 Welcome to class 36 of topics in power electronics and distributed generation. So, we have looked so far at an example of an L filter for a single phase inverter and based on the RMS current calculations, we saw that the ripple that gets injected into the grid would not meet the standard recommendations and we can get a similar result based on a similar conclusions based on a more exact Fourier analysis. So, what we what is shown here is the Fourier spectrum of 3 phase 2 level voltage source inverter, its output voltage is 240 volts RMS which is taken as the base quantity and switching at a frequency 10 kilohertz with a DC bus voltage of 800 volts with sin triangle modulation. So, the modulation index is 0.85. So, if you do the Fourier analysis your fundamental is having a magnitude of 1 per unit, if you look at 10 kilohertz that would correspond to the 200th harmonic and the amplitude at the RMS value of the amplitude at the 200th harmonic is around 0.9. So, if you then look at what the specifications recommendations of the standards would be, you would like the to have the per unit in value of your filter inductor to be equal to V per unit divided by I per unit that is allowable at the 200th harmonic. One thing to notice in the standards recommend even harmonic to be 25 percent of the odd harmonic. So, your standards say 0.003. So, 25 percent of that is essentially what is recommended and the excitation at the 200th harmonic would correspond to 0.9 amps. So, the required value of the of your inductance would be 6 per unit. So, clearly showing that you would exceed what would be realistic for a practical power converter. One thing to keep in mind is, so you can see that the given harmonics is restricted to 25 percent of the odd harmonics. This becomes especially important if you are switching frequency is low of the order of 1 kilohertz which might be more common at a very high power levels. So, it might actually make sense to actually shift your harmonics for example instead of 10 kilohertz if it was 9750 kilohertz then that would correspond to the 195th harmonic. So, you could then eliminate the factor of 0.25 in the denominator over there. So, based on this we can see that having just a inductive filter may not be possible to meet the standard requirements. So, one could consider a higher order filter such as an LCL filter. So, what is shown in this schematic over here is for a voltage source inverter being connected to the grid through a LCL filter. So, you have L1 connected to the inverter side, L2 connected to the grid side and capacitor C and you can have this as a 3 phase 3 wire configuration in which case these dotted lines would not be present or you can have it as a 3 phase 4 wire configuration with capacitor center tab in which case the dotted lines would be connected with physical wires. So, one question that can be asked is why did you go from a L filter to a LCL which is a third order filter why not just a LCL filter. So, we could consider what would happen if you used a LCL filter. So, suppose you have the inverter which can be represented as a pulsed voltage source Vi and if you consider the L filter and then you have the grid voltage Vg. So, the first question is where to place the C you could not definitely place the C on the inverter side because you have to switch from a voltage source to a current source. So, you cannot have the capacitor here. So, you could think about possibly connecting the capacitor here and you know that the impedance if you consider the voltage source ideal you know that the dynamic impedance of a voltage source would be 0 whereas, the dynamic impedance of the capacitor would be 1 by SC. So, whatever ripple would come in the inductor would actually flow through into the grid. So, this is for a ideal grid. So, you can see that there is no benefit of adding a capacitor in a practical grid you would not just have a solid connection ideal 0 impedance voltage source you would have some stray inductance it could be of the lines the transformers etcetera going back to the grid in which case this reduces to a LCL filter. So, this is actually an LCL filter. So, practically you would end up with a LCL filter even if you just connect the C, but having in the LCL filter where one of the dominant impedances be is governed by a stray parameter is not a great idea because the resonances of this particular filter would vary over a wide range depending on what the value of your stray inductances. So, it would be it would not be possible to accurately predict where the resonant frequencies lie in such a situation. So, a more practical situation would be to physically add a second inductor and then whatever non-ideal inductance is there on the grid would be essentially considered part of your inductance L2. So, your L2 can be designed in such a manner that it dominates over the stray inductance. So, whatever variation exists because of the stray inductance in the grid can be accommodated in a not over tolerance range by appropriately designing the LCL filter. So, you can see that there would be a large uncertainty without L and the implication of this is that you would have large variations in the resonant frequency. So, if you look at the circuit that we are seeing for the LCL filter. So, this would accommodate for the configuration for both a 3 phase 3 wire and a 3 phase 4 wire. What one would typically see is that the ripple that is going out into the grid essentially your grid current ripple is higher in a 3 phase 4 wire converter compared to a 3 phase 