 Welcome to class 25 and topics in power electronics and distributed generation. We will start this class by summarizing what we have been discussing which is what we mean by the basic specification of a inverter. So, when we talk about basic specification you start off with something as elementary as a power rating, your AC voltage rating, what frequency you are operating, standard frequencies are 50 hertz, 60 hertz, your nominal range of voltage, what is the range around the nominal and which has implications on current rating, etc. The type of interface whether it is a single phase interface, three phase interface, three wire, four wire, we also discussed about the environmental specifications, temperature, ingress protection, etc. have actually a large impact on the design of the power converter. We also looked closely at the single phase, a simple single phase topology which is essentially consisting of a split capacitor bank and we saw the implication of the AC voltage on deciding what your DC voltage has to be. And the DC voltage we saw is related there are factors relating to the range around you might have 5 percent above the nominal grid voltage which you need to handle, you might have some percentage voltage which gets lost as dead time, on state drops of your devices, etc. You might also have some range of voltage that gets dropped across your filter inductor. And so, for a 230 volt system you might end up having to require a DC bus voltage which might be close to 800 volts and this immediately would imply that now you need a capacitor bank, if you are having 2 in series you are talking about 400 to 450 volts as a value of the voltage that you need for the capacitors. And for the IGBTs or transistors and diodes you need components which commonly would be available at 1200 volts. So, then you can actually look at what given this preliminary specification, what are the implications on your DC bus capacitors and we the starting value point is to discuss what the voltage rating is, then to look at what the current rating is. And we saw that there are a variety of current signals that can flow through the DC bus capacitor bank. And when you are talking about a typical DG application your actual capacitor bank, your voltage sources over here might consist of capacitor banks connected in series. And your actual voltage source is a prime source connected in parallel with the capacitor bank and the capacitor bank is effectively acting as the voltage source on a high frequency and intermediate frequency basis. And what you are trying to regulate is ensure that the voltage across this capacitor bank stays ideally as close to the DC value that you desire. And then we looked at what would be the currents that can flow through your positive bus and we saw that there is a DC current which is essentially what transfers power in a typical DG application. You have a 50 hertz component especially in this center tap topology which flows through your capacitor bank. You have a 100 hertz which would show up in again in a single phase topology or it can even show up in a three phase topology if you have unbalanced in your AC voltage or unbalanced in your AC system. You also have switching frequency components that can actually flow through your positive DC bus and we linked the current in your output eye out to your current in your DC bus. And we derived expressions for the DC current 50 hertz current the second harmonic 100 hertz current and your switching frequency component. And these are relatively simple expressions to actually obtain that can be obtained for your bus capacitor currents. And then the next question is how to actually make use of these currents that you have evaluated in the design of your capacitors. So, the first thing that you when we are looking at say the design of the capacitors are there are a variety of capacitors out there. So, capacitors are commonly classified based on the dielectric that is used in the capacitor. You might have ceramic mica paper, you might have tantalum electrolytic, polyester, plastic capacitors, polypropylene. So, depending on your application for example, mica capacitors etcetera are really good at very high frequencies, but the value of the capacitance that you get is really small. Electrolytic capacitors have a large value, but you may not get it in AC rated as in with AC rating. So, if you look at the common capacitor that is used in the DC bank of a voltage source inverter which is used in a DG type of application you are talking about commonly it is electrolytic capacitors. Once you have decided to say you want to make use of a electrolytic capacitor in your DC bank, then what are the advantages of using a electrolytic capacitor? One thing is you have electrolytic capacitors in hundreds, thousands of micro farads or millifarads range also depending on your voltage rating and a larger capacitance means for a given current ripple you are having smaller voltage ripple or better filtering ok. So, larger capacitance implies you have lesser