 Welcome to class 38 in Topics in Power Electronics and Distributed Generation. In the last class we have been looking at an example of a 10 kVA 3 phase inverter and at 400 volts AC connected to the grid and through a filter inductance and we are looking at different aspects of the problem. We looked at what duty cycle commands would be required at certain instance of time. Then we looked at what would be the value of the filter inductor. We looked at the Faser diagram for operation as a unity power factor active rectifier and then we looked at for short durations of time what the switching functions would be for the three legs, what the common mode voltage would be on the AC side and the DC side and then we looked at the parasitic components of the power converter through which these common mode currents could potentially flow. Then we looked at the expression for the DC bus current and we looked at the different frequency components of the DC bus current and its RMS levels under various operating conditions. So, we use that to then look at what would be the lifetime of the capacitor the power loss in the capacitor bank and the minimum and maximum voltage on the DC bus in terms of looking at the DC bus ripple and we did that for assuming operation under balance conditions. We looked at currents at 5 kilowatt power level and at full rated 10 KVA operating condition. So, in the next part of the problem we are looking at when you have a 3 percent unbalance in the grid voltage and what would be the effect on the frequency components of the currents and what would be the lifetime of the capacitors, the power loss in the capacitor and the ripple on the DC bus under a when you have a 3 percent imbalance. So, under when you had balanced operation the numbers that we calculated was that we are power loss in the or power loss in the capacitor was 5.6 watts and then we use that to evaluate the temperature of the core of the capacitor and the temperature worked out to be 70.3 degree centigrade and we got a life of 2.28 years and our core temperature was 70.3 degree centigrade and the power loss in the bank is 11.2 watts because there are two capacitors in the bank upper and the lower capacitor and then when we looked at the ripple voltage we saw that the ripple voltage was dominated by the ESR effect because the ripple current is at high frequency in this balanced operation and we saw that the ripple voltage due to the capacitive effect had a amplitude of 3.6 volts whereas, due to the ESR the amplitude was 6.3 volts. So, the ripple on the capacitor bank which consists of two capacitors is about 1.3 volts and so we would have a VDC max of 800 plus 1.3 and so this appears at high frequencies as a band around the nominal 800 volts. So, in the next part of the problem we are looking at the 3 percent imbalance and what would be the resulting quantities that would need to be evaluated on a similar manner. So, if you have 3 percent unbalance we are talking about in terms of the ABC voltages we are looking at from a symmetrical component analysis you would be able to say what the voltages are. So, you can write the symmetrical voltage and transformations, symmetrical components transformation V 0, V plus, V minus and V plus is 326 volts which is our amplitude of the line to neutral voltage when your line to line RMS is 400 volts and your negative sequence voltage is 9.8 volts which corresponds to 3 percent of your positive sequence voltage and your 0 sequence voltage is 0. Using this you can calculate what your V a, V b and V c are the 0.1 radian angle difference is from the previous Fischer analysis required for part transfer and V b. So, the signs of 2 pi by 3 is opposite because this is a negative sequence quantity. So, you could then make use of the quantities ABC voltages and we will also assume that the converter still operates with balance currents in its output you could using your the average model come up with your I average would be 12.5 plus 0.375 cos 2 pi 100 T amperes this is at 10 kilowatt and power level of operation at 400 volts. So, you could see what this 12.5 and 3.75 numbers correspond to your 12.5 is essentially 3 into 327.6 into 20.4 divided by 800 which is a DC bus voltage 3 by 2 into 800. So, this is the 12.5 327 is the amplitude of your positive sequence voltage 20.4 is essentially 14.4 amps correspond to 10 k V a operation of a three phase power converter at 10 k V a and the amplitude would be root 2 times 14.4 which is 24 20.4. So, you get that to be your from your DC power transfer and your interaction of the positive sequence current and your negative sequence voltage 9.8 volts times your positive sequence current would give your 100 hertz ripple in addition to that you have the high frequency ripple current. So, you could then evaluate your overall spectrum of the currents that that would be flowing through your DC