 Welcome to class 24 in Topics in Power Electronics and Distributed Generation. In the last class, we talked about transfer of power between two sources and we looked at a efficient way of transferring power from one source to the other with one side being a voltage source and the other side being a current source. And efficient switch arrangement to transfer power in such a condition can be modeled as a single pole double throw switch. And we saw how it can then be extended to a DC to AC inverter which is commonly required for a distributed generation or a power quality or any inverter application. And we looked at the simple case of it where we are looking at it on a single phase basis one leg at a time. And it can also then be generalized to include the actual diodes and transistors to form a real realization of such a converter circuit using transistors and diodes to implement the single pole double throw configuration. So one can then look at what is the output voltage that can be generated in such a system and we looked at the average output voltage that can be obtained. So we looked at the average voltage from the output to the neutral midpoint of the DC bus and this was V DC by 2 T on minus V DC by 2 T off and this is average over duration of T SW which is a switching duration. And we know that the T on can be related to the duty cycle as T on is equal to D times T SW and T off. So it gives the time T on is a time duration of the top switch being on. And we saw that we could actually get a relationship between the duty cycle and the output voltage on an average switching cycle basis. So if your duty cycle is varying between 0 and 1 your output voltage can have a value of minus V DC by 2 at a value of 0 and plus V DC by 2 when duty cycle has a value of 1. So if you look at the grid voltage that or the voltage that gets connected to power converter at the output you would like to get a relationship between what is the actual output voltage at your load or at the terminal of the inverter and the DC bus voltage that is available to you. And that you can get a relationship based on if your grid voltage V G is some A times A amplitude of voltage A V cos omega naught T. A common voltage that we have in a single phase system in India is 230 volts. So A V is 230 square root of 2 would be the amplitude of your voltage that is being applied. And for the positive half cycle your T on max can be as much as your. So in the positive half cycle your T on max can be as much as T SW. So you can apply as much a voltage as V DC by 2. In the negative half cycle T off can be as large as T SW. So you can as apply as much as minus V DC by 2. So you can get a relationship between what is the output voltage that you are applying and what is your DC bus your T on max is T SW when duty cycle is equal to 1. And so you get your average V ON that is being applied can have maximum value of V DC by 2. And if you take V ON average to be equal to the peak of the voltage that you are applying you are talking about 230 root 2 is equal to V DC by 2. So this is about 230 root 2 is about 325 volts. So you are talking ideally of about 650 volts as your DC bus voltage when your input voltage is about 230 volts. We will see that you have to add additional factors over and above this particular DC voltage in an actual application. So before we do that we will actually look at what it means to say you are building a inverter of a given specification. So a typical power converter that will be built you first need to know what power rating it is going to handle. So whether it is 100 watts a kilowatt 10 kilowatt megawatt. So a basic quantity of that is specified when you design a power converter is the power rating. So once you know the power rating the next important quantity is what is the voltage rating especially the AC voltage rating which is what gets connected out of the terminals of your power converter. And common AC voltages are can be 110 volts 220 230 volts 480 volts 560 690 there can be a range of voltages depending on different geographical locations etc. A common voltage for a single phase system is 230 volts for us. So it is not just the voltage rating we also want to know what is the range around the nominal voltage because ideally you get the exact voltage but we know that there is going to be a range around the nominal voltage you might have a positive value above the range and negative value below the range you might have 10 plus 5 percent voltage over the nominal minus 10 percent. So depending on the range around the nominal we also have to know to actually design the specification of the power converter ok. So you are talking about in terms of power rating you are talking about of range of watts could be even as large as mega watts depending on the application. So voltage rating you are talking about your AC voltage rating so you are talking about 110 230 480 volts 690 etc. All these voltages are below 1000 volts so what is typically considered low voltage systems. The frequency rating is also a important factor though it might not directly affect your power semiconductors it might affect your speed of cooling fans etc. Nominal frequency in our case would be 50 hertz. Again the electrical interface required is also a important consideration by electrical interface it is the most basic aspect of it is whether it is a three phase system single phase system whether you have three wires four wires. So the number of AC connections coming into your power converter is a important aspect of the design. So what we have been considering in our basic design is a single phase power converter. Other important factors which are important are environmental specifications. So when we talk about environmental specification the number of factors one is what is the ambient temperature. So higher ambient temperature would be more challenging design for a power converter. Also whether your power converter is designed