 Welcome to class 23 on topics in power electronics and distributed generation. We have been talking about economic evaluation of DG systems and we looked at a few measures for doing that. The simplest is the initial cost which is a short term immediate decision. Payback period is something which is more on a commercial time frame maybe of the order of a year and to make a decision on whether meaningful investment is being made in terms of investing in DG technology. The cost of energy we saw is much longer term time frame which is useful for policy considerations and we looked at examples in these methods and we have we will start looking at looking at effective initial cost and one of the advantage we will see with effective initial cost is that as a power electronic designer we may not know the entire system which is being assembled together. We might know only the power electronic part of the subsystem and we would like to actually make a decision on whether to how to go about doing the design based on a portion of the system which may not we may not be 100 percent clear at the point at which the DG overall DG system is being built. So if you look at the effective initial cost it is a way of summing together different costs with some of which might be upfront and some of which might occur later in time. Some of the upfront costs would be things like material, labor, markup, etcetera for example if you need to build a power converter you need IGBTs you might need capacitors. So these are upfront material costs you might need controllers you might need human input in terms of actual manufacturing process you might also need people to do controller development etcetera and you also have any company would need some percentage as a profit or a markup on whatever system that they are building. Once you manufacture say for example a power converter you may want to ship it to the end location where the power converter is going to be located for example you might be building a power converter may be in Goa may be the wind turbine in which the power converter is going to be sitting might be in Tamil Nadu and you need to actually ship this thing across the country and depending on whether it is a bulky device whether it is a compact this device you have shipping cost you in terms of systems which are used in transportation systems like aircraft space size and weight is a very important requirement. Also once the equipment reaches the end location you need to commission it you need to make sure that the civil work mechanical structures are in place if you are sending it to a location where it is a urban area the cost of area might be different from a rural situation. So depending on many of these things now reflect as initial cost upfront which is fairly well understood. There can also be other cost which can occur when you run this particular power electronic system on an ongoing basis for example you might have operation and maintenance you might have circuit breakers which might need periodic maintenance the springs the contacts might need cleaning. You might have capacitors whose ESR might change as a function of time you might think about replacing the capacitors after so many years of operation. So those are future cost which actually have to be encountered when you operate your system you might also have cost on a running basis for example ideally we might think of power converter as being lossless whereas in a real power converter we know that there are finite loss and in a DG system you might one might consider as something that is being lost in the power converter as something that is not available for you at the load or not available for you as a unit of power which can be sold. So you can think about losses in your converter also as a ongoing expense over the life of the system. At the end of the life of such a system you are talking about power related equipment so you are talking about longer time frames of you are talking of maybe 20 years 30 years etcetera for a large power system. So you might also have cost associated with decommissioning if you have electronic things you might have to look at how to dispose of electronic waste if you are having machines, windings etcetera you might have value in the copper or the iron so depending on what the decommissioning cost are you will have to make a decision. And you will see how all these cost can be effectively rolled up to a up front number to see whether it makes effective design which particular design might be more appropriate depending on your application. So one way of looking at a cost which is going to happen in the future is that if you have some amount C if you have some amount C at the end of a year and if your interest rate is R you will have 1 plus R into C at the end of 2 years you would have C into 1 plus R square at the end of P years you might have 1 plus R to the power of n. So if you have a cost that is incurred at year P and you want to see what is a cost today you are facing because of that you will have the present value of the cost that you are going to incur in the future to be C in the year P divided by 1 plus R to the power of P. So if you have some expense that is going to happen over here you can actually reflect back to the present value. See you might have annualized cost C, A and N occurring over n years. So depending on what this annualized equal cost are you can actually then take this particular number and take over the entire duration over which you are doing your particular calculation and sum it up to get the present value of say for example an annualized cost. So in this example annualized cost may be the losses that you incur in a power converter you might know how much kilowatt hours of loss is there and you know what is the cost per kilowatt hour of for electricity. So in a way you can then reflect back the losses to a upfront number to the initial point to look at what is a effective cost. So many times what can be encountered in the future can be reflected