You can follow along with this presentation by going to nanohub.org and downloading the corresponding slides. Enjoy the show. So the topic this morning is phonon transport. So we've been talking about electron transport and I'm an electrical engineer but when you do electron transport you always dissipate power and the semiconductor heats up and this is a major concern so it's always intimately connected to heat transfer. Heat transport and this is a topic that our colleagues in mechanical engineering and other fields spend a lot of time talking about but even if you're just interested in electrical transport you have to understand something about heat transport and phonon transport. So the goal of this lecture is to present a you know to try to have a look at phonon transport and see if we can establish the main properties and principles and theoretical expressions that people use just by using what we've developed so far and adapting it to phonons and we'll see that that's possible and we can get all of the standard expressions that you find in textbooks on heat transport. So just by way of reminder we've been talking only about electrons so far and we've developed current equations for heat transport but that's the heat transported by the electrons which may or may not be the dominant carrier of heat but these are the expressions that tell us which portion of the heat is carried by the electrons. So the notation I'm going to try to use by the way if you printed this out last night I did some editing so there'll be some minor changes from the copy that you have but you can download the the current one from this website over here. So the notation that I'm going to use is that the superscript little Q means it's the heat carried by the electrons the superscript capital Q means it's the heat carried by the phonons. So this is what we've had to date you know we have heat can be carried the first term pi sigma e we always remember when you see an electric field here if I have concentration gradients then more generally I should replace that by the gradient of the quasi-firmula hole so there's an easy replacement when there's uniform carrier density I can just replace that by the electric field. So the first term is something like the Peltier effect the last term has to do with thermal diffusion down a temperature gradient I can write that in one of two ways they're just mathematically equivalent the second way brings in the Peltier coefficient we've got the mathematical definition of these two thermal conductivities now in metals this is most of the story but in semiconductors and insulators it's not you know there it's the heat is carried by lattice vibrations so this is just a brief look at how we think about phonon transport one of the things I want to do is to discuss the differences so you know in electrical transport you can go from insulators to metals you can go your conductivities your electrical conductivities can go over 20 orders of magnitude the thermal conductivities go over what three orders of magnitude or so you know much less you know why is that what's what's the difference between electrons and phonons so professor Fisher did some discussion about phonon so I don't have to say very much but let me just remind you of a few things we actually don't there's just a few basic considerations that we need to know you know we've been talking about electrons you know and the gas of electrons in a metal or a semiconductor is really a very complicated interacting system but one of the triumphs of condensed matter physics many years ago was that you could treat this complicated system you know the excitations of this kind of this complicated system as a set of independent particles or quasi particles those are the electrons that we've been dealing with and those electrons are described by a Schrodinger equation their waves if we solve the Schrodinger equation in a periodic lattice we get these dispersions so that's our E of K because the crystal is periodic the dispersion is also periodic and it's periodic over this Brillouin zone so in 1D K goes to plus pi over a minus pi over a where a is a lattice spacing and if I go beyond that I just repeat the solutions now to describe a particle a wave is not localized so to describe a localized particle I have to go around a certain K or momentum and I have to add different K's and make a wave packet to try to localize it so that way I can localize my electron particle is really a wave packet of wavelengths that are near some particular momentum and if I want to know the velocity of that wave packet or that's my electron particle the velocity is just given by the slope of the E of K all right now we've been approximating this E of K by a simple parabolic approximation this is our parabolic band approximation this frequently works pretty well for electrons we use it very often I mean the valence band of parabola goes down sometimes we have to do better than that and then we treat the entire E of K as given by a computed band structure and that can be done too now how about phonons so we have the lattice vibrating and we can think of we can decompose these vibrations into their normal modes and we can think of the excitations of these normal modes as particles these quasi particles are phonons quantized lattice vibrations so they're analogous to the electrons okay but those lattice waves are in a periodic crystal so we can compute the dispersion of the phonons also and we'll get something that looks like this and since it's in the same crystal with the same periodicity it'll the solutions will be periodic over a boron zone just like the electrons were you know again I can describe a phonon by a wave packet and if I want to know what the velocity of the phonon is I just take the slope of the of the E of K about that point and so group velocity is gradient of omega so normally when we're talking about phonons we plot the frequency versus wave vector I'll use Q so that we can keep things straight when I say K I mean the wave vector of the electron when I say Q I mean the wave vector of the phonon usually we plot energy for electrons and we plot omega for phonons and the vertical axis but energy is H bar omega so they're really the same thing okay so we have these dispersions now we also have simple ways when we want to analytically compute quantities we have simple descriptions analogous to the effective mass description for electrons and the simple descriptions are the bottom type of waves we simplify by treating it with a straight line this is called the Dubai approximation if I want to do the part of the dispersion that's at higher energy that would be called an Einstein model where I just say the the frequency is independent of the wave vector okay and just remember from your freshman physics or high school physics you know you have this mass on a spring problem if I have some spring constant K and I pull the mass down from its equilibrium position it'll just start to oscillate I can describe the potential energy as parabolic you know so I differentiate it to get the force I could set up the equations of motion and I could solve those and I would find that we oscillate in time and you'd find that the frequency is the square root of the spring constant divided by the mass right now if I wanted to know the energy the energy is going to be proportional classically to the amplitude of the oscillation the more more it's oscillating the more its energy is but I know quantum mechanically everything is quantized and this is one of the first problems you do in quantum mechanics you learn how to quantize a harmonic oscillator and the energy levels then come in discrete units n plus one half times h bar omega so these phonons are going to be something like this problem because I can think of the bonding forces between atoms in the crystal like springs and they vibrate back and forth they have some spring constant they have some frequency and it's going to do something similar to this so the general features of our of a phone on dispersion then would look something like this we have different the displacement of the atoms can be in the direction that the wave is propagating that would be longitudinal so this particular mode would be called a longitudinal acoustic mode because it's like the it's like the propagation of sound in the atmosphere or the distribution you know that my sound waves from my voice that are carrying to you the disturb the disturbances longitudinal in the direction of the propagation of the sound waves so this is called a longitudinal acoustic now I could also have the propagation in the orthogonal directions so there are two modes that correspond to the other two directions transfers to the direction of propagation but there's another set of modes up here these high frequency modes and these correspond to atoms in a unit cell vibrating against each other instead of in the same direction the way the acoustic modes are I think professor Fisher said a little bit about those those are a different set of modes they they have a frequency that's much higher and it is much less sensitive to the wave vector so there's a longitudinal optic mode it's called optic because in a polar material these kind of modes can interact with light and there are two transfers optic modes okay so remember that the slope of the omega versus K is the velocity so the slope of the longitudinal acoustic mode is the sound velocity and this is something you look up on Wikipedia you can find the sound velocity of any material it's frequently measured and easily found you know you'll notice that the optical phonons have a small slope so they it's hard for them to carry very much heat because they don't move very fast so the first order people usually say most of the heat is carried by the acoustic modes all right and depending on the particular crystal you'll see sometimes these lines are degenerate they're at the same frequency if you have a non-polar semiconductor at the zone boundary you'll see this LO mode have the same solution as the LA mode so we'll look at a real dispersion here in a minute so here it is so this is a comparison so these are two different dispersions and this is for silicon so the one on the left is for electrons along a 1 1 1 direction in silicon so you can see the two valence bands down there you can see the minimum of the conduction band out near the zone edge that's that ellipsoid one of those six ellipsoids and the conduction band minimum of silicon and you'll notice that the scale there is electron volts you know this is going over quite a sizeable energy scale and the thermal energy is 0.025 electron volts so it's small compared to this scale so you could run this you go in nano hub and run band structure lab and you can produce a plot like this so this is a computed phonon dispersion and so it looks similar you know it's periodic and the same Brillouin zone you can see the longitudinal and transverse modes you can see this is a non-polar material so those longitudinal optic and longitudinal acoustic modes meet at the edge of the zone boundary you can see that the longitudinal optic and the transverse optic are the same at Q equals 0 so this is what a typical phonon dispersion looks like but there's one very big difference that you can see immediately and that is look at the energy scale instead of going in electron volts it's going in hundreds of electron volts you know so remember the frequency was square root of spring constant over mass you know that the atoms are very massive so the frequency is low so the energy h bar omega is much lower so there's a very big difference between the energy scale of the electrons in the phonon and that ends up being quite important so we're going to say more about that later now you can also compute the wavelength of the phonons and I've moved this into the appendix in this material so it's a simple little derivation but I won't go through the algebra but I'll just give you the answer one of the other differences is if you compute the average de Broglie wavelength of an electron in silicon at room temperature it's about 