 Hello and welcome back to this lecture 12 on Microsystems Fabrication by Advanced Manufacturing Processes. I would like to briefly recap what I taught last lecture. So, we were talking about electrochemical machining, some basics and fundamentals as I would like to just mention here that it is a process where the tool is made the negative electrode, the cathode and the workpiece is made the anode. And then the choice of electrolyte is done in a manner so that whatever dissolves of the anode goes as a precipitate into the solution and it circulated so that there is no deposition back on the tool cathode and this way the whole process is used or is made amenable to machining of materials. So, we also did some estimation of material removal rate and we studied Faraday's law in the process and we were yet to sort of predict what amount of you know material per unit time may be removed by this process when we started looking at the very first important thing in electrochemistry in any electrochemical operation which is the ion transport equation. So, as a part of that equation if you may remember we had actually tried to study what the impact of a positive ion a central ion of interest would be in assembling its negative charge atmosphere around it and we studied that what the potential would be at a function at a point A if we assume that at an infinite distance there is a charge number density represented by n plus 0 and n minus 0 and near this at A it is n plus and n minus. So, and we try to find out the potential arising out of this situation and the xi was represented as A square of k by 4 pi dielectric constant D times of e to the power of minus k r by r. So, that is how xi potential at point A would be as a function of r. So, for electro neutrality the total negative charge assuming the central ion to be positive of the atmosphere about a given ion is of the atmosphere and the charge of the central ion B plus z i epsilon thereby meaning that the atmosphere would have a charge of minus z i e. So, we can assume that let us say if you have a central ion here and you are looking at some distance r from this ion as if there is a shell of thickness D r around this distance r and then you say that there is a distance A or the ion size parameter at which the you know the ion kind of gets truncated. So, this is the radius of the ion. So, we can assume that the total charge which is available because of this positive central ion of interest can be represented as the volume element d v here times of the volume charge density at this point. So, rho times of d v right and this varies between let us say some distance A all the way to when r goes to infinity and that can be made equivalent to the charge minus z i e the total charge in the atmosphere around the central ion of interest. So, d v of course, as you know is nothing but 4 pi r square d r was d by d r of sort of 4 by 3 pi r cube. So, that is how this volume element of the shell would be around this whole central charge of you know interest and. So, therefore, the volume integral A to infinity of 4 pi r square times of rho charge density which has been represented earlier you know that rho has been represented earlier as A k k square by 4 pi d e to the power of minus k r by r. So, this charge here A times of square of k by 4 pi d times of e to the power of minus k r by r times of d r. So, that is equal to minus z i epsilon and therefore, we can just simplify this equation. So, let this be equation 1. So, we can simplify this equation further to obtain A square of k d by I am sorry times of integral 0 to A 4 r times of e to the power of minus k r equals to z i e. So, because there is a minus term which comes from the charge density. So, that is the final integral that we have to solve. So, we have to integrate this expression by parts which as we know that you know the formula of integration by parts is A to B f x g dash x d x equals f x g x between the same limits minus integral A to B of f dash x times of g x times of d x and here as we know that f x equal to r and g x or g dash x if we assume it as e to the power of minus k r. There by meaning that g x can be equal to minus e to the power of minus k r by r. So, we can actually write this whole expression of the by parts as integral A to infinity r e to the power of minus k r as minus r e to the power of minus k r by k varying between A and infinity minus of e to the power of minus k r by square of k varying between A and infinity and this therefore, can be further represented as A e to the power of minus k by k plus e to the power of minus k by square of k and simultaneously we can have A square of k times of d times of e to the power of minus k by k because we are just substituting this result into the equation formulated earlier here the equation 1. So, it has been just the same equation formulated earlier. So, therefore, e to the power minus k by k times of square of k times of k A plus 1 equals z i e or in other words as you already know that the potential functions i is A e to the power of minus k r by r and A from here can be found out as square of k goes away. So, we are left with z i epsilon divided by d times of 1 plus k A e to the power of k A as A