 So, we continue our discussion on metal semiconductor contacts and metal source to injection MOSFETs. Last time we discussed the principle of operation of a metal semiconductor contact. We took a case where the work function of metal is greater than the work function of semiconductor. So, this is the situation in the thermo equilibrium and there is a depletion layer in the semiconductor and a negative charge here plus charge in the depletion layer. So, this is because of the because electrons are transferred from the semiconductor to the metal because the phi m work function of the semiconductor is larger. So, the Fermi level there is usually below the Fermi level of semiconductor. So, electrons get transferred from semiconductor to the metal. So, we get this is the distribution of carriers. This is the barrier phi b n. The electrons which have energy above this barrier can cross there giving rise to current I naught. Now, thermal equilibrium there is no current. So, the electrons on this side which are above the barrier here the barrier is V B I. They are crossing here. So, those two cancel each other and we have shown that current I naught or current density J naught is given by this term which finally is A star. A star is called Richardson constant which depends upon the effective mass of electron with respect to the free mass 120 m that is 120 m n star by m naught t squared means various dependence on the thermal equilibrium. Of the current depends upon the barrier rate. If barrier is more this is less term it will be minus phi b n by k t. If the barrier is low this is large current. Now, thermal equilibrium you saw that they are equal number of carriers crossing each other and that is cancelling as J naught. We also saw the forward bias condition I am sorry reverse bias condition where you make this plus. Now, notice the entire voltage appears across the equation here. So, the barrier height which was V B I on this side has increased by V B I plus V R. So, the potential energy of this conduction band here brought down by an amount equal to V R. So, the energy distribution because the electrons do not have kinetic energy here so you reduce the potential energy. So, kinetic energy is up to that particular distance fixed by the temperature. So, these electrons find a much bigger barrier much more than what was there in thermal equilibrium. Therefore, these electrons from semiconductor are not able to cross to the metal. Now, on this is a very important point we must note on the metal side the barrier heat has not changed. This is the difference between the p n junction and the metal semiconductor contact. In the p n junction the total barrier heat changes potentially variation takes this both on the n side and p side. There is no drop in the metal in the case of there is no voltage drop in the metal in the Schottky barrier contact metal semiconductor contact. Since there is no voltage drop the potential here is same the barrier heat is same thing that is the important thing. So, whenever we apply voltage to a metal semiconductor contact if I make the semiconductor plus the barrier heat on the semiconductor will increase preventing the injection of electrons from the semiconductor to the metal. So, whatever you get the current current direction is from plus to minus like that I not that is due to the these electrons which are having energy above the barrier on the metal side. So, these electrons still can cross these do not get confused from the arrow the arrow is for the direction of electrons a direct current. Electron flow is from here to here by convention you know that negative charge if it moves in x direction current flows in a minus x direction. If plus charge moves from the minus to plus current direction is also same thing. So, I m s the symbol is current is I due to transfer of electrons from metal to semiconductor that is the notation. So, this has not changed because this barrier heat has not changed distributes has not changed. So, I not remains the same thing. So, you get a current whatever you got I not there previously whatever I not was coming due to this transfer of electrons from the metal to semiconductor was getting balanced by the current injection from the electro injection from the semiconductor to the metal. But, since that is removed there is only current component due to the transfer of electrons from the metal to semiconductor that is I m s going in that direction and that we have estimated how much it is a star t square d 2 power minus 5 E n by k t. So, this you can say it is like a P n junction almost the magnitude of current will be different decided by the barrier heat 5 E n in this case. So, the in the P n junction reverse bias means n side positive with respect to P side. Metal n type semiconductor same notation if semiconductor is made plus the current flow is from is less and that is the reverse saturation current. Now, let us see what happens if you forward bias that, but please remember whatever bias I apply to this metal semiconductor contact this barrier 5 B n is not going to change. So, the electrons injected from the metal to semiconductor totally decided by how much is the barrier heat, barrier heat. So, how much if the barrier is very low almost all of them will be available for current flow. If the barrier is high very little will be a current flow in this case. Now, forward bias again