you have an electron going from EF to the conduction band. So this is just phi M minus chi Si can put in the numbers and this is 0.45 electron volts. Then we want to calculate the built in voltage. So V0 is nothing but chi M minus phi Si divided by E. So E is just to convert it from electron volts to volts. It is the difference between the work functions. So this we can substitute and the answer is 0.335 electron volts. In part C, we need to calculate the reverse saturation current and also the current when there is a forward bias of 0.2 volts across the junction. So part C, we want to calculate the current in the junction. So in the case of a short key diode, it is possible to write an expression for the current. This is something we do not see during the course of the lecture. So we can write the current J as some constant J0 exponential EV over KT minus 1. So V here is your external potential, J0 is your reverse saturation current and J0 is equal to BET square exponential minus phi B over KT. So this is called a Richardson-Dushman equation and it is actually used to calculate the current during thermionic emission from a metal. So BE is usually a material constant. It is a property of the interface. So whether you have tungsten and silicon or platinum and silicon, platinum and germanium, the value of BE will change. So BE is a constant that depends upon the materials which is basically the property of the junction. So in this particular case, the value of BE is given. So BE is 110 ampere per centimeter square per Kelvin square. So the value of phi B we calculated earlier. This is nothing but your short key barrier. So from this problem, you can calculate J0. So everything else is known. Temperature is 300. So from here, J0 is essentially 8.5 times 10 to the minus 3 ampere per centimeter square. If you want to calculate the current, you multiply this with the area. So current I is J0 times A which is 8.5 times 10 to the minus 6 amperes or 8.5 micro amperes. We can now calculate the current during the forward bias. So J is J0 exponential EV over KT minus 1. So usually the exponential term dominates. The external voltage is given to be 0.2 volts. So we can substitute the numbers. And J comes out to be 19.3 amperes per centimeter square or the current I is just 0.019 amperes. So this is the current during forward bias. You can see that it is nearly 4 orders of magnitude higher than the current during reverse bias. This is why a short key junction is essentially a very good rectifier. In part D, question says that the experimental short key barrier is actually higher. So phi B experimental is 0.66 EV. So it wants us to do the recalculation. So this experimental value takes into account the fact that the interface is never perfect that you always have some sort of defects at the interface. So if you use this, we can use the same calculations that we just did except that now we now have to use the newer value of phi B. So in this particular case, J0 comes out to be 8.3 times 10 to the minus 5 ampere per centimeter square and current J is 0.19 ampere per centimeter square. So the actual current is slightly lower than what we would expect if you use the theoretical values. But the important fact is that it is still 4 orders of magnitude higher than J0. So that your short key diode, your short key junction still functions as a rectifier. So let us now go to the next problem. The problem 4, we have a platinum silicide short key diode is fabricated on end silicon. So you have platinum silicide on silicon. So how this is usually obtained is by first depositing platinum metal. Usually it is done by some vapor deposition process like thermal evaporation or sputtering or e-beam evaporation. Then the interface is annealed so that we have inter diffusion between platinum and silicon which again react to form the silicide. So depending upon the composition, you can get a single composition PTSI or you could get multiple compositions. Again depends upon the thickness, the platinum layer and the amount of inter mixing. So these silicide layers are usually formed by depositing the metal and then doing some sort of a post annealing treatment. So in this particular case, it is an N type silicon. So ND is 10 to the 16 per centimeter cube. The barrier height in this particular problem is given. So 5B is 0.89 volts. So one of the advantages of doing a post annealing is that that usually eliminates some of the defects so that your barrier is essentially very close to your theoretical barrier. So once again we want to calculate. So the forward current is known. So J is given to be 2 amps per centimeter square and we want to calculate the value of the voltage. So V external is 1, we want to calculate. So we can go back and use the same equation. J is equal to J naught exponential EV over kT minus 1. J naught is nothing but BET square exponential minus psi B over kT. So the value of BE for this problem, we can still take the same value 110 ampere per centimeter square per kelvin. So we can use this and calculate J naught. So J naught is just got by substituting BET square and exponential of minus 5B over kT. Once we get the value of J naught, we can put the value of J naught here. We need to know the value of J. So J is given to be 2 ampere centimeter square. The only thing that we do not know is V external. So once we calculate J naught, we can plug it here and get V external. So I will just write the answer. V external for this particular problem is 0.49 volts, but the calculation is very similar to what we did the previous problem. So let us now go to problem 5. Problem 5, we have a short key diode formed by depositing gold, but now the material is N type gallium arsenide. So we have N gallium arsenide with ND is 5 times 10 to the 16 per centimeter cube. So once again in part A, we need to calculate the forward bias voltage for a current density. So J is 5 amps per centimeter square and we need to calculate the voltage for that. So this is again similar to the previous problem except now the material is gallium arsenide. So the first thing we need to do is to calculate the barrier potential. We are going to assume that it is a theoretical barrier. So we need to know phi B, which is nothing but phi M minus the electron affinity for gallium arsenide. So in this particular case EG of gallium arsenide is given, but more importantly we only need the electron affinity. This has a value of 4.07 EV. So we can still use the other values to calculate the contact potential, but as far as part A is concerned, the only thing we need to know is the electron affinity. So phi M is given to be phi electron volts. So this is the work function of gold. The electron affinity of gallium arsenide is known so that the theoretical barrier potential is nothing but 0.93 electron volts. So once we get that, we can calculate J0. Again for the gold and gallium arsenide interface, we have the values for BE. BE is equal to 45 amps per centimeter square per Kelvin square. So BE is not only a material property, it also depends upon what facet of the material you have. So whether you have a 100 plane or a 110 or a 111 that will also affect the value of BE. So J0 is a number we can calculate. All the numbers are known. So this is nothing but 9.1 times 10 to the minus 10 ampere per centimeter square. So J0 is known, J is known, exponential eV over kT minus 1. So again J is known, J0 is known. The only thing that is unknown is V from which we get V to be 0.58 volts. In part B, we need to calculate the change in the forward bias voltage to double the current density. So J nu which is your new current density is 2 times of the old one. So this should be 10 amps per centimeter square. You can either take the ratio of the old and new J or you could use the same equation J0 exponential eV over kT minus 1 can calculate. So if you take the ratio J nu minus J which is 2 is equal to exponential eV nu over kT by exponential eV over kT. So V is known. We just calculated that in part A. The only thing we need to do is to calculate V nu and V nu is 0.60 volts. So in the case of a short key junction which is essentially a rectifier, we have seen how to calculate the short key barrier voltage, the built-in potential and also the current during both forward and reverse biased. An ohmic contact is much simpler, an ohmic contact is essentially a resistor and the resistivity is usually given by the resistivity of the semiconductor material.