link. So, at 10 kilowatt power level also at 5 kilowatt power level you have 12.5 amps DC at 10 kilowatts or half of that 6.25 amps at 5 kilowatt power level. If you look at your 100 hertz you have 0.37 amps peak or 0.186 amps at 5 kilowatt power level and your high frequency RMS this occurs at the switching frequency and its harmonics this is 8.8 amps RMS and 4.4 amps RMS at 5 kilowatt power level. And you could use these numbers to evaluate your temperature losses in the capacitor bank and corresponding temperature rise. So, the losses in this particular case our loss per capacitor is 8.8 square that is the high frequency RMS current times 71.5 milli ohms resistance ESR plus 0.37. So, this turns out to be the same 5.6 watts because the quantity due to the unbalance is quite negligible compared to the quantity the 8.8 amps flowing at the high frequencies. So, you have 5.6 watts for the capacitor bank and 11.2 for the capacitor and 11.2 watts for the overall bank and because a participation is not changed your core temperature stays the same is 50 degree ambient plus 5.6 watts into 3.64 which was the thermal resistance that we calculated. So, this turns out to be the same temperature 70.3 degree centigrade. So, your lifetime stays the same which is 3000 hours into 1.2 factor into 2 to the power of 95 minus 70.3 by 10. So, this turns out to be again 2.28 years. So, the 3 percent unbalance did not cause a change in the lifetime because the power dissipation stayed the same, but if you look at it from the perspective of the DC bus ripper you have due to 100 hertz you have V 100 the 100 hertz component is 0.37 into the impedance 1 by 2 pi 100 into 1100 microfarads capacitance. So, this is 0.54 volts per capacitor. So, the overall bank would see about 1.1 volt. So, this would be in addition to the high frequency ripple that shows up and the due to the ESR effect. So, your V DC max this is 0.54 into 2. So, about a 2 volts or amplitude ripple riding over the DC bus. So, you can see that in terms of the unbalance there was a effect on the ripple, but not much in terms of participation or lifetime projection. In the last part of the problem you are asked to repeat the again this calculation when the inverter is operating as a STATCOM providing 10 KVA leading war to the grid to a balanced grid. So, you could do a similar calculation and identify the frequency components of the current then use that to calculate the expected life of the capacitor the power loss in the capacitor bank and the ripple on the DC bus. A similar procedure could be adopted and I will just mention the answers in this particular case. So, in this particular case your inverter voltage terminal voltage is higher 253.6 at angle 0 and at 10 KVA your high frequency RMS current turns out to be lower it is 7.2 amps and there is a no 100 hertz ripple because there is no unbalance. And your lifetime of the capacitor in this particular case works out to be 3.66 years and your power loss in the capacitor bank is 7.4 watts in the bank and your ripple is plus or minus 1 volt in the DC bus due to the high frequency ESR of the capacitor. So, you can see that once you have the procedure for evaluating this components you could apply it for a variety of conditions and see how your DC bus capacitor would be operating. In the last part of the problem you are told to evaluate as to evaluate the losses in the semiconductor devices the 3 legs of the inverter consists of 1200 volt 50 amp IGBT modules and the conduction and switching loss parameters are for the IGBT the collector emitter voltage on during on condition is 0.88 volts plus a resistive term of 0.25 times I c. Similarly, for the diode you have fixed on state voltage term of 0.9 volts plus a resistive term of 0.26 times your diode current the switching parameters for the for the loss parameters are evaluated at a DC bus voltage of 600 volts and at 50 amps. And the on state loss of the switch is 5 millijoules under these conditions the off switching off loss turn off loss is 4 millijoules for the diode for reverse recovery it is 3.6 millijoules and we are assuming the scaling factor for voltage and current to be equal to 1. So, you are asked to evaluate the total power loss in the 3 phase inverter operating as a active rectifier under normal grid voltage. So to evaluate this will look at the losses in each component the diodes and the switch. So the first thing to look at is what your duty cycle of each leg would be your duty cycle would be 0.5 plus 328 by 800 times cos 2 pi 55. 