back to the present to see overall what would be the benefit of making one decision or the other way. So we look at an example where say for example you have a 1 megawatt per converter and say the cost of the converter is 50 into the 10 power of 5 rupees and it is used in a machinery with 20 years of expected operational life. And we will assume that in this particular power converter you might have capacitors that might need replacement every 5 years and the cost of the capacitors are given. You might have IGBTs which need replacement every 10 years and your cost of the IGBT is known. So for example of the 5, 10 and 15 year you might replace capacitors, 10th year you might replace your IGBT you are given some interest rate which you find reasonable for your particular application. So the question is when you decide on purchasing this 1 megawatt per converter it is not just the initial upfront cost you will have to also look at what is the cost associated with the operation and maintenance associated with an equipment. So you could then look at what the cost could be. So if you look at your net present value of ONM operation and maintenance so you might have capacitors 10th power of 6 divided by 1.15 to the power of 5 years plus 1 plus 2 into 10th power of 6 divided by 1.15 to the power of 10. So this is IGBTs and capacitors at year 10 plus capacitors at year 15. So if you look at the net present cost of your operation and maintenance in this case it works out to rupees 13.6 into 10 to the power of 5 and then if you look at now your overall cost including your upfront cost which was 15 to 10 to the power of 5 you will need to add this to actually reflect that you need you are making a commitment to operate this particular equipment for 20 years which reflects in the additional operation and maintenance cost. So your effective initial cost including ONM in this particular case is rupees 63.6 into 10 to the power of 5. So we can next ask say suppose you have this 1 megawatt power converter you have say 3 options in technology whether you might have something which is the latest technology having very high efficiency you might have something which is slightly older or something which is now commercially available. So depending on the type of inverter which is being used you might have say a 1 inverter which is 94 percent and its cost might be 45 into 10 to the power of 5. You might have another inverter which is 95 percent, 1 percent more efficient which is 15 to the power of 5 so cost slightly more and then you might have another inverter which is very efficient 96 its cost is slightly more. So then the question is if we know the way in which your particular power converter is going to operate we will assume it is 8 hours per day 365 days a year and for 20 years you are expecting to operate this particular unit and we will assume that the operation and maintenance cost is same as what we calculated in our previous example. We will assume 15 percent interest rate we will also assume the cost of energy is rupees 4 per kilowatt hour. So we could then look at what is now the cost of the losses that you are incurring in this particular equipment and clearly the choice over here is you trade off between paying more for a more efficient converter or paying less for a less efficient and the question is what would be reasonably appropriate decision. So if you look at the example of the 94 percent efficient power converter you can calculate the power loss. So this is P n minus P out so it is 1 megawatt divided by 0.94 P out is 1 megawatt. So this is about 63.8 kilowatts of power is getting dissipated. So if you look at your energy loss per year so this would be 63.8 into 8 hours per day 365 days a year. So this turns out to be 186 into 10 power of 3 units or kilowatt hours per year and assuming at 4 rupees 4 per kilowatt hour this turns out to be rupees 7.45 into 10 power of 5 rupees per year. So then you could then calculate what the net present value of your cost is based on the expression that we just had. So using the expression for the annualized cost you know what is the cost per year you know the rate you know it is operating over 20 years of life. You can then calculate what the net present value of losses so you have the capital cost being 45 50 or 60 into 10 power of 5. We assume the operation and maintenance cost to be 13.6 the respect for all the 3 units. If you calculate the present value of losses for the 94 percent it is 46.7 into 10 power of 5 for the most efficient it is 30 into 10 power of 5 because the losses are a lot lower. So if you then look at what is the overall effective initial cost you find that the what leads to the lowest cost of ownership or the initial cost is actually something which is intermediate not the cheapest or the most efficient you will have to look at what is appropriate depending on whether it is operating 8 hours a day or 12 hours a day. So you can actually go back and calculate according to your particular design and see what would be an appropriate design. In this particular case all the components in this particular system is related to the power electronic converter so you could actually make such a design decision without knowing much about the balance of plant. So you can see that some of these numbers are good as comparison physically the present value of losses is not something that you actually spent out of your pocket it is something that happens. So these numbers are more for comparison rather than absolute numbers which reflect expense that you will be incurring but it can give you a guidance in the direction of what is appropriate or what may not be the most appropriate in terms of making a technology decision. So if you look at the different applications depending on whether you are designing a solar inverter or a wind turbine or maybe a automotive power converter used in electric vehicle depending on the time frame you can adjust your interest what is suitable for your particular application. What might be