60 angstroms if you compute the average wavelength of a phonon in silicon at room temperature it's about a tenth of that so the phonons have very small wavelengths the electrons have much larger wavelengths when you start talking about things like phonon confinement you'd have to go to much smaller structures to see phonon confinement you can see electron confinement in much larger structures okay so that's really all that we're going to need to know about phonons to see if we can figure out a way to compute the heat conductivity using the formalism that we've developed and just adapting it to these other particles okay so here's the model that we've been using from day one you know it's current you know this is something now I hope everyone remembers if you take away one thing from the course you remember that formula and not just memorize the formula but you remember it's physical significance so 2q over h times the transmission now I'm going to put subscripts on these t sub el is the transmission coefficient for electrons because now I'm going to be talking about phonons too m sub el is the number of conducting channels for electrons and f1 minus f2 it always takes a difference in occupation for current to flow and our two reservoirs our two equilibrium reservoirs we described by equilibrium Fermi functions that gives us the population of the states in the equilibrium reservoirs okay so if we were to adapt this model to phonons we would do something like this we would say we have two big chunks of material that are in thermal equilibrium possibly at different temperatures and we have a material in between that we're interested in computing the thermal conductance of and that material in between might be a nanostructure it might be a carbon nanotube we want to compute the thermal conductance of it but it might actually be a big chunk of silicon and I want to compute the thermal conductivity of bulk silicon my channel could be a bulk material so we'll just characterize that if we know the dispersion of that channel then we'll know how to deal with it so we have to be able to compute the dispersion when we start getting down to nanostructures and you have thin structures and you apply boundary conditions then the dispersion that gets say in bulk silicon if you look at the dispersion in a silicon nanowire it will be quite different because there are different boundary conditions to apply so the dispersion will be different but that's okay whatever the if we can compute the dispersion then we can compute the thermal conductance we have two thermal reservoirs and that means we have a thermal equilibrium population of phonons in each of those thermal reservoirs we describe phonons those are Bose particles so we describe them their occupation by a Bose-Einstein distribution just like we describe the electrons by a Fermi Dirac so it looks similar there's a minus one instead of a plus one and another important difference is there's no Fermi level for phonons there is for electrons so I don't have to the Fermi level helps me conserve particles you know if I know what the electron density is then I deduce the location of the Fermi level to give me that electron density and electrons don't you know it's hard to create or destroy electrons unless I add in recombination generation processes but it's easy to I don't need to conserve the number of phonons it's easy to create phonons so I have two different reservoirs both in thermal equilibrium described by Einstein factors at two different temperatures so I would expect an equation like this to describe thermal transport and I just have to figure out how to change it so I can describe the flow of heat instead of the flow of electricity so how would we do that so our current for electrical current is on top and the expression for heat current is right below it so the way we look at it first of all we look at this there's a two Q there and the Q is because I'm interested charges flowing that's why I have a Q there the two is for spin we just chose you know I could have included the spin in the number of channels you know there's I could have embedded it in that but people like to put it out front that's what the two means so if I'm going to adapt that formula instead of charge what I'm carrying is energy so I have to replace Q by H bar omega but then I have to bring it inside the integral instead of energy I'm going to say D H bar omega just so I can remember that I'm dealing with the energy of a phone on it's the same thing I don't have a two because I don't have spin but I have polarization I have longitudinal and two transverse so people conventionally do is they put the polarization in M so that will have to M the number of channels I'll have channels for each polarization longitudinal and transverse so then you can see it's very analogous I still have the one over H inside H bar omega is the energy heat flow for electrons we had energy minus Fermi level but there's no Fermi level for phonons we have transmission now but it's transmission for phonons we have the number of channels but it's the number of channels for the phonons and instead of an F1 minus F2 we have an N1 minus N2 for both Einstein everything's the same okay now we again we're assuming ideal contacts you know and sometimes when people do these nanostructures they sometimes spend a lot of time thinking about do you really have an ideal contact how do you engineer the contact such that the lattice vibrations don't reflect when they go out of the big region into the small region for electrons it's relatively easy to achieve for some of the nanostructures it's as easy to achieve and people account for that there's a factor for the transmission that comes from the phonons transmitting from the contact into the channel the transmission is not necessarily just due to scattering in the channel okay so here's our expression just like the other one and if we're interested in small temperature gradients then we'll be talking about near equilibrium transport so then we're going to be interested in expanding N1 minus N2 for small temperature differences and we do it the same way we'll do a Taylor series expansion in temperature so N1 minus N2 is minus the partial of N with respect to temperature times the temperature difference and now I can just differentiate that Bose-Einstein factor and that's what I get and let's see now I'll just point out that the Bose-Einstein factor with respect to energy I would get something that is very similar in fact the derivative that I want in doing this Taylor series expansion is dN dt and that's just minus dN dH bar omega times H bar omega over TL now why do that the reason is you remember in our expressions for electrical current we had inside the integral dFDE dN dH bar omega is like the minus dFDE so now I have something that looks like what I did for electrons that's the reason for doing it so our final expression this is just a Taylor series expansion just doing that derivative so now I can do near equilibrium transport because we just put that expression into our expression for heat flow and you can see you're going to get a whole bunch of complicated stuff with a minus sign out front times delta T we're going to get heat flow is minus some constant times delta T that's what we expect heat should flow down a temperature gradient and the KL is all of those terms that I collect up together so the KL looks kind of a little bit complicated but the procedure was straight forward we've got a bunch of constants out front we have a transmission we have a number of channels and we have a bunch of things that came from this Taylor series expansion now remember what the electrical conductance looked like the electrical conductance had some stuff out front which ended up being the quantum of conductance and then we learned that conductance comes in discrete chunks it had a transmission for electrons we've got that for the phonons we've got a number of channels for electrons we've got that for phonons and it had a minus D F D E which acts like a window function what it does is it selects the only where minus D F D E is finite that's the only contribution I get to the integral current only flows near the Fermi level that's where minus D F D E is finite so we think of that as a window function without the energies it tells us which energies are responsible for current flow and for electrons it's just something very simple minus D F not D E and if you integrate that it's an easy thing to do just sit down and integrate that from minus infinity to plus infinity you can see that the integral of that is one so it's normalized so it looks like the stuff in the curly brackets there is we're tempting to call that a window function it looks like it's playing the same role for phonons that the minus D F D E did okay so let's look at that window function for phonons a little more we take that out if I were to integrate that I could see whether it's normalized or not and that turns out to be an integral I don't integrate from minus infinity to plus infinity I integrate from zero in the window function for electrons the Fermi level can be above or below the conduction band edge so I have both minus and positive energies for phonons I only have positive energies you do that integral and it turns out to be pi squared over three that can be done so if I multiply three over pi squared inside then that quantity will be normalized and then I bring the pi squared over three out front okay so I'm going to call that my window function and we'll have a look at that it's normalized just like the other one was we now have Fourier's law we have an expression for the thermal conductance that is some stuff out front which is going to be the quantum of thermal conductance so just like electrical conductance is quantized thermal conductance will be quantized we have transmission number of modes times the window function and the quantum of thermal conductance and the window functions look like this so everything maps very nicely electrical current was the same thing for Fourier's law electrical current is g times delta d we had an expression for g we had a quantum of electrical conductance we had a window function for electrons nice symmetry now if you look at those window functions and for electrons we want to plot it as e minus e f so as I said we can have both positive and negative the Fermi level can be either above or below the channel that we're interested looking at if you look at the red line that's minus d f d e just the derivative of the Fermi function you can see it's piqued it has a width that's a few k t the area under that curve is one because it's normalized if I lower the temperature to 50 kelvin the window function gets much sharper the area under it is still one now if you look at this phonon window function we just plot positive energies okay but it looks very similar to the positive half of the electrons if you look at 300k it's got a width of several k t looks like it's just a little bit wider than the electron one the area under it is one because we've normalized it if you cool it down k t is much smaller it becomes much more piqued just looks like the positive looks pretty much like the positive half of the electron one even though its expression is much more complicated okay so the general models very nice mapping they look very similar so we should see now if we could calculate the thermal conductivity and see what happens then so I'm only going to think about the diffusive regime so we'll take the expression that we've just developed and we will insert in for the transmission and we might we could treat ballistic phonon transport if we wanted to we're just going to look at the diffusive limit so in the diffusive limit it will be the mean free path for the phonons divided by the length of the sample and I'm going to expect the phonon modes to be proportional to a cross sectional area just like it is for the electrons we can work it out from the dispersion or the density of states so I have charge I have heat flow is minus kappa times the temperature difference if I multiply by L over A and