So we shall continue now. If you recall what we had learned in the last class in semiconductor devices is that we had learned about doping, things like Fermi distribution, density of states, all those things we are clear. We also had solved some very simple numerical of P and N doping together, right.
One is called minority, one is called majority. So we will continue on the discussion in doping, temperature dependence of your carrier concentration. Things like high doping effect, okay and from there on we will try to move to scattering mobility and other things in the subsequent lectures, okay. So, we will recall what we had last where we had last stopped, okay.
So, we will go to the slide or the page where we last finished in the last lecture and I will bring that up here. So, you can see that in the last class I told you that you know the The doping, whatever you are doping that becomes the majority. For example, if you are doping silicon n type, then you are doping you know with phosphorous or arsenic.
So, that is donor and it is a majority. Suppose you are doping 10 to the power 17 per centimeter cube, that is your say n type doping density, it is called majority. And the minority carrier concentration can be found out by this simple formula. It is the principle of mass action, which is the product of p type and product of n type, the total number of electrons and total number of holes. has to be constant in equilibrium which is equal to square of the intrinsic area concentration.
So intrinsic area concentration is 1.5 into 10 to the power 10 I told you right you know that we derived that. So, if n type concentration is 10 to the power 17 then your number of holes is actually 10 to the power 3 only almost ok. So, that is called minority.
So, it is an n type semiconductor this is a very simple way of actually doing. The finding of the electron and hole concentration. Now another very important thing will come now here is that suppose you have a piece of semiconductor say silicon you are putting some n type dopants you know nd and also you are putting some p type dopants in the whole thing okay. You are putting some n type doping here also you are putting some p type doping in the hole.
So what will be the net concentration will it be electrons or will it be holes because you know an electron in a conduction band. And a hole in the valence band if there is an electron here if there is a hole here they can recombine right they can recombine and they can release the energy right. So, how many electrons will be there how many holes will be there. Suppose I have put tentative power you know 17 doping of your n type doping and suppose I have put you know 1.5 into tentative power 17 per centimeter cube of boron. So, it will give you p type doping.
So, of course, I told you that if I put 10 to the power 17 donors then I will get equal number of electrons. So this also will become equal to electron. If I am giving so many of p type doping or acceptor doping then I will get this many of holes right. So how many of these will be electrons and how many I mean how many electrons and how many holes will eventually survive because many of them will recombine amongst each other know between each other. So to find that we have to understand a very important or we have to you know invoke a very important.
A condition which will be always valid in semiconductor physics and that is called charge neutrality okay. In equilibrium the total positive and negative charges have to cancel each other okay. So the net negative and positive charge have to remain totally 0 that is called charge neutrality.
So suppose I have a phosphorus atom I told you right it will give it has 5 electrons and then 4 electrons it bounce and 1 electron is free that goes away right. When it donates an electron when a donor donates an electron. It leaves behind a positively charged ionized core right. So, I told you that if we have N D, if we have N D number of donors then after giving all the electrons and so N number of electrons will come which is equal to N D, what remains behind is N D plus number of ionized donors which means suppose N D is equal to 10 to the power 16. I have added 10 to the power 16 per centimeter cube of say phosphorus atoms I have added.
So what will happen is that these are donors right they will donor electrons. So after they all donate electrons what remains behind what remains behind is 10 to the power 16 per centimeter cube of ionized phosphorus atoms will remain. I will call them ionized donor atoms right because they have given away the electrons so what remains is positively charged ion. So, this many positive charges remain because the ionized cores remain and also 10 to the power 16 per centimeter cube of free electrons exist now of free electrons exist now because all the atoms phosphorus atoms have given away the electron the extra electron. So, you get you know 10 to the power 16 per centimeter cube of free electrons in the sample these are an electron is a negatively charged particle.
So, you have 10 to the power minus 16 negative 10 to the power plus 16 positive which is the ionized charge core atoms. ions and so the total charge is 0 but electrons will be there because electrons are not recombining with the donor ions there are no holes here okay. So essentially what it means is that ND plus number of positively charged ions will be there right positively charged ions will be there and then N number of negatively charged electrons will be there I mean electrons are negatively charged that we all know right I mean that is we all know.
So, negatively charged electrons will be there and under charge neutrality it is 0, so N will be equal to ND plus ok. So, the total charge is 0, this many free electrons are there that can move and carry current, these are immobile positive ions are there which have come because they have donated the electrons away. Similarly, if I have a P-type material I give N A mine N A of you know acceptor ions acceptor atoms then they will give the holes.