available as an interest rate for a small business may not be the same as what is available as interest rate for a larger business more established business. You can actually again keep in mind that these numbers are not absolute you are making comparisons and when you are making comparisons you can make comparisons at equal level rather than use one rate for one particular comparison and another rate for another comparison. So you can actually make a decision on whether the technology that you have selecting is the appropriate technology. Suppose you have we discussed in the last class that many times because of the addition of a power converter you can have benefits to the balance of the plant. So a balance of plant where you get benefit can be reflected as a negative cost in depending on your particular power converter technology. Also people can refine it further here we are just considered a simple interest rate. You could also include factors like depreciation inflation tax numbers things like that to refine it but those are essentially details. The basic framework is essentially taking cost that can occur on the in the future or on an ongoing basis and reflect it back to the present and people look at this type of design as and people call it creditor to creditor design end to end point design where all aspects of not just your initial cost but also your operational cost are included in making a design decision. And often as a engineer you may not just make one particular design for your customer your payback period might be important for your actual design engineer may be effective initial cost might be important as a company management who might be trying to see whether the technology is going to be broadly useful to society you may want to actually calculate the cost of energy. So you often it is important to calculate all the methods and actually see which particular method gives you justification in making a particular design decision and help you guide through whether the power converter or power electronic technology is appropriate for the particular application. So with this background in essentially making design metric or making a decision on how to actually go about with making a decision on your power conversion equipment we also saw that the role of power electronics is actually increasing quite rapidly in distributed generation applications and your power electronic converter is actually intermediary ideally it is 100 percent efficient at no cost it is compact lightweight very reliable etc. So we will look at what it takes to transfer power so this is an intermediary transferring power from one particular source to another source and what does it take to actually transfer power. So if you look at a typical power transfer system which is based on power electronics what you would want to do is transfer power between sources say here what what is shown is two voltage source one is V in and the voltage source is V out and you want to transfer power between say the two sources and if you use components such as resistors you always end up with dissipation so or if you use switches in active or in the linear range again you will end up with participation. So to get the best possible efficiency in power transfer the method that people use in power electronics is to operate the switch the the the component as a switch which is either on or either off a switch which is on ideal switch does not have any voltage drop across it so there is no participation a switch which is fully off does not carry current through it so again there is no participation so ideal switch does not cause dissipation okay and the question is what possible combinations are feasible for a network of switches between say input and output what is shown over here is that shown is two voltage sources and one thing that we can immediately say is that say in this particular system your switch S1 and S2 cannot be turned on because immediately they would short the voltage sources. Similarly you could actually look at what are the other possible combinations of the switches S3 and S4 cannot be on because the two voltage sources would get shorted through the switch S3 and S4 similarly you can see whether S3, S5 if you turn it on it is again equivalent to shorting V in so you will find that in a switch network between two ports input port and the output port you cannot really operate any of the switch without causing short circuits or we will not be able to transfer power between your input and output in a efficient manner. Similarly we can think about what are the possible ways of exchanging power between different types of sources we just saw that voltage source and a voltage source you can have problems we will see that with even with the current source on both sides you can have problems if you have two current sources such as this you cannot open a switch a current source because it would cause a large interruption in the current and that would cause a large voltage spike so current sources cannot be opened voltage sources cannot be shorted and even if say the current source are connected in series one source will dominate the other and will cause the large stress to be applied from one source to the other. So both voltage source to voltage source and current source to current source would not work well in a efficient manner and it would be quite it would cause a lot of stress on the components in the circuit however we will see that once you have a voltage source on one side and a current source on the other side then it is actually possible to have a fairly efficient good efficient power conversion and in this particular case the voltage source might be battery it might be rectified AC voltage a current source is typically obtained in a power electronic application as a inductor a inductor does not like to change the value of its current rapidly so depending on the design of the inductor you will end up you can end up treating it as a equivalent current source. So if you have