1 00:00:14,750 --> 00:00:23,520 welcome back to this lecture series on pulsewidth modulation for power electronic converters 2 00:00:23,520 --> 00:00:30,240 as we were this is the third lecture here in the first two lectures we were discussing 3 00:00:30,240 --> 00:00:35,570 a few things about electronic switches and d c to d c converters we are in the process 4 00:00:35,570 --> 00:00:45,280 of discussing various power electronic converters here so we have first we have been discussing 5 00:00:45,280 --> 00:00:50,850 numerous power electronic converters before heading towards pulsewidth modulation of power 6 00:00:50,850 --> 00:00:56,020 electronic converters in the process of discussing power electronic converters we have seen that 7 00:00:56,020 --> 00:01:02,110 you know we need sudden switches electronic switches and how we can convert power into 8 00:01:02,110 --> 00:01:04,390 you know using electronic switches 9 00:01:04,390 --> 00:01:10,290 so the first example that we saw was that of d c to d c converter and today we are going 10 00:01:10,290 --> 00:01:15,590 to look at d c to a c converters before looking at d c to a c converters let us have a quick 11 00:01:15,590 --> 00:01:22,240 recapitulation of what we saw about d c to d c converter because d c to d c converters 12 00:01:22,240 --> 00:01:27,310 are the most fundamental converters and the understanding that we developed with these 13 00:01:27,310 --> 00:01:33,640 converters could be extended to many other converters now what i have shown here is a 14 00:01:33,640 --> 00:01:40,250 buck converter which we discussed the other day buck meaning its output voltage v out 15 00:01:40,250 --> 00:01:46,200 the average output voltage v out is less than the input voltage so its a buck converter 16 00:01:46,200 --> 00:01:49,939 given certain input voltage you arrive at any output voltage which is lower than the 17 00:01:49,939 --> 00:01:53,659 input so how do we do that we have a single pole 18 00:01:53,659 --> 00:01:59,740 double throw switch shown here this node p is the pole and there are two throws this 19 00:01:59,740 --> 00:02:08,810 p is either connected to throw t one or connected to throw t two it is connected alternatively 20 00:02:08,810 --> 00:02:15,340 between t one and t two alternately so whenever you connect the pole p to t one the voltage 21 00:02:15,340 --> 00:02:20,799 here is high the entire v in gets separate here and whenever p is connected to t two 22 00:02:20,799 --> 00:02:28,190 what you apply here is zero so either v in or zero gets separate here so you go on repeating 23 00:02:28,190 --> 00:02:36,370 this cycle you go on repeating this cycle so what you get is such a pulsed square waveform 24 00:02:36,370 --> 00:02:44,069 now this has certain average value this has certain average value and this is the average 25 00:02:44,069 --> 00:02:49,769 voltage that we require for the load the average voltage dropped across the inductor is zero 26 00:02:49,769 --> 00:02:54,690 so the entire average goes through the load the inductive filter serves to smoothen the 27 00:02:54,690 --> 00:02:58,819 current as we saw before we can also have an additional capacitance connected across 28 00:02:58,819 --> 00:03:04,890 the load for further smoothening of the current and i mean to smoothen out the load voltage 29 00:03:04,890 --> 00:03:09,300 too so the output voltage here is basically the duty ratio times the input voltage as 30 00:03:09,300 --> 00:03:17,230 we saw if you want a higher output voltage what we do is this width of the pulse divided 31 00:03:17,230 --> 00:03:21,489 by the total time interval we have to increase what is called as the duty ratio we should 32 00:03:21,489 --> 00:03:25,819 increase that you will get higher output voltage if you reduce the duty ratio you are going 33 00:03:25,819 --> 00:03:31,489 to get lower output voltages now so this is a buck converter where you know 34 00:03:31,489 --> 00:03:36,159 which has been shown with a single pole double throw switch generic switch now we have to 35 00:03:36,159 --> 00:03:41,280 realize the single pole double throw