So, P number of holes will be there which are positively charged right they are positively charged but Na will give holes they will take electrons so they are this many number of negatively charged ions are there right and equilibrium P will be equal to Na minus right. this is positively charged holes and this is negatively charged your ionized course okay this is negatively charged ionized course. Now this is only for this is only for p type material right this is only for p type doping right and this is only for n type material n type material the question is what happens when we have both right suppose what I am trying to say is that I have given both electrons and holes know. Suppose I have given N D number of donors and N A number of acceptors together, then what will happen?
Then the charge neutrality has to be written, ok. The charge neutrality condition has to be written first. What is the charge neutrality condition? The charge neutrality condition will say the total positive charge. What is the total positive charge?
One is holes. Holes will be positively charged and N D plus, sorry, N N D plus, this comes from, this is holes. which is positively charged and this is your ionized course of the acceptors right ionized course of acceptors which sorry donors sorry ionized course of donors because they have donated the electrons. So, that remains all remains is basically positively charged I told you that should be equal to the total negative charge which is n plus n A minus because this comes from the ionized acceptors.
This comes from ionized acceptors and this is free electrons of course. So, what I can write is that I can say that total P plus ND plus should be equal to total electrons plus negatively charged ions. This is your charge neutrality condition and from this you find out everything.
If it is n type dope only then the P type doping is not there. So, this will not be there and this will not be there that is true and if you are electrons only p type doping is there then this will not be there and this will not be there right. So that is basically it.
So now we know how to do it. So let us see now I know that so I will write the equation again p plus Nd plus is equal to N plus Na minus but I know that p and in equilibrium is this is all equilibrium I am talking about okay is equal to Ni square right. So what I know I know that p is equal to Ni square.
by n why do not I substitute that back here I will substitute that back here. So, I can write this equation now as p can I can write as n i square by n. plus ND plus okay plus ND plus equal to N plus NA minus. Of course you know ND plus is equal to ND only because if you have given tentative power 16 say per centimeter cube of phosphorus atoms as donors then tentative power 16 per centimeter cube of you know plus positively charged ionized donor will be there. So instead of that ND plus ND minus I can just write it ND and NA this is the same number you know.
We are assuming 100 percent they are ionized. So, what I can write now here is I can multiply both sides by n. So, it will become n i square plus sorry plus n into n d equal to n square plus n into n a, ok. So, what can I write here now?
So, what I can write here now is that n square plus n n a minus n d sorry not plus minus it will be minus ni square is equal to 0. So, does it look like x square plus ax plus b equal to 0, so quadratic equation. So, let us see how we solve the quadratic equation. So, I have n square plus n, n a minus n d plus sorry not plus minus minus ni square is equal to 0. So, n will be equal to minus N A minus N D depends which is the larger here right plus minus the total number has to be positive. So, it cannot be you know the negative it cannot be a negative number N cannot be negative, but the reason I am keeping here plus minus is because this quantity may be negative you know you never know.
So, anyways will be equal to N A minus N D whole square plus 4 N I square by 2 that is your solution right. So, if you see the expression here N A minus N D is there. If one of them is very large, N A is very large then compared to N D then we can just you know suppose N A for example I am telling you a doping situation where I am doping both N and P type doping are coexisting okay that is why we are doing that. If both N and P type doping do not coexist only one type doping is there then we do not care about it because then your N is equal to N D only right.
Only when they are both types of doping this situation this equation is valid right I mean makes sense right. So suppose I am doping holes 10 to the power 17. per centimeter cube P type and suppose I am doping N type only 10 to the power say 15 per centimeter cube then NA minus ND this equation here know this NA minus ND even here for example will be 10 to the power 17 minus 10 to the power 15 which are almost equal to 10 to the power 17 only it is like 100 minus 1 right. So then you can neglect ND and then you can have a simplified expression for example then you will have N equal to minus NA right. will be N A only no there will be nothing there so plus N A minus this will be much smaller this will be 10 to the power 15 or 10 to the power you know square of 10 to the power 20 for example right this will be 10 to the power 24 into 10 to the power 20 approximately and this will be equal to how much this will be equal to 10 to the power 34 because N A minus N B so this is negligible. So it will only remain N A here by 2 this is 0 so almost electrons will be 0 so holes will be.
to NA which is equal to 10 to the power 17 that is easy right. But if both electrons and holes are there then you have to take into account this expression and if you know your doping is low so that your disk cannot be neglected then also you have to use this equation for example okay. So it is important that we use that equation in other words this is a very universal equation which can give you the number of electrons and holes for a semiconductor where both N and p type doping are coexisting okay. That is one thing.
