 Welcome to class 37 on topics in power electronics and distributed generation. In today's class we will discuss a couple of example problems on power electronic component selection analysis of power electronic circuits. So, we will first look at a problem of three phase three wire power converter and it is operating as a active front end rectifier and it is connected to a low voltage AC grid 400 volts with filter inductive filter for interconnection and the filter is 0.1 per unit or 10 percent inductance and the switching frequency of the power converter is 5 kilohertz and the DC bus voltage has a nominal value of 800 volts. So, we are given the grid voltages E, A, G, A, B and C. So, this corresponds to the phase voltages of the grid and it is balanced in this particular case. So, the first question is what is the duty cycle command required by the legs of the inverter assuming that it is light load and you are using sin triangle modulation. So, at light loads the drop across this inductor would be small. So, the voltage at the grid would be close to the voltage at the A, B and C points. So, we could use that to determine what is the voltage the duty cycle required for leg A, B and C and because this is sin triangle modulation the fundamental voltage over here with respect to O would be similar to the fundamental voltage with respect to ground assuming a small drop across the filter L. So, we have V DC is 800 volts and E, A, G is 326 cos 2 pi 50 t and E, P, G is a function of time is 326 cos 2 pi 50 t minus 2 pi by 3 and E, C, G and this corresponds to V line to neutral RMS value of 231 volts and V line to line RMS value of 400 volts. So, if you look at your duty cycle D, A as a function of time is given by 0.5 plus 326 by 800 cos 2 pi 50 t. So, this is actually 0.4075. So, this is because you have a light load. So, your dB of t is phase shifted by 120 degrees and DC phase shifted by further 120 degrees. So, in the next problem you are asked to find what is the value of this particular filter inductor. So, it is 10 percent. So, you can use that to calculate what the inductance is. So, you have P base is 10 KVA 231 volts and I base is 14.5 amps, your Z base is P base by I base. So, that is 15.9 ohms and your L filter inductance is 0.1 times 15.9 divided by 2 pi 50, which is a fundamental frequency. So, this corresponds to 5.1 Newton 5.1 millihenries. So, in the next problem you are asked to look at what would be the actual voltage at the A, B and C terminals with respect to ground when the active front and rectifier is operating at 10 kilowatt unity power factor by drawing a phasor diagram of this particular circuit. So, for UPF operation, unity power factor operation at 10 kilowatts, you have your grid voltage EG phasor, then you have your filter inductance with the current through the filter being IL and you have phase A voltage phasor V A, which you want to figure out. So, we have the phasor diagram would be your grid voltage and you have the converter operating at unity power factor. So, IL is in phase with V A. So, you know that this voltage is 231 volts and the filter inductor is 10 percent. So, the voltage drop across the filter inductor is 23 volts and you are asked. So, you could use that to calculate what your V A phasor is. So, your EG is 231 at angle 0, IL is 14.5 at angle 0 and your V A is EG minus J xl IL and this would be 232 at angle minus 5.7 degrees. So, this corresponds to about 0.1 radians. So, this gives you the information of what the voltages at the terminals of the power converter should be. So, you have this particular voltage at when you are operating at 10 kV A, you can use that to find out what is the actual duty cycle that is commanded to the legs of the power converter. So, the next question is to actually determine the duty cycle and plot the switching functions for the first 200 microseconds. So, your duty cycle for leg A is now given by 0.5 plus 0.4095 cos 2 pi 50 t minus 0.1 and your dB is. So, with this you can evaluate it at time t equal to 0 and you get at t equal to 0. So, this would correspond to the value of duty cycle at time t equal to 0 and assuming that you have a digital controller, this would be the value that is used by the controller to compute what PWM commands to actually provide to control the power converter. So, you would have dA is 0.5 plus 0.4095 cos 0 minus 0.1. So, this is 0.9075. Similarly, you can calculate dB to be 0.2610 and dC to be equal to. So, now you have the dA, dB and dC for the first switching interval. So, for the first 200 microseconds you can then look at what is the switching function, the corresponding switching function and you can actually plot that you have if you plot the carrier and the modulating waveforms dA, dB and dC you have dA is equal to dB is and your switching period is 200 microseconds. So, if you look at the time in microseconds this is 200 and you could then look at your different points of time. So, your dA intersection would correspond to your would correspond to 9.25 going on till 190, your your your B duty or B switching signal would start from 73.9 up to 126.2 and your switching function for phase C would be between 66.8 to 133.2. So, you could actually find the instance depending on your duty cycle and your triangular carrier. So, this would be 200 microseconds, this would be 100 microseconds and you could actually plot your switching functions based on your