50 t minus 0.1 and these were numbers that we got from the facer analysis. We have the switching frequency f s w to be 5 kilohertz which means that your switching period is 200 microseconds. So the number of switching instance in a fundamental cycle is 20 milliseconds divided by 200 microseconds. So, you have 100 switching instance. You also have your phase current I a of t is 10 into 10 to the power of 3 divided by times square root of 2 to get the peak. So at 10 kilowatt power level. So, you could use that to then evaluate the conduction loss. So, starting with conduction loss in the diode. So, top diode d a the energy loss can be evaluated over a switching interval. Your conduction loss is for d a top diode at each switching interval is given by 0.9 times I a. At instant I is 0.26 into I a square at instant I for a duration of d a at the I th instant times your switching duration t s w for I a which is positive or 0 otherwise when I a is negative and then your conduction loss term over a fundamental is given by 1 by 20 milliseconds 20 into 10 to the power of minus 3 summation of the energy loss terms from I is equal to 1 and there are 100 points over a fundamental. So, you have and this number adds up to be equal to 27.8 watts. Similarly, you could calculate the conduction loss in the in the I g b t. So, you have conduction loss in I g b t s a the bottom I g b t at instant I it is given by 0.88 times I a of I plus 0.25 I a square of I and it is operating for a duration of 1 minus d a. Of I times t s w when I a is greater than 0 and 0 otherwise and then to transfer from the conduction to the energy to power over a fundamental cycle 20 milliseconds summation I a is equal to 1 to 100 of e conduction of s a times t s w. And this turns out to be 5.06 watts then the remaining term that need to be evaluated would be the switching loss term in the I g b t and the switching loss term of the diode. So, you have for the I g b t s a bottom at instant I is your e on and then you have at rated condition plus e of at rated condition times your actual d c bus voltage is 800 volts the rated condition specified is at 600 volts to the power of k v which is 1 and I a at instant I and 50 is the rated condition again to the exponent 1. If I a is positive and 0 otherwise. So, you could then transfer from your energy to your power over the fundamental switching loss power loss in the I g b t over a fundamental be equal to the 1 by your duration 20 milliseconds the summation of your loss I 1 to 100 and of e s w s a b of I and this turns out to be 15.06 watts for this particular operating condition the 10 kilowatt operating condition as active rectifier. If you look at the switching loss in the diode the anti-parallel diode you have the reverse recovery term under nominal condition times your 800 by 600 the power of k v again we have taken this as 1 it could be a number less than 1 more commonly and again I a of I divided by 50 again for reverse recovery it is a number typically less than 1, but we have taken it as 1 for I a greater than 0 and 0 otherwise. So, again you calculate your power from your energy over a fundamental cycle the time is 20 milliseconds for a fundamental and this turns out to be 6.25 watts and for the this is per I g b t device or per diode. So, if you have a three phase inverter you have 6 I g b t switches and 6 anti-parallel diodes so for the inverter would be 6 times. So, this would be 328 watts so if you are looking at the efficiency of such a converter your efficiency considering just the losses and the in your semiconductors would be 96.7 percent. So, if you are considering losses in the other components like your DC bus capacitor in your filter inductors the efficiency would come further down. So, this is the maximum achievable efficiency due to the losses in the semiconductor itself. So, in the next problem we are looking at a 3 phase 6 switch SCR current source inverter and we would like to show that it can be modeled as a pair of a single pole triple throw switches and we will label the switches in the sequence of the firing pulse typically given to a current source inverter and we will assume that the input voltages are balanced and sinusoidal phase is phase shifted by 120 degrees. And first part of the problem is to show that the single pole triple throw switch meets the requirement for efficient transfer of power between the input and output of the current source inverter. So, if you look at typical current source inverter you have thyristors or GTOs connected. So, you need reverse blocking devices 3 phase so you have voltage V A V B V C current I A I B I C your DC link has a large inductor which emulates a current source and the output can be a load or it can be a source in when you need regeneration or breaking and you could then write down the expression for the voltages. So, you have a balance set of voltages V A V B is A sin omega t minus 2 pi by 3 and to model it we could so to show that you could model it as a pair of a single pole triple throw switches. So, the previous power converter can be modeled as a pole connected to 3 throws on the top and similarly a pole connected to 3 throws on the bottom. So, you have essentially switch which could link 0.1 to the