If you click in the comments section below, you will be able to ask questions or comment on answers. Week 1: Introduction, Crystal Structures, etc. Q.
In the class, you had been using 'rectangles' as primitive unit cell.
In principle I could use a 'triangle' as a motif as well to tile the
whole surface as well -- it is also smaller than rectangle. Why is this
not allowed? A. An important restriction/requirement of unit cell for a Bravais lattice is that the orientation of the motif can not be changed -- you are allowed only translation, but no rotation. If we use 'triangles', then you will need two motifs -- triangles-up and triangles-down, which is not allowed. Two-motif sets that tile a space completely do exist -- these are called Penrose tiling --- but the these systems are not periodic. Q. You mentioned 7 base units. Can you give me a reference? A. The seven units are time (sec), length(meter), mass (kilogram), electric current)ampere), temperature (Kelvin), amount of substance (mole), light intensity (candela). The other form contains frequency (Hz) substitutes time, velocity (c) substitutes length and time, action (Plank constant), charge (e), heat capacity (k, Boltzmann constant), amount of substance (NA, Avogardo's number), luminous intensity (candala). For a nice discussion, see July 2014 Physics Today article, p. 38. Q. How do you know that the Miller indices will always be integers (e.g. 3,2,5)? Couldn't it be (3.2,1.7,4)?A. The definition of a crystal is that two atoms must be separated by integer multiples of unit vectors, i.e., r12= h*a+k*b where h and k are integers. It follows therefore that the normalized intercepts needed for Miller indices must also be integers. Q.
I read it in my book that Silicon has diamond lattice, while GaAs has
Zinc Blende lattice? You say that they have the same structure, so why
are the names different?A. Zinc Blende structure is generic name for all binary solids (e.g. GaAs) that crystallizes in the same structure. And Diamond structure is the generic name for all elemental solids that have the same structure as Carbon (Diamond is a special form of Carbon). Q.
I do not understand the pairing of Ga and As (black and while) to
make the structure look like FCC. We only had four black balls inside,
yet there were 8 while balls on the corner and 6 on the surfaces. How
can 14 whites pair with 4 blacks? A. This is a terrific question. You can think of it in two ways: First, you should note that WITHIN the box, there are actually 4 whites, not 14 (1/8 of 8 corner whites, and 1/2 of 6 surface whites), so within the box the accounting works out -- 4 whites for 4 blacks. In reality, if you focus on the whole crystals (not only the box) the 14 whites will grab the 14 blacks, 4 from inside the box and remaining 10 from outside the box. If you are still confused, I suggest that you draw a 2D crystal with black and while balls and pair them up appropriately to see how the balls outside the box help keep the balance. Q.
The 8 states/per atom in 3s and 3p states split in groups of 4 and 4.
The bottom 4 becomes valence band and the top four becomes conduction
band. How do we know that it has to 4 and 4, why not 7 and 1 or 5 and 3?
A. The exact splitting depends on the material used. Only those material that splits in such a way that the valence band is full and conduction band is empty is called semiconductor. If the splitting were different, there would be a lot of empty spaces of valence band electrons to move around, giving non-zero conduction even at temperature close to zero -- in fact these materials are called metal. Sophisticated theories based on quantum mechanics can predict these splitting very well. Q. Generally
resistance goes up when temperature is up, but in semiconductors
resistance goes down with temperature because we have more carriers. Is
this correct? A. Absolutely ... In 1870s, it was the first indication that metal (resistance goes up with temperature) is different from semiconductor (resistance goes down with temperature). This issue remained a puzzle until quantum mechanics came along to suggest that the electrons live in energy levels and energy bands, therefore semiconductor conduction has to be treated differently from metal. Q. Sometimes the Winger-Seitz cell looks complicated. How do we calculate the volume of the cell?A. There are many ways -- we could make a replica and dunk it in a bucket of water (Archimedes style) to find the volume! A more practical approach involves putting the Wigner-Seitz cell in cubic box whose volume is easy to compute. Then generate a set of three random number to define a random point inside the volume. If it is inside the irregular volume, we update the counter by 1. Millions of these random coordinates are generated (< 1 sec in most PC). We divide the number of points inside the W-S cell by the total number of points generated to calculate the relative volume of the cell. Then multiply with known volume of the enclosing box to find the volume of the W-S cell. This approach is called a Monte Carlo approach and is often used to define areas. Q. Is there a proof that W-S cell has the minimum volume? A. There is. You should be able to look it up. Q. You mentioned in the class that tetrahedra means four heads? In Greek, the word hedra refers to face, isn't it? A. You are right. When we talk about Platonic solid, we think of faces of the object. Therefore, a octahedra (like BCC-type CsCl) will have to square-pyramids joined at the base; it has 6 atoms at the apex, but 8 faces -- hence the name Octahedron solid. Q. If I have 1 proton, does it mean I have 1 state? A. No, it means that you have 2 1s state, 2 2s state, 6 2p states, etc. Per proton, there are many places the electrons can sit -- although it prefers the lowest energy states. Q. Does [111] means we have a box with 1cm sides?A. No, the 1 in [111] refers to the normalized intercept of the plan with x axis. It is dimensionless. Q. What happens to Miller indices if a plane goes through the origin -- the intercept is zero, the corresponding indices will be infinity?!A. You are right. Remember all points in the Bravais lattice are equivalent -- in this case, one should shift the origin to a neighboring point. After all (100), (200), (300) (001), (002), (003) are equivalent plane with equivalent physical, chemical, and electronic properties. Therefore shifting the origin does not matter. Q. You said that sometimes we need to find crystal planes such as [111] which may be at angle with respect to [001] and [110] surfaces. Presumably those planes have better properties. However, how do we cut something so thin at a given angle? A. It is not necessary to use a SAW to cut at an angle -- we can use anistropic etching to define the planes as well. See the link http://en.wikipedia.org/wiki/Etching_%28microfabrication%29 for a nice example of differential etching across various planes. Q. In the photoelectric experiment, we are told that each electron absorb one photon? Can it absorb 2, or 3, even in principle? A. It can -- this is called a 2-photon process: When the first photon lifts it up in energy, if the second photon strikes the electron while it is up, two photons can be absorbed. If the probability of absorbing a single photon is P (<<1), the the probability of 2 photon absorption is P^2 (<<P). Therefore, unless we are specifically looking for these multi-photon excitation (by amplifying the hf < W region), one may approximate the value to be ~0. Q. 100 years ago, it was classical mechanics; now it is quantum mechanics, who knows how it is going to be 100-years from now. Are we going to see other physics and other devices in the future? A. It is possible that 100 years from now quantum mechanics may be superseded with something else. However, the predictions of quantum mechanics have been tested rigorously with fantastic precision and the predictions have always come out to be right. As engineers, we ask the question if we sufficient physics to predict/control a phenomena. For a Greek catapult or bows/arrow, classical mechanics is enough; to design a IPOD the simple version of QM that will be discussed in this class is enough; to design spintronic devices, a more advanced version of QM is necessary (but it is standard QM). If some other theories are invented, perhaps it will lead to other devices. I would not bet against QM though. Q.
Did you hear you right that the precision of the oxide thickness is
such that "If it snowed 6 inches everyday since the time of Moses, there
would not be more than 1-2 in variation across the whole USA during the
period". Are these numbers real? A. Yes, they are! Total production of wafers are couple of millon wafers/year. If I assume 5 thousand years since Moses and 3 year makes a thousand days, then we have close to 1-2 million days since Moses. Oxide thickness is 10A and wafer is 12 in; the ratio would be the same as area of USA vs. thickness of snow in inches. So yes, the control is that impressive. Q. When we discussed [10] direction for graphene in the reciprocal space, what does it mean? A. After you calculate the reciprocal lattice vectors A1 and A2, you can express any point in the reciprocal lattice by a vector given by K=M*A1+ N*A2. If M=1, and N=0, then the corresponding plane will be [10] plane. Q. We started the whole discussion with the crystal structure of metal/semiconductor material. but how do i get to know the crsytal structure itself (especially of a new material)? Is it obtained simply using some microscope ?"A. The crystal structure is determined by diffraction experiments and by counting how many diffraction peaks you see in the pattern. For example the number of spots for a quasi-crystals is different from that of a periodic crystal (see the link I sent you about Nobel Prize in Chemistry in 2011). The wikipedia may also contain articles about diffraction experiments. Q. Sometimes people mention planes such as (200). Now I thought you said that (100) will do, because you can always shift the axes by one unit and make all parallel planes (100). A. A (200) plane cuts the axes at (1/2, infinity, infinity). In effect, if a is the size of the unit cell of a BCC or FCC lattice, then the atoms going through the body center or face center will have x-axis cut at a/2. As a result, the corresponding axes will be described by (200) plane. You should think if there is a higher order plane such as (300) or (400) for diamond or Zinc Blende crystals. Q. You mentioned something about compact and noncompact packing leading to fcc and bcc lattices? Can you explain one more time. A. Sure. If a wish to arrange a set of balls on a surface, an hexagonal arrangement produce the most compact model. We will call this layer A. We can put a second layer of balls (B layer) on top of layer A, again in the hexagonal pattern, by putting each ball in the center of the triangle formed by three balls of layer A. Now the C-layer offers two options: copy the location of the balls in the A layer, or, shift everything by one spot. If you copy, C is really A -- and the arrangement ABABAB is called hexagonal close packing (hcp); if you shift, C is different from A -- it leads to ABCABC packing -- and this is a bcc lattice. In short, bcc is a closed packed lattice. By the way, you should convince yourself that A, B, C are 111 planes. Examples include: lithium, sodium, potassium, chromium, barium, vanadium, iron, etc. The coordination is 8. All lattices are not close packed. If you arrange balls in little squares (rather than hexagons), then put a 2nd layer again in a square configuration, then the third, etc. and then push everything down so that they are stabilized, you just created a bcc lattice. Notice that the balls are not touching each other -- they are close packed. Examples include many aluminum, copper, gold, iridium, lead, nickel, platinum and silver This following website has great illustrations. http://butane.chem.uiuc.edu/pshapley/GenChem2/C2/1.html .... Also, check http://www.ptable.com/ to see why the metals may form fcc vs. bcc ... Crystal Structure. Patrick Batail discusses a paper by X. Roy where they show a macromolecular counterpart to NaCl by combing C60 (Cl counterpart) and other spherical donors [Co _{6}Se_{8}(PEt_{3})_{6}](Na counterpart) for [Co_{6}Se_{8}(PEt_{3})_{6}][C_{60}]_{2}_{. }http://www.sciencemag.org/content/341/6142/157.abstract http://www.sciencemag.org/content/341/6142/135/F1.large.jpg An updated version of crystal viewer https://nanohub.org/tools/crystal_viewer and the user manual provides a great learning experience. Animated slides: Miller indices: explanation:http://www.doitpoms.ac.uk/tlplib/miller_indices/links.php X-ray diffraction: http://www.doitpoms.ac.uk/tlplib/xray-diffraction/index.php Brillion Zones: http://www.doitpoms.ac.uk/tlplib/brillouin_zones/index.php Dislocation: http://www.doitpoms.ac.uk/tlplib/dislocations/index.php Review Questions. Q1) If I have 1500 silicon atoms, how many ‘chairs’ do I have in the conduction band? Q2) If I increase the temperature of two semiconductors – one Large-gap and other small gap – which one will conduct more? Q3) Effective mass of Silicon is different from Ge. Explain. Q4) How many bandgaps does Silicon have? Which one is most Important?