divide by L over A I'll get the first expression on the bottom on the left the reason for doing that is that delta t divided by the length of the sample looks like a derivative so it's like a dt dx and I know that conductance is conductivity times length times area divided by length so what's in the parentheses there is really the thermal conductivity in watts per meter kelvin so we'll take that expression at the top which is capital K sub L is the conductance and we'll express it in terms of a current density and a temperature gradient and that brings in this thermal conductivity which is the parameter that people are usually interested in and measure so that gives us an expression the thermal conductivity then in the diffusive limit is this quantum of conductance times the mean free path for phonons times the number of channels per square per cross sectional area times the window function integrated over all frequencies very similar to the expression that we get for electrical current instead of a thermal conductivity we have an electrical conductivity instead of a gradient in temperature we have a gradient in quasi-firmly level we have a different sign because electrons are negatively charged the expressions for the lattice thermal conductivity and the expressions for the electrical conductivity that we've seen earlier look very much the same alright now I just want to do a little bit of algebra here if I divide that expression by the integral of m divided by a times window function integrated over frequency that's what I would define as the effective number of channels that are participating in heat flow because the window function is selecting out the ones that contribute to heat flow so that's what I call bracket the effective number of channels now what I'm left with there it looks like I'm averaging a path and weighting it by some quantity m times window function so I'll define that as my average mean free path so it just allows me to write the equation in a simpler form the thermal conductance is quantum of conductance times the effective number of channels that participate in thermal heat flow times the average mean free path and the precise mathematical definition of those average quantities is given over here just a simpler way to remember it so again earlier we wrote the electrical conductance so it looks like everything is pretty much the same between electrons and phonons alright so now we have to get down to the point where we actually have to work this out so to work this out we need to discuss two things we know all about the window function in the integral so we'll put it in we have to talk about the mean free path for phonons scattering I had a whole lecture lecture 6 on scattering of electrons so I'm just going to have a slider to say a few words about scattering for phonons but we have to figure out what the number of channels is for the phonons also alright but I'm going to take a little detour first of all and I'm going to you will often see expressions you will often see in the literature that kappa is proportional to the specific heat so I want to take a little detour and show you where that comes from just so that when you see that you know what the connection is so what's the specific heat so the total thermal energy per unit volume associated with the lattice vibrations is something that would be easy for us to compute the density of states for phonons and that's something that you can calculate from the dispersion the way we calculate the density of states for electrons from the dispersion and if you weight the density of states the number of states at that energy by the energy of the phonons at that point and weight by the probability that that state is occupied and integrate over all phonon energies you'll get the total thermal energy and then the specific heat tells us how much that energy changes per degree Kelvin change in temperature so if I differentiate that I get the specific heat per unit volume and when I take that derivative the Bose factor has an exponential dependence on temperature so if there's any temperature dependence of the density of states it's going to be pretty small and I can just take the derivative of that Bose factor and we get an expression for the density for the specific heat so this is a quantity that's relatively easy to measure you can look this up easily for whatever material you need what the specific heat of silicon is easy to find and find it on Wikipedia ok now so bear with me and you'll see where I'm going so I have a derivative of n and I write that as proportional to minus the n the phonon energy so if I do that then I get an expression that looks like this so bottom line I get an expression that looks somewhat similar to the expression I had for the thermal conductivity I have a density of states times the window function times some things out front so if you compare those two expressions you can see that they look similar now you remember that we discussed how the number of channels is proportional to velocity times the density of states so I could have written the thermal conductivity in terms of that same density of states and then I would start to have some things that look quite similar ok now it takes a little bit of uninspiring algebra and I've included it at the end to show you that you can write that lattice thermal conductivity in this form one third average mean free path this is a little different mean free path average phonon velocity times the specific heat now this is a very common textbook expression you see this in most introductory solid state books probably lots of places simple kinetic arguments you can use to derive this this is a widely used expression so people will estimate the lattice if people want an estimate of what is the mean free path for phonons what they'll do is they know the specific heat that's a well known if it's a large material that's a well known quantity they'll measure the lattice thermal conductivity they will guess at what the average velocity is they might say well the longitudinal acoustic phonons carry most of the heat we'll just use the sound velocity for them and then they'll estimate what the mean free path is so it's a very useful formula very widely used the difficulty is in these simple derivations it's not at all clear exactly what is that average mean free path how would you compute it from detailed scattering processes what is that average velocity is it the sound velocity or is it something else and the point of all of that uninspiring algebra is that when you go through and do this it's in the appendix to this you can get precise expressions so now you know how those averages are defined and I'll refer you to a paper by Changwook with my co-author on this talk this morning he was recently published in JAP where he shows you some specific results and for example that average velocity can be quite different from the sound velocity so the advantage of our formalism is that it tells us exactly what these are notice that that mean free path the capital lambda is the V tau that's what people usually call mean free path if you're using a Landauer approach you want the mean free path for back scattering and we talked about how there's some statistical factors in 3D it's four thirds times V tau so what was the point of doing all of that well as I mentioned you find this expression many places and when you look at this you might wonder how does this expression relate to the ones that you've developed from the Landauer approach it's the same problem so they have to be mathematically equivalent and we've just shown you that they're equivalent but now we also have precise definitions of those quantities alright if you want to see how that's derived there are many many places that you can have a look at that and here are two references that I use okay now when we do electrons we make a lot of use of a simplified dispersion we use the effective mass approximation frequently and sometimes it gets dangerous with students because you get so used to using the effective mass approximation you forget that it doesn't always work sometimes you have to go back and use a better band structure but it works very very frequently we rely on it there's an analogous simple model for phonons called the divide model so let's take a look at that and go back to the effective math why is it works so well remember when I showed you that dispersion the range of the dispersion or the width of those bands the bandwidth went over electron volts you know the electrons have kt of thermal energy or something the Fermi level is usually near the bottom of the band and not too far above it so the electrons are always down near the bottom of the band usually don't get too far away from the bottom of the band and that means this parabolic assumption usually works pretty well we can get by not always but frequently so we rely on it a lot now what happens for well first of all what is the divide model so the divide model is we have these acoustic modes I could fit them on average you know they have two different velocities so I could take a slope of a line that gives me a velocity that I would call the divide velocity which is an approximation to those modes and I could just approximate them by a straight line if you do that you can compute the density of states whenever we have a dispersion we can compute a density of states and I won't go through the details here but it's done the same way with the electrons these are the expressions we get for the density of states it's interesting that the the number of channels is velocity times density of states there's an h over 4 so you can see that the velocity is constant because the slope is constant in the divide approximation so the density of states goes as h bar omega squared and the number of channels is h bar omega squared they have the same dependency so first point is if most of the heat is carried by phonons that are near the center of the Brillouin zone are relatively close where this is a good approximation then we're pretty good now just as a caveat if you go and look in standard textbooks you'll see that the density of states isn't quite the answer that you'll get in the textbooks just because normally it's done as a function of frequency I've done it as a function of h bar omega just because I want to make the analogy to electrons so it's density of states and phonon energy and all that does is it brings in an extra h bar so you have to worry about that ok now now you recall the realistic dispersions the bandwidth of electrons went over electron volts but the bandwidth of phonons was 0.02 or 0.03 electron volts it was on the order of KT also another point to remember is there's a finite number of states if you go all across the Brillouin zone and add up those states there's a finite number there's one for every atom in the solid so if I were to integrate the states well I have a problem here now if I just integrate that density of states across the entire Brillouin zone so if I go and integrate that across the entire Brillouin zone that approximation isn't very good at the edge of the Brillouin zone I would get too many states so you could say well I could do a better approximation I could have a piecewise linear model but what people do is you have to make sure that you get the right number of states so the way you get the right number of states is you integrate up in energy until you get the right number of states and then you stop and you stop that line so what people will do is they'll take that phonon density of states and they'll integrate up to a frequency omega or energy H bar omega and they'll set that equal to the number of states that are there so there's one for every atom in the solid times three for polarization and we'll get the total number of states that gives me a frequency or a wave vector and I can't go above that or else I've got too many states so then the Debye approximation will just look something like this I'll just have to cut it off because that part of the function gives me all the states that are in the solid now I don't have to worry about