top or 0.3 to the top or 0.5 to the top or you could link say 4 to the bottom or 6 to the bottom or 2 to the bottom. Again 4 6 2 1 2 3 you are essentially the switch switches which could be the thyristors or GTO's which would be fired and a typical firing sequence would be 1 2 3 4 5 and 6. So, hence the labeling 1 3 5 4 6 2 for the for the switching devices. So, you could see that in this particular configuration your top single pole triple throw switch would be either in position 1 3 or 5 which means that 1 and 3 is never shorted together because the pole would be only at one of these three points. So, which means that your voltage source would never get short circuited. Similarly, your top throw would be either on 1 3 or 5 and your bottom would be either on 4 6 or 2 which means that your current source has always a path for the current you will never be open circuiting the current source. So, for example, when in this particular configuration your 1 is connected to phase c which means that I c would be equal to I d c and 2 is connected to v a which means that I a would be equal to minus I d c. So, you always have a path for current and you will never short out any of the voltage source and you can efficiently meet the requirements of the voltage sources on the a c side or the current source on the d c side and hence power can be transferred without any shorting of voltage source or opening of the current source and you always provide a path for the source conditions to be met. So, in the next problem we would like to show that the switching states of this particular CSI converter can be shown as by two pair of vectors with three binary values. So, b 3 so the each vector would have a binary value of 0 or 1 and there would be three terms for that particular vector. So, b represents the set of binary numbers. So, if you look at your switching positions your top switch could be considered as a switching vector sp and when sp is device 1 is fired at switch 1 your sp vector could be shown to be 1 0 0. Similarly, when it is at position 3 your sp vector could be 0 1 0 and when it is at position 5 your sp vector could be 0 0 1. So, you could think of sp as a vector with three components sp 1, sp 2 and sp 3. So, you could think of each sp 1, sp 2, sp 3 taking components values of either 0 or 1 and depending on whether it is at position 1 3 or 5 it could take the corresponding values. Similarly, for your vector s n so for the switch s n at position 4 you would have s n would be 1 0 0 and similarly at 6 your s n would be 0 1 0 and at 2 your s n would be 0 0 1. So, similarly s n vector can be thought of as three components s n 1, s n 2 and s n 3 each having a value of either 0 or 1 and the switch provides information about the status of the which switch in the power converter is on at a given duration of time. So, either of the three components of s n can be 1. So, no two of those s n values components s n 1 and s n 2 cannot be simultaneously 1 any one of the three can be 1 at a given point. Similarly, any one of the sp components can have a value of 1 and in the next problem you are told to show the switching space vector diagram for the current source inverter and show that it corresponds to a hexagon and to identify all possible 0 states of the current source inverter. So, we have seen that the voltages V a V b V c are given by this quantity and then we can look at what duration of angle alpha where alpha would be equal to your omega t would each of the switch s p and s n be on for what durations corresponding to say for a diode bridge operation where your firing delay is close to 0. So, under such a condition what would be the values that s p and s n takes. So, for so when it is operating without any delays firing delays etcetera for the thyristor your your s p would be on for 30 degrees less than theta less than 150 degrees and 0 otherwise. Similarly, sp 2 would be on when that particular phase is having carrying the highest voltage. So, this would normal conduction duration would be between 150 and 270 and 0 otherwise. Similarly, sp 3 would be 1 for 270. Similarly, you could write the durations when the bottom switches would be naturally on s n 1 would be 1 for 210. Similarly, s n 2 and s n 2 would be 1 for 90 degrees less than theta less than 210 and 0 otherwise. So, this is when your diode rectifier is essentially operating as a 6 step in 6 step operation assuming constant DC current DC link current. So, if you are operating at high frequencies in PWM you will be using states which are adjacent in this shorter durations of this particular angles. So, you could then look at your transformation from your A B C reference to your alpha beta reference because your state vector diagram is drawn in your alpha beta plane. So, you have your V alpha beta gamma is two thirds minus half times V A B C vector and similarly, you could take your