voltage ripple, a larger capacitance half C v square is the energy stored in a capacitor and large value of capacitance means that you are emulating ideal voltage source more closely it can provide a large amount of energy without getting discharged or changing the value of the voltage ok. You also have now capacitors, electrolytic capacitors with low ESR, ESR is equivalent series resistance. So, low ESR again we will see has a advantage in terms of dissipation, life, temperature rise etcetera and today commonly you get electrolytic capacitors with low ESR capability which has implication in the amount of ripple current that can actually flow through these capacitors. Another negative point of a electrolytic capacitor is that it does not have AC voltage rating, but AC voltage rating is not required in the DC bus of a voltage source inverter. So, even though it is a drawback it is not a major drawback for a electrolytic capacitor in a VSI application. So, depending on the nature of the ESR you can also have different types of electrolytic capacitors, you might have electrolytic capacitors where the ESR might be minimum at lower frequencies and you might have electrolytic capacitors where your minimum of your ESR might be at this the switching frequencies tens of kilohertz or few kilohertz. So, if you are having a specific capacitor which is designed to have lower ESR at may be 100 hertz 150 hertz you might be having a rectifier grade electrolytic capacitor which is used in standard diode bridge rectifiers where you know most of the ripple is going to be at lower frequencies. Sometimes people refer to high frequency electrolytics as SMPS grade electrolytics which are optimized for the ripple now flowing at may be many kilohertz tens of kilohertz etcetera. You also have other parameters of the capacitor which is important if you look at a capacitor which is acting as a equivalent voltage source in bus of a inverter you also have a parasitic inductance which can actually come in series with the capacitance and this parasitic inductance actually can cause a significant problems because the switching speed of your transistor it can actually have very large d i d t d i d t's of the order of kilo amps per micro second to tens of kilo amps per micro second. So, even a few tens of nano Henry's of inductance can actually cause a fairly significant voltage spike of the order of 100 volts 200 volts etcetera depending on the inductance. So, it is very important to actually minimize the ESL of your capacitors. One thing that people might do is instead of just having a electrolytic capacitor you might actually put snubber capacitors in parallel with the electrolytic capacitors. So, many times one might wonder why is there a 1000 micro farad capacitor and in parallel with that you are putting a one micro farad capacitor or a 4.7 micro farad capacitor it is from a ESL perspective where the ESL of your snubber grade capacitor would be much lower and for higher frequencies the ESL is also a significant factor in ensuring that you do not have voltage spikes that come across your capacitors. Also another aspect to keep in mind is when you just draw a power converter you might just connect source to a transistor often because your parasitic inductance is so critical you might actually decide to make special DC bus connections you might have parallel structures plate structures in your printed circuit boards or copper buses with a parallel structure to minimize these inductances. So, as to ensure that even though your DC bus might be sitting at 800 volts and your device is 1200 volt if you do not pay close attention to what these stray inductances are given the capability of modern day transistors you can end up causing voltage spikes that can damage these devices. So, once you actually have adequate margin between your voltage in your DC bus and you know a preliminary value of the voltage then you can ask what are the constraints that can happen on a capacitor bank and we saw that the DC the voltage rating is a basic constraint your DC voltage rating is something that we just discussed. In addition to the DC voltage rating we know that we are having low frequency currents 50 hertz 100 hertz also switching frequency currents those currents can actually induce and cause ripple voltage and you need to ensure that the ripple voltage is also adequately addressed. So, we also saw that the 100 hertz ripple might become so significant that you might actually want to add additional DC to DC converter etcetera to prevent this ripple voltage from flowing back into your prime source and cause losses in your overall dg system. Also you have current rating which is a important factor in any component many times your terminals that you connect a component is specified by what current that terminal connection can handle. If your current becomes too large then your terminals your screws your whether snap in connections or your PCB insertion points they can overheat and can melt. So, current rating is a important aspect of terminals