that for electrons because that parabolic dispersion if I integrated that over the entire Brillouin zone it would give me way too many states but I'm always right down near the bottom and I'm only occupying a tiny fraction of the states anyway so I never worry about that for electrons but for phonons we have to and we'll see why in a minute so that gives me a Debye frequency I could convert that frequency KT is an energy that's the maximum energy that I allow that linear dispersion to go up to I could convert that into a temperature KT is an energy that would be a Debye temperature and that's a number that is frequently used you can look it up and find out what it is in particular materials it's just a different way to express that cut-off energy cut-off wave vector and we're going to see a little later in this course when the lattice temperature is much less than the Debye temperature ok so now I could easily go in and I take my general expression for the thermal conductivity and I use my Debye approximation to approximate the dispersion that gives me the density of channels or the number of channels M of omega I insert that in the integral I only integrate up to omega and I cross the Brillouin zone and these are classic models these are two really good papers they're still very readable and very well worth reading and still highly cited we'd all hope that we could write a paper 50 years later is still highly cited but these folks did so I'd encourage you to have a look at those they're very readable and still very good so let's take a look at this it's possible to take that actual computed dispersion and to compute the number of channels rigorously from that don't make the Debye approximation this is what the number of channels versus energy looks like for that silicon dispersion that I showed you earlier right pretty messy now if I look at what the number of channels looks like in the Debye approximation that's that red dashed line remember it went as h bar omega squared so you're seeing that parabola for low energies you can see that that fits but it only fits for low energies now if I look at my window functions my window functions are going to tell me which of those channels are occupied and here's the big difference at room temperature the window function is very broad you can see that all of those channels are occupied even the ones for which the Debye approximation is a terrible fit at 50 degrees kelvin the window function is much more peaked near zero energies the the blue dashed line which is going off scale there is the window function at 50k and you can see that it just populates the states in the part of the density of modes that the Debye approximation works so I would expect the Debye approximation to be pretty poor at room temperature but to be very good at 50 degrees kelvin now why does that happen it's because the bandwidth of this is only about 20 or 30 milliolectron volts the width of the window function is on the order of KT so all of the states across the Brillouin zone for the phonons are occupied for electrons the bandwidth is on the order of electron volts only a few states near the center of the Brillouin zone are occupied that are well approximated by a parabola so that makes it very difficult the Debye approximation is widely used but it's not nearly as good as the effective mass approximation that we use for electrons now so here's a situation for electrons this is the same calculation for electrons we showed you the electron dispersion earlier on later and you can calculate the number of modes for electrons that's the red dash line there's a solid black line underneath it that's calculated from the rigorous band structure in 3D the number of channels versus energy for electrons is linearly proportional to energy so that's that straight line you see now if you look at you can see I'm only going up to .2EV if you look at the window function at room temperature the solid blue line you can see that the dashed red line which is the effective mass approximation is falling right on top of the actual dispersion and only states right near the bottom are being occupied and at 50 degrees Kelvin closer to the bottom so effective mass works really well for silicon but the bi approximation doesn't work nearly as well for phonons we're lucky for electrons because many times we can avoid dealing with having to do everything numerically and having a table of E of K over the entire brillant zone we're much less lucky with phonons a lot of these in order to compute thermal conductivities rigorously we need to do the integrals numerically because the simple analytical approximations just aren't very good for phonons but we also have a mean free path inside that integral so let's talk for just a little bit about phonon scattering so I'll remind you lecture 6 I talked about electrons can scatter from defects charged impurities a neutral impurity might perturb the crystal potential and they have a weak scattering cross section also crystal defects and other things electrons can scatter from phonons they frequently do that's frequently a strong mechanism electrons can scatter from surfaces and boundaries surface roughness scattering is a big deal for silicon MOSFETs and electrons can scatter from other electrons but that's usually not so important because it might destroy the momentum of one electron but it's given its momentum to another so on average the ensemble the momentum is conserved and we compute those scattering rates from Fermi's golden rule how about phonons so phonons can scatter from various types of defects it doesn't matter whether they're charged or not but there could be some kind of impurity there could be different isotopes of silicon in the same piece of silicon these things occur in nature there could be dislocations and various other lattice defects that scatter phonons phonons can scatter from other phonons too this turns out to be a more important process and I'll show you why later phonons can scatter from surfaces and boundaries so this is one of the things when people make thermoelectric devices out of nanowires and if you roughen the nanowires you can get a lot of phonon scattering and kill the thermal conductance that's one of the reasons people got excited about their thermoelectric properties because you could reduce the thermal conductance and phonons can scatter from electrons so there's a phenomenon that you'll hear called phonon drag that if electrons and phonons are scattering strongly then if I'm pushing electrons or if let's say I have heat flowing in one direction then that momentum can be transferred to the electrons because if they're scattering so I can push the electrons along with the heat flow if the two systems are interacting that's called phonon drag usually it usually is important that lower temperatures when we don't get other scattering processes so what's phonon phonon scattering is actually quite important now if the crystal potential is purely harmonic then that's what leads to the dispersion and if it's purely harmonic phonons just travel through without scattering but there are higher order terms to give us anharmonic terms here which act as scattering potentials and which knock you from one state to another and the picture could be something like this so here you can see I'm not conserving phonon number but there's no need to I have two phonons with two different wave vectors and two different energies coming in and scattering and what goes out is a third phonon with a different momentum and different energy but I have to conserve momentum and I have to conserve energy so this is a scattering process and you can compute these scattering rates from Fermi's golden rule the same way we do for electrons now this process again we would argue that this process shouldn't have much effect on thermal conductivity because the momentum of the whole ensemble of phonons is conserved so no big deal this is called a normal process so this is the picture in the Brillouin zone two of them coming in momentum is conserved but if I think about the whole ensemble on average nothing has happened now there's another process called a umklap process one's normal you would think the other one would be called abnormal but it's not it's called umklap and it looks like this what if those two incident phonons that interact and scatter this third phonon that conserves momentum what if they had larger momentum to begin with now momentum conservation would say that the red arrow should be bigger but the red arrow bigger means that it's outside the Brillouin zone it really physically corresponds to a lattice vibration with a wavelength that's shorter than the lattice spacing which is unphysical so we have to map it back inside the Brillouin zone with a reciprocal lattice vector G meaning it really belongs over there where the red solid line is now you can see that that has changed the momentum that's flipped it around and that has destroyed momentum that's an umklap process and those umklap processes can lower the lattice thermal conductivity these can occur for electrons too but they don't occur as often because the electrons tend to be near the center of the Brillouin zone they have smaller wave vectors and these kind of processes don't happen as often ok so we need a large population of these large q states in order for this scattering to occur so that means we need a high temperature so that the window function is broad and we're populating these high q states near the zone boundary and if you just go through the Bose-Einstein factor and if I were to say well at high temperatures then the argument of that exponential is going to be small so I can expand e to the x is 1 plus x I guess and that means that the ones will cancel out and the number of phonons is going to be kt divided by h bar omega that kind of makes physical sense kt is the thermal energy h bar omega is the energy of the phonon so the number is just kt divided by h bar omega the scattering rate would be proportional to n which means it would be proportional to temperature so you would expect these processes to get more important at high temperature ok so the scattering summary we have the scattering rate is going to be the sum of the scattering rates due to defects due to bounds scattering off of boundaries due to the umklap processes the mean free path is velocity times that so I could have a mean free path due to defects scattering a mean free path due to boundary scattering a mean free path due to umklap processes if you look at how those work out we would have to go to textbooks and people work this out from a defect the scattering time goes as omega to the fourth that's called rally scattering that's the same kind of scattering scattering light in the atmosphere off a dust particle or something the boundary and surface scattering you think would have something to do with the time it takes for a phonon to get to the boundary so that's velocity divided by the thickness of the sample the umklap processes have something to do with how hot it is and how many of those large wavelength I think large wave vector phonons I have that can lead to those I gave you a simple argument people use a more complicated fitting to describe that generally okay so then we can wrap up by seeing if we can understand some measured characteristics so these are some calculations comparing so this is if you put in the theoretical scattering rates for these different processes work out these integrals compare it to the experimentally measured thermal conductivity of silicon you can see that the theoretical expression that we've developed works well fits the measured data and the measured data has the thermal conductivity dropping at low temperatures reaching a peak and then dropping at high temperatures and if we want to understand that the thermal conductivity has two things it has a number of channels and a mean free path if I look at the number of channels when I'm at very low temperature that window function is very sharply peaked and just a few of the channels near the center of the Brillouin zone are occupied so there's a