a combination such as this the number of switches that you need can be reduced to just two S1 and S2 and to prevent the voltage source V in from getting shorted it means that S1 and S2 cannot be simultaneously on so either S1 is on or S2 is on but cannot be simultaneously on similarly you need to provide a path for the current I out to flow it means that S1 or S2 has to be on you cannot actually prevent the current from flowing through the I out which means that S1 or S2 has to be on. So if you look at this particular model of two switches S1 and S2 this corresponds to a single pole double throw switch you can think of it as we will look at the next slide you can think of it as a switch where you either connect to the bottom or the top so your throw is either connected to your one end of your source or to the other end depending on how you want to transfer power and this can be used in a variety of configurations. You can actually see that the basic buck converter boost converter etc are actually variants of this particular configuration of having a voltage source on one side and a current source on the other. So if you look at an example the realization of this particular circuit you in a DC to DC converter form you have V in so your pole is connected to 3 and because you are either in location 1 or location 2 you always have a path for the current I out and because 1 and 2 are not simultaneously connected anytime you will never create a short circuit in V in okay. So it satisfies both the conditions for the voltage source and the current source. If you then look at what could be a realization of this particular circuit if you think about it in terms of a practical switch along with diodes or to form a power converter one realization would be your buck converter. So a buck converter is a simple realization of this particular circuit where you have a voltage source on one side and your filter inductor along with whatever load is connected acts like a single pole double throw switch. Your boost converter is also equivalent to this particular configuration. So if you have a boost converter so the boost converter also can be thought of as a realization of this single pole double throw configuration and you it is possible to have efficient transfer of power between your input and your output whether it is a buck or a boost as we are familiar from our Paul electronics courses. You could also have bidirectional power flow if you have with a buck boost type of configuration. So depending on your control you can actually send power in either direction and depending on how you send your power you can have a buck or a boost effect. So then so the next thing that we would be interested in is what how to actually make use of this rather than on a DC to DC basis how to actually generate a AC voltage how to put together a elementary single phase power converter using the same configuration. So building a simple DC to AC converter is straight forward again assuming the SPDT configuration for your power converter. So what is shown over here is essentially your DC input to be consisting of two voltages V DC by 2 and V DC by 2 and the midpoint N which you might one might consider to be the neutral and you have again the SPDT which can either be into positions. If the SPDT is on the top position so if throw is at P so your V ON is now plus V DC by 2 whereas if your throw is at N at the negative DC bus your V ON is minus V DC by 2. So depending on whether it is connected to the top or the bottom it is possible to get an AC voltage and value V DC by 2 or minus V DC by 2 depending on the position of your throw. So the next question is how can we then generate a sine wave from a configuration such as this because many times you are interested in interconnecting to the AC grid and the AC grid voltage is a sinusoidal quantity. So again similar to what we saw in the buck converter boost converter example converting this single leg single phase single leg inverter using transistors and diodes is possible and we could make use of pulse width modulation of the switch by appropriately controlling your switching action of your switches in the positive half cycle and for short durations. So when you are talking about generating a fundamental output your fundamental voltage is of the order of time duration you are talking about 50 hertz or 20 millisecond the switching times that you are typically talking about this of the order of 10s of kilohertz. So you are talking about 10 kilohertz you are talking about 100 microseconds. So if you look at in this particular case your fundamental frequency is 50 hertz so your T naught your fundamental period is 20 milliseconds whereas your switching frequency depending on your particular application you might have switching frequency which might be a few kilohertz to maybe 100s of kilohertz. So if we take a example of 10 kilohertz switching frequency so you are talking about TSW switching period of 100 microseconds. So you are talking about maybe 200 durations of switching period per fundamental in a 20 millisecond fundamental time frame. So when you operate such a power converter as a simple inverter on the one side you have the voltage source which is shown over here as Vdc by 2 and Vdc by 2 with the midpoint being considered the neutral. The output is essentially again a filtering action provided by a inductor we are considered over here as simple RL type of load and the inductor provides the filtering. So the question is how quick how fast you need to switch when you are looking at a PWM action and essentially what you are looking at in terms of the switching period is in terms of your switching frequency has to be much higher than the frequency of the poles of your system. So from your filter perspective or from your equivalent physical system perspective what is seen by the filter is essentially an average value rather than the instantaneous value. So larger the value of the filter you have more integrating effect so the ripple will be reduced. So