and connection points and also components and when it come to components the current rating reflects itself in terms of the thermal effects which eventually leads to overheating ok. So, if you are having overheating in a component like say the DC bus capacitor one thing is you could directly say you are passing too much current through a component ok. If you are passing too much current one way to prevent say large current is to parallel components if you have a given amount of current and you want to actually you find that the current that is flowing is too large you could then connect components in parallel and then the currents take the parallel parts and ensure that the current per device comes down ok. The other aspect which can lead to higher thermal effects over overheating in components is the ESR itself is being is too large. If you select a capacitor where the ESR is large your I square R the R part of it is given a current the R part is too large which means that you are having higher power loss and which in turn results in too much temperature rise. So, you need to ensure that whatever capacitor or component that you are selecting has a ESR that ensures that the temperature does not go too high and which was one of the criteria in modern day electrolytic capacitors having low ESR capability. Low ESR which is actually advantage. Another thing your IRMS capability might be your RMS current might be small your ESR might be small, but given how a small a power dissipation is having is occurring within a component. If the component is not being cooled then it is an integration process where you are adding thermal energy into a component where the thermal energy does not flow out. So, if the cooling is not adequate then again you are going to get overheating which means that you need to consider say whether your capacitors how is it packaged. Is it packaged in such a manner where you allow say natural convection if your capacitors are too tightly placed there may not be possibility for air flow around it. If it is kept in a horizontal configuration rather than a vertical configuration it will determine whether you can have natural convection around a capacitor bank or you might be as simple as maybe you need to add a fan for the capacitors. You might have this capacitor sitting in a sealed cabinet and you might be having poor air exchange and you need to have some air circulation within whatever cabinet you have to ensure that your components does not get too hot. And these so there are multiple aspects when you consider the thermal effect it is not just a question of just the current being too large it can be ESR it can be cooling so all these factors have to be considered together. Also if you are looking at the constraint another constraint on a capacitor bank is how much energy is there in the capacitor bank. So, for example, if you have a short term outage your energy that is there as half CV square might be available to provide for a short term requirement within the power converter. So, this has implications on hold up time of your converter in many dg applications people are talking about low voltage right through or 0 voltage right through which means that you need some energy storage capability which is becoming more and more important in dg applications. And out of many of these constraints the hold up time thermal current voltage rating many times when you eventually come up with the design the thermal effect is actually a major factor which needs to be considered. And so close attention to thermal effect would also incorporate the current rating issues and the voltage rating issues because the current causes ripple the current causes overheating. So, many times when you get to the detail of the thermal design you are in fact, covering many of the other constraints in the design of your capacitor bank. So, then you could ask if you have say different temperatures what is the constraint that what is causing problems within your capacitors ok. So, if you look at a typical component say you are talking about a electrical electric capacitor and any chemical process if you are looking at temperature effects on a chemical reaction say you are having some A plus B reacting to form C. And we know that the rate of a reaction can be expressed at as some R is dependent on some rate constant K which is a function of temperature it is proportional to the concentration of your reactants A and B. And these are things that we have looked at studied in maybe a high school chemistry class and we know that these rate these rates of reaction is actually a function of temperature. So, if you think about the chemicals within components like the electrolytic material etcetera they degrade at a higher rate at higher temperature. And often your K of t there is people express it as a adrenous rate law some A exponential activation energy gas constant your Kelvin temperature. So, with factors such as this you can see that at a higher temperature you end up with a higher rate constant. And