and potential functions i can be represented as z i epsilon d to the power of or d times of 1 plus k A times of r times of e to the power of minus k or let us say plus k A minus r. So, that is how the potential functions i can be calculated as. So, I would like to now just sort of bring your attention to what would happen at the surface of the charge central ion which is at a distance A from the center of the ion. So, we can assume that this influence or domain of the charge varies between r equal to 0 and A although the charge is kind of super concentrated at the center of the ion. So, that is the concept of a point charge the charge is concentrated at a point, but then it can have a radius 0 to A. So, at A the potential because of the charge which is somewhere here or hidden here can be represented as psi at r equal to A is z i epsilon by d times of 1 plus k A times of A r equal to A and when you put r equal to A here this exponential term goes away. So, this is equal to 0 sorry 1. So, therefore, that is how the potential at a point A on the surface of this point charge can be recorded as. We can further try to sort of split this into a partial fraction and try to have a look into what all components would go in formulating a potential at the point A. So, here let us say this I therefore, at r equal to A can be written down as z i E by d times of 1 divided by A times of 1 plus k A meaning thereby it can be z i E by d times of 1 by A minus k by k A plus 1. This is how could happen when it is resolved into partial fractions and we can represent this as z i E by d A minus k times of z i E by d times of 1 plus k A and this can further be expressed as plus minus z i epsilon by d A minus plus depending on the value of the charge k z i epsilon by d times of 1 plus k A. So, just by the look of it it appears as if this can be the effect of the central positive ion and this component because it is exactly the opposite of it can be the effect of the charge atmosphere which is principally negative charge. So, therefore, you know the potential contribution at the surface of the charge would come due to the ion itself and due to the atmosphere of the ion and the equivalent radius of this ion atmosphere can be treated to be 1 plus k A by k. So, this is the from this term here this is the equivalent radius of the ion atmosphere and this let this be called equation 2 here and this would be of some relevance because we will try to utilize this in order to find out what is the potential of such a surface which would get which is nothing but an assemblage on an assemblage of the charges of one type with respect to the charges of other type which is present lined up as the Helmholtz layer near the surface as a dual layer. So, this ion atmosphere would become very relevant in calculating the zeta potential of the surface at that stage, but you do understand that the contribution of the ion atmosphere on to the ions surface also is of equal amount of significance and relevance in this particular illustration. So, let us now look at some of the very basic fundamentals of what happens when an electrode is dipped in a solution which creates an assembly of the charges cut. So, let us now look into a slightly different aspect of the double layer as I was mentioning before. So, how does the double layer get formulated? Let us say this is the electrode surface that we are looking at and this zinc and this is immersed in just a brine solution N S E L plus H 2 O salt water. So, there is an automatic tendency of the zinc to formulate zinc ions. So, there would actually be ions of zinc which would come out dissolved manner from the electrode and there would be water molecules which would be immediately encircling the zinc so called. So, this is the water molecule encircling the zinc and that is how this whole orientation would be. So, each of the zinc atom is surrounded by a bunch of water molecules and that is true for all the zinc atoms which are there in the solution and correspondingly there are negative ions which are there on the electrode side. So, there are positive ions in the solution and the negative ions in the electrode side. So, they should be equal and opposite if there is no external field that is influencing the system because whatever number of zinc atoms are liberated as ions into the solution should have charge wise the similar amount of electrons on the electrode surface. So, what is important and worth here to mention is that there is a thin layer of water which would come between the positive charges and the ions separating them from each other and thus the zinc plus 2 cannot get deposited back on to the electrons and this becomes electron rich and there is a field which exists between the you know the first layer of zinc ions which are here. This is known as the Helmholtz layer and the layer of electrons which are on this other side here. So, there is an electrostatic interaction between the electrode charge and the solution charge that is number one and then ions from the solution may approach the electrode surface only so far as there inner solvation shells. So, solvation shell is this shell which is formulated because of the water molecules around the zinc plus 