switch electronically and how do we do that so we 36 00:03:41,280 --> 00:03:47,840 look at the two states of the d c to d c converter lets call this as state one and lets call 37 00:03:47,840 --> 00:03:54,739 this as state two what happens in state one pole is connected to throw one and in state 38 00:03:54,739 --> 00:04:00,780 two this pole is connect to throw two in the so called state one there is current flowing 39 00:04:00,780 --> 00:04:09,069 from throw one to pole as indicated in this direction then this pole to throw two you 40 00:04:09,069 --> 00:04:13,220 know what happens here is it is blocking certain voltage 41 00:04:13,220 --> 00:04:19,580 what is the polarity of this the pole p is positive with respect to the throw t two that 42 00:04:19,580 --> 00:04:25,500 is the polarity of voltage which it is required to block now when you go to the state two 43 00:04:25,500 --> 00:04:34,400 what happens is pole p is connected to throw t two so here the conduction is from throw 44 00:04:34,400 --> 00:04:39,110 t two to pole p because what is being flowing is the load current the load current flows 45 00:04:39,110 --> 00:04:43,300 in this direction so the load current flows in this direction so the load current is free 46 00:04:43,300 --> 00:04:51,660 wheeling now and that conduction is from t two to pole p so here what happens p to t 47 00:04:51,660 --> 00:04:57,080 one is now blocking certain potential and that potential you know is the polarity such 48 00:04:57,080 --> 00:05:03,390 that t one is positive with respect to p this is the kind of potential that you have here 49 00:05:03,390 --> 00:05:08,590 now how are we going to realize this so these are the requirements that we have we need 50 00:05:08,590 --> 00:05:13,750 a switch between pole and throw one you know which can conduct in one particular direction 51 00:05:13,750 --> 00:05:20,280 namely from throw one to p and we need a switch which can block a potential which is where 52 00:05:20,280 --> 00:05:26,310 you know t one is positive with respect to p that is what we need and you see that a 53 00:05:26,310 --> 00:05:32,870 transistor perfectly fits the bill so it can conduct in this direction in state one the 54 00:05:32,870 --> 00:05:37,370 transistor conducts in this direction and similarly a diode what fits the bill for the 55 00:05:37,370 --> 00:05:43,060 other one so in the so called state one a transistor conducts in this direction and 56 00:05:43,060 --> 00:05:51,331 the diode blocks with this polarity and in the so called state two you have the transistor 57 00:05:51,331 --> 00:05:55,789 blocking like this on the diode conducting like this 58 00:05:55,789 --> 00:05:59,740 so you have a transistor that is connected between p and t one and you have a diode that 59 00:05:59,740 --> 00:06:04,100 is connected between p and t two whenever the transistor is turned off the diode gets 60 00:06:04,100 --> 00:06:09,910 turned on that is because of the energy stored in the inductor it gets turned on and whenever 61 00:06:09,910 --> 00:06:14,010 the transistor is turned on the diode is turned off so you basically dont need to have any 62 00:06:14,010 --> 00:06:18,080 other control device here an uncontrolled switch will do this point now this is what 63 00:06:18,080 --> 00:06:22,880 we saw the other day so we are recapitulating it today because it is important for us to 64 00:06:22,880 --> 00:06:27,699 do so because we are going to extend this to d c to a c converters now so this is what 65 00:06:27,699 --> 00:06:33,569 we found with d c to d c converter i mean a buck converter the next thing is a boost 66 00:06:33,569 --> 00:06:37,530 converter so this is again a buck converter which is 67 00:06:37,530 --> 00:06:43,180 shown here now you give such kind of a getting signal to this transistor you give such kind 68 00:06:43,180 --> 00:06:49,070 of getting signal so that t one gets connected to p and when this getting signal is low t 69 00:06:49,070 --> 00:06:54,110 two gets connected to p and it goes on like this and you get a pulsed waveform here and 70 00:06:54,110 --> 00:06:57,780 