value of the duty cycle and the carrier. So, in the next problem you are asked to evaluate what would be your common mode AC side voltage and the common mode AC side voltage is V a with respect to ground plus V b with respect to ground plus V c with respect to ground divided by 3 at the inverter terminals and you can evaluate this on the AC side of the inverter is. So, we have from the the converter circuit it is a three phase inverter. So, from the circuit of the power converter you have the voltage at V is equal to the voltage from the grid source plus L di by dt the voltage drop across the inductor. So, you can use that to evaluate V a g. So, your common mode AC side voltage is. So, the first term would be this term would be 0 if you are considering a balanced grid operation and the second term over here would be 0. If you are assuming a couple of things one is that the this is three phase three wires. So, the sum of the currents should add to 0. The second thing that you are assuming is that there are no parasitic paths to ground. So, your I a plus I b plus I c is equal to 0 under AC side. So, your common mode AC side voltage would be 0. Next you are asked to plot the common mode DC side voltage and your common mode DC voltage is V p the positive bus with respect to ground plus negative bus with respect to ground by 2. So, you can use the definition of that to find the common mode DC voltage and we know that V p with respect to ground can be written as V p with respect to your negative plus V n with respect to ground. And so, your common mode DC is V p n by 2 plus V n g. So, if you take that as 1 and you could then write your a b c voltages as V a g is the voltage at like a with respect to ground is V p g into s a plus V n g 1 plus s a 1 minus s a plus. So, because it is a complementary function. So, we should consider this as 2. So, you could write this as V p with respect to n s a plus plus V n g V p g into s a plus minus V n g into s a plus is V p n. Similarly, you can write V b with respect to ground is and V c with respect to ground can be written as V p n into V c plus plus V n g. So, if you sum 2, 3 and 4 you can get essentially V a g plus V b g plus V c g which we know is 0 from the previous derivation from the common mode AC side voltage. So, we have. So, you could make use of this expression and substitute that in the expression 1 for the common mode DC bus voltage to get V DC common mode. So, you it can express it in terms of your DC bus voltage and the switching functions of the individual legs. So, if you plot it you will get a function that looks roughly such like this. So, you have V DC common mode is equal to V p n into 0.5 plus minus. So, if you plot that when all the switches s a s b and s c are high then you have 0.5 minus minus 1. So, that would correspond to the duration over here you will end up with minus 0.5 V DC when you have all the switches sitting at 0 which would correspond to say this particular duration these 2 durations you would have V p n is equal to 0.5. So, this would correspond to plus V DC by 2 and then you have steps of one third V DC in depending on whether you have just one switch being high or two switch being high. So, you have 0.166 V DC or minus 0.166 V DC and the timing instance where these transitions occur. So, 0.5 V DC would correspond to 400 volts, 0.166 V DC would correspond to 133 volts for a 800 volts DC bus and these points of time are the same intersection points if this is in microseconds here you have 9.25, 66.873.9 and by symmetry it would continue on the other side. So, you could actually plot the common mode DC bus voltage which has a 6 step structure in every PWM cycle and one could then look at the edges when this sharp transitions occur and depending on what is a parasitics with respect to ground at each of those points those parasitic capacitance can get charged or discharged causing ground currents in the overall system. So, the next question is to identify what are the parasitics to ground for this active front end rectifier you have parasitics to ground you have device heat sink to ground capacitance. So, you have the device heat sink to ground capacitance. So, you have capacitance between your collector of your IGBT chips or the cathodes of your diode chips with respect to ground. So, for the top chip and for the bottom chip. So, for all the 6 devices you would have capacitances you could have capacitances of the DC bus capacitor the actual capacitive foils with respect to the frame which connect to ground through parasitics. So, you would have parasitics of the DC bus capacitance to ground also the bus bars with respect to ground another path could be your inductors are wound the windings are wound on a core and the core might be connected to the cabinet. So, you would have parasitic capacitances going to ground from the inductors to through the parasitics to ground you could have then the converters sitting within a frame. So, these ground currents would couple to the converter frame and the frame itself might be grounded and then there would be parasitic current paths going into the ground ground point of the of the source and then flowing back through the power lines. So, you are you can have a variety of parasitic paths. So, device to heat sink capacitance bus plate and capacitor package to frame capacitance you have inductor winding to core capacitance and you have your actual