Q5) Of the four lattices (simple cubic,
bcc, fcc, diamond) which One has the smallest volume fraction,
which one has the largest.
Chaikin
Physics Today article: Thomas Hales: 2005 paper on Densest packing
shows that Kepler conjecture that no packing of congurent balls in
Euclidean tree space density greater than that of the FCC packing. This
does not however mean that this density can be achieved in only one
way: FCC, Hexagonal closed packed, double HCP (structure found in
Lanthanide elements), random HCP in colloidal crystal all have FCC
equivalent density.
8000-9,000 species of ants -- 1 quadratrillion in number (1e15). Britannica, 1993, 437. Number of leaves in a tree (3e14) E.J. Wilson. 1e6 wafers, 700 per 300 mm IC, 1e9 per IC. ~ 1e18 number of transistors. If you add all other chips, that may be another factor of 10.Memory $50 per Gigibyte, $50B annual market .... 8 *1e9 * 1e9 ~ 1e19 transistors in MRAM per year Hard drives: 7e19 bytes per year shipped (Stuart Parkin's presentation) (Chip Making's Singular Future By Rajendra Singh and Randhir Thakur)There are more transistors than ants and tree leaves in the world. Com. 2 A glass of water spread of the oceans across the world will still contain 100 original molecule per glass. (Number of Waterdrops in all oceans: 1e25). http://answers.google.com/answers/threadview?id=543278 Q. I read in the literature that the carriers
of graphene has the massless characteristic. All the paper mention this
problem but they didn't prove it, would you mine clarifying it for me?A. It is easy to define a mass of a classical object (e.g. a stone) by using Newton's law -- it is a positive, finite number. We can do the same for an electron in vacuum -- and the value is listed in all textbooks. In a solid, whenever possible, we define something called an EFFECTIVE mass for an electron. When a electron moves in the solid, they are continuously scattered by the atoms. The right way to handle the transport problem is to treat the electron as having a standard positive mass, and account for all the scattering with the atoms by Schrodinger equation. The second, simpler way is to say the following: it takes a certain amount of time for the electrons to cross a finite distance in the solid. What would be the mass of an electron-like object (without the atoms) to cover the same distance at the same time.? We will call this an effective mass -- because this mass includes all the scattering within it. This mass can be positive, zero, or negative. Since this contains information about the original electron and the scattering with atoms, effective mass is different for every material. It is not always possible to use the second, simpler way for all materials. In graphene, the scattering is different that one cannot define an effective mass. So depending on how you define it, it can be zero or infinity: In graphene E=v*k, v is a constant, k is the wavevector. dE/dk = v and d^2E/dk^2=0 so m* ~ (1/d^E/dk^2) = infinity, on the other hand, E=v*k relationship looks like that of a photon and photon has zero mass. All these confusion arises because we should solve the full problem, not try to approximate it. There is nothing fundamental about it. Notes. Holes - a short history. The concept of a hole as a positively charged counterpart to electrons was first proposed by Frenkel in 1926-1927. Remarkably, Dirac was inspired by this idea when he was searching for an interpretation of the negative energy states of an atom predicted by the relativistic Dirac equation. In analogy -- he called his interpretation a hole theory, eventually it led to the discovery of the positron. The Fermi-Dirac distribution traveled the other way -- originally used to interpret the energy distribution of electrons in stars was eventually used to describe carrier distribution of electrons in a solid. Topological States at the surface. A K-P model applies to infinite periodic crystal with repeating units of identical potential wells. The first and the last atomic potential of a finite system are different. The surface state (Tamm state) is slightly higher than the rest, because one side of the surface atom sees larger potential than the other side. This surface state cannot contribute to band-formation, but does serve as bandgap state that contribute to surface recombination. Instead of 1D system, a 3D system will have the surface states on all six faces. These surface states however will have exactly the same energy, which can create a surface states -- unmixed and protected from the bulk states. If the bulk is ferromagnetic, then the surface states will feel an effective magnetic field and the surface states will have split for up vs. down spins. Shockley discussed the surface states in a 1938 paper; a modern version of it involves topological insulator. Hofstadter Butterfly -- A K-P model of a super-lattice. The Brilliouin zone associated with a lattice of spacing 'a' is given by pi/a to -pi/a, and the process folds large free electron parabola into multiple bands separated by bandgap discontinuity. If a second periodic potential -- with spacing 5a, for example -- is superimposed on the original K-P model, the BZ will now reduce to pi/(5a) to -pi/(5a). And each of the bands in the original BZ will now split into several minibands. In general, it is difficult to produce a super-lattice with multiples of original lattice spacing. However, if two graphene sheets are rotated at a small angle, the underlying phase come-in and out of phase, producing another hexagonal supercell with large lattice constants (Moire pattern). One can view this as 2D equivalent of superlattice K-P model. * The Hofstadler Butterfly involves the plot of the energy vs. magnetic field plot in such a system that produces solution that appear like a butterfly. A nice set of questions and answers about various aspects of solid state physics is given at http://www.researchgate.net/topic/solid_state_physics?ev=tp_pst_dtl_xkey Week 3: Doping, carrier density, etc. Q. How can you have 2 atoms at one lattice site of a FCC/Diamond
lattice?A. Its a way of simplifying things. For a compound semiconductor like GaAs, just pair the atoms on the edge with atoms which are 1/4th of the body diagonal inside the cell. This way a diamond lattice can be visualized as a normal FCC lattice. Q.
Regarding amphoteric doping of GaAs by silicon -- I do not understand
why you said that there are Purdue professors who spend their life in
trying to put the atom either at Ga or As sites. If it goes to Ga site,
it will donate an electron to conduction band, if it goes to As site, it
will help park an electron from the valence band. In either case,
conductivity increases. A. This is very good question because it shows that you understood the concept of doping very well and can apply to various situations. In the presence of amphoteric doping, things are a little complicated. Often what will happen is that the extra electron from the donor site will see that there is vacant spot in the acceptor site -- it will drop down and fill it up. The extra electron is gone, so is the extra parking spot -- the donor-Si canceled the acceptor-Si and we are back to intrinsic semiconductor. So we want atoms to end up either as a donor or as an acceptor, but not both. Q. I know that P has 15 neutrons and 15 electrons -- looks pretty stable to me. Why would they bother giving away an electron? A. If P was on its own, it would not give up its electron. In silicon however, the binding between the proton and extra electron is considerably weaker. Recall that force between two charged particle q1 and q2 in inversely proportional to the square of the distance and the dielectric constant. Since the extra electron of P orbits the neutron at a radius which includes many silicon atoms, it will see dielectric constant of Si -- as a result the forces will be considerably weaker and the extra electron will float away. The other electrons of P are pulled tightly close the nucleus and they do not see the full dielectric constant of Si matrix and they are safe. Q. I hear
that people are reducing operating temperature of computer to make them
go faster, yet you say that carrier concentration becomes smaller as
reduced temperature because ionization of donors is difficult. A. The electron current depends on the number of electrons and how fast they are going. When the temperature is lowered, the number of electrons in a semiconductor is reduced, however at the same time electrons go faster at lower temperature -- therefore under certain circumstances the current is higher at lower temperature. Since circuit speed depends on the ability of current to charge load capacitors, higher current means faster computer. The reason why electrons go faster is this -- at the temperature is raised, the atoms begin to vibrate with increasing amplitude and the electrons are repeatedly scattered as they try to go from one contact to the next -- these scattering reduces the speed of the electrons just as speed bumps reduce velocity of car. At lower temperature, the 'road bumps' are milder -- and net velocity is faster. Q. What is the difference between EF and Ei? Both
are Fermi levels: Ei is the special name of EF when the material is
intrinsic. When the material is doped (extrinsic) by donors/acceptors,
one should only care about Fermi level, EF; nonetheless, people still
draw the Ei levels to indicate where the Fermi level would have been,
had it been a intrinsic semiconductor (just for reference). You see, one
can say if a material is donor doped or acceptor doped simply be
noticing if EF is above Ei or not (you do not need to know where the Ec
or Ev is). Other uses of Ei level will be discussed later. It
is a fundamental requirement of statistical physics that at EQUILIBRIUM
conditions (no voltage applied -- no one pulling electrons in and out
of the system), EF be constant throughout a material (regardless the
complexity of the doping). I will show in a few days that if this were
not the case and if EF changed with position at equilibrium, current
would flow even at zero bias (which is impossible). Q. Why do you say EF is constant? Q. What is the difference between
degenerate and non-degenerate conditions? How does it relate to
Boltzmann and Fermi-Dirac distribution? Simply
put, degenerate means crowded, and non-degenerate means 'not-crowded'.