3 wire converter because you have now additional parts in the 4 wire case for the 0 sequence components to flow. So, if you have a filter design that meets the standards for the 3 phase 4 wire case it would typically meet the requirements for the 3 phase 3 wire situation also. So, we will consider the 3 phase 4 wire filter to be a typical filter which we would then look at in detail. So, if you look at the LCL filter now for the 3 phase 4 wire power converter we can then look at it on a per phase equivalent circuit basis. So, on one side is essentially the inverter voltage V i between the output of your phase leg and the midpoint of the DC bus followed by the LCL filter. On the grid side you essentially have the grid voltage V G which at frequencies other than the fundamental can be considered a short circuit and whatever stray inductance impedances are there in the grid can be accommodated along with L2. So, the grid side can be considered a short circuit for looking at frequencies other than the fundamental frequency. So, with this simplified single phase equivalent circuit one could go about analyzing the LCL filter. You can then take a look at typical magnitude response plot of the grid current injection divided by the inverter side voltage. So, essentially you are looking at the transfer function I g of s divided by V i of s and you are looking at the Bode plot. And if you look at the low frequencies below your resonant frequency your attenuation is corresponds to minus 20 dB per decade. So, this would be similar to the attenuation characteristics of a simple inductive filter beyond the resonant frequency the attenuation of the LCL filter would correspond to minus 60 dB per decade which would correspond to a third order LCL filter. So, if you take what is shown over here is the frequency in radiance per second and the magnitude of I g by V i transfer function which is essentially a admittance transfer function. If you had essentially L filter it would continue all along to the higher frequencies at the same minus 20 dB per decade. So, if you are considering say a frequency such as say 10 kilohertz. So, this would correspond to around 6 into 10 power of 4 radiance per second. So, you are looking at essentially this point on the LCL filter and this point on the L filter you have attenuation difference of about 40 dB. So, you get a improvement in attenuation by a factor of 100 for the LCL filter compared to L filter. So, you know that you could actually meet it is feasible to meet the attenuation requirement of suggested by the specifications with the LCL filter. So, what is shown over here is not an ideal LCL filter where the resonant peaks would actually go off to a very large value. This shows a damped LCL filter where the gain at the resonant frequency is managed to a lower level. One could then use the to look at the transfer function i g by V i of s we can actually get it from the simple equivalent circuit of the LCL filter. So, from the LCL filter network one can actually calculate what is i g as a i g of s divided by V i of s and that can be done with just looking at the impedance values. So, i g of s divided by V i of s is 1 by SL1 plus the series value of SL1 plus the impedance of of c in parallel with L2. So, essentially you have SL2 by SC by and then whatever current flows through your inverter side splits between these two branches depending on the ratio of the imprudences. So, you can write it as 1 by SC divided by this is equal to 1 by this can also be written as 1 by SL1 plus L2 into 1 plus S square L1 L2 c divided by L1 plus L2. So, the quantity L1 plus L2 can be thought of as L. So, essentially the inductor is split into L1 plus L2. If you had a first order filter then essentially the capacitor connection would not be there you would have essentially just L1 plus L2. Now, that you add a capacitor in between you get the additional attenuation provided by this term over here. The this term over here is essentially the parallel L1 and L2 in parallel. So, LP is and your resonant frequency is 1 by root LP times c. So, these relationships can be used then to obtain this transfer function i g of s by V i of s. So, this is the ideal LCL filter without any damping. So, you have very high gain at the resonant frequency and you have the definition of what the resonant frequency and the L value is. We can actually make use of this to actually determine what would be one of the limits for this particular filter design. So, if you look at essentially this particular expression over here we can write this particular term as i g at your dominant frequency and your omega dom we saw from our Fourier spectrum is essentially 2 pi switching frequency and this would correspond to 1 by sl min 1 times 1 plus omega dom square by omega resonant. So, we know that your i g has a value of 0.003 on a per unit basis and your V i can be obtained from your Fourier inverter harmonic spectrum. So, one could then rewrite this particular expression in terms of we know what this quantity is and assuming at this point we know what the resonant frequency is. We could make use of this particular expression and put it in the form of determining L minimum in terms of quantities that are known. You know what the dominant frequency is we will discuss on how we could select the resonant frequency. So, you have essentially minimum limit for what your inductive term is. You also have a maximum limit for what L max is what is the maximum value of L1 plus L2 and this is based on the DC bus voltage requirement. So, if