remember this would not change. If a forward bias thermal equilibrium this 5 a thermal level was the same level see the previous case the thermal level has come down what is the thermal level because entire energy levels have come down, but here on the thermal equilibrium you have made this plus on the metal side this is minus. So, the potential barrier across this which was plus minus because of the plus minus on opposite side that is reduced by B applied. Now, what happens earlier this distribution was just coming up to this end. Now, because of the forward bias the potential energy at the conduction bandage has gone up. So, this distribution remains the same thing. So, some of the electrons which are above this particular line dotted line see dotted line above this barrier can go from left to right electrons whatever electrons are there from the dotted line up to the peak they can cross from the right. So, that there are more electrons now crossing from semiconductor to the metal. Now, how will this current vary? This current varies as J is current due to the transfer of electrons from semiconductor to the metal actual direction is from current flow direction is from metal to semiconductor electron flow is from semiconductor to the metal that is why this is given like that. That is number of electrons on surface into velocity, velocity thermal velocity root K T by some number that we have seen already. So, only n s number of electrons actually will be whatever carrier concentration here is there into e to the power of the potential difference and n 0 is the carrier concentration here where neutral region into e to the power of minus Q barrier now originally it was V B I. Now, it is V B I minus V because of forward bias by K T. So, what is n 0 e to the power of minus V B I that was electron concentration which was present under thermal equilibrium situation at this point. Now, that is raised by an amount equal to V to the power of V B I 2 P by V T. So, what you have to remember is whatever thermal equilibrium carrier concentration was there here it has raised by e to the power of V by V T. So, current has gone up whatever thermal equilibrium J naught was there has gone up by an amount of e to the power of V by V T. So, what you have got is this current due to transfer of electrons from the semiconductor electrons transferring from the semiconductor to the metal that is I m s that remains the same thing, but I s m that is current due to the transfer of electrons from the semiconductor to the metal transfer is like that current flow in that direction that has gone up exponentially. How much it has gone up depends upon how much is the voltage. So, what you say me imply is this current is J naught e to the power of V by V T like in the case of forward bias P n junction diode whatever J naught is there e to the power V by V T and this current is due to that that is presented in the opposite direction. So, total current is from the metal to the semiconductor in fact from the plus side it comes like this battery is driving the current that is J s m current density is J s m minus J m s J s m is J naught e to the power V by V T and current due to transfer of electrons from the metal to semiconductor is J naught itself you subtract that. So, the current is J naught into e to the power V by V T minus V T plus 1 this is what I have written here like exactly like the case of diode there is no difference. The only thing is the J naught term in this case will be different from the J naught term in the diode case we will see what it is. Once again remember this barrier heat has not changed because of forward bias because of forward bias the current flow has increased because of increase in electron injection from the semiconductor to the metal. In the case of p n junction the current increases because of whole injection from p to n and electron injection from n to p. But in this case whatever electron injected remains the same thing that is this quantity remains the same thing J naught. So, it is the this component which is actually a kind of due to increase in the electron injection from semiconductor to the metal. In the reverse bias direction only that J naught is there which is due to the injection of electron from the metal to the semiconductor there is nothing from semiconductor injected. Please note if I have to have electron injection from the metal to the semiconductor whatever electrons are there above this barrier can only inject. If I want to increase that electron current injected this is important for MOSFET action. In the MOSFET n channel your n plus source which injects electrons here your metal if it has to inject electron to the channel n channel this barrier must be small because only those electrons which are above the barrier can get injected whatever is here is going in the opposite direction. Let us see that now if it is a reverse bias if the potential you reduce here this cannot inject electrons back this is the theme of basic ground for the MOSFET. Now just see the I V characteristics J naught depends upon J naught depends upon H dot T square root minus 5 B N by K T. These are constants for a given temperature it depends upon 5 B N. So, I V characteristics it is a rectifying diode J naught is small if the barrier head is large number of electrons which are having energy above that barrier is small. So, high barrier head which