for indoor or outdoor application the environmental degradation experienced by a converter meant for outdoor can be more severe compared to indoor. So you will have to pay attention to where exactly is the location of your power converter. You also have what people refer to as the ingress protection of your power converter. So people talk about IP rating of your converter cabinets. IP ratings like IP 00 means that there is no protection you can put your hand in and you will potentially get a shock. There is no water proofing whereas a larger number like IP 66 means that it is fully sealed and your dust will not enter the cabinet. Also you can have water jet sprays under which some of these cabinets can be immersed and you won't have damage to the electronics. The electronics would function even when the ingress protection number is at the higher side. Again the cost of equipment goes up with higher IP numbers ingress protection numbers. So if you have a enclosure which is very tightly sealed it means thermally cooling it is more challenging whereas something which is open frame is easier to cool but there is more risk that some dirt might fall on it might get damaged. So there is always a risk between how you handle your IP ratings. Typically when we develop something in an academic lab we might develop things in an open frame basis but finally when you have to send it to a customer aspects such as IP protection is important. Also there are other important factors such as vibration or whether your equipment will survive an earthquake depending on what level of impact or shock is being experienced in the cabinets whether there are capacitors in the cabinet which might fall off from your mounting. These are important aspects of specifying a power converter. We will start with the basic single phase power converter and if we take a single phase power converter and assume that it is going to be operating is going to be operating as a distributed generation unit will assume that it is going to be at unity power factor. We will assume that there is unity power factor operation sinusoidal AC voltages and currents. So your power is VRMS IRMS and so your I rated is P rated divided by your V and one thing that we just mentioned is the V depends on what range around the nominal you want to actually operate. So for example if you have a range around the nominal as plus 5% minus 10% it means that for a given power rated if your voltage is 10% lower it means that your current has to be rated higher which means that your component costs are going to be higher. If you have a system which is now going to be used in an application where the power quality is going to be poorer it means that your voltage range may be wider it has impact on the cost because it means that you have to have higher current ratings. So you need to make sure that there is a trade off between how much power quality situation you are willing to encounter and to what specification you want to maintain your power requirement. So a larger voltage range would imply that greater current rating or the components given P rated. So we will again look at what is the voltage that we would we need to have on your DC bus given AC voltage and these voltages are related to your nominal AC voltage variation. So if you think about the grid voltage as a phaser V grid you can have a number of factors one thing that we mentioned is that the grid voltage might have some tolerance which might increase the value of the actual Vg. The other factors that can actually affect us between the power converter and the your output voltage and your inverter you have a filter inductor. So you might end up with a drop on the filter inductor. So you need to also incorporate how much drop you might expect on that particular filter inductor drop. So depending on the power factor if your I out is going to be lagging for some reason if you want to actually provide capacity watts you might end up with further increase in value in your terminal voltage that is required by your power converter. If you take a typical filter starting assumption might be your filter inductance might be 10 percent. So you are talking about here a AC voltage variation of say 5 percent addition over here you might be talking about a 10 percent addition over here. You also have factors such as dead band which is will explain what the dead band is. Typically when you have a leg of a power converter we saw that to the one of the condition is that we cannot short both the tops and the bottom together or you will cause a shoot through across the voltage source. The other is that your current always has a path. So to prevent the positive and negative being shorted you always have a duration where both S plus and S minus is simultaneously kept off for a short duration called the dead time. Now if you look at the effect of this particular dead time the effect of the dead time depends on the polarity of the current. So if you look at an ideal sine triangle modulation whenever your duty cycle is greater than the triangle V TRI your S plus is on and S minus is on whenever S plus goes low. But in the actual signal that is provided to the gate of the devices your S plus and S minus your starting edge of your signal is always delayed a little bit. So as to prevent both top and bottom switches from shooting through and this delayed duration is called the dead time. Now if your current polarity is positive turning off S1 will automatically cause the complementary diode to turn on. So there is no error being caused during the positive polarity current when your when your current is in the positive polarity but when you are trying to actually turn on the top switch any delay in the turn on of the top switch will increase the amount of time for which the bottom diode is still conducting which leads to an error in voltage and this error in voltage depends on the ratio of your the average value of this