you look at essentially typical power converter that is connected to the grid. If you consider power converter operating with at 20 percent impedance filter impedance value of L being 20 and operating say at unity power factor i j then essentially at fundamental frequency your voltage drop is j omega L times i and i is 1 per unit say Vg is 1 per unit. So, you have a drop voltage drop of 0.2 per unit assuming L is 0.2. So, then if you look at what would be the inverter voltage V i this would be actually 1.02 per unit. So, it is not very different from Vg if you are operating at unity power factor. If you are operating say the power converter as a STATCOM say supplying reactive power then essentially you would have Vg and i g and the voltage drop would now be having a value of 0.2 and your inverter voltage required would be now 1.2 per unit which would be at the higher on the higher side. Suppose that you are not operating as a either as a STATCOM or even though you are operating as in a unity power factor mode typically for a DG you might consider say some variation in the power factor when you are operating this particular power converter say you have the grid voltage Vg and say you have your i g that you would ideally like to inject in unity power factor and say you have a range of i g going say 30 degree lead to 30 degree lag and then if you look at what would be the say the voltage that would be required at the terminal of the power converter. So, when you are trying to inject leading watts you would need a high terminal voltage when you are trying to inject lagging watts you would require a lower terminal voltage. So, you could you could require say about 10 percent more DC bus voltage when you are operating in say a range of such conditions you might be required to operate in some such condition under dynamic situations when you have phase jumps frequency shifts etcetera your controls take time to actually react. So, you might need some margin to actually accommodate the filter. So, if so, keeping in mind what would be a upper limit for your DC bus you would then be able to obtain a range of value of the inductance between say l min and say l max. So, your value of the inductance should lie in this particular range. We could also think about say in this particular case we are looking at what l is, but what we are really interested in is what l 1 should be and what l 1 l 2 should be. So, the question is how should we split the l between l 1 and l 2. So, one thing that you could then do is look at the expression omega r square is equal to l 1 l 2 c by l 1 plus l 2 inverse and then look at l to be equal to l 1 plus l 2 and define say l 1 to be equal to a times a l times l 2. So, for example, if a l is close to 0 it means that the entire inductance is l is equal to l 2 if a l is equal to 1 it means l 1 is equal to l 2 and if a l tends to infinity it means the entire inductance is l is in l 1 and l 2 is negligible. So, you could then write this particular expression for the resonant frequency in terms of l in this particular expression and a l you would then get an expression of the form l is equal to 1 plus a l the whole square divided by omega r into a l c. So, you can then look at this particular expression which is a constraint between the value of l a l omega r and c and look at a few minimizations. So, one thing that you might ask is what is the requirement for minimum l a l is equal to 1 keeping omega r and c constant. So, if you take the previous expression and differentiate l with respect to a l you could then equate it to 0 either at the minimum or the maximum take that second derivative to see whether it is the minimum or the maximum and look at what are the relationship which you could get and you would see that the requirement for minimum l occurs when a l is equal to 1 keeping omega r and c constant. Another property that you would observe is that if you have the ripple in i g is minimum for fixed l and c when a l is equal to 1. So, keeping other parameters like l c and omega r constant you can then take the expression for i g and take the derivative with respect to a l and you will see that the ripple in i g is minimum at this particular point of a l. Another point that you could observe is that for fixed i g ripple and l comma c requirement for l the. So, keeping i g ripple fixed and keeping l fixed you are allowed to vary a l, but keep l 1 plus l 2 equal to l as a fixed parameter you will then see that c requirement is minimum when a l is equal to 1. So, from all these properties it is actually seen that l 1 is equal to l 2 is equal to l by 2 is desirable and in a practical situation you may not be able to exactly make l 1 is equal to l 2 because l 1 might be a physical inductor that you might add l 2 might be a the inductance of a interconnection transformer you might try to make the values close, but you may not be able to make it exactly match, but you know what is the desirable direction. So, you can see that keeping things like you could actually look at the minimum of these functions and keeping things like the value of l to be small is desirable because in a filter if you look at the cost of energy storage if you look at point number 1 that which we just wrote keeping the l valued l minimum given other parameters to be constant would be actually important from a cost perspective the cost per k v of capacitors etcetera is actually much lower than that of a inductor. So, it is actually desirable to actually keep your inductive components to be as cost effective