If the doping exceeds 1e18-1e19, then for every 1000 Silicon atoms,
there is a dopant atom -- the free electrons are beginning to meet each
other more frequently -- a crowded, degenerate situation. Fermi-Dirac
distribution is used to describe the 'seating-chart' for these
electrons. On the other hand, if the doping is relatively low, then electrons are few and far between -- and the simple Boltzmann distribution (originally developed for gas molecules) is sufficient to describe their 'seating' preference. In this course, we will mainly focus on non-degenerate situation. Q. You promised the link to a simulator that physically interprets Boltzmann distribution. Q. Why is it that the electrons at the bottom of the band can be described by an effective mass, while elsewhere it cannot? A. First of all, let us be clear that electron at the bottom of the band may not be defined by an effective mass. Recall the band-diagram for graphene, where E=a*abs(k) where a is some constant. The second-derivative does not exist, neither do effective mass -- in the classical sense. Nevertheless, we can calculate the position and wavevector of the particle at all points at all time by using the relations discussed in the class. Now it is true that for many semiconductors, one can define an effective mass at the bottom of the band. Here is why: in many semiconductors, there is a mirror symmetry between +x and -x direction, therefore the solution of the Schrodinger equation must also be the same in the +kx and -kx directions. Therefore, mathematically we can write E-k diagram only as follows: E= a * abs (k) , E= b*k^2, E = ck^2*abs(k), E= d*k^4 , or some combination thereof. Now, IF a material happens to be described by E=b*k^2, the second derivative will be a constant, it will appear as if it is a classical mass. Indeed for many traditional semiconductor, this turns out to be case. If the E-k relationship is more complicated as it goes away from the conduction band or even close to the conduction band, the second derivative becomes energy-dependent (if it exists at all) and life becomes more complicated. Finally, there is experimental convenience answer. When cyclotron data comes in for measurement of effective mass, experimentalists sometime find it convenient to FIT the data as E=b1*k^2, even though the data might be better fitted with E=b*k^2 - d*k^4. Here b1 is an effective parameter that interpolates between b and d in some approximate way. When this parameter is given to you, you can only use the information the way it was extracted. This is another reason in the books you will find tabulated values of effective mass from cyclotron experiments, even though the theory may not support an energy independent effective mass. Q. I have one doubt
that's been bugging me: In silicon the band gap is around 1.1eV at room
temperature, with only a few
electrons in the conduction band. At room
temperature the energy that electrons could obtain from the lattice is 26meV. How is it possible for even a few
electrons to overcome the bandgap of 1.21eV (when they go from
valence band to conduction band) with as little as energy as
26meV?A. When I was a student, I had the exact same question: How is it
possible that with only 26 meV of average energy, an electron can
cross the 1 eV bandgap? The answer lies in probability and large numbers, similar to a
lottery. Remember the number of electrons constantly jumping up and down in the valence band are fantastically large, of the order of 1e21-1e22 per cubic cm, equal to the number of atoms in the solid. a electron absorbs 26 meV energy from the lattice vibrations (called phonons), its electron energy goes up by that amount. A few ps later, it is highly likely that the electron will give the 26 meV energy back to the lattice, and return to the original state. Now some of the electrons may be luckier, and can catch two 26 meV phonons in a row (go up by 52 meV), and bit later, give them all up. The number of these lucky electrons catching two phonons in a succession is smaller, just like getting two sixes in a row in dice throw is smaller than the probability of getting just one. In other words, if p is the probability of going up by one phonon (26 meV), p^2 is the probability of going up by catching two phonons, etc. To jump 1.1eV bandgap, an electron has to catch 42 phonons in a row, a near impossibility -- in other words, the probability is p^(42) ~ exp(-1.1/0.26). However, the original number of electrons trying to make this transition is so large (1e21-1e22) that a very very small fraction (1 lucky one in a trillion, 1e12) still makes it, giving the intrinsic concentration of 1e10. Bottom line, yes, the probability of making it is very very low, but then the number of electrons playing the lottery is so large that some of them are guaranteed to make it. Q. What is the physical significance of the Fermi level?I mentioned in the class that Fermi level is uniquely related to carrier concentration. Therefore, a carrier density uniquely determines the Fermi level. And I joked that it is set to a value of 1/2 because Fermi was born ahead of me, he did the math --defined the function, and named the variable. More seriously, however, you can see for donor level distribution, ED=EF does not give you 1/2, but 1/3. Of course, if you incorporated the gD=2 in the energy level information ED* and redefine the function so that it looks like Fermi function, then the probability is again 1/2 -- but this is cheating! Finally, the HW based on metal should give you another interpretation for the Fermi level -- at T=0, the band is filled upto the Fermi level and empty beyond it. Q. I was going through lecture 4 again on nano hub and you mention in
slide 3 that valence electrons are the ones closest to the potential and
hence the most difficult to displace and there are electrons higher up.
I thought it was the other way round and its the valence electrons that
hop around. This happens at 52 seconds into slide 3. I just wanted to
bring that to your attention and clarify that.A. Recall that a semiconductor has many bands. The bands below the Fermi level are called valance bands and the bands above the Fermi level are called conduction band. Also, recall that the bands with lowest energies also have large effective mass. These valence band electrons want to stay close to their parent atoms and do not hop from one site to another (e.g. 1s states). As we go up in energy through various valence bands, the electrons are held less tightly to the nucleus and they can move around a bit more. The electron in the top-most valence band (just below the Fermi level) can move quite easily. Finally, the electrons in the conduction bands are highly mobile as well. Therefore, you should distinguish between two types of valence bands -- the ones close to the nucleus that does not move, and the one close to the Fermi level, that move easily. Q. I calculated the Mn* to be 6^(2/3)*(h^2/(0.6(a^2))). I'm not sure if this is right or wrong. but i'm confused with this equation in terms of dimension. the dimension of h is "ev" and a is "meter". This seems to be more like mass/unit area.
A. Start by recalling that 1/2 *m*v*2 = E gives you an eV. v=cm/sec, therefore a unit of mass is eV*sec^2/cm^2. Now h^2 is (ev.s)^2 and a is meter. So you have the same unit. Q. I am confused by the idea of effective mass tensor? How is it that when I apply a force in one direction, the material moves in another direction? A. Every child remembers sliding down a slide -- the gravitational force is vertical (say in z direction), but the child moves both in z and x direction. Assume that the slide is made of invisible material -- then for someone standing apart, things will look strange -- force in one direction, but the mass moving in different direction, defined by the constraint of the potential. Replace the child by the electron and replace the slide by the atomic potential in different direction -- you get the idea of tensor nature of the mass. *Rolling down the ball in an inclined plane is the experiment that Gallieo did when he was imprisoned and helped him understand the nature of gravity. Thank goodness, he did not have facebook access during his captivity. Q. You mentioned several times in the lectures that the relationship np=n_i^2 holds only in equilibrium. How do we know that. A. Our current derivation involves multiplication of n=N_C exp(-(EC-EF)/kT) and p=N_V exp((EF-EV)/kT) -- showing that it equals a constant N_C N_V exp(-EG/kT) which we called n_i^2. The notion of equilibrium hides -- as we will see later -- in our assumption that both electron and holes are defined by the same E_F. At nonequilibrium situations, E_F for electrons (F_n) will be different from that of holes (F_p), so that the generalized relationship will be np=n_i^2 exp (F_n-F_p)/kT. There is another way to establish the relationship that makes the kinetics of the problem explicit. Assume that there are n_v number of electrons in the valence band (total number of states is NV), and n_c number of electrons in the conduction band (total number of states is NC). At equilibrium, the up and down transitions must balance, namely, NC*K_vc*n_v=n_c*(NV-n_v)*K_cv. Here K_vc and K_cv are transition rates from valence to conduction band, and vice versa. Now K_vc~exp(-EG/kT) because going up is difficult, but K_cV=1 because going down is easy. And assuming n_v~NV for the left-hand side, we find NC*NV*exp(-EG/kT)=n_c*p_v, or np=n_i^2. This approach of finding concentration by balancing the rates is quite common in chemistry. If Na+Cl= NaCl, then we write a similar equation [Na][Cl]/[NaCl] = constant. which involves balances the rates the same way as we did in the previous paragraph. The rule described above is completely general and arises is broad range of system. Assume that you have a half-filled bottle of water -- initially the air is dry, but soon the humidity reaches 100%. The water molecules in air are free, with zero cohesive energy; while the molecules in water are surrounded by other water molecules, the hydrogen bonding reduces energy. Now we have two energy levels, E_air and E_water. Again write the detailed balance probabilities -- with probability of a water molecule going from air to water is 1, while that of water to air is exp(-(E_water-E_air)). The detailed balance fixes the humidity to 100%. If you put some salt, the stronger attraction between salt ions and water molecules, reduces the energy level of the water further. This in turn reduces the amount of water in the air. A saturated solution provides unambiguous RH level. Bivalent salt reduces the RH even further. In fact, you can connect up the air exposed to different salt solution to produce any arbitrary RH. One final approach to derive np=n_i^2 showing that the results shows up in multiple areas of physics. Assume two sets of boxes: one set contains NV box and the other set contains NC boxes. Now if you wanted to take n number of electrons from the NV boxes and throw them out to infinity, you have S1= NV!/(n! (NV-n)! options. The entropy is S=K_B ln S1. Now bring these electrons back from infinity and put them in NC boxes, therefore S2= NC!/(n!(NC-n))!) and the entropy is S=k_B ln S2. Total free energy is F=Eg-T(S1+S2), minimize F with respect to n. And you will get back n=sqrt(NC*NV)e^(-Eg/2kT). In fact, this is how the material science people derive the relationship of forming an interstitials, see Kittle. Q. I hear the term Fermi-level pinning. Can you explain? One technique to keep the conductivity invariant under accidental contamination, is to use a small amount of midgap defect levels. Now any accidental donor will donate the electrons to the midgap defects, rather than contributing to conduction. To a large extent, the Fermi-level remains pinned close to the defects, insensitive to donor concentration. A chemist uses the same trick when he/she wants to hold the pH fixed. A buffer solution acts like a 'midgap trap' that gobbles up any change in pH, and would not let the pH change dramatically. Q. You mentioned in a recent class that a single dopant may make a Nanocrystal degenerate. What did you mean? A. A 4-nm NC contains approximately 1000 atoms. Even a single impurity dopes the NC to 1e19 cm-3, a degenerate limit. In addition, the confinement of NC quantizes its own state, and the levels of Cu atoms introduces extra energy levels close to the quantized levels. In an Science article, D. Mocatta et al. (Science, 332, p. 77, 2011) demonstrates this effect beautifully. ------------------------------------------------------------------------------------------- Review Questions. Q1) Electron energy at room temperature is kT = 25meV, Which donor (Li, ET=33eV, Bi=45 meV) will ionize easier? Q2) What is the difference between effective density of states vs. Density of states? Q3) How many electrons do we have at Fermi level of intrinsic Si? Q4) In intrinsic Si, if I have 396 electrons, how many holes do I have?
Q5) If I have same number of acceptor and donor atoms in a sample, How would the conductivity change with respect to intrinsic Si? Q1) How many Bravais lattice do we have in three dimension ? Q2) What are the units of effective density of state vs. density of state ? Q3) When should we use Fermi-Diract statistics vs. Boltzmann statistics? Q4) Is Zn (column 2) in GaAs an acceptor or a donor ? Q5) Your ipod will not work if you visit Mars, do you know why ? Electrons from
both conduction and valence bands could be hit by a photon (a small
packet of energy), however since there are many more electrons in the
valence band (remember that there are 2N chairs in each band for N atoms
and that below the Fermi level, the valence band is almost full)
compared to conduction band, that the probability of getting hit is
orders of magnitude higher for valence band electrons compared to
conduction band electrons. Week 4: Recombination Generation ProcessesQ. When a photon hits a semiconductor, you always show that an electron is lifted from valence band to conduction band leaving behind a hole? There are electrons in the conduction band as well? Why couldn't they get hit ? Q. You derived a capture formula like -cp*n_T*p, but then approximated it as -cp*N_T*p, that is replaced n_T by N_T. Why?When a material is n-doped, it is the minority carriers (holes) that control the overall recombination -- they are too many electrons compared, so the availablility of holes dictates everything. Now n_T means the number of traps that have already captured electrons and are waiting for a hole to come by and with so many electrons electrons around almost all the traps are full and ready and waiting, so N_T=n_T+p_T ~ n_T. That is why I made the replacement. Q. In the recombination formula,
sometime you say (delta_p)^2 ~ 0, but I still see (delta_p) floating
around in the final formula. What is going on? What we can drop in formula depends on the relative sizes of the terms I compare. While I can drop (delta_p)^2 in the numerator COMPARED to n0*(delta_p), because n0>> delta_p, I can not drop a term when delta_p is sitting alone and there is nothing to compare it with. So when I set sometime to zero, it does not mean it zero by itself, rather it is negligible compared to others. Q. When you say something is
in equilibrium and n0 and p0 are time-independent, do you mean that stay
in their respective bands forever?Even in equilibrium, inside a semiconductor a violent exchange of particles is going on. The number in conduction and valence bands do not changes in equilibrium for sure, but if you marked an electron in conduction band red at t=zero and came back a milli-second later, you will see that that red electron may have recombined with the hole and is now swimming deep down in the valence band -- forever forgetting that it was ever in the conduction band. This does not mean, we have lost one electron in the conduction band -- because within the same millisecond, a electron from valence band has climbed up to the conduction band after great difficulty (40 kT high -- it will be like your jumping from here to the the moon) leaving behind a hole. Yes, the red electron is long gone, but a new one has taken its place -- keeping the total number unchanged. Per second, this rate of exchange is more than 1e10 events/sec. Analogy: Just like when you look in the sky and the air may seem still and nothing seem to be moving and yet you realize that there is tremendous fluxes of airflow going all over the place. Q. Why are we spending so much time thinking about recombination? Shouldn't we try to avoid it altogether?A. Semiconductor materials can never be made 100% pure, there are always trace amount of other impurities. If these impurities are Cu or Fe within silicon host, you will recall that Cu or Fe provide energy level somewhere in the middle of silicon bandgap, and therefore they can be very efficient recombination centers. The calculations we are doing helps us decide the at what level of impurity can we accept without affecting device performance. For example, the transistors in your iPAD are purer than that of a solar cell. In addition, we will see later that sometimes we will intentionally introduce trap level (e.g. Au in Si) to reduce minority carrier lifetime so that the diodes can be turned off fast. As we will see later that when minority carriers have to get out of a region on their own, they do so by diffusion -- by randomly bumping along -- until they reach a contact; in contrast, traps allow these minority carriers to get out by first recombining with a majority carrier and then getting out from the device with dielectric relaxation time (we talk about it in a few days). Q. I did a optical trap experiment last semester? Are the electronic trap related to optical traps?A.