2 atom. So, as much as this solvation shell can allow only that much distance the ion can have in proximity to the electrode charge, electronic charge which is on the electrode. So, the surface array of the ions is the cushion from the electrode surface by a monolayer of solvent and the line drawn through the center of such ions at this distance of closest approach marks boundary and this is known as the outer Helmholtz plane is indicated here ok. So, this is an assemblage of all the zinc plus 2 atoms hold on to a certain line and the size of ions formulating the outer Helmholtz plane are larger and the total number of ions needed to do a complete charge balance cannot all be fitted on this plane. So, obvious reasons that electron here in the electrode is a part of the electronic structure much more densely packed in comparison to the ions which are in the solution which are positive ions. And so therefore, not one layer, but subsequently a set of layers would be needed for. So, for example, let us suppose this is the electrode surface and you have negative charge assembly here in the electrode surface. So, the first set of Z n plus 2 ions would not be able to take care of the total electronic charge which has been emanated in this surface this is mind you are. So, it is sort of atomically impregnated into the surface these electrons and these are full size ions in the solution. So, there has to be subsequent layers which would be responsible for whatever field escapes this first line of ions and then slowly the field gets neutralized and the ion layers get thinner and thinner as you go radially from the electrode into the solution. So, typically this layer in such a case is known as the first layer of charge and so it is the Helmholtz layer and this is demarcated by this outer Helmholtz plane. And this layer here right here is sort of a diffuse layer of charges and therefore, in electrochemistry mostly there is a Helmholtz layer and then there is a diffuse layer which has its own contribution in terms of capacitance. So, this is CH and this is CD. So, the capacitance of the Helmholtz layer is what generates because of this solvation shell of individual ions separating the ions from getting back into the electrodes back to the electrons and the diffuse layer which is there in formulated inside the solution. So, that is what the orientation here is. So, therefore, the remaining charges are all held as you know with increasing order disorder in the outward direction into the solution and this is known as the diffuse layer of charges. So, that is why double layer. So, you have one layer which is the Helmholtz layer and then another layer which is the diffuse layer of the charges. So, that is how in a nutshell the you know the whole electrode is oriented with reference to the solution. So, let us actually now look into how the potential zeta would vary with respect to distance from the electrode surface. So, if you plot the potential zeta with respect to x the distance into the solution as you can see the potential varies linearly like this all the way up to this value here which is actually as I will later on tell you the zeta potential of the surface and then after that the as a moment it crosses the outer Helmholtz plane the potential starts varying of the diffuse layer in a different manner. It is not a it is a non-linear behavior in this particular zone. So, we have to somehow be able to find out what this potential function is with respect to x and as obviously as we have done before at x equal to infinity the potential function zeta is actually equal to or xi is actually equal to 0. So, this calls for the solution of the Debye Huckel theory in three dimensions as we did just about few slides back and in a ion solution to a 1D case a 1D is only the dimension of interest from the array of ions which are somewhere here on the electrode. So, this is the array of ions and with respect to that array of ions maybe we can consider this all focused on to a single place with respect to this one let us say central ion as an available as an array format on the electrode what is the distribution of the charge in the solution and thereby the potential function would be recalled in that case. So, it is simply a one dimensional approximation of the Poisson's equation we can just write d 2 xi by del x 2 is actually equal to square of k times of xi as you already know from earlier equation this square of k is basically the function which comes in terms of sigma n i z i square square of epsilon by k t times of so on so forth. So, that is what the k square term is here also what is of interest and significance is that as we found out last time that with respect to the central ion the there is an effective radius of the ion atmosphere and this effective radius was given by 1 plus k a by k in other words it means it was 1 by k plus a. So, that was the radius of the effective surrounding ion atmosphere with respect to the central charge of interest. So, we will consider the similar value here and try to see what this function results in. So, by this theory of ion association I would just like to just simply state it without really getting into the proof of it, but by the ion