So the next thing we will learn today is that you know I told you that suppose I have only an n-type semiconductor let us not make it both n-type and p-type. Suppose I have only n-type semiconductor and Fermi level is here you know the and you know the expression if you do not remember you should remember the total number of electrons is given by n equal to nc exponential of Ef minus Ec by KT right. So Ef minus Ec is basically I keep telling you D spacing the negative of D spacing okay EF-EC is the negative of D spacing and NC of course depends on you know effective mass like 2 pi M star KT this is the Boltzmann constant by something like H square into 3 by 2. Of course this depends on effective mass of the material and each material has a different effective mass for example gallium arsenide will have a different effective mass, silicon will have a different effective mass, gemini will have different effective mass.
So if effective masses are different then NC also keeps differing that is one thing we should keep in mind. So this expression comes from the you know Maxwell Boltzmann equation if you remember call it Maxwell Boltzmann approximation to Fermi Dirac and that is how we get it. It takes into account both Fermi Dirac distribution as well as density of states.
Now what are we going to tell you you know here is that suppose I take. You know the donor ionization energy level is very close up here I told you right ED when you add say you know if you take a silicon and I am telling you that if I would say put phosphorous here then one electron is free that goes away you know that is loosely bound that is why that is free and it goes away and that loosely bound coulombic energy which the electron overcomes and goes away to become free is called the donor ionization energy. A similar thing also exists for acceptor ionization energy for p-type I am talking only of n-type here. So donor ionization energy is basically the energy required for the electron to let loose from this phosphorus atom and go and become an n type you know it contribute to n type conductivity.
So this donor ionization energy I told you is very small typically in the range of or smaller than around thermal energy KT. So that room temperature energy thermal energy KT is sufficient to break this you know bond and make this electron free and go away okay. So the donor ionization energy is typically very very small and the spacing and energy different energy level.
is positioned as ED very close to the conduction band it is know that it is basically EC-ED okay that donor ionization level is there is very small it is in the range of KT it can be in the for example KT 0.026 so this can be say 0.010 for example E log 10 volt which is very small this is much smaller than 0.026 room temperature. So at room temperature the thermal energy is sufficient to excite the electrons from the donor ionization energy. I mean you know this to loosen the electron from the host atom. Actually you can imagine that this donor ionization energies is also it is a donor level which is filled up with electrons and it is so close to conduction band it is so close to conduction band that the energy 10 milli electron volt is very small with which it is bound.
So, any room temperature fluctuation of energy which exist all the time will be sufficient to excite this electrons up here and so all the electrons will come here right and that is how they populate the conduction band in lot of electrons. and that is how conductivity changes and that is how doping actually changes your conductivity or conductance of the material that is why you know semiconductor is very awesome because materials and insulators cannot do that. So this donor ionization energy level exist very close to the conduction band, Fermi level of course can be close to that or even below that for example and then this is the valence band right.
Now we all know this things and the reason I am telling you this is again because let us zoom it very well so this is conduction band. Suppose this is valence band, this is silicon only okay and suppose this is your Fermi level and drawing it in a very zoomed way and suppose this is your conduction sorry this is your donor ionization energy level which is say E d of course it will be constant okay there is no slope here my drawing is not very good but you know it is there is no slope here. So your what is it called your donor ionization energy level is very close to conduction band very close okay it is close to almost you know less than k T. Now at room temperature at room temperature I told you.
That your energy the thermal energy Kt is almost equal to 26 millelectron volt and this donor energy I call it Ec-Ed this donor energy is say 10 millelectron volt. So room temperature energy is sufficient to actually excite the electrons from the donor level to the conduction band that is how you get conductance but the question is what happens this is the important question that we want to answer now what happens you know when for example. temperature is cool down you know temperature is lowered for example or a sample is cool down. So, see you see room temperature energy Kt becomes 26 milli-Lt only 300 Kelvin but if I go to say 100 Kelvin or if I go to say 50 Kelvin or 10 Kelvin then your Kt becomes very small much smaller right because the temperature is reduced no much smaller whereas your E c minus E d this quantity donor ionization energy level. actually does not change the temperature much so it will always remain a 10 MeV.