you get when phi m is larger than phi of s for n type semiconductor you get a rectify you get I is equal to I naught into a V by V T and I naught will be low will be low because I B N is large that is the J naught and the current is exponentially increasing. Now if the barrier head is low the reverse saturation current is large I naught is large if I naught is large I naught into that quantity is large. So, this is the characteristic of this diode with a smaller barrier head. So, if a larger barrier head is there you get good rectifying junction very large barrier head you can get the current reverse current close to 0 or none over to pico. High barrier head there will be large current in the forward direction exponentially increasing reverse bias current will saturated J naught a large value of J naught. I naught is high when phi B N is low which is the one that you may be looking for in a MOSFET action because I naught in the reverse bias condition is due to injection of electron from the metal to the semiconductor. Now, let us have a more detailed graph there are number of graphs I have plotted here large phi B N I naught is low as increase the phi B N reduce the phi B N reverse current keeps on increasing in the reverse direction this keeps on increasing on the positive side even the smaller v applied voltage will give large current smaller phi B N the current is like that in that same graph I shown number of them keep on reducing phi B N you get more and more current and you see ultimately when phi B N is very low if you look at the characteristic at the source end that is a most linear there that is voltage drop is very little you can have large current flow that is close to an ohmic contact. So, phi B N very small indicates that you get a ohmic contact phi B N large gives you rectifying contact. Now, what you have said is phi B N is phi M minus chi. So, if phi M higher you must get lower barrier head more closer to rectifying ideal rectifying characteristics. So, now, let us see the next slide comparison between the P N junction and the metasemical contact this is a very good comparison property I V characteristics you take both of them have the same type of same type of expression J equal to J naught e to power v by v t minus J equal to J naught e to power v by v, but what is the difference in the characteristics? J naught in this case is S star reticence constant is 120 into M star by M naught in silicon it is almost equal to 120 or 120, gallium arsenate is about 8 t squared into exponential this quantity. Now, in the case of semiconductor it is proportional to N i squared. So, I have got a constant which it is depend is decided by the doping etcetera, but the exponential term N i squared is e to the power of minus e g by k t. Now, let us see the mean contributor to the reverse current is this exponential term saturation current in the case of P N junction it is exponential e to the power of minus e g by k t and for silicon e g is 1.1 electron volts for gallium arsenate it is 1.43 electron volts. So, you can immediately say that the J naught is between the silicon and gallium arsenate junction gallium arsenate junction will have much lower much lower J naught because e g is larger there, between metal semiconductor contact and this P N junction if you take silicon e g is 1.1 and barrier height will be let us go back and see thermo equilibrium if you see the barrier height pi B n is much smaller than the band gap pi B n is this much band gap is you can see from here to here. So, even if this permeability comes down here it will come closer and closer to e g, but always less than e g. So, if you go to this particular expression where we have comparison this is 1.1 electron volts this is something like 0.6, 0.7 electron volts. So, this will be this particular quantity will be larger compared to that e to the power of minus 1.1 by k t e to the power of minus 0.7 by k t. So, this will be larger. So, always the reverse saturation current in the metal semiconductor contact will be higher than that in the case of P N junction. Now, current in forward bias condition in the case of P N junction current in a forward bias diode is due to injection of holes from the P type to N type, injection of electrons from the N type to P type. The barrier height for both holes and electrons will be reduced in the case of P N junction. In the case of metal semiconductor contact the current flow in the forward direction is dominated by injection of electron from the n side region to the metal. So, 5 B N does not change. So, the electron injection from the metal to semiconductor does not change it corresponds to the J naught term. Now, if you want to see the different this is the thing that I can you can see it one shot metal N type semiconductor thermal liquid will be like that. Forward bias more electrons are injected from semiconductor to the metal. Whereas, the electrons injected from the metal to semiconductor do not change that is this arrow to that smaller arrow. And when you go to reverse direction bias you make N region plus nothing no electrons are getting injected from the semiconductor to the metal because they do not have the energy to go above the barrier. All that you have current is due to injection of electrons from this metal to semiconductor that is J naught itself. So, J naught J naught e to power V by V t no current. Now, before