often when you look at many electrical systems further simplification that people might make is to say that when you are increasing your temperature by say a 10 degrees you might lose your lifetime by a factor of 2 ok. So, these are again simplifications given a given temperature activation energy, but this can be some simplifications which give you some insight into how to actually select a component for a given lifetime or what are the implications of temperature on the operation of a given component. So, with this we will look at say a design example of a capacitor bank and we will look at an example of say a 2 kilowatt power converter single phase center tap capacitor topology the topology that we have been looking at at 230 volts it is operating as a unity power factor converter. And the switching frequency is 10 kilohertz say it operates at a ambient temperature of 40 degrees and maybe there is a temperature rise between your ambient air temperature and the temperature of your cabinet this power converter might be sitting in a sealed cabinet and your ambient may be 40 your inside cabinet temperature might be 50 and say you are operating maybe it is a solar inverter it may be operating during the daytime 8 hours a day and around the year 365 days a year. And the question is how could you go about selecting a DC bus capacitor given a simple topology as what we have discussed ok. So, one thing that we can start off with is what would be the voltage and we know that the peak voltage that we are operating is 230 root 2 is 325 volts and based on the discussion that we had about the additional margin of due to dead band voltage variation 5 percent voltage variation in the grid again making assumptions relating to the filter design we might talk about VDC roughly about 800 volts. And then given that we need 800 volts we might select a capacitor which is maybe 450 volts and we go to manufacturers data sheet and take a look at what are the range of capacitors that are available and as a starting point we might say let us start off with maybe a 150 micro farad 450 volt capacitor and this I picked off from manufacturer Cornel Dublier CD.com you have many of manufacturers Chemette, Vishay, Sanu, Epcos many of these manufacturers provide data sheets of capacitors on their websites. And when I when you look at the capacitor they provide values of what years are is there a common frequency that people might specify might be at your second harmonic which might be a dominant frequency in such capacitor designs depending on whether it is 50 hertz, 60 hertz people might consider different second harmonic frequencies we will assume second harmonic frequency of 100 hertz. And there is a 0.8 ohms resistance at ESR resistance at 100 hertz. From the data sheet we also have current multipliers. So, the data sheet says that at 100 hertz you are able to carry one ampere of current, but at 50 hertz you can carry only 0.8 amps current at 10 kilo hertz you can actually carry 1.4 amps current. So, when you are talking about different current levels which can be carried at different frequencies you are consuming assuming that you get same dissipation now at with these different current levels in a given capacitor ok. So, you are talking about the fact that you are now your I square r effect because of the current at different frequencies are actually not the same for a given capacitor. Also in the data sheet it is provided that when the ambient temperature is increased for example, if it is 85 degree centigrade I can pass 1 amp of current and expect about 3000 hours of load life in the capacitor. But if I now reduce my ambient temperature to 55 I can actually carry 2 amps and get the same 3000 hours of life. If I reduce it to 45 degrees I can pass 2.25 amps and still get the same load life. So, you can see that you if you are able to reduce your temperature you can actually get improved life or on the other way looking at it in another way you can actually have higher current capability depending on how well you have cooling. So, the first question that you could ask is this capacitor appropriate for this application from both maybe current ripple and a voltage ripple perspective. And if it is not appropriate then what should it be maybe should it be 300 micro farads should it be 600 should I put many of them in parallel. How about how can I go about deciding on what would be a appropriate capacitor to choose. So, the first thing that you can look at is then your current rating. So, AC rating for this particular system your IAC is it is a 2 kilowatt system 210 power of 3 and 230 volts nominal. So, you have 8.7 amps RMS or 12.3 amps peak. So, the first thing that you could immediately calculate is your 50 hertz component because it is a center tapped DC bus topology. So, your I in your C DC at 50 hertz is 4.3 amps. We know your based on your DC bus voltage we have V DC times I P of T or I DC of T is 230 volts root 2 cos omega t and 12.3 amps peak cos omega t and we know your V DC is 800 volts nominal. So, we have