the current is smoothened by the inductor and its so one end you can also have an optional 71 00:06:57,780 --> 00:07:03,871 capacitance here right so now what we are going to look at here is this is a voltage 72 00:07:03,871 --> 00:07:11,139 buck converter the output voltage is less than the input voltage now lets look at this 73 00:07:11,139 --> 00:07:19,220 we have a supply we have a load so between the supply and load we have basically 74 00:07:19,220 --> 00:07:26,370 switches and the passive elements for filtering lets say these are all ideal the power that 75 00:07:26,370 --> 00:07:33,210 flows into this converter is v in times i in that is the input power what is the output 76 00:07:33,210 --> 00:07:38,150 power it is v out times i out if the converter is ideal that is if the switches are ideal 77 00:07:38,150 --> 00:07:42,780 and the passive elements are all ideal there are no losses then you have v n is equal to 78 00:07:42,780 --> 00:07:48,560 v out times v in i in is v out times i out that is power balance the output the input 79 00:07:48,560 --> 00:07:55,479 power is equal to the output power if you rewrite this v out by v in is equal to i in 80 00:07:55,479 --> 00:08:03,699 by i out so v out by v in if this is less than one then i in by i out should also be 81 00:08:03,699 --> 00:08:08,930 less than one so if you have you know if if the voltage 82 00:08:08,930 --> 00:08:18,810 is bucked then the current gets boosted you have v out lower than v in you will have i 83 00:08:18,810 --> 00:08:24,139 in lower than i out or i out greater than i in so if its a voltage buck action this 84 00:08:24,139 --> 00:08:28,180 because of this power balance here if the output voltage is less than the input voltage 85 00:08:28,180 --> 00:08:32,419 the output current will be greater than the input current therefore a voltage buck converter 86 00:08:32,419 --> 00:08:37,330 what we commonly call as a buck converter is actually a voltage buck converter and this 87 00:08:37,330 --> 00:08:45,430 is actually a current boost converter now if you want basically a voltage boost converter 88 00:08:45,430 --> 00:08:50,500 what you need is actually a current buck converter so this is what i have drawn now so this is 89 00:08:50,500 --> 00:08:54,920 a current source here and there is a current buck converter now there is a constant current 90 00:08:54,920 --> 00:09:00,500 flowing through this line whereas if you look at this i out what flows through this t two 91 00:09:00,500 --> 00:09:05,300 it it only flows the there is a current equal to i in zero this current is equal to i in 92 00:09:05,300 --> 00:09:12,090 this is equal to zero or i in zero you have such a pulsed current flowing through this 93 00:09:12,090 --> 00:09:17,459 now this pulsed current flowing through a resistive load will produce a pulsed voltage 94 00:09:17,459 --> 00:09:21,360 waveform you can consider the you know you smoothen the voltage waveform considerably 95 00:09:21,360 --> 00:09:25,470 by putting in a capacitance across this as we discussed the other day anyway what we 96 00:09:25,470 --> 00:09:28,589 are worried about is the output voltage v out now 97 00:09:28,589 --> 00:09:33,810 so now what we are trying to do is here the average current is lower than the average 98 00:09:33,810 --> 00:09:38,230 output current is lower than the input current therefore you would expect the output voltage 99 00:09:38,230 --> 00:09:43,260 to be greater than the input voltage therefore this is a boost converter now so lets go on 100 00:09:43,260 --> 00:09:47,580 to the next step of replacing this current source with a voltage source and a series 101 00:09:47,580 --> 00:09:53,880 inductance now this is a voltage boost converter now there is no average voltage that is dropped 102 00:09:53,880 --> 00:09:58,230 across an inductor if you have a positive voltage during the during one of the switching 103 00:09:58,230 --> 00:10:02,399 intervals or during the on time or off time it will be negative during the other time 104 00:10:02,399 --> 00:10:07,210 on over switching interval this voltage will get will