you go further into this I just want to show how the currents in various devices will be. We draw the voltage versus current. If I compare P N junction silicon with P N junction with the metal semiconductor junction rectifying P N junction with silicon will be going like this exponential reverse very little very little going to 0. And in the case of metal semiconductor contact what will happen? Barrier is low. So, current is more reverse current is more I am sorry reverse current will be more passing through that it will go like that and it will go like this. So, these two are not same scale forward and reverse this is the metal this is the P N junction. Suppose P N junction silicon this is silicon this is metal silicon. Suppose I take gallium arsenide the gallium arsenide P N junction that will be like this. The E g is larger therefore, reverse saturation current will be even smaller that is the gallium arsenide P N junction. So, if I make a P N junction in gallium arsenide it will have more cutting voltage cutting voltage of almost about you know silicon it may be about 0.7 volts gallium arsenide it may be close to about 1 volt 1 electron 1 volt. So, you will have the that is voltage. So, this is gallium arsenide. Now, if I make a metal semiconductor to contact in gallium arsenide that will be somewhere higher than this because you can get higher barrier height in that compared to the silicon. So, that will be m s contact with gallium arsenide. So, in the gallium arsenide you can make metal semiconductor to contact much more effectively than in silicon. If you want to make J-fred they use metal semiconductor to contact in gallium arsenide because you can get cutting voltage something like in this case V gamma cutting voltage for that case will be something like about 0.6 volts close to that of cutting voltage of a silicon P N junction this one this is a gallium arsenide metal semiconductor contact that is because V g is higher you get a barrier height which are higher than what you get in the case of silicon. So, that is the idea that you can think of. Now, let us get back to this. So, having said that if you want to inject electrons into N channel you must have smaller barrier height that will go into that N channel MOSFET if you want to make. In that case of conventional jet on a schematic diagram in the case of conventional MOSFET this is a channel blue is the channel red is the oxide above that I have the metal and P type substrate I am talking of silicon junction silicon MOSFET N plus source N plus strain N channel when I apply plus voltage you can invert it. Now, when I apply plus voltage to the gate there is a depletion layer below this and this potential is at plus twice phi f when it is inverted surface potential is plus phi f that mean between the ground and this channel beginning there is plus twice phi f this junction is forward biased that means the barrier whatever barrier was there originally I am just showing it still as vertical, but it will go that barrier will be whatever built in potential is there minus twice phi f. So, originally the barrier was built in potential and if there is no inversion if there is no voltage appearing between the source and the channel barrier height between the channel and the source with built in potential. Now, because there is a plus twice phi f voltage here because of gate voltage at the channel potential difference there is V B I minus twice phi f forward biased as I was mentioning earlier when you invert the channel the source is getting source junction is getting forward biased by twice phi f. So, barrier height is V B I minus twice phi f that means the whole electrons can easily get injected across that I am just showing still that distribution of electrons from the permeable up to that like that in N type region. So, you can see a lot of electrons get injected here if I increase the twice phi f what a bit of electrons can be injected into channel. Now, from here to here when I apply a drain voltage V D S whether there is depletion layer at the drain end or not there is a drop or there is from this source end to the gate end there is a voltage rise. There is gate end is at plus voltage with respect to source end that means if I draw the energy band diagram that will be going down like this plus is at lower end compared to minus which means this is plus this is minus I put it at lower electron lower energy in electrons because plus and minus here electrons can be easily go from the minus to plus it can attach to plus. So, that is another way of looking at that. So, whatever electrons are injected here go down into the along the channel the drift electric feed from minus to plus and they collected. How much current flow is there here depends upon how much charge is present here because of the gate voltage also how much is the barrier heat here the barrier heat is reduced by phi f that is quite a large reduction in the barrier heat. So, entire current is controlled by the gate voltage more gate voltage more charge and whatever is collected here can be supplied from the source end because there is a reduced barrier heat is there. Now, let us take a look at the metal semiconductor contact on the right hand side see it is very interesting to compare this this is what makes makes it difficult for the technologies to realize metal source metal source drain metal source source metal