small number of channels but as you increase the temperature you start occupying more and more of those initially the number goes is t cubed and then finally you saturate because you occupy all of the channels so that t cubed is similar to the is it t cubed or t to the 3 halves it's the same as the specific heat whatever that is but it's the product of the number of channels and the mean free path so if you look at the mean free path if you're at low temperatures and you have very little phonon phonon scattering you might be limited by the boundaries as you get a little bit higher some of these defect scattering mechanisms become important and as you get a little bit there you start getting more short wavelength phonons being populated they are more strongly scattered by those small defects that's why they become important and then when you go to higher and higher temperatures you have more and more of these phonons that can undergo these unclip processes and your mean free path starts to drop so you can understand this curve by the fact that initially you're dominated by boundary scattering because the mean free path is very long you scatter off of the boundaries but the thermal conductivity is increasing because you're occupying more and more channels then you get the channels all occupied and you're dominated by scattering from various types of defects and then as you continue to increase the temperature the path starts to suffer it turns around and goes down ok just then quickly about going back why all of the expressions look so similar electrons and phonons we can go back and forth why do we get such differences why do thermal conductivities vary only over a few orders of magnitude factor of a thousand or so they vary over 20 orders of magnitude well if you look at the mean free paths they're actually relatively close they're not too different so it all comes from the modes the difference is in the average number of channels for phonons it doesn't take much temperature the dispersion is very small on the order of KT all of the modes in the Brillon zone are occupied for electrons you can occupy the modes by positioning of Fermi level you put the Fermi level way below the conduction band and you have virtually none occupied you put them close and you have orders of magnitude more modes participating you push it into the band and you have even more so you can vary the number of modes that carry electrical current by orders of magnitude you can't do very much about that you have different dispersions heavier atoms have lower vibration frequencies which means they have lower energies and lower bandwidths so there's some variation in the bandwidth but it's not huge mean free paths can vary but again they're not 20 orders of magnitude so you get this fast difference between the range of thermal and electrical conductivity now the other thing that I will mention you know I don't have a nice I had this nice beautiful picture of quantized electrical conductance where people did experiments and it just you control the width of a channel and the conductance just went up in units of 2 q squared over h it's hard to do that kind of experiment for phonons you know to vary the width of the channel electrically say and watch those things happen but people have done experiments and have measured this quantum of thermal conductance if you go down to very low temperatures where the transport is ballistic so it's the phonon transmission at low temperatures so that T would be 1 and if you have a very small nanostructure in this particular experiment they had some thin little beams on a suspended nanostructure and they computed the phonon modes and they had 4 of them and when they went down to low enough temperature that they observed ballistic transport they measured for their thermal conductance was 4 times this quantum of thermal conductance so it too is a well established experimental fact that thermal conductance is quantized alright so let me just summarize you can easily extend the model that we had for electrical conduction and you can get all of the same mathematical expressions that people that work on heat transport have worked with for a long time just as for electrons phonon transport heat transport is quantized but this difference in bandwidth has important consequences it means we have to deal with the entire Brillouin zone for phonons and the Dubai model just isn't very good and the fact that we have no Fermi level means you can't control the population of phonons over orders of magnitude so there are lots of good books the first two are textbooks the second one by Gang Chen is more recent and talks about a lot of the recent work on nanostructures and thermal electric effects you might find that interesting these two classic references that I mentioned earlier this is the reference from the group that first measured quantized thermal conductance and all of the results that I've shown here are described in more detail in this paper by my student Chang-Wook who is the co-author with me on this talk here this morning ok so we'll end there and see if you have any questions thank you do we have a simple model to understand the number of models for phonon I mean we're not quite sure about what we're going to study but do we have a simple model to understand the number of models for phonon I'm not as familiar with what people I mean this is a standard thing people do a lot of calculations imposing correct boundary conditions on a beam or a slab or a metal wire or something and there's been a lot of work and there are even standard packages for computing phonon dispersions Chang-Wook which one did you use to compute those phonon dispersions so you might see Chang-Wook at the break but there's there's been an awful lot of work done about how you get the right phonon mode and when you go to nanostructures you know the modes change quite a lot in terms of in terms of our formulation it just means that we have a different M of E that we have to put in you can try phonon it doesn't make any sense to look at phonon as it's that phonon at a scale that's basically a new movement they could have a way to this is in a nanometer in some of the connectors and you said also I'm glad you're today that confined phonon is being important at basically the dimensions of the structure is much more than that and so could you maybe comment on how when does it become important to look at confined phonon band structure as opposed to like a both phonon band structure well ok so there's a question about confined phonon band structures you know I think so my point there is if you think about electrons and you want to know when quantum confinement becomes important let's say you have a well and you want to know whether the energy levels are quantized what you would do is you would look at the de Broglie wavelength of an average electron and you would look at the dimension of the well and if the two are comparable then you would think well I have to solve a particle on a box problem and figure out what they are and as I pointed out the average wavelength of a phonon is much shorter so that would lead you to believe that the quantum confinement effects would occur in much smaller dimensions now on the other hand you just do elastic theory and compute the dispersions and you'll see changes in the modes at much larger dimensions you know you just do conventional elastic theory and let's say you have a slab you impose boundary conditions on it you can you know you have different types of modes propagating in a beam or in a thin sheet or in a rod and these are the kind of calculations that people do on these small structures so they're basically using bulk elastic continuum theory but they're just imposing the proper boundary conditions on a maturel you know column what happens when the main repath of the phonons gets close to important device dimensions and also when do you expect that to look at them yeah so again when I'm responding to these questions you have to realize that I'm a double lead trying to figure out what this you know how I can understand these papers that I read on phonon transport I don't have a lot of experience with this but you know I know ballistic phonon transport is something that people worry about even when you look at those temperature dependent characteristics even when you have a large chunk of material that you're trying to measure the thermal conductivity of bulk silicon and you go down to very low temperatures the mean free paths can get very long and they attribute part of that drop to boundary scattering you know we can estimate phonon mean free paths people like in thin SOI structures I think the lattice thermal conductivity would be considerably less than it is in bulk silicon I think there's a lot of data on this I don't know if it's an order of magnitude less or what and then you have these simple expressions you know you can estimate just as I went through some examples where we estimated the mean free path for electrons in a MOSFET if you know the measured thermal conductivity and you know the specific heat you can estimate what the mean free path would and then you know what that would I guess what that would tell you is that you know your question is what would you expect to see then you start expecting to see boundary roughness scattering so you start expecting to see the mean free path to drop which I guess is probably why SOI films have a lower thermal conductivity than bulk silicon because you're scattering off of the boundaries and another possibility could be that their dispersion has changed so I'm not exactly calibrated on which one is the most important it's probably the roughness scattering you know these nanowires that people the Majumdar group in Berkeley measured they had roughened silicon nanowires I think these were on the order of diameters of 10 to 20 nanometers and the roughness there lowered the thermal conductivity by a factor of 10 to 20 nanometers so at those dimensions the boundaries were enormously important let's see I think so is there any confusion can you add another question yeah there is and I don't know anything about it really I know that professor Fischer here does NEGF calculations of phonon transport so there is a quantum theory of phonon I don't know very much about that work but you know we could go over here to the Burke Center and talk with professor Fischer but I know there are people that do these calculations yeah so and you know also you'll frequently hear people talk about a phonon Boltzmann equation you know and just as I talked yesterday that you can do these things with a Boltzmann equation it's more common to see these kind of calculations done with a phonon Boltzmann equation and then just like there's there is a NEGF to do a quantum version of the Boltzmann equation for electrons there's an NEGF to do a quantum version of the phonon Boltzmann equation yeah can you turn that to slide number 30 which one slide number 30 30 okay this one here whoops was that the one yes talking about this e h e 1 that was are we speaking about the acoustic phonon Boltzmann equation well yeah so that's a good question and that's the difficulty when you go through standard textbook descriptions this is done very simply and you get an argument like this and you know when you do those kind of things you usually think that well this is probably the longitudinal acoustic sound velocity but if you want to know what it really is you know this is what it is and it means you actually have to do this integral you know over the density of states at any energy it's the average velocity of all of the phonon branches that have a density of states at that particular energy and you have to do that integral numerically and you know if I that reference that I gave you at the end the paper by Changwook shows a calculation of this for silicon and compares it to the longitudinal acoustic sound velocity and it actually is quite different you know so it's something that you know it's something that you need to be aware of it's common to use an expression like that and say well I've measured my lattice