association theory the value of 1 by k in the ion atmosphere comes out to be equal to the diffuse layer thickness let us call this thickness delta. And as you already know square of k times i in this equation del 2 xi by del x 2 is actually equal to minus 4 pi rho by d where rho is the charge density at a particular point d is the dielectric constant of the medium that is how the Poisson's equation was initially illustrated. So, for this part of the equation the simple solution for this part xi x comes out to be equal to the standard solution a e to the power of minus k x plus b e to the power of k x that is how the standard solution to this problem would look like and we again apply in the same manner the boundary conditions. So, we know that at x equal to infinity in this particular case the potential at infinity is 0 meaning thereby that b automatically becomes 0 because otherwise at infinity this term is very large the only way you can satisfy the solution is by assuming a very small b negligible or 0. And simultaneously the xi x comes out to be equal to the potential function at point x comes out to be equal to a e to the power of minus k x simultaneously again the rho which is actually represented by minus square of k times of xi d by 4 pi looking at these two parts of this equation come can be written down here as minus a times of square of k times of d by 4 pi e to the power of minus k x. So, that is how rho is 0. And of course, there is a d term here the electric constant and we will have to somehow determine what this k is in an identical manner assuming a charge distribution and principle of electron neutrality at within the case of a one dimensional ionic form. So, let us assume that the charge density on the surface of the electrode is actually equal to sigma per unit area and somehow because of the principle of electron neutrality the electrode and solution charges should be the same as far as the magnitude of the charge goes. So, sigma therefore, of the solution or the electrode can be represented as minus of the volume charge density of the solution times of d x because the only variation available is in the x direction we do not consider any other direction for comparing the sigma value. So, we assume that there is a homogeneity of distribution of charges as we move along each plane along a certain value of x equal to x 1, x 2, x 3, x 4 so on up to infinity. So, if you put the value of rho here in this equation and try to solve we actually get you know and this charge density here being positive this is of course, the negative charge density the negative sign goes away. So, we are able to see this as a to infinity times of a square of d sorry square of square of k times of d divided by 4 pi e to the power of minus k x times of d x and this can further be expressed as a d k divided by 4 pi e to the power of minus k x 1 k term goes here and therefore, a can further be expressed as 4 pi e to the power of minus sigma divided by d k e to the power of k x positive at x equal to a which thereby means the start of the dual layer this value would be 4 pi sigma by d k e to the power of k a. So, that is about the expression regarding a would be and because the term potential functions i of x is related to this a e to the power of minus q x I am sorry minus k x meaning thereby that it is 4 pi sigma divided by d k e to the power of k times of a minus x where obviously, the expression a the constant comes from this particular term here. So, therefore, we can fully write the potentials i x in this case as 4 pi sigma by d k e to the power of k a minus x other words add x equal to 0 x equal to I am sorry x equal to a indicative of the surface of the electrode the value of potential function also known as the zeta potential of a surface with respect to a solution can be represented as 4 pi sigma by d k e to the power of 0 1. So, it is 4 pi sigma by d k in other words the zeta potential is basically represented as 4 pi sigma delta by d if you may remember delta was the diffuse layer thickness with which had been earlier said to be predicted by the ion association theory as 1 by k. So, that is how the zeta potential of any surface comes into existence and this is a very critical value which matters particularly in electrochemical machining processes because every surface would have a characteristic zeta potential with respect to the electrolyte of interest that you are going in and that zeta potential has to be somehow crossed over by the applied potential external potential in order to be able to furnish the electrochemical machining operation and so the zeta potential is absolutely important to be kept in mind and to estimate things like material removal rate etcetera which happens normally during the process of electrochemical machining. So, if we look at the diffuse layer theory where we talk about let us say a surface here followed by a inner solvation shell of the ions which are there in the solution followed by let us say a positive ion outer plane here and then subsequently a diffuse layer which is starting from the Hall-Moltz outer plane here all the way to the to infinity. We say that if supposing the capacitance here is capacitance of Helmholtz