as possible. You also have one more constraint on in the LCL filter. So, if you look at the LCL filter you while operating the power converter if you now look at the operation at the fundamental voltage you have say 240 volts over here and 240 volts RMS over here. The voltage across the capacitor is would be actually close to 240 volts. So, we are talking of approximate value of the voltage in the filter network. So, there is actually reactive power which is being drawn by the filter capacitor and you do not want that reactive power to be excessive. So, that would actually form another constraint on the filter design. So, we know that q of your capacitor is equal to v naught omega naught c and if you keep this to be a limit q max say 20 percent 20 percent power q would correspond to a power factor of 0.96. So, under operation you do not cause a large say offset in your power factor because of the introduction of the capacitor. So, you have c max and we know the constraint that omega r square is equal to 1 by c max and the corresponding L would correspond to L min and we will call this L min 2 by 4 because we know that L 1 is equal to L 2 is equal to L by 2 is essentially the value and if you then look at L p which is essentially the parallel combination of L 1 by L 2 this would be L by 4. So, the corresponding constraint would be L min 2 by 4. So, on the minimum side there are two constraints one one constraint is about the ripple current being injected into the grid. The second constraint is on the amount of reactive power that is being consumed by the filter and you could then say your L min may be the upper limit of your L min 1 and L min 2. So, by selecting this you are ensuring that both the constraints are being met and then your final inductance would be belong to the range between L min and L max. So, the another question is what would be how to actually go about the selection of the value of the inductor in this range between your minimum and maximum and the actual choice of the value of L between the minimum and maximum could be based on minimizing the power loss in your filter. If you look at the power loss in a filter in your value of the power loss at fundamental frequency plus the value of the power loss at the switching frequency is what you are trying to minimize. And if you look at the trend of power loss at the fundamental frequency it would actually increase if you have a larger in value of inductance because you need more turns more length of the winding maybe a larger core. So, you would end up with fundamental loss which tends to increase and so it is desirable to operate closer to L min from the fundamental loss perspective. If you look at the switching loss in the filter inductor your switching ripple current tends to reduce as you increase your inductance. So, the losses also would reduce as your ripple current through the filter reduces. So, from your switching loss perspective it is actually desirable to operate close to L max. So, then you look at the overall picture to actually see what point would correspond to the lower loss and we will look at this issue in the subsequent slides to actually determine optimum value of your inductance between the L min and L max range depending on analyzing the loss components within the filter inductor. So, one item that we still need to figure out is what the resonant frequency is and if you look at a typical filter designs you are you are essentially having a filter where you have a pass band and you have essentially a stop band and you essentially would like your filter to transition from your pass band to your stop band when you are actually going between this particular range. So, the question is what belongs to your pass band your pass band might have frequencies you definitely have the fundamental frequency see in your pass band you might have say your fifth seventh harmonic etcetera your harmonic frequencies in your pass band especially if you are trying to build a active filter which is trying to inject and control the harmonics in accurate manner. You might also say for example, try to reject the harmonics that the grid is putting on your power converter by actively ensuring that your band width includes your harmonic frequencies. So, your pass band need not just be your fundamental frequency it can actually extend further out depending on your band width targets of your power converter. Your stop band would correspond to your switching frequency and its side bands and harmonics and you are trying to ensure that all that is in the stop band and then you could actually select your resonant frequency to be a geometric mean between your pass band and stop band. So, with this choice now you have essentially obtained all the components of your LCL filter you know your resonant frequency you know the choice of L you know L 1, L 2 and once you know your omega r knowing L 1 and L 2 you automatically fix your capacitance because it is related through your resonant frequency. So, the next thing one could consider is then what does this design procedure address there are multiple constraints in the filter design. So, you have essentially at this point preliminary one issue that it addresses is it definitely addresses the attenuation requirement it looks at the reactive power constraint it looks at the DC bus voltage constraints and it looks at your pass band and stop band targets. So, the actual value of now that we have your preliminary L 1, L 2 and C the next question is what is the actual value of L 1, L 2 that can be