thermal conductivity I know my specific heat I'll assume that the average velocity is the sound velocity and then I'll deduce what the mean free path is but it's not so simple to know what the average velocity is you've got to do a numerical calculation well so for acoustic it's what five six thousand five six thousand kilometers per second is it and what hundred times smaller just one second here five thousand meters per second or so something on that order pardon me so the optic phonons are very small because their slope is very low but they have some finite slope but you know this would be for the longitudinal there are transverse also they have lower velocities they participate in heat flow also okay let's see slide number nine so you see that the velocity that people would usually think of is the velocity that the heat that the phonons that are carrying the heat is that for the longitudinal acoustic because that's the highest velocity but those transverse acoustic will be carrying some of the heat too and even the optical phonons might have a little bit of a roll okay in the screen blue cough scatter the g vector which takes you between three of these is that going to be transferred to the crystal as a whole or this as a whole yeah does that where did that momentum go I'm not sure that's something we can think about over the coffee break I guess it is preserved but you might lose it in a photo yeah I'm not sure how to answer that question let's think about that a little bit slide 45 this one okay you know that's because you know this is the way I calculate my dispersion and the dispersion is like considered a perfect crystal corresponding by spring you know when you do those calculations of dispersion this is this is the potential that you assume so under those conditions you have a perfect crystal and the phonons just travel through there's nothing to scatter them the dispersion tells me how carriers propagate in the absence of scattering now I've neglected these higher order terms and you know we think of those higher order terms then as perturbations and they knock them from one state to another in that dispersion yeah same way you know electrons travel through the electron dispersion without scattering the scattering potentials are something else


Hello, and welcome to this session in which we would look at a consolidation problem that you could see an advanced accounting or a CPA exam. And on the CPA exam, you may not see a complete comprehensive consolidation problem like the one I'm going to be going over today. You might see a mini-consolidation problem. Nevertheless, the concept is the same. So in this session, I will explain the consolidation process and specifically subsequent to the date of acquisition, because in the prior session we looked at how do we consolidate on the date of acquisition. Now we're going to look a year after the acquisition, how do we consolidate? So it's very important as a CPA candidate to understand this whole problem for two reasons. You could see this in a simulation or from this problem, I can generate 20 different multiple choice questions. So on the multiple choice questions, they may ask you about one thing in the consolidation process. So if you understand the consolidation process, it's very easy to understand that small thing. And sometime you just need good basic. I'm going to repeat, good basic. So you don't need to know advanced consolidation as long as you know your basics, you're strong in your basics, you'll be able to answer many multiple choice questions that deals with consolidation because they only scratch the surface when it comes to the multiple choice. Now whether you are an accounting student or a CPA candidate, I strongly suggest you take a look at my website farhatlectures.com. Now most likely you have a CPA exam course, review course, that's fine. I'm not asking you to get rid of your CPA review course. I'm asking you to add me as a useful addition. I explain consolidation differently than your CPA review course. How so? Well, your CPA review course will assume you know the material and they would review it with you. I don't assume this. I'm going to teach you consolidation and once I teach you consolidation, your CPA review course will do a great job reviewing that information with you. Your risk to try me is one month of subscription, your potential gain is understanding and pass in the exam. If you're willing to take this trade off, go ahead and sign up, try me for a month and see how it works. And if not for anything, take a look at my website to find out how well or not while your university are doing on the CPA exam. I do have lectures and resources for other courses as well. If you haven't connected with me on LinkedIn, please do so. Take a look at my LinkedIn recommendation, like this recording, share it with other, connect with me on Instagram, Facebook, Twitter and Reddit. Now we're going to look at a consolidation problem and I'm going to go over this problem step by step. I'm going to try to slow down. I'm going to try to tell you what questions you could be asked about in a form of a multiple choice. But again, as a simulation, you could see this as a mini simulation, not as comprehensive as the one I'm going to go over, because I'm going to go over the whole process. Starting with, we're going to be buying a subsidiary. We have the subsidiary's book value as of the beginning of the year. We have the fair value as the beginning of the year when we are making the purchase. Current assets, book value 320. The fair value is 320. Therefore, there's no difference. Trademark, we have a trademark on the books are worth 220. The fair value is 220. So the trademark, they have an additional 20,000 and fair value. We have to account for patent and technology on the books. It's worth 320. Fair value is 450. The difference is 120,000, 130,000 and they're telling us this has a 10-year remaining life. Now why are they telling us this has a 10-year remaining life? Because we are going to amortize this.  Therefore, it's not amortized. Equipment on the books are worth 150. Fair value is, I'm sorry, on the books are worth 180. Fair value is 150. They are $30,000 less. Again, they're giving us the life because what we're going to do, we're going to amortize this reduction in depreciation, because notice the fair value is less than the book value. Therefore, when we transfer this asset to our books, we're going to have less depreciation. Liabilities, 420 and 420, they're the same. The net book value, now how do you find the net book value? Sometimes the question is, find the net book value for this company. That's as simple as that. You could have a CPA question that tells, find the net book value. What's the net book value? It's asset minus liabilities, net. Okay, sometimes it's called net book value. Sometimes it's called net asset. Well, simply put, all what you are doing, all what you are doing is taking the assets. If you add up the assets and subtract the liabilities, you'll get the 600,000. Okay, asset minus liabilities. Now also, the net book value is the same thing as the equity. And let's take a look at the equity. 200,000, 20,000 and 380. 380 plus 20 equal to 400,000 plus 200 equal to 600,000. Notice the equity is 600,000, the net book value equal to 600,000. And that could be a question on the CPA exam. They could give you a lot of numbers and saying about assets liability and they'd say, what's the net book value for this company? And that's the net book value, 600,000. Now, why is the net book value important to understand? Because you need to understand the net book value of the fair value. So the company is worth fair value, net book value, fair value, net fair value, net identifiable fair value, 720. Why is that important? Because when you buy the company, you're going to say, well, how much should I pay above this amount? 720. The difference between the net book value and the net fair value is 120,000. So the company is worth 120,000 more on the books. Now, the first thing we're going to do is I'm going to take all these assets that I identify that they have either more value or less value and make a record of them. For example, the trademark, I'm going to have a $20,000 more to allocate to my trademarks.  The annual access amortization is zero. Why?  We don't amortize something that will have indefinite life. The patentant technology 100 and 130,000, it's going to last 10 years. So we're going to amortize it using the straight line, 13,000 per year. Equipment is $30,000 less, less. We're going to amortize this less depreciation over five years, and that's $6,000 less. Now, we're going to assume for this illustration that we paid for this company 900,000. Now, if we paid 900,000, what did we purchase? Well, we purchased the equity. The equity is 600,000. So paid 900 minus the 600, we have $300,000 remaining. What are we going to do with this 300,000? Well, we know what we're going to do with some of it. We're going to allocate 20,000 of this 300,000 to the trademark. We're going to allocate 130,000 to the patentant technology, and we are going to reduce this by 30,000. Because why? Because we purchase something that's worthless on fair value. When we do all of this, we're going to have a remaining. When we account for all of this, reducing this asset, increasing this asset, increasing this asset by allocating 300,000, we're going to have $180,000 remaining. Since this $180,000 remaining cannot go anywhere, in other words, we cannot connect it. We cannot allocate it. The correct term is allocate. We cannot allocate to any particular identifiable asset. Well, guess what? What do we call this? Hopefully we know this. We call this good will because we identified all the assets that we can. Everything else is good will. Therefore, we're going to have a good will that's worth $180,000. Remember, good will will have an indefinite life. Therefore, there's no annual access amortization. Now, when we net out our amortization that we're going to be booking, we're going to find out that every year we have to book an additional $7,000 in amortization because we purchase this company. And because we purchase this company and the fair value is more for two assets and less for one asset, well, we're going to have an access amortization to book every year because we have more assets on the books. Therefore, more assets that's being subject to depreciation, more amortization expense we have to book. Now, also on the CPA exam, they could give you this problem and ask you what is the annual access amortization as a result of this purchase. That's all what they can ask you about. Okay? Or they can ask you what is the new, what's the consolidated patent and technology balance? Well, it's going to be, for example, 450. It's going to be on the consolidated paper 450. So they could ask you many questions. But now we understand where all these numbers are coming from. Now we said we paid 900,000. Let's go through the journal entries when we purchase this company. When you purchase a company, you're going to have an investment in sub 900,000. You're going to credit cash 900,000 to record the acquisition of the sub. On August 1st, the company declared $40,000 of receivable. So the subsidiaries declared it, and we purchased the whole company. By the way, we purchased 100% of this company. We have no minority interests, no non-controlling interests. We're going to debit dividend receivable and credit the investment in son because what happened is the investment account goes down when they declare dividend. In other words, every time the company pays dividend, our investment goes down because we're taking out the money, the dividend, not the money, we're taking out. Actually, it's going to be the money soon. We're going to be taking the money out of income. Okay? Then they pay the cash. Then they pay the cash. We debit dividend receivable, credit cash. Now the company, the subsidiary made $100,000 in earning. That's going to increase our investment because as the company earned money, we're using the equity method as the company earns profit and meant to