layer C H and here is C D the overall capacitance which would come into existence is actually 1 by C H plus 1 by C D or in other words C can be represented as C H C D by C H plus C D. Here if supposing C H dominates C D, C H is dominating C D let us say. So, then C becomes equal to C D and vice versa if C D predominates C H C becomes equal to C H. So, therefore, this particular model is known as the Helmholtz model it was the predominance of the diffuse layer capacitance and this is known as the Gai-Chapman model when you talk about the capacitance modeling of a diffuse double layer with respect to any surface or in solution. So, let us look at some aspects of mass transfer and electron exchange process involved therein. So, in considering the electron exchange reaction at the electrodes we are concerned essentially with the layer of solution very close to the electrode surface as you have just seen and model this before and there should be some means. However, available for oxidant or the reductant to reach the electrode surface and if you look at this interface formulated with this so called double layer diffuse and Helmholtz layer. So, this layer whole has to be crossed over for any electrochemical reaction to happen and the way that they are principally done are through three processes or three steps one of them is called migration another is diffusion and the third is convection. So, the migration is the movement of cations or anions through a solution under the influence of an applied potential between electrodes placed in that solution and migration automatically starts happening when the electrode just about touches the solution and this is the reason for the double layer and then you have of course, the electrode reaction which depletes the concentration of the oxidant or the reductant at an electrode surface and produces a gradient therein. So, this gives rise to the movement of species from higher to lower concentration occurs in both charged and uncharged species and then of course, we have convection which relates to thermal or stirring effects of the solution can arise extra honestly through vibrations shock and setting up of temperature gradient. So, these are the three principle mechanisms in which the ions can move back and forth between the electrode and the solution and cross the double layer that is established and that is the mechanism of mass transfer which happens mostly in the even the electrochemical machining processes. So, let us look at the fundamental of the ECM process here. So, when a metallic body cut so just for the sake of repetition in an ECM process the following events actually take place it starts with when a metallic body is submerged in an electrolyte metallic atoms leave the body becomes ions, ions move to the body and becomes atoms and the process goes on continuously and an equilibrium is maintained. The potential difference exists between a point on the surface of the metallic body and an adjacent point on the electrolyte and this difference is known as the electrode potential and the electrode potential varies depending on the electrode electrolyte combination. So, this actually is the nothing but the zeta potential that they are mentioning of a surface and if two different electrodes A and B are in turn immersed together in the same electrolyte and the potential difference is given externally between these electrodes. First of all there will be a potential difference will automatically exist and then you can externally give a potential difference also. There would be some kind of transport of ions from the solution into the electrodes or vice versa and the total amount of electrode you know difference of potential which would does arise is basically the potential of the electrode A minus the potential electrode for B and that would be the net EMF which is available which would create a condition where ions start moving if supposing we are not giving any external field or external potential to these set of electrodes and in the process supposing if you have two electrodes copper and iron as mentioned here in Brian's solution or solution of kitchen salt in water and you already know that the standard electrode potentials of both these reactions where iron gets into ferrous state releasing two electrons is minus 0.049 volts and copper getting into cupra state and releasing two electrons is 0.304 volts. The difference between electrode potentials of the copper and the iron electrode can be represented as the potential of the anode minus the potential of the cathode. So, it is basically potential of the copper Brian solution minus the potential of then this basically is anode it is placed higher on the electro activity series potential of Fe with respect to the Brian solution. So, therefore this becomes equal to 0.304 plus 0.4090.713 volts and the nature of electrolysis process depends on the electrolyte used and supposing if you were to do ECM then whatever comes out here for example, if Fe plus 2 state comes out of this electrode here and being the cathode and this basically gets converted into immediately into FeOH2. This ferrous hydroxide is not dissolvable anymore in the solution. So, it precipitates down it