particular error in voltage depends on your DC bus voltage in this particular case the error is negative it depends on your dead time it depends on your switching duration TSW and you can think of it as a voltage which is like a square wave and if you want to convert it back to a sine wave of some particular fundamental it will you can add a factor of 4 by pi to get the fundamental effect of the dead time and this gives you an additional factor which you need to have in your AC voltage output to generate your your required PWM output. So, we will assume that this is also having a factor of plus 0.5 percent. So, overall if you then look at the relationship between your DC bus voltage and your AC output you will end up with now your 230 times root 2 plus a factor for your dead time plus a factor for your delta Vg your Vg variation and you might have a factor for filter drops some of these factors might be not exactly in phase for example, the filter drop might be 90 degrees lagging whereas your dead time might be in phase, but we also need to consider that you have a margin between what is the actual dynamic voltage available to control your output and the actual voltage that is there at the terminals of the inverter. So, overall keeping in mind factors such as this we will see that the actual DC bus voltage required is not just 325 it might has it would have or 650 volts it has to be greater than 650 volts a common voltage you might end up requiring maybe you multiply these factors you might be getting a voltage close to 800 volt. So, you can see that these factors do add up and you have to actually get adequate margin between your actual DC bus voltage and your operating voltage for your AC side of your system. Also keep in mind that we are assuming basic single phase sine triangle modulation for your pulse width modulation of the switches this voltage now has an implication on your selection of components within your power converter two parts are there to at one is we know that in the actual power converter your DC bus is which is over here shown as 2 voltage sources of V DC by 2 is actually built with capacitors. So, the first implication is on the rating of the capacitors that are used in your DC bus. So, if you have to have a DC bus voltage of 800 volts it means that your electrolytic capacitors have to be maybe 850 900 volts etc. and because it is not very common to have high voltage capacitors greater than 500 volts you might have to end up now having series connected capacitors in a power electronic application where you are having to have say 230 volts AC. So, in this particular case maybe because your DC bus voltage is 800 volts you might use two capacitors each capacitor having 450 volts. Another important implication of this particular relationship of AC voltage with the DC voltage is the rating of the switches and the diodes. So, the switches that you would typically use it might be some transistor to some extent maybe you could even find a FET which can operate at that voltage commonly people might use IGBTs which are insulated gate bipolar transistors. And again the voltage ratings of these components these transistors are not continuously varying you might have say FETs at 100 volts 200 volts 400 volts 600 volts then it may jump to 1200 volts there may be a few components which are available at 1000 volts. So, quite often you might end up selecting the voltage rating of your switches to be 1200 volts commonly when you deal with a 230 volt system because it now directly links your AC side voltage to what is required on your DC and we will see that you also need adequate margin in your transistors because there is large di-dt's during switching of the transistors and you need adequate margin between your actual voltage rating of your DC bus and your voltage rating of your transistor devices. So, now that we have a feel for what would be the preliminary voltage rating required by some of the components we can then take a look at what could be some of the currents that are flowing through this particular circuit especially on the DC bus. And we will look at the capacitor currents closely and there are actually in a circuit such as this different frequency components of currents that can actually flow through the circuit. The first component that we will look at is the fundamental component of the capacitor current because this is a single phase system your I out is actually a 50 hertz AC quantity and we saw how I out is related to your P rating and your AC voltage rating. So, your I out in this particular case can flow through your source which is your AC grid, but then it has to flow back through your neutral and come back into your DC bus. And if you take your current I out to be AI cos omega naught t your your IRMS is AI by root 2 taking the output current to be a sinusoidal quantity which is well filtered. We also know that you can relate your grid we had taken it as A v cos omega naught t where the instantaneous grid voltage is related to the amplitude and it is a sinusoidal function. So, you can relate your power rating your P rated is A v AI by root 2 root 2. So, you know what your your AI has to be in terms of your power rating and your terminal voltage rating. And so, you know what your current is that is actually flowing into your neutral point. So, this neutral point current is actually a 50 hertz current and we are assuming that the switches S plus and S minus are being modulated in a symmetric manner. So, during the positive half cycle what is happening to S plus it is identical to what is happening in the negative half cycle to S minus. So, there is symmetry between the operation of the top and the bottom leg of the switches in this particular power converter ok. So, whatever current is flowing in splits between the capacitor C dc C dc as i cap going in one direction into your top bank and i cap going into the bottom direction in the bottom bank. And one