drain drain junctions metal contact here. Here as you already have seen whatever I do on between the channel and the gate source that barrier heat whatever was there phi b m is not going to change that is going to remain in the same thing. Whatever I applied minus twice phi f I do here see in this case the minus twice f was reducing entire barrier in this case if I do minus twice f that is reduced here only this barrier heat is reduced here on the n side. So, this is reduced plus minus here and that is across this depletion layer there that is small depletion layer that is whatever twice phi f is appearing across that nothing here. So, that means number of electrons available here are fixed by the barrier heat phi b m this distribution is fixed by temperature. So, on the barrier heat if it is a rectifying contact the barrier heat is high. So, this number of electrons available for transporting from the metal to the semiconductor only in that small region. So, here whatever charges are here this portion is same as this there is a voltage drop from here to here because of current flow. So, the energy band diagram is like this. So, electrons reaching here will flow in that direction get collected by the range that no issue on that thing because there is downhill there you can be collected here. So, now the point that you would see here is if this is not able to supply whatever charges are inverted here it is getting collected, but the rate at which the current can flow is ultimately limited by how much is this can supply. If the barrier heat is let us say almost like there very high nothing can collect. There is no injection of carriers because of barrier for electrons in the source region is limited by the barrier heat and I am sorry number of electrons that can cross this barrier depends upon the barrier heat more the barrier heat more electrons are available. So, the moral of the story is if I want more number of electrons available here to be injected I must have this barrier heat reduced 5 B N must be lower and 5 B N is as you keep on getting lower and lower ultimately what you saw was you get here. If I keep on reducing the barrier heat if you see this graph here I will be characteristics I move along this line ultimately you get ohmic contact. So, ideally what you would like to have here would be you would like to have this as a ohmic contact or 5 B N as low as possible 5 B that is the barrier heat with respect to the channel not with respect to P region what we are concerned is the barrier heat between the metal and the N region which is the inertial region. So, what we are looking for is metal semiconductor contact at the source end which will which will give 5 B N as low as possible. Now, let us see whether you can get that by changing the work function. If you look at the this characteristics here, here I just go back to those slides see here. Here you had taken 5 M much larger than 5 of us that is the metal which has got large work function means the Fermi level is at a much lower level compared to the Fermi level of the semiconductor. 5 S small means the work function of the semiconductor is smaller between between the vacuum level and the Fermi level in semiconductor the energy difference is less compared to the energy difference between the vacuum level and the Fermi level in the semiconductor. So, now let us see what happens instead of I want to reduce the barrier heat. So, that there is supply is there what will happen if I get 5 M less than 5 of us. See, if this 5 M were large that would have been here that would have been here then the electrons would have got transferred from the semiconductor to the metal giving a depletion layer that give to rectifying. Here 5 M is small compared to what is 5 M from the vacuum level to the Fermi level that is smaller compared to this difference between the energy of the vacuum level and the Fermi level in semiconductor. So, when you connect them together in the system what you will have will be electrons will get transferred from the metal to the semiconductor till the two Fermi levels are equalized. So, that is the situation electrons have got transferred from the semiconductor to the metal. In metal there are already large number of electrons. So, if you transfer electrons to the semiconductor to the metal there will be an accumulation layer. See more number of electrons have got transferred from the metal to semiconductor there will be a small charge sheet that is no depletion layer. If it has to be depleted the electrons have to be transferred from the semiconductor to the metal. But now because of the higher Fermi level here electrons get transferred from the metal to the semiconductor. So, this surface gets more negatively charged this surface gets more positively charged that is the situation. So, you have got plus charges here and those plus charges you can put it as the electron density here are plus charges. But electron distribution is like this in the metal electron distribution semiconductor is like this. So, that is in thermal equilibrium there is no barrier here because this is accumulation. Accumulation means very little band bending you know that accumulation we used to take phi s is equal to 0. But it is not 0 slightly negative here. So, very small barrier if the electron very small bending. So, barrier heat only this much phi b n can be made very very small again