formulates as crystals here. So, it does not get re-deposited or anything else as a matter of fact does not get re-deposited on the copper. So, this can be the tool and this can be the work piece and this can be a basis of electrochemical machine. So, that is what that is how the electrochemical machining is realized these are the reactions for example, the iron getting removed and the hydrogen getting split up by the two electrons that is released by the iron getting into H2 plus 2 H minus and the resultant FeOH2 which is non-dissolvable is formulated and this is the precipitate which can be later on removed and it furnishes the machining operation. So, basically let us look at some of the mathematical details or estimation of the MRR material removal rate using the Faraday's laws that we had discussed before. And as you already know that by the Faraday's laws the amount of weight that is removed cut the amount of weight in grams of a material dissolved or deposited is also given by the is proportional to the ambient current. The total time duration for which this current is flown and epsilon which is actually equal to the gram equivalent weight the gram equivalent weight of a material and by the definition of it the equivalent weight is 1 mole of the material which is the weight of the 1 mole of material which is also equal to the atomic weight per unit the valency of the material Z ok Z is the valency. So, we can call it A by Z and essentially the amount of charge which would have to be flown in the system is related to this quantity i into t as you know i already is dq by dt the amount of you know the rate of change of charge or rate of flow of charge. And so, this is the time integral of i meaning thereby that this is the total charge and when we are considering 1 mole in one atomic weight the total charge that would be involved in removing this 1 mole is basically equal to 1 mole electrons right. And this 1 mole electrons is 6.023 10 to the power of 23 times of charge of 1 electron which is 1.6 10 to the power of 19 minus 19 would formulate the term 96500 Coulomb which is also known as a Farad which is also which is also described by the term F ok. So, this 96500 Coulomb is the unit for 1 mole charge or 1 mole electrons. So, therefore, the mass rate of removal the rate at which the mass or the weight gets resolved or deposited would be equal to the rate at which you are flowing charge that is the current times of the equivalent weight that you are removing that is a by z times of how many moles of current is flowing per unit time represented by the F. So, i by F is that term right the moles of electrons which are flowing and so, many moles with 1 mole removing a by z weight of a particular material at a certain rate which is determined by you know this current term here rate of flow of charge would be the m dot or the rate of flow of dissolved dissolution or rate of dissolution or rate of deposition or rate of flow of material in fact. So, basically the mass removal rate of an ECM process is therefore, given by this term m dot ai by z F just for the sake of repetition a by z is the atomic weight of 1 mole of a certain ion and exactly how many moles per unit time is being flown is given by the quantity i by F. And so, you are essentially looking at what is the mass rate which is coming out from the system because of the flowing of this particular charge. So, if we assume further that the density of the anode material is rho there can be a estimation of the volumetric removal rate. So, it can be given by m dot by rho. So, rate of change of rate of change of mass per unit volume of a per unit density giving thereby the volume which is being emanated or the volume rate which is being emanated. So, this can be represented as ai rho z F ai by rho z F and units of this could be in centimeter cube per second. For the sake of repetition again a is the gram atomic weight of the metallic ion i is the current in amperes which you are trying to flow and rho is the density of the anode in gram per centimeter cube. There is a valency of the cation of interest which is coming out or being deposited and F is of course, the amount of charge corresponding to 1 mole electrons given by F the quantity F of arid 96500 cooler. So, that is how you can sort of get an estimation of the volume rate of removal q dot ai by rho z F. So, this is only if predominantly one metallic system is existing right, but however most of the real engineering applications come when there are alloyed components or many such metals are participative in the system. And so therefore, we somehow need to tailor this mass removal or volume removal rate in a manner. So, that it can be applied to alloyed systems and so today we are kind of at the end of this lecture, but then in the next lecture we will take this up and try to see that if there is an alloyed system with more than one metals participating in the process of this electrochemical removal, then what would be the overall rate based on a sort of average between all the alloys which are there in the system of different densities, different atomic weight, different valencies so on so forth. So, we will look at that problem in the next class. Thank you.