designed and the procedure to design the L 1 and L 2 is based on the area product calculations that we had discussed previously. And we need to choose L 1 and L 2 from keeping in mind the power loss in the filter inductor and we have seen that the power loss in the filter inductor consists of the core loss component and the winding loss the core loss depends on the type of material that is selected the flux level the temperature because of loss parameters would and the properties of the material would change with temperature also the frequency of excitation the winding loss would again depend on the material conductivity the material conductivity can again change with temperature the winding length the geometry geometrical parameters number of layers etcetera and frequency of excitation. So, both these terms the core and winding loss terms can be then evaluated at the fundamental frequency excitation and the switching frequency excitation to look at what would be the components on the loss components in L 1, L 2 and the other parts of the filter and we can we will actually do that. So, if you look at the filter inductor L 1 this is an example of 10 kba 415 volt inverter switching at 10 kilohertz what is shown in these figures are essentially what is plotted is the power loss and the inductor L 1 the first figure over here shows the winding loss as a function of L which is L 1 plus L 2 ranging between say 1 percent to 14 percent and the next figure shows the core loss again when L is the varied between 1 and 14 percent and you are looking at the loss in the core with a variation of L loss in the core of L 1 when L 1 plus L 2 is varied. So, you can see that if you compare these two figures this is actually for a ferrite inductor. So, you would expect it to be used at very high frequencies it is being used at 10 kilohertz. So, the core loss is actually much lower than the winding loss. So, if you look at the winding loss components you are looking at numbers of the order of 20 whereas, if you are looking at the core loss you are looking at numbers the range of this particular plot has a maximum of 4. So, you can see that the winding loss in this particular due to the selection of this particular material the winding loss dominates over the core loss potentially if you had used steel laminations rather than ferrites you could get much higher losses in the core rather than in the winding. So, you could then look at then the trends between the core loss and the winding loss if you look at the core loss in this particular case you can what is plotted along the two y axis one is the 50 hertz loss. So, this is the 50 hertz loss is shown in this particular axis you can see that the maximum value of the 50 hertz loss is in the range of 0.04 and then you look at the core loss at the switching frequency you have a value of 0.4. So, if it was actually plotted on the same axis the core loss at 50 hertz would be almost a straight line at the very bottom. So, it would not affect the overall loss characteristics significantly, but you could see the underlying trend the 50 hertz component of the losses as a increasing trend as you increase the value of l total and if you look at the switching frequency term at low value of l the switching ripple is large. So, you have having higher losses, but as you increase the value of l the switching ripple reduces. So, your losses come down as your l is made goes to a larger value. If you look at then the winding related loss again you have a similar trend if you look at the 50 hertz loss in the winding it has again a increasing trend whereas, if you look at the switching ripple related loss you would end up with decreasing trend as you are increasing the value of l and so, the overall trend for the losses would be to select a value which tends to give you the lowest losses which would lie in this between this range. You could you can see that in this particular example the trend is can be seen from the curve the actual value of the loss is having a up and down characteristic this is because each of these points corresponds to a different inductor design. So, l being 1 per unit would correspond to 1 particular inductor l being 0.2 or 0.3 or 0.4 they are different inductors. So, if you go beyond say a value of l you might need to go to the next larger size score or you might need one more layer in your windings. So, the these jumps that you see is because of discontinuities because of you are changing your physical inductor from one particular value selection to a next selection which makes this curve have a have a non uniform non uniformities in it, but the overall trend can be seen under underlying this non uniformities where at the very low values of l your switching related terms would dominate and at the very large values of l it would be your fundamental related terms that would dominate. So, once you so this is for loss in inductor l 1 you could then do a similar analysis for inductor l 2. If you look at the inductor l 2 the we know that are your ripple current in the grid that is being specified I g ripple at the switching frequency is 0.003 per unit. So, in terms of the ripple current flowing through inductor l 2 there is almost no ripple. So, if you if you look at the high frequency loss in the grid side inductor it can actually be neglected you can think of it as essentially a fundamental frequency which is flowing through the inductor l 2. So, then if you look at the fundamental frequency loss term in the inductor l 2 again you can see the underlying