your I DC of t to be equal to 2.5 amps 1 plus cos 2 pi 100 t. So, your so you know your DC current is 2.5 amps and your 100 hertz component of your current I C DC 100 is 2.5 amps peak or 2.5 amps by root 2 equals 1.8 amps RMS. So, then so another thing that we could maybe from the design you might start off with a switching frequency of maybe 10 kilohertz that is used in this particular application. So, we would like to calculate what is the 100 hertz component of your current flowing through the DC bus. So, to calculate I C DC 10 kilohertz you know your I out because if you are switching at 10 kilohertz will consider it as N switching instance. So, 200 switching instance in one fundamental cycle. So, you can think of it as your output volt current is 12.3 amps cos 2 pi 50 N T SW. So, depending on each switching instant you know what your output current is your output current is a sine wave which you are trying to control to be a sine wave. You also know your duty cycle duty cycle again depending on which instant you are operating is given by close to 0.5 plus 230 root 2 by 800 cos 2 pi 50 N T SW. Again this is assuming that your ideally ideal grid voltage is reflecting back to your pole voltage. You might have some small changes in the value because of the voltage drop across your filter, but it will not be majorly different. So, you could take this as an approximation and you also know that given FSW switching frequency is 10 kilohertz and F naught is 50 hertz so your there are 200 switchings per fundamental. And then you could calculate what your 100 10 kilohertz component is. So, your ICDC 10 kilohertz is 1 by 200 summation i from 1 to 200 i square of N T SW into D of N minus D square of N the whole thing under a root. So, you could evaluate what this quantity is you know what your D is what your I out is and you can calculate the current you end up in this case of around 3.1 amps. So, if you look at your total RMS current in that flows through your capacitor is now your 50 hertz 100 hertz and your 10 kilohertz component. So, I RMS in your capacitor. So, we see that there is 5.6 amps and this is much greater than the current rating of the capacitor which was roughly around 1 amp. So, definitely 150 micro farad capacitor that we started off with is not adequate so this is much greater than rating current rating of. So, we need multiple capacitors in parallel and we will also we will now take a look at what the what the ESR implication of having the different current level at the different frequencies mean. So, if you look at the capacitor data so, that it can carry 1 amp at 100 hertz 0.8 amps at 50 hertz and 10 at 1.4 amps at 10 kilohertz. So, if you look at what does that mean. So, at 50 hertz I rated equals 0.8 amps. So, if you look at your 0.8 is your ESR at is your current I square R ESR at 50 would be similar to now your I square R at 100 hertz and we know at our ESR at 100 hertz is 0.8 amps which was there from your data sheet. So, you know that now your ESR at 150 hertz at 50 hertz is actually 1.25 ohms. So, similarly you can do a calculation at 10 kilohertz your you know your I rated is 1.4 amps. So, 1.4 square ESR at 10 kilohertz you can evaluate that into 1 square into ESR at 100. So, you get ESR at 10 K to be 0.41 ohms. So, in a way giving your current multiplier at different frequencies is a giving you the same information as ESR as a function of frequency. So, some manufacturers might actually give you a plot of ESR at different frequencies. So, in this case you might have a plot that looks like this where ESR is reducing with frequency. So, you had 50 hertz you have some value at 100 hertz and some value at 10 kilohertz. So, the other parameters that we can actually look at from the temperature multipliers that we have provided to us is that we were told that at for 3000 hours of life you at 85 you have 1 amp at 55 degree C ambient for the same life you can carry 2 amps etcetera. So, now with that information we can actually calculate what is the thermal impedance between the core of your capacitor and your ambient and also what is the actual core temperature at which this particular number is being specified ok. So, if you look at the number at 1 amp at 85 degree centigrade your life is 3000 hours and 2.25 amps at 45 degree centigrade you have the same life. So, your T core of the capacitor is similar under these 2 conditions because the degradation within the capacitor is happening at the same temperature because it results with similar lifetimes. So, if you look at your thermal impedance or thermal between your core of your capacitor and your ambient. So, you have T core minus 85 by your participation is 1 square into point 8 ohms which is your ESR is equal to T core minus 45. Now you can carry actually instead of 1 amp you have 2.25 amps into again the resistance is still point 8 ohms it implies that you can solve for what your T core is. So, the T core that is probably being used by this particular manufacturer is 94.8 degree centigrade and the thermal impedance