thing that you could immediately note about this current. So, if there is a current flowing into the bottom bank this is actually charging this bottom capacitor whereas, the same current which is flowing out into your top bank is actually discharging the top capacitor. So, the current is actually a common mode current it does not affect the differential voltage between your positive and negative dc bus. Even though it affects the individual capacitor voltage the total voltage between your positive and negative is not affected by this common mode current because one is charging when the other is discharging and the other way around depending on the instant at which the current is flowing. So, the next component of the current that can flow through this capacitor bank is we will see that it is has a dc and a 100 hertz component. And for this again we are assuming your the voltage to be a V cos omega t. So, V o n is a V and we can we can look at what is the power going out of your part of the ac leg and the power coming in. And we will make some assumptions we will assume that the inverter is very efficient which means it is a lossless and the filter is purely inductive. So, this again no losses in the inductor. So, we will also assume that the voltage at the input V dc is quite stiff typically by design you will try to maintain that voltage to be a constant. So, assuming we have p out I p is the current flowing through your positive dc bus and so, this is p out is a V a i cos square omega t. So, you can write what your I p of t is a V a i by V dc cos square omega t. So, you can see that by looking at the power balance between your input and your output you can see that there is a dc quantity that is flowing through your dc bus which you would naturally expect because you are exchanging power from one source to the other. But in addition to that now you see that you have a 100 hertz component which is coming in the dc bus and has a value of a V a i by 2 V dc and then there is a second harmonic component and its amplitude is also a V a i by 2 V dc. So, if you then look at where is the this particular second harmonic this particular current flowing this current is actually flowing through this particular circuit back through this particular loop and this is actually a component which is circulating in the dc bus. So, it shows up as a differential voltage across your dc bus. So, this component will so, if you actually measure the voltage across the dc bus the 50 hertz would not show up across the dc bus as a total across the total voltage whereas, the 100 hertz component will actually be measured across the dc bus. And some part of it because if you are thinking about say an application like a photovoltaic inverter you might have a prime source which might be your your PV panels it might be your batteries if it is a storage system it might be a fuel cell stack depending on whether what application it is. So, one thing that a 100 hertz component can actually do is it will split between your capacitor c dc and your prime source depending on the impedance of these two sources at that particular 100 hertz. So, if you have a higher impedance for c dc it means that most of your 100 hertz might flow through the prime source and many times the effects of this 100 hertz ripple in the prime source is actually negative in the sense that in a photovoltaic system the 100 hertz component would just cause heating of the elements not any real generation of power also the 100 hertz voltage which would mean that your actual operating point is deviating from your MPPT point. So, in multiple ways you are actually losing out on power if you are taking a battery your effective capacitance of the battery can be quite large which means that a fair amount of 100 hertz can actually flow into it and this will cause losses in the electrolyte the electrodes etcetera again in things like a fuel cell it is it will cause cause heating of the stack etcetera. So, these are undesirable components to some extent that people might even think about adding a power converter over here to prevent that 100 hertz ripple from actually flowing into the prime source and affecting the performance of the prime source. The other item of interest when we are looking at the 100 hertz component is that if you look at the path for the 100 hertz component the output current I out consists of largely the fundamental. So, this is 50 hertz it might have some switching frequency ripple, but you are assuming that your filtering is being designed in a efficient manner that it is largely 50 hertz and you are not having much of switching ripple and also your controls are ensuring that you are not having low frequency harmonic distortions in your output voltage. So, if you look at where is the path for the 100 hertz component it is flowing through the DC bus it is flowing through the switches. So, it is an interesting path that even though we as we know that the both the switches do not say conducts simultaneously because of the modulation effect of the power converter the non-linearity involved in modulation you end up with a 100 hertz current through the switches through your DC bus and you could actually explicitly look at these waveforms in a time domain simulation to convince yourself that the path for this 100 hertz component is actually through the inverter. The third component of the current through a power converter such as what we have just discussed is now the high frequency component. So, we looked at the DC 50 hertz 100 hertz. So, you also have because of the PWM switching action a high frequency component of the current flowing through the switches and to determine what these high frequency currents are in the switch we can look at the switching functions that are being generated by the power converter. So, if you look at the power converter we have essentially we had the switching function of the stop switch being S plus. So, whenever