determined by the difference between the phi m and chi f s phi f s phi m minus chi f s is negative. So, barrier heat in this much small only. So, the electrons here all so many electrons have energy above that here so many electrons are there. If I apply a voltage to this either way if I apply either way voltage there is very small movement here. So, there can be large current flow in the it is difference between the two is pre-escape this there was a big barrier here. Here there is no barrier for either case very small barrier. So, if I apply plus here or minus here there will be large transfer of electron from metal to semiconductor to semiconductor. So, this is the way to go. So, what you are telling is why do you choose metal with high work function choose a metal with low work function. For example, if I take aluminum work function is 4.05 and phi f s in semiconductor is 4.05 practically no barrier. So, you can be very happy if everything is fine you will say that the you are you can have ohmicont at if I have aluminum on to a surface of the n type semiconductor or as a source of the n channel MOSFET. So, now let us see if I have p type substrate because after all if you take n type substrate you get when invert you get p type surface. If I take p type substrate you get this type of n channel. Now, let us see what about the p type substrate. In fact, whatever we have discussed I am not getting down to this supposing the phi m is less than that of phi f s like in this case in the p type if it were p type what would happen formula will be even below phi m is less than phi f s. Electrons will get transferred from the semiconductor metal to the semiconductor if it is p type thing if electrons get transferred from the metal to the semiconductor there will be depletion layer. So, if phi m is less than phi f s you will get a depletion layer here and that will be rectifying. So, thing to remember is if you if when you make the contact if there is a depletion layer in the semiconductor you can change that depletion layer width and the potential barrier in semiconductor you can go you can change the barrier height in semiconductor that is rectifying. If I have a phi so, if there is no depletion layer if there is a accumulation layer then that is the only contact. So, exactly by the same argument I am not getting down into that you can determine figure in this case it is transferred from this metal to semiconductor introducing depletion layer that will give us the depletion layer. So, by applying a voltage you can change the barrier height so, it will be rectifying contact. If phi m is greater than the chi of s if phi m is greater than phi f s in the case of p type material this is way down you will transfer from there to there or you will get a you will get just like exactly reverse in the n type semiconductor if phi m is greater than phi of s you have got electrons transfer from the semiconductor to from the metal to semiconductor causing depleting. The opposite is true in the case of p type semiconductor if the phi m is larger than the phi of s you will have accumulation layer in the case of p type semiconductor. So, the sum up to summary of this discussion phi b n is phi m minus chi that is the ideal and if phi m is chi phi m minus chi is chi so, rectifying contact in n type semiconductor. What is phi b p I do not know whether I pointed it out to you let me go back to that and show you this is phi b p you take a look at this from if there is a hole here there is a electron is absent there to take the hole above this to the other side that has to cross this barrier height. So, the important thing to understand is if an electron is at this end to take it here you have to spend energy. So, a hole at the near the fermi level if you are talking about taking it to the above this point up to this point is equivalent of taking an electron from this point to that end that means to take an electron from here to here you have to you have to spend energy that is higher energy for electrons same thing if I have to take a hole from here that is fermi level to this edge of the gap barrier height you have to spend energy. So, this is barrier for holes and here the barrier for electron is that. So, when you go back to other diagrams here this is the barrier for the electrons and that is the e g band gap band gap minus phi b n band gap minus phi b n that is phi b n is that total is band gap that minus phi b n is phi b p. So, that is if you do not get it see if I have the energy band diagram like this the energy band diagram is like that that is n type. Now, that is the barrier height that is the barrier height for electrons and this is phi b p barrier height for holes because it is difficult take you have to spend energy for plus charges that is this is the higher energy for electrons that is higher energy this is higher energy for holes. So, this is the phi b p and phi b p you can see is the band gap e g phi b p expression electron volts minus phi b n. So, you can see that if phi b n is high phi b p is always low if some process gives you higher phi b phi b n that gives you lower phi b p this is something which you must remember. Now, let me go further down. So, phi b n if phi m is larger than chi phi b n is high at the same time you can see phi m is larger than chi phi b p is low because phi b n is high phi b p is low. So, you get phi m is larger than chi