under the measured measured condition your RTH from your core of your capacitor to your ambient is about now 12.3 degree centigrade per watt dissipation in the capacitor. Now you based on the current currents that we know is that are flowing through the capacitor you could then say what is a equivalent 100 hertz current that might be flowing in the application that we are considering. So, the currents in the capacitor is 4.3 amps at 50 hertz 1.8 amps at 100 hertz and 3.1 amps at 10 kilo hertz. So, equivalent 100 hertz current 100 hertz current in the capacitor from a thermal perspective. So, in terms of what gives the same temperature rise or in your capacitor is say some I prime square 100 into 0.8 is 4.3 square into 1.25 plus 1.8 square into 0.8 which is a ESR at 100 hertz plus 3.1 square into 0.41 which is the ESR at 10 kilo hertz. So, you are talking about I prime 100 we are talking of something like 5.4 amps and if you want to say target in your application 3000 hours roughly 3000 hours of life and we know that 2 amps of ripple can flow if your ambient temperature is 55 degrees which is the multiplier given to you at 55 degrees. So, targeting 3000 hours of life assuming 2 amps ripple at say 55 degrees because your cabinet temperature is 50 degrees not too far away from 55 ambient. So, one would need about 5.4 amps divided by 2 amps. So, you are talking of 2.7 capacitors rounded off to 3. So, you need roughly 3 capacitors in parallel. So, instead of 150 maybe you could selecting a 450 micro farad capacitor might be a good starting point ok. So, then you could ask now if you take 3 capacitors in parallel what would be the power dissipation in each of these 3 capacitors that you are going to put in parallel ok. So, the power dissipation per capacitor in this parallel combination of 3 banks is 4.3 divided by 3 square into 1.25 plus 1.8 divided by 3 square into 0.8 plus 3.1 by 3 square into 0.41. So, you have 2.63 watts plus 0.28 watts plus 0.43 watts. So, you have about 3.3 watts dissipation per capacitor in the parallel combination of 3 capacitors and then you could immediately say what your core temperature is given that your cabinet is sitting at 50 degrees centigrade. So, your core temperature in this particular case would be your 50 degrees which is your inside cabinet ambient temperatures given for your application plus 3.3 watts power dissipation per capacitor and we know at your thermal impedance from ambient to core which is 12.3 degree centigrade per watt. So, your actual core temperature if you use say 3 capacitors in parallel will be about 91.1 degree centigrade ok. Then you could actually calculate what would be the lifetime of this particular bank of 3 capacitors. We know life is approximately 3000 hours for your nominal core temperature. So, it is 3000 hours into we will again use a simplifying assumption of decrease in life by a factor of 2 for 10 degree temperature rise. So, it is 2 to the power of 94.8 which is your nominal specified temperature at which we made those specifications were being made. Your actual temperature is 91.05 divided by 10 to reflect the factor of 10 change having a multiplying factor of 2. So, you are talking about now 3900 hours of operation of this particular bank and then if you look at what it is in terms of 8 hours per day 365 days per year. So, this implies 3902 divided by 8 into 365. So, you are talking about 1.3 years is what you could roughly expect from this bank ok. Then you could ask what is the power dissipation in this bank you have 3 capacitors on top 3 capacitors at the bottom and the power dissipation in the capacitor bank is 3.3 watts per capacitor into 6 capacitors. So, there is about 19.8 watts of dissipation happening in this particular capacitor bank ok. So, you can see that you now have a way of linking the selection of the capacitor to your current ripple. I mean there are many other things that we need to calculate that the voltage ripple, but we could start off with some simple design tradeoffs. What say maybe a basic question could be instead of why did we select 3 capacitors in parallel we could have put 4 in parallel ok. So, what would be the implication of putting 4 capacitors in parallel rather than 3 ok. So, if you do your power calculation for 4 capacitors in parallel your power dissipation per capacitor is 1.25 into 4.3 by 4 square. So, instead of 3.3 watts per capacitor now you get 1.88 watts per capacitor. So, your power dissipation has come down because you have gone to a larger number of capacitors your core temperature now comes down from the 91 to about 73 and your life now comes goes up. So, it is 3000 hours into 2 to the power of 94.8 minus 73.1 by 10. So, you are talking about 13500 hours by 8 into 365. So, you are talking about 4.6 years rather than the 1.5 1.6 years that we previously looked at when you had 3 capacitors. So, you can see that your lifetime has increased because now you are able to bring down the temperature you could also now calculate your power loss in your capacitor bank when you