S plus is on your current in IP gets connected to I out. So, there is a you can make use of the switching function to relate IP to I out we also make can make use of the switching functions to relate your output voltage to your DC bus voltage which was what we did to look at what the average output voltages. We will assume that all the switching frequency effects occur at the switching frequency rather than spread around a range of band around the nominal switching frequency. We will also assume that the ripple in AC current is small and we can make use of the switching functions to obtain your voltage relationship we saw that VON is S plus VDC by 2 plus S minus minus VDC by 2. So, this is O with respect to the neutral if you look at voltage O with respect to the negative bus you would have S plus VDC to be directly giving your V O with respect to your negative DC bus. Similarly, you can write an expression for what is your positive DC bus current is S plus your times I out of T. So, this gives the relationship between your actual current that is flowing in your positive DC bus to your AC output current and the AC output current you know what it is based on the specification of the inverter and we saw how the specification of the inverter is linked to the power rating voltage rating etcetera. So, what is plotted over here is this black waveform over here is I out what is shown over here in red is your V O with respect to your I P of T this is I out of T and we can actually then calculate what the average value of I P is in the different durations. So, if you look at over one switching period interval say from one peak to the next peak we will take that your current is not changing by much because you have about 200 switching cycles when you are talking about a 20 hertz I mean 50 hertz fundamental and 10 kilohertz switching frequency. So, I out times D because D is the duration D times T SW is your on duration. So, if you look at your I out your on duration over here is D times T SW. Now, if you look at the I P average over and what we have averaged it is over the duration of T SW the actual if you look at it over multiple such cycles you will see an average which has a 50 hertz buried in it it has 100 hertz buried in it has DC. So, it is not just a flat quantity it is a it is something which is changing slowly on a cycle by cycle basis ok. You could also then look at what is a RMS value of this I P current which is flowing on a flowing in your positive DC bus. So, the RMS. So, you can calculate what your I P RMS is and we know that I P square of T can be written as it is on only during the T on duration. So, it is T on divided by T SW into I P square. So, T on by T SW is your duty cycle. So, you can write your I P RMS to be equal to the magnitude of I P times square root of D ok. So, if you look at the RMS component of the switching frequency in I P during this particular T S duration the RMS switching frequency component because I P average contained also the low frequency 50 hertz 100 hertz effect. We want to then calculate what is the switching frequency effect component we will get we can get it as. So, you can take the I P RMS square minus the I P average square and take the square root of that we will get the I P switching frequency component during that particular duration T SW. So, this is equal to I P I out of T times square root of D minus D square where T the D term is because of the RMS and the D square is because of the average effect and we know that your D belongs to the interval 0 to 1. So, D minus D square will always have the appropriate sign. Now, we know that this is on per switching cycle basis. So, if you will want to look at the switching frequency effect over the whole fundamental cycle then you need to take this individual high frequency harmonics and sum it up appropriately on a sum of square basis over the longer time frame ok. We have N times T SW is 1 by F naught W is 1 by F naught F naught is your fundamental frequency. So, using that we can get a relationship for your output for your output if you are for your RMS current as I P FSW RMS is 1 by N summation K is equal to 1 to N over the N points I out of K T S. So, it depends depending on what your AC output is at the particular instant square. So, you do the RMS calculation over the number of cycles. So, if you will look at. So, if this is duration say K is equal to 1 K is equal to 1 to N over the N points I out of K T S. So, it depends depending on what your AC output is at the particular instant square. So, you do the RMS calculation over the number of cycles. So, if you look at. So, if this is duration say K is equal to 2 K is equal to 3 and you know what your I P values are at these different instance. You can go through a fairly simple calculation just a summing over say 200 points for fundamental frequency of 50 hertz and a switching frequency of 10 kilo hertz. So, you will be able to calculate what your high frequency current is going to be through your capacitor bank. Now, once you have the value of the currents calculated at the different components frequency components then we have to make a decision about what type of capacitor we would actually be using in the DC bus and that the type of capacitor depends on many factors of one important factor is the type of dielectric that is being used. And depending on the type of dielectric you can have ceramic capacitors, mica capacitors, paper capacitors these are high frequency type of capacitors. You could have tantalum type capacitors or electrolytics you can have poly propylene or polyester if it is AC application. And in the DC bus of a voltage source converter a common capacitor that is be that would be used is a electrolytic capacitor. And we will see that now once you have this effects how next can we actually go in and see what would be the appropriate electrolytic capacitor that can be used in the DC bus of power converter. Thank you.