say money when the company earned income, we increase our investment and we credit equity and subsidiaries earning, which is an income account. Then the last thing we have to do, remember the $7,000 here, access amortization, we have to book that access amortization. We're going to reduce our earnings, basically like reducing earnings by $7,000 and reducing our investment by $7,000. Why did we do this? Because when we purchased this company, we had additional assets with additional values. As a result, those assets would reduce our value. So those are the entries that you have to do as far as purchasing the company accounting for the receivable. You're going to see this $40,000 again. Just write it down. You're going to see this $40,000 again in receivable. Okay? You're going to see this $100,000 again in earning and you're going to see the reduction. So 100 minus 7 is 93,000. You're going to see this earning again soon. Okay? But those are the journal entries that you have to make when you purchase this company. And what I'm going to do next, I'm going to show you what's going to happen at the end of the year when we're going to consolidate those two companies together, the parent and the sub. Okay? So remember, this is one, one, 2022. I'm going to move to the Excel sheet. I'm going to show you the balances for these companies, 12, 31, 22, which is the end of the year. It doesn't matter which particular year. Let's take a look at the Excel sheet and start the consolidation process. So this is the parent and subsidiary as of December 31st, 2021, I'm sorry, 2022 at the end of the year, 2022. And we have the parent, the sub, the consolidated entry and the consolidated balances. We have the income statement. We have the income statement for each company separately. And by the way, when it looks negative, those are credit balances. So revenues for the parent company, 1.5 million cost of goods sold 700,000, amortization expense 120, depreciation expense 80. Then they have additional income from the subsidiary equity and subsidiaries income 93,000. Remember, this 93,000 is 100,000 minus seven. Remember, I told you, you will see this because we are the parent company. The income comes to us and it's 93,000. Therefore, the parent company has net income of 93,000. The sub company has net income of 100,000, just following the same concept. The statement of retained earnings, the beginning retained earnings is 890, the net income, 693, dividend declared by the parent company is 120, retained earning as 1231, 1,413,000 credit balance. The same thing for the sub, we have beginning retained earning, giving net income of 100,000, dividend declared 40,000. You remember this dividend, I hope you do. Then we have the assets. We have 940,000 in current assets. Now we have an investment of 953,000. Now how did we come up with this 953,000? Let me just show you where this number is coming from. We started with 900,000 investment. Then we deducted 40,000 when we had the dividend. Remember, the dividend reduces our investment. Net income increased investment by 100,000, then the access amortization reduced it by 7,000. Therefore, our investment balance is 953,000. We have the trademark 600,000. Patent and technology equipment, those are the parent, book value, those are the subsidiaries, book value. Then we have liabilities, common stock for each one. And by the way, retained earnings goes from here to here. Retained earnings goes to the balance sheet, goes to the balance sheet. Now once again, you know, they could ask you a question as simple as, tell me what's the investment account as of 1231 before the consolidation. Well, 953, I showed you how to do so. So that's another question they can ask you. Now we're going to start the consolidation process. How do we start the consolidation process? The first thing you have to understand about the consolidation process is only the equity of the parent company will survive. So that's a very important concept on the CPA exam. They always ask you about this. You have parent and you have subsidiaries and you have the consolidated amount. And the consolidated amount, only the equity of the parent because the parent purchase the sub. Well, if the parent purchase the sub, there shouldn't be any sub equity. Well, what does that mean? It means let's start by eliminating all the equity of the sub. The sub has 200 in common stock, 20,000 in paid in capital. Let's do that. So what's going to happen now is I'm going to start to zero this down. So I'm going to debit common stock for the sub debit additional paid in capital for the sub debit retained earning for the sub. Okay, and let me show you the entry. So we're going to debit all of those. Let me just go ahead and show you the journal entry, then I will show it to you on a worksheet. And this is what the worksheet would look like. Okay, let me just change the colors here. There we go. Let me make it white. Okay, so what happened is this, I debited common stock for the sub. I debited additional paid in capital for the sub and I debited retained earning for the sub. Now, what do I credit? I credit them against investment. Therefore, I'm going to debit investment 600,000. Now, investment is debited 993. Don't worry about this. This is the 600,000 because I'm going to be using this investment account more, but this is the first entry. And the first entry is to do what? Is to eliminate the sub equity because the only thing that should survive, the only thing that should survive is the parent equity. The only thing that should survive is the parent equity. What does that mean? It means the 600,000 of the parent should be on the consolidated. The 120,000 of the parent should be the consolidated. And the dividend, which we're going to close the dividend soon, because I'm going to close the dividend separately, the only thing that's going to appear here, I'm going to show it to you up front before we even get there. The only thing that's going to be 120,000. I'm going to show you how I'm going to be eliminating this 40,000 in a moment. This 40,000 in a moment. And I'm just going to tell you that this is what's going to survive. Now, the second thing I'm going to do, do you guys remember we prepared this schedule? We said we're going to allocate more to the trademark, 20,000. We're going to allocate more to the patent and technology. We're going to allocate less to the equipment and we're going to add goodwill on the books. You remember we said this? We said we're going to have to do this, right? Let's do this. Now, we're going to adjust the fair value, which is recognize the allocation. Well, starting with the trademark, trademark, we're going to add 20,000. We're going to debit the trademark. We're going to notice here that we're going to debit the trademark account 20,000. We're going to debit the patent and technology. We're going to credit equipment. We're going to credit equipment. We're going to add goodwill of 180,000. And guess what? The difference will go into investment. Give me, give me investment in the sub. I'm going to credit the investment in the sub. Let me just make this bigger so you can see the whole entry. So I debited the trademark patent and technology to do what? To allocate this. And as a result, I'm going to credit. This is the credit of 300,000 up here in the formula. Okay. Remember I had 600,000 credited investment 600,000. No, I'm crediting investment 300,000. Remember when I bought the assets, I bought them with my investment account so I cannot have the assets and the investment. I cannot have both. Simply put, I'm going to go ahead and tell you the investment account has to be zero at the end. The investment account has to be zero because I cannot buy the asset, add the asset to my consolidated total, then keep my investment. Then I have the investment, which is a financial asset, and I have the physical asset on the books. You can't have both. So when you add the physical asset, you have to remove them against the investment. I hope this makes sense. So that's done. Now I have to eliminate any intercompany income. Well, intercompany income. What's the intercompany income? I earned 93,000 from this company. Okay. I have to eliminate this. I cannot have this 93,000 because this is intercompany income. So I'm going to debit 93,000 and I'm going to credit 93,000 of the investment. Now my investment is credited 993,000 in total, which is 60,000 to eliminate the book value of the equity of the sub 300,000 to add the assets and 93,000 to eliminate the intercompany income. Okay. So that's the, let me show you the entry. This way you can see the entry as well. Again, I don't want to keep repeating this. On the CPA exam, they can ask you about one particular transaction and all of this. They can ask you a question as a simple question or a tricky question as what is the investment account at the end in the consolidated balance sheet? It's zero. You cannot keep the investment account. Okay. Or what's the equity consolidated balances? Remember, the equity consolidated balances are only those of the parent company, which is 124 dividend, 604 common stock, 124 paid in capital. Also I have to eliminate any intercompany dividend again, because it's between the parent and the subsidiary. Well, as a result, I'm going to have to credit dividend, credit dividend, 40,000 to eliminate this debit. And I'm going to have to reduce my investment by 40,000, reduce my investment by 40,000. So this is the eliminate the intercompany dividend. And that's done. Now, if I take 953 plus 40, which is 993 minus 993, my investment account is zero. So this is important. So this is important. The investment account will be zero when you consolidate because you cannot have the investment account plus you want to have the assets because you already are adding the assets. You cannot have both. You cannot have both. Okay. Last thing I'm going to do in these entries is recognize access amortization of fair value. Remember, I had access amortization of net was 7,000. Let me show you the journal entry. I'm going to have to add 13,000 and access amortization. That's what's not showing amortization expense because the background have to change the background. There we go. 13,000. I'm going to have to debit equipment, debit equipment. Well, let's keep going with the expenses. Debit amortization expense, credit depreciation expense. So this is the 7,000 difference. I'm going to have to credit depreciation expense 6,000. I'm going to have to debit equipment in additional $6,000. Let me do that. I'm going to have to debit equipment $6,000. Where's the equipment account right here? Debit equipment. Why? Because this is against the depreciation expense and I'm going to have to credit patent and technology 13,000 because I'm reducing the patent and technology every year after year. This $130,000, it's going to be in chipped.  