get average to zero the average voltage 105 00:10:07,210 --> 00:10:12,130 across this will always be zero now so you have nothing and so you have the same 106 00:10:12,130 --> 00:10:17,010 v in available here and you will have a v out which is going to be greater than that 107 00:10:17,010 --> 00:10:20,760 which is going to be because there is a current buck action and there will be a voltage boost 108 00:10:20,760 --> 00:10:26,820 action now so here you have also you also have a single pole double throw switch as 109 00:10:26,820 --> 00:10:33,060 we had in a buck converter but it is connected a little differently now right so lets move 110 00:10:33,060 --> 00:10:41,000 on further and see how this single pole double throw switch in a boost converter be realized 111 00:10:41,000 --> 00:10:46,890 the approach is similar to what we saw in case of a buck converter so so called state 112 00:10:46,890 --> 00:10:54,740 one we have this pole p connected to throw t one and this is the direction of connection 113 00:10:54,740 --> 00:11:03,070 then this is the voltage blocking requirement on the part of p to t two now if you move 114 00:11:03,070 --> 00:11:12,140 over to state two the direction of conduction is from p to t two and the potential block 115 00:11:12,140 --> 00:11:19,890 is p positive with respect to t so this these are your requirements you have 116 00:11:19,890 --> 00:11:25,110 to find out devices which will satisfy this requirements now as we can see very easily 117 00:11:25,110 --> 00:11:31,730 if you have this state two you have a diode here the diode can conduct like this and the 118 00:11:31,730 --> 00:11:38,329 transistor can block a potential like this now if you want to look at the state one you 119 00:11:38,329 --> 00:11:43,310 have a transistor and the transistor can conduct like this and the diode can block as shown 120 00:11:43,310 --> 00:11:47,610 here so you can realize the single pole double throw switch using a transistor and a diode 121 00:11:47,610 --> 00:11:54,680 as shown here so we have all of them put together so essentially d c to d c this buck as well 122 00:11:54,680 --> 00:12:00,151 as boost converters have one single pole double throw switch now and you see that this pole 123 00:12:00,151 --> 00:12:04,010 is always in series with the inductance because the inductive current should not be opened 124 00:12:04,010 --> 00:12:08,389 and you should always provide a path for the inductive current flow through 125 00:12:08,389 --> 00:12:13,150 so p is either connected to t one or t two so the current through l or the current through 126 00:12:13,150 --> 00:12:17,730 p always has a path to flow through either through t one or through t two the same story 127 00:12:17,730 --> 00:12:23,740 here also it is connected that will differently now so here because of this chopping action 128 00:12:23,740 --> 00:12:28,339 we know you turn on and turn off the average output voltage is less than the average input 129 00:12:28,339 --> 00:12:33,200 voltage in the case of a buck converter now if you look at the voltage polarity whenever 130 00:12:33,200 --> 00:12:38,290 the transistor is off this t one is positive with respect to p or the collector is positive 131 00:12:38,290 --> 00:12:44,680 with respect to emitter so you have certain average voltage v t with this polarity that 132 00:12:44,680 --> 00:12:51,610 is dropped and your output voltage is v out is basically v in minus v t v out is equal 133 00:12:51,610 --> 00:13:00,200 to v in minus v t is what you have and there is a buck action 134 00:13:00,200 --> 00:13:04,320 on the other hand if you look at the boost converter there is no voltage dropped across 135 00:13:04,320 --> 00:13:10,019 the inductor now whereas if you have a diode a diode blocks voltages of this polarity and 136 00:13:10,019 --> 00:13:16,579 when it is on ideally the drop is zero therefore the average voltage dropped has this polarity 137 00:13:16,579 --> 00:13:22,220 the cathode positive with respect to anode now what happens here is your v out is the 138 00:13:22,220 --> 00:13:27,140 sum of this v d plus