characteristics the core loss is in the range between 0 and 0.03 the winding loss is in the range between 0 and 20. So, if you look at the overall losses the winding loss is the dominant term because we have selected a ferrite core inductor and the over the losses in the total inductance would be the sum of the losses in l 1 plus l 2. So, the underlying trend in both these cases is actually increasing losses as you are increasing the value of l. So, one could then look at what would be the total loss in the in the LCL filter. So, this actually includes not just the losses in the l 1 and l 2 it also includes the losses in damping terms in the circuit. So, if you look at the loss term overall as a function of l. So, what is shown in this green dotted line is a total loss as a function of l 1 plus l 2 where it is varied in the between the 1 percent to 14 percent range. And we are looking at a three different frequencies one is at very low switching frequency. So, this corresponds to 8 kilohertz. So, this would correspond to a low switching frequency and this would correspond to a high switching frequency. So, this is 16 kilohertz and what is in between is for the 10 kilohertz term. So, you could see that when you your loss in the filter design is having this tradeoff between the losses at the low value of l and the losses at the high value of l even for the different frequency ranges. But it is also interesting to see the point at which the minimum occurs is for the low value the high value of switching frequency a desirable range of your minimum loss is occurring around 0.4 around 4 percent value of inductance. Whereas, if you look at the case when you are operating at low switching frequency the desired value of l the total inductance l 1 plus l 2 shifts from a low frequency to a high frequency of higher value of l t to about 0.9 percent. So, at 16 kilohertz you are having a minimum loss around 4 percent at around 8 kilohertz switching frequency the minimum loss point shifts up to a larger value of inductance. So, this is what you naturally expect if your switching frequency is low you need higher value of inductance in your filter and if your switching frequency is high you need a lower value of inductance in your filter. The other thing to observe is that at the low value of switching frequency the losses in your filter inductor is actually about 50 watts. Whereas, at the high value of switching frequency your losses in your filter is lower about slightly lower than 20 watts. So, you could then consider not just the losses in the filter you could also consider the losses in the filter filter plus the semiconductor devices because you know that you get lower losses in the filter, but your losses switching losses in your IGBTs and recovery losses in your diodes is going up with switching frequency. So, you could not just look at it from the filter perspective you could actually look at a minimization from the overall perspective. So, depending on the frequency you have chosen you could then consider value of l. So, you could select a value of l that leads to minimum loss that could be a preferred filter design. There are multiple perspectives one is you have high efficiency because you have selecting a filter which leads to lower losses you have low temperature rise, low temperature of the filter components which improves the reliability. So, if you look at it from your net present value basis your cost of loss is actually reducing your net present value and sometimes if you just go by the rule of thumb you might have ended up selecting a larger value of filter inductor, but you know that physically because you have done the analysis you might actually be able to select a filter which is actually having a lower value than initially thought which actually reduces the initial cost too. So, from multiple perspectives you can actually end up with a better filter by doing a close analysis of what would be the power loss and the thermal rise in your filter. So, we can also see that. So, we some of the points to note are is a function of switching frequency the second thing is at high f s w lower value of l is preferable. So, we saw that at 16 kilo hertz your preferable l was closer to the 4 percent range and the power loss in the filter reduces as the switching frequency increases. So, your overall switching frequency in your overall design can because in many designs you often ask what is your switching frequency and you might say 10 kilo hertz or 20 kilo hertz or some number. So, later on you can actually justify why you selected a particular switching frequency through such a low loss calculation which leads to what is the best switching frequency that the power converter can operate. So, you could select your frequency to minimize the sum of the losses in the semiconductor plus filter. So, if you look at the overall characteristics you can think about the total power loss is a function of what your l per unit is and it is also a function of your switching frequency essentially you are looking at what is the point which gives you your best desired loss subject to constraints on your total value of l lying between some minimum and maximum range. So, at this point we have actually now your selection. So, of l 1 comma l 2 comma c and we can actually look at non idealities of your power converter and primary non ideality which you automatically get once you go into a higher order filter design is how to deal with filter resonances. So, this is one aspect that we would need to consider. Thank you.