In someways they are. In an optical trap, two laser beams are crossed
to create a spot with intense electric field. When a beam of atoms are
injected in this region with slow enough velocity, then they might be
trapped by the attractive fields of the laser-trap for a while.
Similarly, a Cu or Au atoms creates an electric field where a slowly
moving electron can be trapped for a certain period of time. However,
the recombination process has a second step -- that of an hole coming
along and allowing the electron to drop down and disappear in the
valence band. There is nothing comparable for the optical traps. A.
The concepts are very similar -- for nuclear reaction, we talk about
nuclear cross-section -- this is the target of the size of a nucleus
that a fast moving neutron sees during impact. Similarly, trap
cross-section is related to the target-size offered by a Cu or Au atoms
to an incoming electron. You may remember that the atoms in Si are
roughly 5A apart. Therefore, when a Cu atom replaces a Si atom, it has a
region of influence of the same-size (this is the size of the spider
web). Therefore, the cross-section simply is pi*(5e-8)^2 ~1e-14 cm-2 -
which is typical trap cross-section. If the trap has already caught an
electron, its potential is repulsive because it does not want any more
electron -- so its cross-section decreases. See this is easy, no need to
have nightmares!Q. I took a class in nuclear physics and the word 'capture cross-section' gives me nightmares. But seriously, is the capture cross-section of traps in anyway related to capture cross-section of nuclear reaction? A.
Yes you can. There are phototoconductive materials, photo-conductive
transistors etc. that uses light to control electron transport. Q. You say E_T is approximately at midgap? Does it mean it is at E_i? Q. Instead of voltage, is it possible to control current flow by light? Q. What happens when light has energy that is not sufficient to get absorbed? Does it simply pass through? Or does it interact with the atoms? A. This is a terrific question! See if the light passed right through without acknowledging the atoms, then the dielectric constant of the material at this wavelength should have been equal to free space. But you know this is not correct, because you remember from your high school experiment that a light passing through any material unabsorbed bends a bit due to refraction. Refraction occurs because even though light is not absorbed at low energy, they do put the atoms in vibration. When they re-emit, the phases are such that they bend a bit. In short, therefore they light are absorbed and re-emitted by the atoms of the material, even if an electron can not go all the way from conduction to valence band. Q. In a p-doped
material, Auger recombination involves scattering of two holes followed
by recombination with a electron and the other hole being pushed way
down in the fermi sea. When the holes collide, why do they not become
one bigger hole? A. They would if you took my hole=bubble analogy too seriously! Assume that you have 100 states in the valence band and 98 electrons to occupy them -- so you have two holes. After the 'holes' scatter (actually the 98 electrons did something that we are accounting for by holes), then after the scattering you still have 98 electrons, because two electrons can not combine to make a bigger electron. Therefore after scattering you still have two empty places. In other words, just like electron numbers have to conserved, so are the holes numbers. Q. Even for indirect recombination in Si, I need a vertical transition to the gamma point-- wouldn't photon be emitted during the transition? And if so, why can I not make a laser out of Si? A. Interesting question. Indirect recombination can occur in one of two ways -- a (low-energy acoustic) phonon-photon combination (light will be emitted) and a combination of acoustic phonon plus a series of optical phonon emission. The second process is called cascade capture and is dominant in Silicon. The energy is dissipated by lattice vibration as optical phonon decomposes into acoustic phonon, but there is no light emission. Review Questions. 1)What is an exciton? What is the charge of an exciton? 2)How strong the field should be for impact ionization? 3)What is the difference between direct and indirect recombination? 4)What is the difference between a trap and a donor? 5)What is the difference between surface state and a trap? Q.
If I do not have any generation, but only recombination, then the
system is supposed to be in equilibrium. But how can it be ? Wouldn't
everything will recombine and be lost from the conduction band due to
recombination? A. You have to distinguish between optical-generation (the term we include in the minority carrier equation) and thermal-generation (the term we do not include anywhere explicitly, but the term which is hidden in intrinsic carrier concentration). At equilibrium, thermal generation balances standard recombination, so that there is no net change and the product of electrons and holes equals ni^2. Q.
You made an approximation during derivation of R-G rate when you
calculated the rate of (minority) hole capture is proportional to
-cp*nT*v ~ -cp*NT*v ? How do I understand this approximation? A. Recall the analogy that majority carrier electrons are like job applicants in USA, while minority carrier holes are like jop opening in Europe, for example. And the traps are like Placement agencies (NT). You realize that when there are many job applicants, all the agencies will be full of applicants waiting for a job opening to come around, (nT~NT). Therefore, I am allowed to make the approximation. Detailed mathematical derivation will also give the same result. Practical application of Photogeneration. Here is an interesting example of how photons create e-h pair to wash out corona deposited positive charges for applications in photocopiers. http://www.physics.udel.edu/~watson/scen103/xerox/ Q. In one of the lectures on nano hub you describe how absorption of a
photon with the right energy causes an electron that matches that energy
to absorb it and jump to the conduction. For me this is a possibility
of creating very efficient solar panels made of a single layer of
graphene atoms. If we can draw the Ek diagram for it, we know what
spectrum of photons need to be made incident on the sheet to cause a
current to flow efficiently as we are dealing with just a 2D structure.
Since we are working in 2D it makes it easier to define the energies of
the electrons in the valence band from the Ek diagram. Does this hold
any basis?A. That is a very nice observation. Indeed, people have been trying to to make graphene as a photodetector and a solar cell for sometime now. The challenge is that graphene is so thin that it allows 98% photons to pass through without having time to absorb them, therefore the efficiency of the solar cell is very low. However, we are currently working on transparent electrodes based on graphene -- we will discuss this briefly in the class. Q. In Q4.13 of the text book, it's asked to calculate "degenerate doping limit as a function of Temperature". I'm not clear on what exactly is "degenerate doping limit" and how is it supposed to be calculated. Please explain a bit.A. When a material is undoped, the Fermi level is close to the middle of the bandgap. As you begin to add donors, the carrier density increases and the Fermi level begins to move towards the conduction band. At low doping density (non-degenerate limit), you can calculate the carrier density by assuming Boltzmann approximation of Fermi-Dirac statistics, and you get excellent results. At high enough density (degenerate limit), the Fermi level begins to get within 3kT of the conduction band. This high density is known as degenerate limit, because Boltzmann approximation is no longer valid. Therefore, you need to calculate effective density of states N_C at each Temperature T. And use the equation n~ND and n=N_C *exp(-(EC-EF)/kT) where (EC-EF)=3kT. By equating two sides, you will get the value of ND which makes the semiconductor degenerate. Week 5: Carrier Transport: Part 1 Actually
I cheated a bit for now -- something I would own upto in a few days'
time. If you look carefully at the graph, you will find that the slope
of ln(rho) vs. ln(NA) is not really -1, as it should have been. Rather
is a bit different -- and this difference reflects the mobility
correction. There are two ways to fix it:Q. You said that people measure doping density by measuring conductivity, but mobility depends on doping as well (through impurity scattering). How do you separate the two? -- Assume a fixed mobility appropriate for low doped sample, get an estimate of doping. In the second iteration, use the value of the doping to find a new mobility (from the graph), and recalculate doping. In 2/3 trials, you will find a self-consistent value. -- There is special experiment called Hall measurement that can be used to measure doping density directly. And then you can use the conductivity formula to get mobility at that density. This is how the mobility-doping curves in your book has been created. When
a donor atom loses it electron to the conduction band, leaving behind a
positive charge -- other electrons can not leave it alone, but crowd
around it in a manner similar to zone defense in basketball or
body-guards for celebrities. Other electrons coming in the neighborhood
finds not an isolated positive donor, but a donor whose potential has
been 'screened' by other electrons. At higher temperature, the screening
is more effective -- the defense is more agile -- therefore, the
visiting electrons are scattered less. As a result, mobility due to
impurities increase at higher temperature (opposite to that of phonons).
Q. How can the mass of an electron change in GaAs as a function of field, that would give us negative differential conductivity? Q. Can you explain why mobility due to doping increases with temperature? Donor and acceptors
both scatter electrons and holes equally well. So the total scattering
will be given by the sum of the impurities, not their difference. Q. If a material is co-doped at 1e17 (both donor and acceptor densities are 1e17 cm-3), then should we calculate mobility at net doping of 0 or sum of the doping at 2e17? You
see it does not matter -- because only thing we are really interested
in the potential difference, not the absolute potential (no such thing
anyway). So we can take the reference level anywhere we wish, while the
absolute value will be different, the difference in potential will
remain unaffected. Q. How do I know where to put the reference energy that is needed to calculate the potential profile of a semiconductor? Q.
Today you calculated the distance between successive scattering of an
electron by two approaches, first by using mobility and then by using
doping density. What were you trying to say ? Assume
that ND=1e18 cm-3. At this density, mobility vs. doping curves say
that the scattering should be dominated by Ionized impurity scattering.