converter frame cabinet your grounding wires your earth back through the power lines. So, this would form a loop through which you could have common mode currents flowing which would lead to phenomena's interference phenomena. The next question is to look at what is the DC bus current obtain an expression for the DC bus current in terms of the duty cycle of the signal that can be used to obtain the average and RMS currents in the positive DC bus. So, if you consider your power converter as the three phase power converter you have the three legs. So, a quick schematic. So, this is IDC of your positive DC bus. So, you have switches S A plus S B plus S C plus and you have currents I A and I C and what you would like to evaluate is what is the average and RMS currents in the DC bus plate on a per cycle basis. So, to evaluate this in the duration 0 to 200 microseconds we have you can look at your duty cycle signals D A D B and D C. So, if you look at the duty cycle signals D A D B and D C. So, we had the values of D A D B and D C in a previous slide D A is the maximum. So, this is your D max the D B signal is the minimum D min and D C is the midpoint in the middle between D A and D B. So, we will call it D mid. So, if you look at the switching signals similarly you could corresponding to the phase which is at the maximum and the minimum you could call this as your S max and the one which is having the narrow pulse we could call it the S minimum and the pulse which is in between would be S mid. So, you could then identify what would be the currents that would flow through the DC bus positive DC bus in each of these durations. So, if you look at the first duration over here when all the switches are low then if all the switches are low then essentially your DC bus current would be 0. Similarly, when all the switches are high all the some of the input currents would be 0. So, the currents over current over here would add up would add up to 0. So, you get 0 current in the duration corresponding to the first duration when all the switches are high are low and when all the switches are high you will have 0 current. If you look at the duration when corresponding to S max between S max and S min then essentially you have one phase which is connected to the top and the other phase which is connected the other two phases which are connected to the bottom. So, if you consider the current in phase corresponding to D max to be I max then the current that would flow in this particular duration between the point at which your S max rises and the point at which S mid rises would correspond to the current in the phase which has D max. So, we will call that as that current as I max if you if you look at the duration between the midpoint and the between the mid the switching of the mid switching signal and the narrow switching signal then you have two switches which are connected to the top and you have one switch which is connected to the bottom. So, if you have two switches which are connected to the top and one switch which is connected to the bottom then the what that particular switch would connect to the bottom DC bus which would circulate and come back to the positive DC bus with a negative sign. So, you would have this level corresponding to minus I of the phase which corresponds to the minimum duty cycle. So, you have identified the durations of the duty cycles and you have identified the magnitudes you can use that to calculate your average and the RMS currents in one PWM duration. So, this is this would correspond in our case to 200 microseconds. So, an expression for so, in this particular case we had dA to be the maximum and dB was the minimum and dC was the middle value. So, if you look at the expression for the DC bus current you have if S max is equal to S med is equal to S min and all of them would be 0 then you have IDCP would be 0. If S max equal to 1 and S med is equal to S min equal to 0 then IDCP would correspond to I of the D which is at the maximum level. If S max is equal to S med is equal to 1 and S min is equal to 0 then your IDCP would be minus of I D in the phase corresponding to the minimum duty cycle. And if S max is equal to S med is equal to S min is equal to 1 then DCP is 0. So, in our particular example if you evaluate the time instance your I of D max is I a which is 20.4 amps and I at D min is your B phase current which is minus 10.2 amps. So, you could use that to calculate the DC bus current and as you are moving along the sinus sine wave the value of what phase would have the D max and D min and the D min D med would actually interchange as you proceed with time. So, you could actually evaluate your DC bus currents at each of the switching instance. So, you could write an expression for your average current. So, IDCP average at the nth instant would be I of D max at the nth instant into D max at the nth switching cycle instant and D med. So, you could write an expression for your average DC bus current and similarly you could write an expression because these are square wave pulses you are ignoring the ripple in your phase currents and you are assuming the switching duration to be sufficiently long that your AC currents can be assumed to be approximately flat. So, your IDCP RMS square at instant n is I D max square at n into. So, once you have the average and RMS DC bus current in the bus plate you could then calculate what is the high frequency