So if your KT becomes less much less than you know 10 MeV then your room then the temperature the thermal energy at low temperature will be so low that they will not be sufficient to excite the electrons from the donor level to the conduction band what will happen then then the fraction then you know then the fraction of donors that is ionized fraction of donors that is ionized will also reduce. We will reduce what it means is that what it means is that suppose you know I have a conduction band I have valence band it is n type dopes of Fermi level is up here and I told you that you know there is a very small you know there is a very small difference there is this donor ionization energy level ed I told you that if I have tantality power say 17 phosphorous atoms I am giving then I am expecting to get tantality power 17 electrons only why because all the tantality power 17 phosphorous atoms will give electrons here in the conduction band because at room temperature KT is 26 milli electron volt is actually larger than the ionization energy say 10 milli electron volt but at temperature equal to say 10 Kelvin your KT is much lower In that case not all not all 10 to the power 17 you know donors would be able to give electrons which I call donors may not be ionized or will not be ionized ok. All the 10 to the power 17 donors that I have added here will not be ionized because the temperature K T has become much smaller and that smaller energy is not sufficient to excite the carriers.
So the fraction of donors the fraction of donors that is ionized. And that is able to give electrons will reduce with lowering temperature. What will also it will result is that your free electron concentration will reduce.
Your free electron concentration will also reduce because your number of donor atoms that are ionizing also reduces. And why does it happen? Because your temperature is reducing say 50 Kelvin, 20 Kelvin, 10 Kelvin.
So your KT is also decreasing. So it does not have energy. The thermal energy does not have energy to actually excite. So, this process the carriers reducing at lower temperature is called carrier freeze out okay.
They are freezing out at lower temperature that is what is happening. It is carrier freeze out it is basically the lowering of free electron concentration or free hole I mean anything is fine free carrier concentration is getting lowered when your temperature is reduced right. When your temperature is reduced then your free carrier concentration is reduced that is called.
You know it will go exponentially reduction actually it will go exponentially so if I have a the carrier concentration is suppose n I plot on the y axis and suppose on the x axis I have temperature 1 by temperature okay so it is Kelvin inverse 1 by temperature I'm at love this is very high temperature okay this is lower this is even lower lower lower lower lower right. So at very low temperature okay at very low temperature what will happen very low temperature means this region right very low temperature your carrier concentration reduces temperature reduces so it will decrease like this and this is by the way log of depth so it is a log scale. So log scale is linear so exponentially actually reduces so what is happening is that this is your carrier freeze out that is happening okay carrier freeze out what is happening is that your carrier concentration is reducing with reducing temperature because fewer and fewer carriers are able to get ionized okay fewer and fewer carriers. are able to get ionized as you are reducing the temperature, okay. As you are reducing the temperature fewer and fewer carriers are getting ionized.
As you are increasing the temperature which means you are coming this side, this is increasing temperature, right, 1 by T, this is increasing temperature. If you are increasing temperature more and more fraction gets ionized, eventually approaching room temperature or even lower than that, what will happen is that all your 100% of that will be ionized. So it will be limit.
flat. What is this flat because this is defined by N D the doping that you have given if you have given there is no P type only N type suppose P type can be talked about separately suppose you have given 10 to the power 17 per centimeter cube of doping to the number this is remember number of free electrons. So number of free electron also will be 10 to the power 17. number of electrons will be also 10 to the power 17 that is the number of donors which is this is equal to number of free electrons which is the same 10 to the power 17 so it remains flat ok it remains flat ok. It remains flat for a some reason of temperature for some range of temperature including room temperature it remains some range of temperature it remains flat ok and this region is called the extrinsic region. Because, this is the region where your number of free carriers decided by the doping density N D, so it is flat and you want to operate the device in this regime.
Because here you exactly know the number of free carrier which is equal to N D which you pre decide, so all conductivity resistivity current voltage everything that we have to do in a device we know exactly the carrier concentration here, lower than that gradually drops down because the carriers are freezing out they are not able to ionize. Although the number of doping as 10 to the power 17 or whatever you have kept is fixed, but out of this 10 to the power 17, not all 10 to the power 17 are ionizing. So, the fraction of free carrier available is reducing. That is what is happening here.