phi of s or chi you get rectifying contact with the n type material or n channel, but then for the same work function difference phi b p because if phi m is large this is e g minus that large quantity will be small phi b p will be small that will be omicontact in the p type substrate. So, if you get very easy to remember if you get rectifying contact with the p type material n type material or n channel device you will get omicontact in the p type material or p channel. So, low phi m phi m minus chi low value of phi m and chi because low phi b n that gives omicontact on n type material rectifying contact on p type. So, now what will you say? So, if I want to make a this ideal theory tells us that I would choose a metal with a lower function to make omicontact with the n channel because that can inject lot of carriers. If I have a p channel I will make a high work function material here that is ideal. Now whole thing is under the assumption that phi b n is phi m minus chi that was the most ideal case. Now let us get down to our requirement experimental results on phi b n for metals with different phi m. You can see this slide you did not see that measurements on phi b n there is some ways of measuring by the I v characteristic etcetera or C v I phi b n showed that the first order theory whatever we have mentioned is not correct phi b n equal to phi m minus chi is not satisfactory. That is the phi b n all these thoughts are measurements the crosses are the measured phi b n by making the measurements on metal semiconductor contact either C v or I v it will give you that you can understand that. So, what they did was they had silicon which was freshly cleaved you can just cleave a silicon material you can just apply some pressure it will cleave you cleave along a particular plane and freshly cleaved means what if I have the thing which is cleaved I have got the surface which is free I have the surface of the semiconductor which has been just cleaved that means there are a lot of dangling bonds there that will tell you that that surface has lot of surface states it is not ideal ideal case there are no surface states when I discuss the ideal metal semiconductor ideal metal semiconductor contact theory assumption was that there are no dangling bonds there are no surface states when you have an oxide layer there it is the same surface state which you call it as call it as a interface state because it is the same surface state manifests that as the interface state many of the dangling bonds are satisfied by the oxide there are still some of them which are not satisfied that is called interface state. So, freshly cleaved absolutely see this is one the x axis here the work function corresponding to aluminum, silver, copper and gold all this cross x. So, the phi B n is about 0.8 almost about two thirds of E g. So, that means once you have fixed the E g for silicon that was fixed that is freshly cleaved thing, but it gives a clue that there are large dangling bonds which were responsible for that the surface states were responsible for this. So, if you are able to remove those then you are in business. Now, what they did was let us see that let us cleave the surface give a chemical treatment for example, if you clean it with H n H c l H 2 O 2 solution and ammonium hydroxide and H 2 O 2 solution then what you get will be a surface which has a layer of oxide a thin layer of oxide will be there even when you put it in H n like that R c 1 R c 2 because of that H 2 O 2 action there will be some amount of oxide. Now, when a very, very thin layer may be less than a nanometer 2 angstroms, but nanometer is 10 angstroms couple of angstroms that is sufficient it passivates some of those dangling bonds. So, when you do that and put a metal you can see that where it has gone down is changed and as you increase the work function you expect phi m minus chi, chi is same for silicon phi m minus chi increases at increase and that increases linearly. We are very happy that it is obvious the law phi minus chi, but the change in the 5 B n say from 4.2 you change it to 4.7 or 4.75 around that for gold. Let me take 4.2 to 4.7 that is that is about 0.5 I should have got as per our theory the 5 B n should have increased by the same extent that is 0.5 electron volts, but if you see here it is about 0.5 here it has gone up and become 0.8 it should have increasing by 0.5 electron volts it has gone up only by 0.3 still not satisfactory it changes with phi m as we are happy that first order theory is ok, but it does not change as much as phi m. Let us see other materials is it characteristic of silicon alone same thing they did with gallium arsenide gallium arsenide results on cleaved 1 1 0 gallium arsenide cleaved there are slight dependence phi m varied from about 4 right up to about 4.8 it change hardly 0.2 electron volts. So, when it when you change it by about 1 electron volts closed 1 electron volts phi m your 5 B n changes hardly about 0.2 electron volts, but again the indication is there on the dependence of 5 B n on phi m, but not as much as you expect chemically prepared thing see here this is just freshly cleaved. In the case of silicon it was flat here it is slight dependence over there may be the substrate number is less than that 1 1 0 ok. Now, another surface 1 1 1 1 or it is a family there they chemically prepared chemical treatment you give ok. Then when 4 2 