Is this true? Method 1: The mobility curve says the mu=300 cm^2/V.sec at ND=1e18. We know m*=0.5x9.1e-34 gm, q=1.6e-19, so that the time between successive scattering is tau~ 1e-12 sec. (mu=q*tau/m*). If the field is 1e3 V/cm, then the mobility field curve says that the velocity is 2e6 cm/sec. Therefore the distance between scattering is 20 nm. Method 2: The distance between the neighboring ionized atoms is obtained by simply 1/(1e18 )^1/3 cm = 1e-6 cm = 10 nm. Therefore we expect the electrons to scatter roughly at the same distance interval. And the calculation above says that it does!If you still did not understand, perhaps the following will help. Assume you are James Bond and you are being taken blindfolded to a secret location in a car that is going at 100 miles/hour. Although you can not see, you can feel the effect of speed-bump every 10 sec, therefore you figure the distance between speed bumps is 100 miles/hour x 10 sec. This is Method 1 of the calculation, the 'blind' electron being bumped every 1e-12 sec traveling at a speed of 2e6 cm/sec. Next day, you are free and then check to see which parts of the city has speeds bumps 100 miles/hour x 10 sec apart. This is method 2 -- you are finding the separation between speed bumping by physically seeing where the scattering atoms are. When they two methods match, you know where you are. Q. Why is it that +trap and -trap have the same effect on electron with the same sign? Q. Where does the T^3/2 dependence of mobility come from? Week 6: Carrier Transport: Part 2Q. Why is fermi level constant in equilibrium even though there is band bending?A. Derivative of Fermi level is related to current -- at zero voltage, current is zero, therefore the derivative of the Fermi level is zero -- that is Fermi level is constant. Band bending involves electrostatic potential -- this is related only to the drift current, but has no influence on diffusion current. Bands may bend as they wish and change the drift current in the process, but at equilibrium diffusion current will exactly counter-balance the drift current and Fermi level will flat. Q. What does it mean by minority injection current. Do the injected carriers not recombine with the majority carriers present ?A. Certainly they do. Assume that you inject some of the professors to a piranha filled river (Here professors are minority and piranha are majority population) -- there number will rapidly diminish as they try to swim further into the river. And you can model their disappearance by solving the minority carrier differential equation that is given in the text book. Before they completely disappear however, there will be flux of professors from the shore to their eventual endpoint -- hence the minority carrier current. Q. Can kinetic energy of particles be zero in non-equilibrium conditions?A. Sure it can be form some electrons that are sitting at the very bottom of the conduction band. On average however, the kinetic energy of an electron is given by 1/2 m v^2=k_B T. Q.
In solving steady state problems, you said that concentrations at the
left a boundary should be the same as that on the right, that is
delta_n(x=0-)=delta_n(x=0+). But this is a diffusion problem, if the
concentration is the same, then how can there be any diffusion?A. When we say that concentration is continuous across a boundary, we mean two points across the boundary are so close together that they are essentially the same point, therefore continuity of concentration is appropriate. Diffusion involves distances that are separated by distances that involve scattering of electrons. Q. Ok, but then why do we need delta_n(t=0-)=delta_n(t=0+) for transient problems? What will happen if they were discontinuous? A. Do you remember that voltage across a capacitor can not change instantaneously and current across a inductor can not change instantanteously. If they did, then P=1/2 CV1^2 - 1/2CV2^2/dt with dt going to zero, would have translated to infinite power supply. Same is true here -- dn can not jump instantaneously because that will require infinite supply of flux. Q. Are the statements 'no thermal R-G', 'no traps', 'tau=infinity', etc. all equivalent and should be treated as same?A. Yes, since tau= 1/(velocity*trap number), then zero traps translates to infinite lifetime. It takes the electron forever to recombine because there are no traps available for recombination. Similarly, since R-G term is -delta_n/tau, no R-G means that you put tau=infinity and thereby remove the recombination term. Q. What is the difference between traps and surface states?A. Both types have empty bonds that allows electron capture and help in the recombination process. The empty bonds in a trap comes from a impurity atom like Cu or Au in Silicon, on the empty bond in a surface state arises from the fact that at the surface, the neighbors have gone away leaving behind states that are empty. Therefore, electrically they behave similarly -- but their physical origin is very different. Q. What happens if you have a problem that has both transient excitation as well as multiple region? A. There is no analytical solution for the problem. Those have to solved numerically -- we will have nanohub exercise. Q.
When you solved Poisson equation in the last class, the solution looked
similar to the 'half-illuminated bar" problem you did today? What are
they so similar?A. Because both involve 2nd order differential equations: For p-n junction and band-diagram, the equation was eps*d^V/dx^2= charge , while for minority carriers in illuminated bar problem is described by D d^2 n/dx^2 = G-n/tau. Since the math is the same, the solutions look similar -- the physics obviously is very different. Q. Why does not dielectric relaxation response as soon as the minority carrier is injected in a diode? Why does it need to wait till minority carrier recombination? A. This is a beautiful question and shows that you have a deep appreciation of the related physics. My guess is that it has to be a majority carrier perturbation (which happens after recombination) is a pre-requisite for dielectric relaxation. Just any perturbation of potential by minority carrier injection will not do. I will need to think more. Resources. An excellent review of the physics of drift diffusion is given by Bill Frensley. Another article by Frensley explains theory of band alignment http://www.utdallas.edu/~frensley/hlts/hjalign.html In an email, Frensley also provides a fascinating historical background of the Haynes-Shockley experiment:The Haynes-Shockley film was made as a part of the Semiconductor Electronics Education Committee (SEEC) program at MIT in the early 1960s. I had seen this film in the one semiconductor device course I ever took, in a Freshman elective at Caltech in 1969 taught by Carver Mead. The film is the best source I know for an intuitive understanding of semiconductor transport. I had to ask about it on some internet forums, and it turned out that Karl Hess had made a videotape from the film, and he loaned me a copy. I had thought that the film must have been made by Bell Labs, but they disclaimed any knowledge of it. When you go back and look at it, and compare it to the original experiment (Phys. Rev. 75, 641 [1949]) you'll see the MIT fingerprints all over the experiment. Dick Haynes did not have a flashlamp in 1949. They were all still in Harold Edgerton's lab then. What he used were square waves generated by an "electromechanical interrupter" (more commonly known as a buzzer), and he measured the delay and shape of the leading edge of the detected pulses. Also, I seriously doubt that any experiment ever conducted at Bell Labs was subject to all the calibration checks that Haynes shows. That was the MIT view on how to do experiments. Note that Streetman and everyone else who treats the Haynes-Shockley experiment describes the SEEC version. The Haynes-Shockley film seems to be the only one from SEEC that anyone found useful. I don't know if any others were made, but my guess is that there were. I had always found the SEEC materials to show a bit idiosyncratic view of the subject: Ken O recently told me that Paul Gray had explained to him that they wanted to "simplify" semiconductor devices by eliminating all the band diagrams and presenting everything in terms of equivalent circuits. If that was the MIT view of simplification it certainly validated my decision to go to Caltech rather than MIT. Carver Mead of course presented everything in terms of band diagrams (though to be fair, in 1969 he had to explain the MOS inversion layer). Carver certainly prepared me to understand exactly what Herb Kroemer was talking about in our first meeting in 1974. PART 2: Diodes, Transistors, etc. A.
In typical situation, the left and rightmost regions are most heavily
doped -- therefore, you the standard definition will work. If the
regions in between have high doping, then there will be the standard Vbi
as defined in the class, but additional Vbi due to internal barriers.
The context of the problem will make it clear which Vbi to use. Week 7: p-n junction and other diode characteristics ... . Q. You said that Vbi should be computed from the leftmost and the rightmost doped region. However, what if the region in the middle have a high doping -- higher than the region on the right? Then the barrier height will be maximum in the middle of the device. A. I will show it in the class -- it is very important topic and it should be understood clearly. Q. How do you draw band diagram at the metal-semiconductor interface ? Q. You mentioned that the class that the 'electron affinity model' that we use to draw band-diagram involves a bit of cheating. What did you mean? A. The 'electron affinity model' was originally developed for Metal-semiconductor junction by Schottky and it has been adapted to semiconductor heterojunction since then. Often it gives good results, however, the main limitation of the approach is the following. The 'affinity model' is based on bulk workfunction, however, the surface of a semiconductor is different than the bulk due to formation of surface dipole. When an interface is formed, the surface dipole from the two semiconductor will contribute in opposite direction. If they cancel each other, the 'affinity rule' will give good results -- but if they do not, then one must additionally account for this interface dipole to obtain correct band-discontinuity. A simple but power procedure is given by ( Ruan/Ching, An Effective Dipole Theory of band lineup in Semiconductor Heterojunction" JAP, 62, 2885, 1987.). Note however that if you use a quantum mechanical simulator that accounts for the wave-functions spillover explicitly, then you need not account for it during the calculation of band-discontinuity. There are poor man's version of the calculating or estimating the dipole such as 'common anion rules' which explains how bands move when atoms are changed. For example, for Al_(1-x)Ga_x As/ Ga As system, As is the anion and anions dictate valence band, while cataions dictate conduction band. Therefore, if we wish to change the conduction band without affecting the valence band, we will change Al concentration, but not touch the As concentration, etc. Some review questions ... 1) What is the very
first step of drawing a band
diagram?
2) What is the third
step (hint: it has to do with vacuum level)?
3) What is Vbi ? What is the Vbi for a homogenous
semiconductor?
4) If left side of a
junction is more heavily doped than the right
side, which side will have more depletion? More voltage drop? 5) What is difference
between a homojunction and heterojunction?
For which type of junction do you expect a notch at the junction? Q. I know I can differentiate Avanlanche and Zener breakdown by noise, but I still do not understand the difference between Avalanche and Zener tunneling from a physical perspective? Zener tunneling on the other hand involves tunneling of valence band electrons directly into conduction band. It is like opening the floodgate and creating flood of electrons in the conduction band in the process. (very different from avalanche) Review Questions (Lecture 18) 1)The electric field in
the minority carrier region is defined by majority carriers. Is this true? Why or why not?
2) What is the law of
mass action for a biased junction? When do we use the law?
3) Minority carrier
flux is not a constant when traps are present. How is the current continuity
preserved?
4) What is the
difference between Avalanche multiplication and Zener tunneling? How do we differentiate?