And as you keep increasing temperature more and more, very high, what happens is that you have an Ni that intrinsic carrier concentration in silicon is 10 to the power 17. 10 approximately at room temperature at 300 Kelvin. So, that is much lower that is much lower somewhere here right that is much lower, but as you increase the temperature your N i has a exponential dependence on EG by. K T. So, as you keep increasing temperature this N I will keep increasing at sufficiently high temperature may be 5 600 degree Celsius or so for silicon your N I actually becomes so high actually N I keeps coming like this from here it will take over.
it will take over here, Ni is actually this one, it keeps it is very low, but with higher temperature Ni keeps increasing, increasing, increasing, increasing at some point may be five six hundred may be seven hundred may be four hundred that you can calculate at some temperature degree Celsius, Ni will become so large that it will even more be more than 10 to the power 17 or whatever you are doping and it will go crazy high like this, it will go exponentially like this. This region is called intrinsic region except that in intrinsic region excuse me. The carrier concentration the intrinsic carrier concentration Ni has become too high, the too high means it is much higher than end even. So, that here you know at very high temperature your carrier concentration has become so high that your sample becomes almost like metallic, it has become too many it has too much too many electrons too many holes Ni electrons Ni holes are there, it is much larger than 10 to the power 17 almost like metallic it can carry much higher current and you lose the control of conductivity. by doping which is what makes semiconductor very ideal candidate you want the semiconductor to have conductivity that you define in this region as this one but if it is higher temperature it will go crazy because Ni has increased.
If you go too low also Ni I mean the Ni is not there but free carrier will also freeze out. So that is this is how the temperature versus log plot looks like so if I can draw it fresh again so essentially log of N you are plotting right versus 1 by temperature. At very high temperature this is 1 by temperature so you know this is increasing temperature increasing temperature.
So, at very high temperature it will blow up then it will stay flat eventually it will fall down. This is called freeze out ok region this is called extrinsic region where the conduct the number of electrons is almost fixed and you want to operate in this regime only and then this is called intrinsic regime ok. Here carrier concentration will go exponential is e to the power minus e g by k t. That is what is happening.
So you see silicon for example has a band gap of 1.1 so the carrier concentration is 1.5 into 10 to the power 10 per centimeter cube. Suppose gallium arsenide has a higher band gap it has like 1.4 eV band gap so the carrier concentration is around 10 to the power 8 or 10 to the power 6 I do not remember exactly it is probably like 10 to the power 6. So you see silicon will basically have it has a higher intrinsic carrier concentration. So, at higher temperature it will blow up even faster this will withstand little higher temperature it can withstand till little higher temperature because the carrier concentration is low. So, you have to go to a little higher much higher temperature than this case in order for this to blow up which means because wide band gap material if you have wide band gap middle of the band gap is large I told you in one class that Ni is much smaller there right because it goes inversely as band gap the exponent of that.
So, you can heat a wider band gap temperature to higher temperature you can heat up a wide band gap material to a much higher temperature and expect the Ni to remain small. But for small band gap material like silicon or germanium this Ni is already very large I mean 10 to the power 10. So, if you heat up the material then Ni will become even much more larger. So, wide band gap materials can be used for high temperature electronics okay for if you want to use high temperature electronics you want to measure some something in the fume hood of a missile or a satellite fume with lot of high temperature 1000 degree Celsius or any other application industrial automotive vehicle industry where you know 5600 or 1000 degree Celsius processes are there and you want electronics to do some sensing some measurements and some integrated circuits and so on. Then high temperature electronics can be used for that you need a material with wider band gap because wider band gap materials will have lower Ni and hence this blowing up of the Ni with temperature will be much delayed.
Because, smaller band gap material has very high Ni and so smaller band gap material will quickly exceed this Ni quickly, that is why we go for wider band gap material for high power electron high temperature electronics like gallium nitride it has a band gap of 3.4 eV, silicon carbide it also has a band gap around 3.3 eV. So, this kind of materials are used for high temperature electronics because they can withstand much higher temperature okay. Still the reason is because their Ni is very small Ni is very small you want very small Ni otherwise it will blow up okay. So, that is very important thing that we have studied.
So, now we have considered the temperature variation of doping we have also talked about charge neutrality if you remember we talked about charge neutrality in the beginning of this lecture okay. We talked about charge neutrality right which is P plus ND plus is equal to N plus NA minus the total negative charge and the total positive charge has to be you know equal right. And then another thing we have talked about is temperature dependence of carrier concentration, temperature dependence of carrier concentration.
How carrier concentration changes the temperature we talked about that also right of carrier concentration okay. So we will end the lecture here, so these are two very important things. So in the next class we will talk about the following up the temperature dependence what is the next thing. The next thing is actually something called incomplete ionization okay.
We will quickly touch upon the facts of incomplete ionization, whatever you are putting need not all get ionized even at room temperature. So how will things look like in incomplete ionization? We will talk about incomplete ionization and then we will move forward to the basic concept of drift, conductivity, diffusion and so on and eventually we will talk about mobility and other things maybe in the subsequent classes okay. So thank you for your time and we will meet again for the next class.