RMS currents which would essentially flow into your DC bus capacitors. So, your I high frequency RMS is you subtract the squares of these and take the square root you would get the high frequency RMS currents that would essentially flow into your DC bus capacitor and then you could actually. So, this calculation is on a per cycle basis. So, if you want to calculate over the entire fundamental cycle you have I cap high frequency RMS to be given by which is given by your number of switching instance on a switching cycle. So, this would be F naught divided by F SW summation I is equal to 1 to n. So, we have considered this particular term to be equal to n of I high frequency RMS square at the instant I and taking the square root. So, if you evaluate it for the active rectifier at 10 kilowatt 5 kilohertz switching frequency. So, you have 100 switching cycles per fundamental. So, you get high frequency RMS current in a DC bus capacitor to be about 8.8 amps. So, for the active front end rectifier operation you have about 8.8 amps flowing through the DC bus as high frequency RMS currents. So, if you look at the frequency components of the currents that are flowing through the DC bus what we calculated over here was the RMS high frequency RMS. So, if you look at it at different power levels you can change the different power levels would correspond to change in current levels on the AC side current you can evaluate that these expressions you will find that your for operation of a 3 phase power converter the average balanced operation the high average of n stays constant at for a power level of 10 kilowatts it stays constant at 12.4 amps for all n at 10 kilowatt and if you look at the 5 kilo watt power level this would be 6.2 amps. So, essentially the average current stays flat as a function of time there is no low frequency riding on top of the average current. So, if you look at then the high frequency positive bus current it is essentially a DC plus high frequency components. So, high frequency AC. So, if you look at the current waveform you would find that the geometry of the waveforms if you do a time domain simulation would have the geometry would repeat every 60 degrees. So, just looking at it you would think that there is a 6 harmonic, but under ideal switching conditions taking ideal switches there would be no low frequency component it would essentially be a DC plus high frequency component and there are no harmonics of 50 hertz considering a ideal power converter. Of course, if you have non-ideality such as dead time on state drops etcetera those could actually introduce harmonics in the system, but in an ideal case it would essentially be DC when you are having a balanced 3 phase operation. If you look at the period of repetition you would have essentially you would have the switching period would be the period corresponding to when you have similar waveforms they are not exactly similar because the duty cycles are changing over the sinusoid, but you could think about the high frequency component repeating at 5 kilohertz or TSW and it would not exactly be FSW would have sidebands because your duty cycle is varying as the sine wave proceeds and your duty cycle levels are actually varying with time. So, if you look at your high frequency RMS currents it is 8.8 amps at 10 kilo watts and 4.4 amps at 5 kilo watts. In the next problem you are given details about the DC bus capacitor. So, we have in our problem two capacitors that are connected in series. So, you have two capacitors in series and their value is 1100 micro farads at 450 volts and you are given lifetime parameters at 3000 hours at 85 degrees centigrade ampian at 5 amps RMS current at 100 hertz and the ESR at 100 hertz is 110 milli ohms and you have current multipliers and the ambient is reduced. So, at 55 degree C ambient you have a current multiplier of 2 at 45 you have current multiplier of 2.24 and you have also a ripple current multiplier. So, this at you can have 5 amps at 100 hertz you can have only 0.82 times of that at 50 hertz you can have 1.24 times of that at 5 kilohertz and 1.27 at the higher frequencies. We are told that the ambient temperature for the active rectifier is 50 degrees. So, this is the temperature within the converter cabinet and your actual rating of the capacitor is 450 volts your DC bus voltage is 800 volts. So, your capacitor voltage is 400. So, you have improvement in the lifetime factor by 1.2 because you are operating at a reduced voltage. So, you are asked to calculate the expected capacitor life, the power loss in the capacitor bank and the ripple on the DC bus voltage. So, to do that we will start with looking at the thermal properties of this particular capacitor bank. So, you have your thermal resistance RTH from your core to ambient of the capacitor and we are told that at 85 degree centigrade you have this would be the temperature rise between the ambient and the core and you can pass 5 amps. So, your power loss is I square R I square is 5 amps square times 110 milli ohms and you are told that you can pass twice that current if your ambient temperature is reduced to 55. So, Tc minus 55 is 10 square into 110 milli ohms. So, this is the I square R. So, you could then use this particular expression to calculate what the core temperature is and the thermal resistance from core to ambient. So, your core temperature is 95 degree centigrade and your RTH from