5)If Esaki diode must
be degenerately doped, what statistics
should we be using? Q. You said that the I-V region with slope of 2kT/q involves traps? What is a trap? How do we differentiate it from a dopant? A. If you have a high-powered microscope (electron microscope for example), you can see individual atoms. Among rows and rows of Silicon atoms, the atoms of Cu or Au (traps) will be indistinguishable from the dopants Phosphorous or Boron. So you can not tell them apart by looking at them physically. However, you can tell them apart by looking at the location of their energy level with respect to conduction or valance bands. Atoms with energy levels very close to the conduction or valence bands are called dopants, while atoms with energy level in the middle of the band-gap (almost equidistantant from conduction and valence bands) are called traps. Since traps energy level is in the middle of gap, they can help ferry the electrons from conduction band to valence band easily -- and therefore act as great recombination center. Q. In one of the HW problems, they say that in a given region the trap density is infinity and therefore the lifetime tau is zero. Does it mean that the excess carrier density is zero as well? How does it compare to metal contacts?A. You remember that the solution of the diffusion equations with recombination is delta n = A exp(-x/LD) + B exp(x/LD), where LD= sqrt(tau*D). We set B=0 so the solution does not blow up. Now as tau goes to zero, so does LD, so that exp(-x/0) = exp(-infinity) = 0, so delta_n=0 in this region. Physically, it means that there are so many traps that the minority carriers disappear by recombining with majority carriers as soon as they enter the region, so there are no excess minority carriers in that region. In some sense, the metal contact behaves exactly the same way. As soon as the minority carriers enter metal, the lattice vibrations swat them left/right until they lose all their excess energy and join the majority carriers -- here again the minority carrier lifetime is zero. Q. You say that the depletion region does not have mobile charges. Does it also mean that it does not have any electric field? I am confused, because I thought depletion region does have electric field. A. Remember the formula that dE/dx= (p-n+ND-NA)/eps. Even if mobile p and n are absent, ND and NA are still there. When you integrate over the charges, you will find that there is an electric field. Now if you assumed an extreme case, where ND and NA are also zero (e.g. in the intrinsic region), the equations above says that right hand side is zero, and that there will be no CHANGE in the electric field. If the field was zero when it entered the intrinsic region, it will stay zero -- but if it were non-zero before entering the intrinsic region, it will retain the same value. Q. For a pn junction, the band of each side with be bent and p/n side will have the same fermi levelfinally in equilibrium. Suppose I have two different metal, like Au and Co with different work function separated by a thin Al2O3 tunnel barrier. The Al2O3 barrier is thin so that electrons can tunnel through. In this case, can I follow the same rules as before. A. Yes. Recall that in a metal, the Fermi level is high-above the associated conduction band. Therefore, if you follow the same rules as you did for p-n junction, you will find a bending in the conduction band of the metal at the Au/Al2O3 interface, and Co/Al2O3 interfaces. In the absence of Al2O3, this metal/metal discontinuity is essential to interpret properties of thermo-couples. Q. Related to the previous Question. In equilibrium state, if I connect a voltage meter across the junction, should I detect a voltage signal (Any difference from pn junction)? A. No, because voltmeter does not measure voltage,different, but the difference in Fermi-levels (or electro-chemical potential). At equlibrium, the difference is zero, and so will be the reading of Fermi-level. There are however ways to measure the build-in voltage by ejecting electrons by X-ray from various depths in the sample. Q. Input to a diode is I*V. Assume purely radiative recombination, so that output of the cell is I*Eg. In some cases, I*Eg > I*V, output is greater than input. A. This is a beautiful problem. The extra-power comes from the environment. It is opto-electronic cooling, similar to thermo-electronic cooling. It is really a fundamental principle of forward-biased diode -- it involves cooling of carriers moving against the field (a topic that Landauer discussed in great detail -- I will find you a reference). Please see the article below http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.108.097403 Q. I am reviewing the previous materials, and on slide 26 of Week7-L18-20 there is a question concerning dead space. Is it true that if W (depletion width) is less than the mean-free path (dead space), then avalanche breakdown will not occur?A. Yes. Some of the modern transistors have very short depletion width and breakdown voltage can increase due to dead space effect. Recall that Silicon has a bandgap of 1.1 eV. Therefore, in order for the avalanche process to start, the first electron rolling downing the hill must acquire at least 1.1 eV of energy (actually 3/2Eg~1.5eV), before it can kick one electron from the valance band to conduction band (i.e. you have to acquire a certain amount of money before you give someone in charity!). Until this energy is achieved, no impact ionization can occur -- hence the name 'dead space'. Therefore, when we do calculation for impact ionization, the limits of integral should be not from zero to L, but lower-limit is dead-space, the upper limit is (L -- dead-space). Q. Also there are several journals on the internet that work on reducing noise in APD. They say that the dead space effect will introduce noise into the diode. Why would dead space introduce large amount of noise in avalanche breakdown? Is that the same noise you have mentioned in the class, which actually differentiates avalanche breakdown from zener breakdown?A. Yes, APD is more noisy than Zener diode, but the noise from dead-space could be an additional contributor. You see, because the electrons must have at least 1.5 eV of energy before avalanche. Different electrons acquires this energy at slightly different times, therefore the initiation of the avalanche becomes a statistical process. This could lead to additional noise -- but I thought this was not big. Perhaps I was wrong. Q. You have mentioned that at high forward bias, the diode current
saturates either due to ambipolar transport and due to series resistance
drop. How do I know which one is dominant? A. It depends on the doping of the n and p regions. If the doping is high, the voltage drop due to series resistance will be small; the ambipolar region will show up before the roll-over due to series resistance. If the doping is low (and the resistance high), the series resistance drop will push ambipolar transport to higher voltage. A. The recombination dominated n=2 regime occurs at low voltage -- here the electrons need to climb only halfway up the barrier before recombining. The n=2 ambipolar transport occurs at high voltages, this reflects the reduction of minority carriers due to electrostatic effect, when injected electrons and holes become equal. Week 8: AC Response and Schottky DiodeFew review questions. 1) What is the difference between AC and DC response of a diode? What is the reverse bias ac conductance of an ideal diode? 2) What carriers are responsible for diffusion capacitance (minority or majority?) 3) What is dielectric relaxation time? What does it explain? 4)Which side of a P+/N
diode will most of the voltage drop. Which minority carrier diffusion should
we worry about?
5) Shottky diodes are called majority carrier device. Can you explain why? 1)Is there any
diffusion capacitance in Shottky barrier diode. Why or why
not? What about junction capacitance?
2)Some say that Shottky diodes are faster
than p-n diodes
Can you explain why? 3) We have used thermionic emission theory for Shottky barrier diode, and diffusion theory for diodes. How would I know which one to use in a given situation? 4)The I-V
characteristics of a Shottky diode and classical
diode looks very similar. How would you know the difference?
5)What do you need to
do to a diode in order to make it like a Shottky barrier diode (Hint. Think minority carriers)
Q. The symbols of Shottky and Zener diode look the same. What is the difference? A. I could not recall the difference during the class. Now I remember that there is an extra-turn in the Shottky barrier symbol. See the symbol list at http://en.wikipedia.org/wiki/File:Schottky_diode_symbol.svg and compare with Zener diode. Q. When explaining the band-diagram of MS junction diode, you said that it is similar to P+N or N+P junctions (heavily doped side have negligible depletion and voltage drop) and you drew the electric field continuous at the M/S junction. However, given that these are two different materials, should the field be continuous? A. I am glad to see that you have sophisticated understanding of the band-diagram. Indeed, since M-S junction is a heterojunction (abrupt discontinuity in conduction and/or valance bands) with two different material on two different sides, therefore it is the displacement field that will be continuous, and the electric fields will change by the ratio of the dielectric constant. However, since the voltage drop is neglible on the M-side, we do not need to know either the carrier concentration (except the fact that it is very high), nor do we need to know the dielectric constant (poorly defined anyway). But you are absolutely right that conceptually at least, the fields must be different. Q. What do you mean by the statement 'there is no minority carriers in Shottky diode -- as soon as they are injected, they RELAX and join the majority carriers." How do they relax? A. -------------------------------------------------- First reread the answer to the question that was discussed above: "In one of the HW problems, they say that in a given region the trap density is infinity and therefore the lifetime tau is zero. Does it mean that the excess carrier density is zero as well? How does it compare to metal contacts?" You remember that the solution of the diffusion equations with recombination is delta n = A exp(-x/LD) + B exp(x/LD), where LD= sqrt(tau*D). We set B=0 so the solution does not blow up. Now as tau goes to zero, so does LD, so that exp(-x/0) = exp(-infinity) = 0, so delta_n=0 in this region. Physically, it means that there are so many traps that the minority carriers disappear by recombining with majority carriers as soon as they enter the region, so there are no excess minority carriers in that region. In some sense, the metal contact behaves exactly the same way. As soon as the minority carriers enter metal, the lattice vibrations swat them left/right until they lose all their excess energy and join the majority carriers -- here again the minority carrier lifetime is zero. ----------------------------------------------------- Now getting back to the original question: The loss of electron energy by being swatted by lattice vibration is called 'energy relaxation' for the electrons. Once they join the majority carriers, their response is dictated by dielectric relaxation times -- as discussed in the class. Q. If Schottky barrier diode is faster than classical tubes, why is it that we do find more widespread use of these kinds of diode? A. Controlling variability of Schottky barrier diode is a challenge, making their integration in large scale circuits difficult. Therefore, most of the applications are generally limited to small or medium scale circuits. Q. During discussion of thermionic emission current, you said that we only have to worry about electrons that can go over the barrier. Electrons can tunnel below the barrier as well. Why do we not account for them? A. This is a wonderful question. You are right that electrons can tunnel, especially those with energy just below the barrier. What people do however is that they find the expression of the current assuming there is no tunneling. Then the adjust the 'effective barrier height' slightly lower than theoretical barrier height to match the experiment. The reduction is barrier height accounts for the fact that the top of the barrier becomes transparent to the tunneling electrons and therefore should not be viewed as contributing to the barrier height. Q. I used to build Ge diodes for low-resistance circuits? How does it fit in with what you are discussing here? A. Very simply. If you write out the expression for diode, i.e. I=I_0*(exp(qV_A/kT)-1), you will notice that I_0 depends on n_i^2 directly. Since n_i^2= NC*NV*exp(-Eg/kT), therefore lower bandgap material like Ge has higher n_i^2 compared to higher bandgap material like Si. Therefore, for the same applied voltage, V_A, Ge will have higher current compared to Silicon, for example. Now since conductance R_FB=R_s + m*kT/ q(I+I_0), it should be clear that Ge with higher I and I_0 should have significantly lower small signal resistance than Silicon. Q. You said something about the sign of x_n = const* sqrt( V_bi - V_A) formula. I did not understand what you were talking about. A. I tried to distinguish between polarity of the battery (V_A) applied to a terminal and the sign of V_A used in the formula. You remember that we always ground one of the terminal of the diode. If we want to forward bias the device, then we use the terminal of the battery that will reduce the barrier height from Vbi to something smaller. In doing this, we recall that the band goes up when you touch to the device the negative terminal of the battery, while the band goes down, when you apply the positive terminal of the battery. Therefore, if p-side was grounded, then to forward bias the diode, one must connect the negative terminal of the battery to the n-side. Conversely, if n-side had been grounded, positive terminal of the battery should be connected to the p-side of the device. The