use 4 in parallel is now 8 capacitors dissipating 1.88 watts. So, this is 15.04 watts. So, previously you had 19 watts when you had 3 capacitors it is come down to 15 watts when you are putting 4 capacitors. So, you can see that you have now a framework where you could address the what is the effective initial cost of selecting a single component just a capacitor because between 3 and 4 capacitors you know what the cost difference is. You have a way of evaluating your operation and maintenance life and you have a way of looking at your net present value of power loss which is going to be happening in your capacitor bank. So, a framework where you can actually do a such an evaluation on a per component basis of each component that goes into a power converter can actually then be rolled together to do a overall design of your bigger system or your overall power converters power conversion system. So, if you look at again the factors that we looked at in the life calculation of this capacitor we had essentially your life at a given temperature is related to some nominal operating life given by a manufacturers and using a simplistic assumption of 2 degrees per different I mean factor of 2 in life for 10 degrees change. So, you have a TCO minus T actual by 10 degrees into you also have a multiplying factor based on your voltage suppose you have a DC bus capacitor bank rated at 450 volts and you are using it very close to 250 volts then your you have a factor which imply effects on your life because you are operating very close to the rated voltage as you bring down your voltage of your capacitor you can have some small improvement in your life. So, you could incorporate factors that address different issues like temperature, a margin your thermal margins etcetera as we have just calculated and just for simplicity maybe you could take your voltage factor to be roughly 1 as long as you are ensuring that your voltage does not exceed your rated voltage during operation of your power converter ok. So, you can see that with this you could also ask a question what does it mean by having a capacitor which is going to last 4 years or 4.6 years or whatever number you have evaluated and that typically is reflected in terms of what is the failure rate of a typical component and people talk about mean time between failure of a component failure rate or when you look at a typical component essentially at the point at which it is initially built you might have large failure rates you might do some initial tests within your factory to ensure that some initial tolerance which is not being met in the component does not get shipped out of your factory. So, you might do some initial tests to ensure that you do not see what people talk about early mortality or infant mortality of your components in your power conversion systems. So, once you pass your qualification test within your factory whatever gets shipped out has fairly low roughly constant failure rate it is not that the failure goes to 0 it might be you might have a random failure depending on what the parameters the complexity of your component etcetera is, but you would have a low roughly constant failure rate. But after a while so, after a while it start the failure rate starts picking up it no longer stays that the low value what it was because of your operation because of the usage of your component you might end a reach a end of life value at which point your failure rate starts increasing. So, in the case of a capacitor what would one consider to be a end of life. So, your physical capacitance value might actually change with time your leakage current in your capacitor might change with time after a long time your leakage current through the capacitor might suddenly start increasing. Or if you look at your ESR of your capacitor initially you might have a ESR of 0.8 ohms, but after some 5 or 6 years you again measure the ESR of the component it may not stay the same 0.8 ohms it might increase to a different value. So, manufacturers have different specifications of what it means for a component to be reaching its end of life in case of say a capacitor a common specification might be the ESR goes up by a factor of 2 at the end of life ok. So, if a ESR goes up by a factor of 2 given the same RMS current that is flowing through the capacitor it means that now your participation has doubled which means that again your temperature is going to accelerate up. So, your degradation is going to now become more and more rapid at the end of the life you are quite quickly going to reach a point where you might end up with a failure of the device. So, you can see that it is possible to actually link the operation of the components to the operational life and try to incorporate these factors in the design. We have still to consider what is the impact of looking at the voltage ripple because we have looked at the current ripple. So, we also have to look at the voltage ripple to see whether the capacitor selection would be adequate for a for the design. Thank you.