core to ambient is 3.64 degree centigrade per watt. Your current multiplier at 5 kilohertz is 1.24. So, you can use that to calculate your ESR at 5 kilohertz. So, you have 1.24 into 5 square into ESR at 5 kilohertz is your I square into R the participation at 100 hertz. So, you could use that calculate your ESR at 5 kilohertz. So, your ESR at 5 kilohertz is 71.5 milli ohms and your power loss in your capacitor is I square R. We calculated the high frequency RMS currents to be 8.8 amps and your R is 71.5. So, you have 5.6 watts power loss in each capacitor. So, you can calculate your core temperature is 50 degree centigrade which is the ambient plus your participation which is 5.6 into your RTH which is 3.64. So, this is 70.3 degree centigrade. You are also asked to calculate the participation in your capacitor bank. So, there are 2 capacitors. So, you have your participation in total is 11.2 watts. So, you know your core temperature which can be used to calculate your life. So, you have 3000 hours at the nominal core temperature and you have a factor of 1.2 because you are operating at reduced voltage provided from the data sheet and you have nominal temperature of the core is 95. Your actual operating is 70.3 and considering the simple lifetime model of doubling in life for a 10 degree reduction in temperature this would correspond to 19.9 to 10 power of 3 hours or 2.3 years. So, you have the power loss in the capacitor bank your expected life. So, this is running at rated temperature 24 hours a day around the clock. So, the next thing you could calculate is the ripple voltage in the DC bus capacitor bank. So, one thing we can we will do a some simplifying assumptions. We will assume that your 8.8 amps which is a high frequency RMS current at the switching frequency would have underlying sinusoidal value of 8.8 amps into root 2 to be the envelope of this high frequency RMS current. So, due to the capacitor capacitive effect your V high frequency ripple would be 8.8 amps into root 2 divided by 2 pi into 5 kilohertz which is your switching frequency into your value of your capacitance. So, you have about 0.361 volt as the capacitive voltage ripple you could also calculate the voltage ripple because of a ESR. So, that would be high frequency due to the resistive ESR effect would be 8.8 root 2 times the ESR at the high frequency which is 71.5 milli ohms. So, this is 0.632 volts per cap. So, if you look at the cap bank you have 2 capacitors in series. So, you have a total voltage drop of about 1.3 volts. So, your V dc max is 800 plus 1.3 and V dc min is approximately 800 minus 1.3. So, it is you could think of it as a noise which is around your nominal DC bus voltage assuming that your controllers are working and ensuring that your DC bus is related to 800 volts. In the next problem you are asked to look at the case where you have an unbalance. So, we are essentially repeating this problem for this analysis when the grid has a 3 percent unbalance caused by a negative sequence in the grid voltage. So, the first thing is to evaluate what this negative sequence voltage is and then make use of that to find what your ABC voltages are and use that to evaluate your duty cycles. So, we have V ABC is and we have our V plus to be 326 volts. It is continuing to operate under the nominal positive sequence voltage your negative sequence voltage is 3 percent. So, that corresponds to 9.8 volts and your 0 sequence you are assuming that to be not present which is 0 volts. So, you could actually now calculate what is your AB and C voltage we did that previously for the balance condition. So, we could now add the unbalance term which would correspond to the 9.8 volts the ABC voltages can actually got be obtained through the phase to the sequence to phase transformation you have V a of t to be 326 cos omega t minus 0.1 plus 9.8 cos omega t. So, your negative sequence is rotating in the opposite direction compared to the negative is rotating in the opposite direction compared to the positive sequence. So, you have the signs to be handled appropriately. So, you using this you could then calculate your duty cycles your duty cycle for example, for phase a d a would be V a of t divided by V d c which is 800 volts plus 0.5 and you could then calculate your d b and d c you know your I a and I b and I c will assume that the converter control still provides provides balance currents. So, by appropriate compensation it is ensuring balance currents even when the grid voltage is unbalanced. So, you can you know your I a I b and I c which is the same as what you had for the balance condition. So, you could then calculate your I average as a function of time would be your I a times d a plus I b times d b plus I c times d c and you could evaluate that to be it happens to be 12.5 plus 0.375 cos 2 pi 100 t at 10 kilo watt power level. So, if you look at the value of 12.5 this correspond to your 3 by 2 into 326 close to 327 times 20.4 this is your 10 kilo watt the peak current corresponding to your 10 kilo watt power level that would be your 12.5. So, while working out this particular simplification you could actually see that this is indeed the case and your 3.75 would correspond to 3 by 2 into 9.8 which is your unbalanced voltage times again 20.4. So, it is your interaction of your unbalanced voltage times your balanced current which gives rise to your 100 hertz ripple on your DC bus. Thank you.