opposite is true for reverse bias. When using the depletion formula however, we should forget about which terminal was connected to which region. Now, if the barrier was lowered because of the battery connection, VA in the formula is positive; if the barrier went up, VA in the formula is negative. If you keep these two concepts distinct, you should have no problem. Q. I can not recall if you answered this question before. You said in the class that in a Schottky diode, the metal plays whatever role the semiconductor side wants it to play, i.e. if the semiconductor is n-type, metal will p+; while if the semiconductor is p-type, the metal should be n+. Now I thought metal has a lot of electrons, shouldn't it be always n+? How can it be p+ because there is no acceptor doping?A. A metal has equal number of electrons AND protons -- it is charge neutral. When it comes in contact with a n-semiconductor, some of the electrons in the metal are pushed out -- now you have fewer electrons than the number of protons, so metal becomes p+. In contrast, if the metal was in contact with p-semiconductor, electrons will be attracted to the junction, metal will borrow some extra electrons from the contacts. Now you have more electrons than protons, so you have a n+ material. Q. Why is it that image-force lowering plays such an important role in defining the the reverse bias current in Shottky barriers, but has no effect on p-n junction diode? A. The answer lies in the 'doping' asymmetry -- MS junction is one sided junction, therefore anything that happens at the interface affects the absolute barrier height. A p-n junciton is more symmetric, a phenomenon that occurs between the bulk and the depletion region does not affect the barrier height. The effect in question is attration between electron and its impage force, which creates energy levels below the single partcle spectrum (just like excitons). The phenomenon occurs in all junction; however a slight reduction in energy of the electron as it is entering the depletion region is much smaller than that of the one sided junction. Hence the difference. Week 9: Physics and Design of Bipolar transistorQ. When we used transistors with a given beta in ECE255, we assumed that it can not be changed. Are you saying that beta can be tailored by transistor design? A. Of course. That is the whole point of the course. You can get any gain you want by changing base width, emitter and base doping, etc. We will discuss these in greater detail in next two classes. Q. You said that no current from base contact can surmount the reversed-biased base-emitter junction. But didn't we learn that in reverse bias depletion region there are generation? Wouldn't that contribute to current?A. You are absolutely right -- I was postponing this discussion till next week. Today I am focussing of transistor characteristics based on ideal diode behavior (no R-G in depletion region). Q. If beta is defined by transistor geometry, why is it that they vary so much even for transistors from the same batch?A. As you will see small variation in thickness (e.g. 10% variation in the base -- in the order of the size of your DNA) translates to 10% variation in gain. Which means that even if the nominal gain is 500 -- the actual gain may vary from 450 to 550. This is why we need various emitter-feedback resistor or base-feedback resistors to stabilize against fluctuation in gain. This is something you may have heard about in ECE255. Q. When you say 'common emitter' configuration has emitter as a common terminal between input and output -- does it also mean that emitter is grounded? A. It often is ... but as you realize that it need not be. Ground defines a reference point for all potentials -- in principle it any terminal may be grounded, it will not change the input/output configurations or the designation of something being common emitter vs. common base. Q. I am confused -- I understood how single diode works, but when you put two diodes in series and put all these extra-voltages, are the electrons coming from emitter and collector collide. The holes are going to knock off the electrons.A. Consider the analogy of planes -- when small planes and jumbo jets fly, they do not collide because they operate at different altitudes. If someone was looking down from outer space, they may be worried because the planes might seem to be operating in the same region, but once the concept of difference in altitude is clear -- everyone can rest easy. Same is true for electrons and holes -- electrons live in higher-altitude conduction band, the holes in lower-altitude valence band. They can fly in the same space without ever colliding with each other. Which transistor configuration would you use for high current gain? Common Emitter or common Base? Why isn't there a common collector configuration? Will inverted active operation of common emitter transistor can be viewed as forward active model of common emitter configuration? Q. What is difference between single and double HBT transistor? Was the first transistor single or double HBT? Why are HBT transistors fabricated in mesa configuration while double-diffused homojunction transistor fabricated in planar configuration? Q. Common Emitter configuration has smaller breakdown voltage than common base configuration.Is this true? Do you understand why? Do you expect any gain for inverted active operation of bipolar transistor? Why or why not? 1) How does BJTs amplify the input voltage by cutting the connection between electron current and hole current in half? A. By the phrase 'cutting the connection in ... half', I assume you wanted to ask "How does separate control of electrons and holes allow voltage amplification?. Let us start with a diode: You may remember that the electron current in the p-side of the diode is I_n=qD_n*n_i^2/N_A ( exp(qVa/kT-1). Similarly, the hole current in the n-side of the diode is I_p=q D_p*n_i^2/N_D (exp(qVa/kT-1). The ratio of the current -- I_n/I_p=N_D/N_A .... and if N_D > N_A, then I_n > I_p. In other words, if the doping of the two sides are asymmetric, the electron current is 'amplified' over the hole current by the same ratio. Consider the analogy of two children (named I_n and I_p) playing on a see-saw where the pivot point divides the length by the ratio (N_A/N_D). Obviously, the child on the longer arm I_n will have his efforts multiplied by the ratio of N_D/N_A compared to the child I_p sitting on a shorter arm. The current amplification of a diode is similar. While there is amplification of electron current over the hole current in a diode, a two-terminal device like a diode can not keep the electron current separate from hole current to make the amplification useful. The sole purpose of adding the third contact (i.e. collector terminal) is to take away electron current, while keeping the door shut for holes (in a n-p-n transistor). Now only the hole current comes out of base and the electron current comes out of collector. And we can see the effect of amplification in a collector current vs. base current ratio. 2) Does the equations for Emitter efficiency and base transport factor changes based on if the BJT is npn or pnp?A. Expressions are the same. Only thing that has to change is that for n-p-n transistor, base transport factor involves the ratio of ELECTRON current at the emitter and the collector junction, while for p-n-p transistors, this involves ratio of HOLES fir the corresponding junctions. Q. Why is the breakdown voltage of common emitter configuration smaller than that of common base? A. For any transistor configuration, the bias current at the input terminal must be held constant. Therefore for a CE configuration, IB=constant, but IE is not; while for CB configuration, IE=const, but IB can change. Let us assume a n-p-n structure. When impact ionization is initiated in the base-collector reverse biased junction, transistors configured in CE and CB respond differently. For CB, the extra hole current created by impact ionization can easily escape through the base because base current in CB configuration need not be constant. As a result, the CB breakdown voltage is simply the breakdown voltage of base-collector junction (and is typically high). On the other hand, for CE configuration, the holes can NOT flow out of base contact (IB=const), so it must flow out of emitter contact. However, to get out of the emitter contact, the E-B junction must be further forward biased -- this leads to more electron injection at the emitter contact, more collector current, more impact ionization, etc. and a positive feedback is established. As a result, the breakdown voltage is reduced. If this is not completely clear, you may want to study Fig. 11.9 carefully along with the comments I made. Q. You said something regarding 'Army and Rebels' regarding Kirk effect? Can you explain one more time? A. I was trying to explain Kirk effect by analogy. Assume that Collector doping is equivalent to an Rebel formation -- the length of the formation is equal to the depletion width and the number per-row is equivalent to doping density. Also, assume that the Base doping is like a Army formation. For a given base-collector bias, they are well balanced with each other. Now comes the electrons (like planes) from the emitter-base junction -- they help the Army in the base region, but decimate the Rebels in the collector region. At some point, given enough electrons (planes), the Rebels will vanish and the Army will no longer be balanced by the rebel and Army will push all the way to the subcollector. It is just an analogy to keep you awake, so do not take it too seriously! Q. During the analysis of the Kirk effect, should I focus only on the collector doping? Or, is the base doping also important? A. Here we should only worry about collector doping and the possibility that the electron flux from the emitter (for a npn transistor) is overwhelmed and the junction disappears. Yes, if you want to know the junction widths in the base region before the complete base pushout, you will assume that the net doping of the base has increased to (NB+JE/q*vth), while the effective collector doping has been reduced to (NC-JE/q*vth), where JE is the emitter current coming from emitter-base junction and vth is the velocity of those electrons in the base-collector junction. However, at the base pushout point, the effective collector doping vanishes -- NC-JE/q*vth=0, and this depends only on NC and nothing else. http://www.allaboutcircuits.com/vol_3/chpt_4/6.html http://assets.openstudy.com/updates/attachments/505a04d8e4b03290a4150095-nick67-1348396716776-lec_9_ccandcbdesigns_08.pdf A. My use of the term was confusing. When I used the term "common collector", I assumed a circuit in which you take a common emitter configuration and then keeping everything the same, switch the collector and emitter terminals. In the inverted mode operation, this configuration will not have any current gain. On the other hand, the "common collector" defined in the links is configured exactly the same as the common emitter, except that the output is taken from emitter resistance. Here the transistor operates in the normal mode, and there will be certainly a current gain -- the same as in comment emitter configuration. I know this circuit being as a "emitter follower". Week 11: Introduction to MOSFETA terrific resource: History of Transistors -- PBS documentary -- search for 'Transistor Documentary' and then you can view the 6 videos (1/6 ... 6/6). Also at youtube: 'From sand to chip - How a CPU is made' ,'Get Inside an Intel 45nm Chip Factory' Appropriate for undergraduate students. Lecture 29 Q. When you refer to accumulation of p-doped MOS-C, which accumulation are you referring to? The metal side accumulates electrons, while semiconductor side accumulates holes -- which one should we consider.A. In all cases involving MOS-capacitors, the terminology refers to semiconductor side. When we say that p-MOS has been accumulated, it means there have been accumulation of additional holes in the substrate. Same is true for depletion and inversion. Notes. We did not consider the transition from subthreshold to inversion properly. A rigorous solution is possible, and is given by the F function, seehttp://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-720j-integrated-microelectronic-devices-spring-2007/lecture-notes/lecture23.pdf Q. How does polarization of charges affect inversion or accumulation of charges?A. There are two types of polarization relevant for MOSFET and both do not matter explicitly (they are hidden in dielectric constants). For the semiconductor, polarization affects dielectric constant (depletion width, etc.) and dielectric response time -- the time constant with which majority carriers respond to changes in potential. For the oxides, the dielectric constant dictates the discontinuity of electric field at the oxide/semiconductor surface. - Name one or two important differences between MOSFET and
Bipolar transistors.
- Why is there no operation mode equivalent to inversion in BJTs?
- What is the difference between a normally on and normally off transistor?
- BJTs have three terminals, but MOSFET has four. What is the extra terminal called?
- It takes a long time to invert a channel, but the accumulation and depletion is quicker. Can you explain why?
- What is the difference between surface potential and gate voltage? How are they related?
- If (Ef-Ev)=0.1 V, then at what surface potential will the transitor invert? Are the inversion charges electrons or holes?
- When the transistor inverts (or accumulates), which region sees most of the potential drop?
- What should be the polarity of the voltage to invert a n-type semiconductor? Give simple physical reasoning.
- The word accumulation, inversion, depletion applies to semiconductor side or the metal side?
- MOS transistors are often called MIS transistors. Can you explain why?
- There is a discontinuity between electric field in the semiconductor and insulator. Why?
- Why should the oxide must be amorphous? What would be the problem if it was crystalline?