4.8 when you change it increases from instead of 0.8 electron volts increase in 5 B n it increases something like about 0.3 electron volts. So, still almost a linear variation is there with phi m, but not as much as you expect everything points out that if there are interface state densities are high the 5 B n will become independent of phi m, but if the interface state densities are reduced to some extent by some passivation it will become proportional to phi m, but not as much as there is a proportionality constant a factor which is not equal to 1, but than 1. Now, let us see. So, we have seen silicon we have seen gallium arsenide let us see germanium ok germanium if you see why we are taking a look at germanium is this is the one material which they are trying to use for making the MOSFET with metal semiconductor contact. They are using germanium because of its much higher electron mobility compared to silicon much higher whole mobility compared to silicon. If you go to gallium arsenide electron mobility is much higher compared to that of silicon whole mobility is not it is almost same as that of whole mobility silicon, but in germanium you get both of them by a factor of 2 to 3 higher. So, let us we will have made metal semiconductor contacts with germanium varying the work function from right from about 2.5 to about 5.5 or more platinum from cesium metal to platinum. You can see there all over the squares are the measured see put the thing together it does not follow any rules, but I can draw or at least if it is 5 m minus chi this should have been the dependence taking a chi of the minimum that is also about 4.05 close to that of silicon and if I had 5 m you can see 4 and 4.5 means that should have been somewhere 0.5 electron volt this point 0.5 it was corresponding that 4.5 minus 4.05 that is 0.5 would have been the barrel head you do not get that. If I take 3 electron volt I must have got it somewhere down here, but you get almost like this. So, just some quick thing we will get back to this later on to examine that I can draw a straight line like this then it is almost flat compared when I am using hoffnium titanium nickel etcetera it is almost flat close to about slightly less than 0.6 electron volts. Remember the band gap of germanium is 0.66 electron volts in the case of silicon we had 1.1 electron volts and you had the barrel head flat coming to about 0.8 electron volts that is about two thirds of that in the case of germanium band gap is 0.6 electron volts. So, what you get is slightly less than 0.6 electron volts that is below 0.6 electron volts, but I can draw a flat line like this or if I am more optimistic let us say it is dependent on phi m I can draw this line leaving out hoffnium and nickel and tantalum I can draw this line straight line and say there is still some phi m dependence. So, this is the phi m here and this we will come back to this is the band gap e c minus e v and phi m is there for that case see very high this thing here. So, this we will discuss there is a neutral level neutral level is very close to valence band that is indication of that. So, these two this is actually the ideal theory the dotted line that theory was given by Bete and Schottky and barrier height was called Schottky barrier, but whatever phi b n we are talking of was called as Schottky barrier because of name of Schottky who named it. So, that is the Schottky limit you would have got phi m minus chi means it is Schottky limit and Bardin said it will be independent of phi m it will be having certain value decided by a neutral level in the permeable in the permeable surface. So, that is the Bardin limit due to strong pinning limit surface pinning permeable pinning limit all that what is there we will be able to discuss later. So, from these things one can fit in a graph like this I will just discuss this how this comes up and what are the implications of that more details we will take up in the next presentation. So, what we have concluded from these experimental results is phi b n can be expressed by a relation ideal case it is phi m minus chi in the practice I can express phi b n as gamma times phi m minus chi plus 1 minus gamma time p g minus phi naught because we saw that depends upon e g minus phi naught. So, gamma equal to 0 or gamma equal to 1 second term is 0 gamma equal to 1 phi m minus chi that is the ideal case first order theory. When gamma equal to 0 first term ideal theory is gone and 1 minus gamma equal to 1 e g minus phi naught. So, we can say that when high freshly glued surface gamma is equal to 0 therefore, first term went off and you got a term which is independent of phi m e g what is that phi naught we will see in my next discussion and gamma in between the two. So, what we said is ideal theory it is phi m minus chi non ideal theory worst case gamma equal to 0 and it is depends it is independent of phi m and you get e g minus phi naught in between you get gamma is less than 1. So, something 0.5 times phi m minus gamma 0.5 times e g minus phi naught like that. So, in summary that is what we are telling you can express it in this fashion the details of all these things how it comes up due to the surface states or interface states we will just discuss in our next presentation complete details of that how to overcome that how to overcome some of these problems and how to use it we will see in the next presentation.