the natural averaging effect of physical systems comes into picture and you get an averaging effect by the PWM operation of the power converter. So if you look at essentially a filtering effect because of a inductor if you look at the input to be a step waveform going through integrator which might represent a inductor or a capacitor or some electromechanical time constants in a motor type of system the output is essentially going to be a ramp when you are having a step input if you are having say a pulsed input essentially you will have something which steps up and stays flat as you proceed through the filtering circuit if you have something which is having a pulsed input with an average of 0 essentially what you would have is in this particular case the output of such a waveform would be the integral of that would be a ripple the ripple would be riding on top of whatever is the fundamental that you are trying to apply to the load and the magnitude of the ripple is can be reduced by selecting a larger inductor or a larger capacitor for improved filtering. So many physical processes naturally have integrators in it and they lend itself to PWM action and the constraint is that your switching frequency is assumed to be a lot larger than your maximum pole frequency of your plant. So in such a situation what you get at the output of your power converter is essentially the average effect rather than the instantaneous effect so the averaging can be considered. So if you look at the average voltage your average voltage in the leg of your particular power converter essentially when switch is on you are connecting your output O to P will S 1 is assumed to be the S minus is assumed to be complementary to S plus. So S plus takes a value of 0 or 1 and S minus is actually the complementary of that. So you can then calculate what is the average voltage V o with respect to n. So whenever the when you are doing say a sine triangle comparison will assume that say you are having a triangular waveform we try and you are comparing that with say a duty cycle D and whenever the duty cycle is larger than the triangle then you switch S plus is considered to be turned on and when the D is below the triangle then we will assume the switch to be off. In this particular case we will assume that the triangle is between 0 and 1 the duty cycle D is a continuous signal which belongs to the range say 0 to 1 and it is a continuous signal where S plus is 1 if D is greater than V triangle and 0 otherwise. We will assume that the bottom switch S minus is essentially the complementary of S plus. So whatever is when S plus has a value of 1 then S minus is 0. So depending on whether your S plus is on or S minus is on you can actually calculate the average value of the voltage between O and N averaged over a duration of TSW where TSW is your switching period. So when when switch S plus is on you have plus V DC by 2 for duration of T on when your duty cycle is below your triangle you have minus V DC by 2 for a duration of T off. Also T on plus T off is your switching period. Also T on can be related to your switching duration TSW through your duty cycle D and T off is 1 minus D times TSW. So if you look at then what is the this average voltage that you are applying over the period your V ON is V DC by 2 into D minus 1 minus D substituting for T on and T off. So you can see that this is equal to V DC into D minus half. So you can think about your single phase inverter as a amplifier where essentially you are having a gain of V DC on an average basis that is getting multiplied by the duty cycle with some offset depending on the value of your duty cycle being in the range 0 to 1. So that half correspond to the mid value of that particular range. So you can see that essentially in this particular situation your V ON the D is generated internally by your controller it can be PI controller or some other type of controller or it can be some open loops type of structure which is giving a waveform which belongs to the duration 0 and 1. So this is actually a signal that can be thought of as generated by the controller and if you look at then the waveform your S plus S plus has a value of either 0 or 1. If you look at the average value of your S plus over at switching period the signal S plus is also referred to as a switching function the average. So over a duration S plus to TSW S plus is ON for T ON it has a value of 1 during T ON by TSW. So you can see that the switching function and the and the duty cycle give information back and forth. So knowing the duty cycle you know what the switching function is and knowing what the switching function is you know what the duty cycle is. So you can go back and forth between your switching function and your duty cycle. Also if you look at the average voltage expression V ON average and you look at your duty cycle between say D belongs to the range 0 to 1. So when your duty cycle equal to 0 you have output voltage which might be minus V DC by 2 when your duty cycle is close to plus 1 you are having a value of plus V DC by 2. So depending on your duty cycle you can actually think about an average voltage which is being applied and essentially your single phase leg of the converter acting essentially as an amplifier from your voltage perspective. Now once we have a feeling for what the voltage can be generated in a single phase basis in such a power converter we could then ask the question what would be the voltage level that would be considered appropriate for a given application often we are interested in connecting to the AC grid. So we will see that important factors is factors are what is a nominal voltage and what is the range around the nominal voltage that one has to consider. What is the filter voltage drops that one has to consider and also effects such as dead band plays a important role in determining how much margin you need to have over and above your typical voltage when you are selecting a given DC bus value in your inverter application. Thank you.