Q. You say that MOSFET is analogous to BJTs in many respects. There is however no gate current for a MOSFET; does it mean that current gain of a MOSFET is infinite?A. You are right that unless the oxide is exceedingly thin when tunneling current may appear, for all practical purposes MOSFET gate current through the oxide (which is an insulator with 9 eV bandgap) is zero. Therefore, in the strict sense of the word -- current gain is indeed infinite. However, MOSFET is a voltage controlled device and the quantity of interest is voltage gain -- ability to change the output voltage by a certain amount with small change in input voltage. That voltage gain is obviously finite. In the class, I will give example of an inverter to illustrate the point. Q. When you draw band-diagram of a MOSFET, the oxide region is shown to have a constant slope. How do I know that this should be case?A. This is because the insulator has very large bandgap, and since ni=sqrt(NC*NV)*exp(-Eg/2kT), therefore there will be only a few electrons in the conduction band. Since there are no mobile charge and since the oxide is not doped, it means that there is no charge, therefore whatever electric field enters from one side of the oxide will exit the other side unchanged. When you integrate this constant electric field, you get the a linear potential drop -- which is what I drew in the diagram. Q. Can the bandgap be so large that the conduction band is higher than the vacuum level? A. It can not be. Otherwise, it will mean that the electrons from the conduction band will float away from the material into the vacuum, destroying charge neutrality. Therefore, although the affinity can be small, it will always be finite. 1)Why is the small signal conductance and diffusion capacitance absent for MOS capacitors? 2)What is the expression for inversion capacitance? Why isn’t there inversion capacitance in a diode? 3)What is the difference between flatband voltage vs. threshold voltage? 4)When would you use deep depletion formula vs. small signal formula? 5)Explain why there is a difference between low frequency response vs. high-frequency response for a MOS-C, but there is no such distinction for MOSFET. Q. When you were calculating sub-threshold current (in analogy to BJTs), the band-digram was almost symmetric with respect to Source-to-channel and Drain-to-channel. How can diffusion current flow in a symmetric junction? A. Very insightful question. You are correct that had the junctions been symmetric, there would have been no current flow, because any flux from source-to-channel would have been exactly balanced by flux from drain-to-channel. If you look carefully at the band-diagram however you will see that I have the drain-to-channel voltage larger than source-to channel voltage by at least several 10s of mV. In this case, the current flow will be from source-to-drain, as described in the class. Q. When there are different ways to calculate threshold voltage, which one do we take? A. There are at least three ways to determine threshold voltage: First, in CV technique, threshold voltage is determined by transition of capacitance from its value at depletion to its value at inversion. Second, you could look at the linear plot of ID vs. VG for very low values of VD and pick up the point where the current begins to rise linearly as the point for threshold voltage. Third and finally, you can look at the long-linear plot of subthreshold current (ln ID vs. VG) and pick up the point where the exponential curve (characteristic of subthreshold operation) gives way to superthreshold (high-injection) regime. The values you will obtain will be slightly different, but generally consistent with each other. Often you will be given only one, so there is no question which one to pick. If you are given two, you can take either one -- you will get extra credit for checking for the consistency. Q. While I was listening your ece 305 lecture, I came up with a question. MOSFET uses the inversion state when the current between source and drain is going to flow. Then, how about accumulation state? I mean when the source and drain are P+ doped and the body substrate is just normal p doped, the accumulated hole concentration near the oxide-semiconductor junction can enhance the current between source and drain. A. Your analysis and conclusions are correct. The type of transistor you are describing is called a "normally ON depletion mode MOSFET" which is different from the traditional "normally OFF inversion mode MOSFET" that I am discussing in class. In a normally on p+/p/p+ transistor you describe, the source-channel barrier is very small (draw a quick band-diagram to convince yourself), therefore when you apply a negative bias and accumulate holes, the transistor is fully turned on (as you write above). On the other hand, when the transistor is biased in depletion and then subsequently in inversion, the transistor is turned off. Before mid 1980s, most PMOS transistor was like this -- depletion mode devices. It works fine, but it consumes more power and is somewhat slower. Q. In the simplified bulk charge model, you use the body factor m to account for the extra depletion in the drain size. Since m=1+ C_s/C_o, therefore its value is different below inversion (~1.1-1.3) vs. above inversion (~10-15) because the inversion charge changes the semiconductor capacitance. Which m should we use? A. We should use m calculated for below-inversion condition. This is because the goal of the correction factor
in the bulk charge theory is to account additional depletion near the
drain (reflected in the below-threshold m), because the inversion charge has already accounted for in Co(Vg-Vt-V). If we use the value of m above
inversion, there will be a double counting of inversion charges. Week 14: Advanced MOSFETAfter going through the lectures, you should be able to follow general article like this and understand the basic ideas of what they are talking about. Transistor Wars http://www.serkantopaloglu.com/archives/271 source: http://spectrum.ieee.org/semiconductors/devices/transistor-wars/0 Q. You say that the electric field at the oxide/semiconductor junction should be discontinuous by the ratio of the dielectric constants. And yet, you did not say much about oxide/metal interface. Shouldn't there be similar discontinuity. A. It is difficult to define dielectric constant of a metal. Dielectric constants is related to shift in atomic wavefunction so that the electron-wave functions and the protons are no longer co-centric. A tiny shift gives rise to dipole charge and the dielectric constant. For an insulator, the field from outside easily penetrates into the material and dielectric constant is easy to define. For metal at low frequencies, fields are screened by surface charge and cannot penetrate the bulk. In such a case, it is difficult to define a dielectric constant. Therefore, we can not determine the exact value of the field inside the metal (although the oxide side is completely known). Not knowing the field inside the metal does not matter for us. The area under the field is very small and does not contribute anything to the voltage. This inability to define the field in the metal is the reason we always start integrating the field from the bulk semiconductor side, not the metal side. Q. While discussing short channel effects, you mentioned that one of the ways to control short channel effect is to reduce the doping. With reduced doping, the WS increases as well. That should make the transistor less controllable, not more. Moreover, if I change doping, would it not change my VT. A. You are correct on both counts. First, recall that reduced doping will not only increase the WS, but also WT (the bulk depletion). Therefore, the total area of the trapezoid is now larger. Therefore, although WS does eat into the area of the trapezoid, the fractional change is smaller and hence the threshold control is better. Second, yes, VT will change, but there are other ways to manage VT (gate workfunction, pocket implant, backgate bias), that can alleviate the problem. VT is important, but control of VT is even more important. Q. You say that various non [100] directions are being used for higher mobility channels in modern transistors. Why were they not used before? A. You may recall that the number of bonds in [100] surface is the smallest, therefore managing interface defects are the easiest in this direction. Other surfaces increase mobility, but also increase reliability concerns. Q. For the NBTI simulation HW problem, I find that Ni(t) vs t is linear with a change of slope at around 0.11 secBased on our derivation in class, we should expect the a plot that goes by a power of 1/4 throughout duration of t. I couldn't understand why the slope changes. A. This is a very good question and I am glad that you noticed it. The analysis I did in class is a asymptotic analysis, i.e., applies to longer time. The approximation dN/dt~0=kf(N0-N)-kR*NH*N. does not apply for short time. At short time, dN/dt ~ kF(N0) because N is too small initially. Only when N builds up to some quasi-equlibrium values, will the 1/4 exponent emerge. If you really want to know more, you could read my paper http://www.sciencedirect.com/science/article/pii/S0026271404001751 Q. I am often confused/surprised by the approximations you make, such as flat quasi Fermi levels, setting dNit/dt=0 in the NBTI calculation above. How do you know when and how to approximate? A. Once you have explored a region very well, you begin to notice that there are shortcuts that you take to go between two points. You often discover these shortcuts accidentally -- while solving a problem numerically, you may notice that for some applications some quantities are much smaller than others; you simplify the problem by dropping the terms and see if you still get the final result. Once you have solved a lot of problems, you begin to spots these shortcuts more easily and it becomes a habit. So there is no rule of discovering these shortcuts, with time it comes to most people naturally. For now, I hope you will accept it as gifts from generations past. Notes and comments on Strain Mobility: Q. You gave a generic description of lattices and strain. Give us some specific numbers and illustrative examples. A. Take Si and SiGe as an example. Lattice constant for Ge .. 5.64 A and Si 5.43 A. Si on SiGe substrate has biaxial tensile strain. SiGe S/D with Si channel has unixiaxial compressive strain. Electron mobility for uniaxial tensile strain: mu_<100> better than mu_<110> Hole mobility for uniaxial compressive strain mu_<110> better than mu_<100> Q. How are the node numbers defined.A. The actual channel length is defined by the S/D distance distance just underneath the channel. On the other hand the node number is often associated with edges of the gate defined by lithography. Actual channel length is always smaller than the node number. Details of how the numbers are defined are found in ITRS roadmap. Some more notes are given below. From ITRS 2007 Roadmap: Mobility Enhancement Factor due to Strain (ratio). ratio = [enhanced mobility]/[reference mobility], where [enhanced mobility] is the actual mobility including the enhancement due to strain, and [reference mobility] is the mobility in the absence of strain. Following the literature, the value of ratio is limited to a maximum of 1.85. Mobility enhancement was implemented in product in 2004 to meet the required saturation drive current, and hence the coloring for extended planar bulk is initially white. However, there are numerous approaches in the literature for mobility enhancement (including global strain using thin silicon epitaxial layers on SiGe epitaxial layers, different process induced strain approaches such as strained thin overlayers of SiN, selective epitaxial SiGe in the PMOS S/D and selective epitaxial Si:C in the NMOS S/D, hybrid orientations, etc.). As we continue to scale MOSFETs, it becomes more difficult to maintain the strain, and it is unclear what the optimal approach(es) will be and how to integrate them into the process flow. Consequently, the cells are colored yellow in 2009, when Lg=20 nm and the doping approaches 4.5E18 cm-3 according to the MASTAR modeling. For UTB FD and DG, the color is red because strain and enhanced mobility are less well understood in these structures. I think all items that are marked red the ITRS report (link : http://www.itrs.net/Links/2007ITRS/2007_Chapters/2007_PIDS.pdf) Means "MANUFACTURABLE SOLUTION ARE NOT KNOWN". This is mentioned in the legend on the pages 22,24,26,28). MASTAR is a modelling tool used by ITRS to obtain various device paramaters, given their structure. More details can be found at: http://www.itrs.net/Links/2007ITRS/LinkedFiles/PIDS/MASTAR5/Mastar5_ITRS_2007/instructions.pdf. The tool can also be downloaded from ITRS website. The following is from the user manual of MASTAR: Mastar 4 is a computing tool especially conceived for the calculation of the electrical characteristics of advanced CMOS transistors based on different technologies such as planar bulk, Double Gate (DG) or Silicon On Isolator (SOI). The calculation is based on analytical drift diffusion equations, which depend directly on the major technological parameters, such as gate length, channel doping, oxide thickness, etc. This application allows the user to evaluate immediately the impact of these technological parameters on the main transistor characteristics such as the threshold behavior, performance values or time delay. Moreover, the influence of physical secondary parameters such as mobility, poly depletion and dark space can be visualized giving a deep insight in the physics of CMOS devices. An extension towards ballistic transport is planned and will be included in one of the following releases. ------------------------------------------------------ Some suggestions 1. There are so many symbols, it will be helpful if they were all consistent and we could have table. -- Okay. Symbol Table Errata
The reverse bias conductance expression should have a 1/4, rather than 1/2 (after taking the derivative). A |