So the following year, I will chip away from the 130,000, 13,000 every year. That will be the case. So this is the to recognize access amortization. Those are the entries. Now we're going to have to do the whole consolidation revenue. The parent company $1.5 million. The subsidiary $400,000. We have no debit, no credit. Therefore, the balance will be $1.9 million consolidated revenue. There is no debit, no credits. Cost of goods sold, $700,000 for the parent, $232,000 for the sub. Together is $9.32. If there's anything here, we'll have to net it out. Amortization expense, $120,000 for the parent, $32,000 for this sub. Then an additional $13,000 of amortization, which is what we're going to add. Therefore, the net amount is $165,000, which is parent plus sub plus debit minus credit. The same thing for depreciation. Depreciation for the parent company. Depreciation for the sub plus zero debit minus the credit will give us a balance of $110,000. The equity in the subsidiaries has to be zero because this is intercompany transaction. And now the net income for the consolidated company is, sorry, the net income for the consolidated company, $693,000. Immediately, once I have net income here, you can plug it here. No, not retained earnings. You can plug it here. You can plug it here. So net income goes from here to here, the consolidated net income, because the only thing that's going to survive is the consolidated net income. Again, the retained earnings, the only thing that's going to survive is the parent company retained earnings. Again, this is one of those from the equity account. The only thing that survived is the beginning company retained earnings. So this is gone, $380 and $380. And we have dividend, the only dividend that survived, $120 plus $40, minus $40 is $120. That's the only thing that survived. Therefore, the ending equity, retained earnings is the retained earnings of the parent company. As I told you, the only equity account that survived consolidation is the equity of the parent company. For retained earnings is $1,413,000. Simply put, if you have a simulation, you can fill out those quickly, especially if you are pressured under time. See, retained earnings, even without knowing what you need to do. $840 will survive, $693 will survive in the consolidation, and the dividend will survive. And the retained earnings should be that of the parent company. Now we're going to do the same thing with current assets. Current assets, $940 plus current assets, $500,000. We'll give us a total of $1,440,000. Now down the road, we're going to learn we might have to debit or credit. It doesn't really matter. Investment in the subcompany should be zero. We already know this. We already did this. Trademark, $600,000 plus $200,000 plus $20,000. And that's going to give us a total of $820. Patent and technology, $370 plus $288 plus $130 minus the $13,000 amortization, it's going to give us $775. Equipment, same concept. Parent plus the sub, plus the $6,000 addition minus the credit of $30,000. We'll give us $446. Goodwill, $00180. Goodwill is $180. Now we have total assets. And total assets are $3,481. Liabilities. Liabilities of the sub plus the liabilities of the parent company, $1,345. Additional paid in capital. Again, those are equity accounts. The only thing that's going to survive is the parent equity. And retained earnings, obviously it's going to go right from up there. Right from the statement of retained earnings. Right from the statement of retained earnings. And make sure your liabilities and equity equal to your assets. And what I did, I went through this whole process. Once again, will you be expected to solve something like this on the CPA exam? I doubt it. They're going to give you all of this. They might only give you to do those entries, for example, or maybe two or three entries like this. I doubted that they're going to give you a complete consolidated worksheet, or they might give you all the numbers that you need and you need to do the debits and the credits. Or they might take this, I will take this consolidated problem and give you 15 different multiple choice questions. I can ask you, what is the Goodwill? That's the only question I could ask you about this thing. Or what's consolidated net income? I can ask you again, what's the ending retained earning? Well, that's easy. It's the parent company ending retained earnings because the only thing that survive with the equity balances are only the parent company. So basically going over this consolidation, over this consolidation process using the equity method will help you tremendously understanding the concept for the CPA exam.  I don't intend to do so, but I strongly encourage you to add me as a supplemental tool to your CPA review course because I explain the material differently than your CPA review course. I teach you. I don't review it. What is it? I did not review. I explain everything in detail, every single step in details. And this is what differentiates me from your CPA review course. Look, make this investment for a month. You don't like it. You cancel. You like it. It's, you feel it's benefiting you. You keep it. The CPA exam is worth it. It's a long-term investment. It's 20, 30 year investment.


Very large and important work for me again a family painting the Sherkill family from my mother's side Painted it over two years Took a lot out of me in some way the scale produced a certain Formality a certain kind of distance science of fire a forest fire the whole inside a blaze Overlapped as an image mixed media and collage The planes of Punjab burning a series of charcoal drawings started off on a visit to Auschwitz The quality of the landscape a kind of subterranean a very dark and very bleak journey taking place Many of the images quite abstract didn't have any sense of location Could feel images about war War in some sense the second world war images of nuclear holocaust of violence has scorched our own situation and once that is over The paintings again form part of the desire to return to a very direct human Sensuous palpable experience It's such a large You've had a long history of friendship of doing in the sense of exhibitions together But not actually working together I thought that working on a canvas We all won't be able to participate the way we did in glass painting So we thought we'll be able to forget our egos and work together Baroda is a small town Compared to Bombay. I've been working in Baroda since last 28 years. I think Temples, mosques, churches are very much part of our living They form essential part of my painting These gods and goddesses are almost next door neighbors or someone who stays upstairs Naturalistic painting has not interested me My interest is mostly in the narrative painting the time sequence in painting interest me in 1971 I started doing paintings based on professions. I Used the colors that were used by middle-class people Particularly shocking pink green and blue. I'm happy with these paintings Though I did not sell these paintings for a long time. I Used the things around us. I used colors of posters garish and vulgar At present I'm interested in play of light in my painting The way light falls on objects The way things get eliminated. I've been collecting this oligarch since 1962 right in the beginning the things which I had collected are from the cinema posters The gods and goddesses posters and also very late miniatures All these things had interested me These things have been ignored because they were considered as things which are on periphery of art the earlier work Like a tailor watch repairer is a satire on the existing work being done Certain things which I thought I had no courage to say. I needed some courage to say So I'd use totally different way of approaching the problem Slowly I could gather the courage and say these things things Which interested me started coming? the problems of Transvestite problems of homosexuality problems of so many things which I was not able to depict on the canvas Slowly I could do it in the later work Although I do paint things about people who are on this street I don't go down onto the street and sketch What I like to do is observe as much as I can or at least Incorporate in my mental space much that I see around me and I have been seeing around me for the 12 years that I have worked in this studio and Then come back to my studio and recall all that I have observed And this is how I actually work. I mean I like the stress of recall I like the stress of you know the strain of recalling from memory pulling it through various Screens almost and then bringing forth a figure. I had started working on the watercolors in 1983 and the idea started Because I had to vacate my studio Which was under repair and had a very small space to work in at my home and It was really starting in a very small way in a very small format So it was at this point that I started to do the watercolors After having done a whole series I showed them to a friend in Ima shake and she introduced me to the works of an artist the Turkish artist called see a column I wish she felt it was important for me to see at that point and I think that was really something fantastic See a column is a series of pictures done by Turkish nomadic artists of the 15th century Wandering in the deserts and recording the nomadic lives of the tribes of that area and With a poverty of means It's simply washes of CPR and Bansi and Persian blue even little flecks of gold at times I guess when they had little money, they could just put a little bit of golden With just this they were able to record the most marvelous things You know dervishes and fakirs and then I think that was for me a revelation series about Conversations with artists of the past who have played a vital role in my art perceptions The works entitled rethinking Raja Ravi Varma and Marsha and the other Vermeer women are from this series the picture entitled of monsters and angels a Fable is about reinventing figures from the miniature painting traditions and placing them in post-colonial times Signified by the appearance of the Occidental Angel another set of pictures manifest the city my city Bombay The areas and locales that I encounter every day Beneath the mundane skin lies an alternate world Which reveals itself only when it's not being looked at That is when it is absorbed in itself and does not rise to meet your gaze I Am curious about the possibility of painting states of one's mind. I Do not wish to be mystical at all by states of mind I'm referring to emotions that grip on an ordinary context of one's life Like oppression anxiety self-absorption anger when I paint this figure I put in the posture all at once with a loaded brush like a mark and That mark has in it the charm of the stoop the gesture the facial expression all of it incorporate in that very instant I do pictures of the human face out of a passionate obsession These are faces of people I have been meeting for many years. I Walk down my street where my studio is situated in Lohar Chal It is the party while I'm waiting in the rain for a quick assignment For the boys who run errands for the ODP restaurant Or the old widow who I meet going up and down the stairs Who lives in her memories of being an actress at one time? I do not know whether this is truth or fantasy I'm obsessed with getting into these people that have secret lives that I will never know a series of pastels called journeys Very joyous works The whole thing about journeys see journeys leaving one shores exploring other people's other cultures a lot of them 19th century fantasy travels a range of possibilities that excites me These journeys started off as abstract works, but the latter ones Concretized having a certain geography is a place a reference to a certain time Going to the opposite of starting with abstraction and then making a kind of story or a narrative I'd like to go out fantasy journeys And then come back home and then leave again Many of the paintings that I do I think about for sometimes for years I get get an image in my head and that image I ruminate about I make sketches and it might be Two or three years later that actually come about to do that painting because many changes take place But the painting at the moment is a constructed Artifact it is something that I put together from this kind of long History that I have of this image or images that make the painting The landscape in Kasoli has become a very central feature of my work By landscape I also include the human beings and in fact over the last 20 years since I've been coming here many Local people that I have established a very close personal contact have become a very important feature of my work The way that the colors come in the way the quality of the light the kinds of textures that are there Where the landscape then becomes a grand metaphor? 1980 did a portrait of my father a Turning point in my work that being Considerable crisis the crisis about bringing the human figure in From that point a series of paintings evolved Set the pattern which still exists today In the oil paintings that I'm doing The desire to have as large a spectrum in terms of my subject matter Where I do allow the personal characters Friends of mine to be subject matter of painting which has been the modernist movement Anatma you don't paint your friends or your beloved or your family members