switching vectors s alpha beta gamma to be two thirds of this same matrix times your switching vector s where s corresponds to s 1, s 2, s 3 and this in turn corresponds to your s p vector minus your s n vector where you are taking your s p to be your 100, 100, 100, etcetera and adding them as real numbers to get your switching vector s. So, then you could look at the durations say for duration between say for theta belonging to say theta say 30 degrees to 90 degrees. If you look at this particular duration then we could see that s p 1 would be having a value of 1 with the others being 0 and similarly, your s n 2 would be having a value between 30 degrees and 90 degrees this would be having s n 1 would be having a value of 1 and the s n 1 and s n 3 would be 0 and s n 2 would be having a value of 1. So, you could write your s vector to be equal to 1 minus 1 0 and you could use this transformation to look at your s alpha beta to be equal to 1 minus 1 by root 3. Similarly, you could look at your other switching positions from 90 degrees to 150 degrees your switching vector is 1 0 minus 1 and your s alpha beta would be equal to 0 1 by root 3. Your third position would be for theta belonging to the range 150 to 210 your s vector would correspond to 0 1 minus 1 and your s in your alpha beta plane would correspond to 0 2 by root 3 and your vector 4 would be for the duration 210. Then 270 your s would be minus 1 1 0 and your s alpha beta would be minus 1 1 by root 3 between 270 and 330 degrees your switching vector would be minus 1 0 1 s alpha beta be equal to minus 1 minus 1 by root 3 and your 6 vector would be theta during the duration 330 to 30 degrees. So, if you plot these vectors you will see that the 6 vectors you would have the corresponding positions of 1 2 3 4 5 6 there are 6 points around a hexagon. So, you would get a similar space vector diagram and 0 vector corresponds to the center unlike the voltage source space vector diagram this is actually rotated by 30 degrees. So, this is very similar to what you would experience in a for a voltage source inverter you could make use of the modulating states and show that for a current source inverter also you have a hexagon corresponding to the state vector diagram. The 0 states would correspond to the condition when say for example, if both the the throws are at the same point then essentially the voltage seen by your DC source would be 0 because 2 switches would be connected to the both the throws would be connected to the same pole and your voltage sources would all be open circuit at which means that your I a I b and I c would be equal to 0. So, there is no energy transfer during the 0 states and there are 3 possible 0 states you could either be connected to 1 and 4 or you be connected to 3 and 6 or 4 and 2. So, you have 3 0 states. So, you could write down the 3 0 states as so there are 3 possible 0 states and unlike in a 2 voltage source inverter where there are 2 0 states in a current source inverter there are 3 0 possible 0 states. To look at the input-output relationship between your AC and AC voltage and the output DC voltage and also the DC current and the input AC current you could write down the relationship. So, you have your I a is equal to S p 1 minus S n 1 times I d c is equal to S p 2. Similarly, your I b is S p 2 minus S n 2 times I d c your I c is S p 3 minus S n 3 times I d c. You could also write it as in a more compact form we have I a b c vector to be I a b c vector to be I a I b I c and S p 1 S p is equal to S p 1 S p 2 S p 3. So, you could write it as I a b c vector is equal to S p minus S n times I d c. So, this relates your AC side currents to your DC current of your CSI. Similarly, you could write your where essentially your S p vector is S p 3 and S n vector is similarly you could write your output voltage V out which is the voltage across your positive and negative poles of the of the switch is S p 1 V a plus S p 2 V b plus S p 3 V c minus S n 1 V a plus S n 2 V b plus S n 2 V b plus S n 3 V c. So, this could be written as essentially S p vector minus S n vector transpose take the dot product times V a B c vector would give you output voltage vector. So, you could make use of your switching states vectors to actually obtain your input output relationship. So, here you get the output voltage in terms of your input voltages and then you get your input currents I a b c in terms of your DC current. So, this gives you your input output relationship of this particular power converter. So, we have looked at now the example problems in the next class will continue with where we had left off for the filter design. We had looked at LCL filter and looked at how to determine the values of L and C. We will start with looking at how to determine what would be the damping components required to damp out the oscillations in such a LCL filter in the next class. Thank you.