v n and you can see that the output voltage is greater than the input 139 00:13:27,140 --> 00:13:32,190 voltage and you have boost action now so we have you know both the buck and the boost 140 00:13:32,190 --> 00:13:35,209 converters available for us here both have single pole double throw switches as shown 141 00:13:35,209 --> 00:13:39,810 here and the single pole double throw switches you look at the conduction and the voltage 142 00:13:39,810 --> 00:13:45,500 blocking requirements between every pole and every throw and you come up with an appropriate 143 00:13:45,500 --> 00:13:50,500 electronic realization which is one transistor and a diode as shown in these two cases now 144 00:13:50,500 --> 00:13:56,199 so here we find certain rules here what do we find about how switches are to be connected 145 00:13:56,199 --> 00:14:02,529 what do we find we find a single pole double throw switch is used why is it used to provide 146 00:14:02,529 --> 00:14:07,089 path for inductive current there is a inductive current and it has to you know be provided 147 00:14:07,089 --> 00:14:11,750 with the path it cannot be opened therefore you provide for the path or you know for the 148 00:14:11,750 --> 00:14:15,480 inductive current flow throw i throw one throw or other throw now so you have a single pole 149 00:14:15,480 --> 00:14:21,600 double throw switch now the pole is always connected in series with the inductor the 150 00:14:21,600 --> 00:14:26,380 pole is connected in series with the inductor or a current source and you look at the throws 151 00:14:26,380 --> 00:14:30,459 there are connected across the capacitor ah voltage source now 152 00:14:30,459 --> 00:14:35,790 so this is what you commonly find so you once again you know you connect the pole in series 153 00:14:35,790 --> 00:14:43,680 with the inductor or current source so the inductor current or you know current source 154 00:14:43,680 --> 00:14:48,480 it has a path to flow through t will either flow through t one or through t two this is 155 00:14:48,480 --> 00:14:54,800 what will happen now if you look at more complex cases than buck converter or a boost converter 156 00:14:54,800 --> 00:14:58,580 what you may find is you may not find a single double single pole double throw switches you 157 00:14:58,580 --> 00:15:04,300 may find a number of single pole double throw switches and also there are cases where you 158 00:15:04,300 --> 00:15:08,430 will find not single pole double throw switch but single pole multiple throw for example 159 00:15:08,430 --> 00:15:13,850 single pole triple throw switch you look at those examples today a little later now 160 00:15:13,850 --> 00:15:19,410 so you this is how you basically connect a switches in a converter so you can realize 161 00:15:19,410 --> 00:15:24,490 you know we know for one of our first attempts would be to realize a converter using a network 162 00:15:24,490 --> 00:15:30,410 of single pole multiple throw switches and then go about electronically realizing each 163 00:15:30,410 --> 00:15:35,399 single pole multiple throw switch lets that could be an approach to coming up with any 164 00:15:35,399 --> 00:15:41,240 power circuit that may be necessary that may be useful for us e are going to look at this 165 00:15:41,240 --> 00:15:46,040 simple example for a d c to a c conversion which is our main topic for today this was 166 00:15:46,040 --> 00:15:50,940 something we briefly discussed on the you know in the first day or in our first lecture 167 00:15:50,940 --> 00:15:57,639 so lets say we have d c and we need an a c that is we have a load which needs fifty hertz 168 00:15:57,639 --> 00:16:01,699 a c this is very very common we like for example when you power supply is failed you just have 169 00:16:01,699 --> 00:16:05,709 a d c battery you have a load which requires and you want to feed it now what do you do 170 00:16:05,709 --> 00:16:10,470 you use an inverter that s what we call as an inverter i mean which is part of a u p 171 00:16:10,470 --> 00:16:15,009 s now lets look at what is a simplest possibility 172 00:16:15,009 --> 00:16:24,370 what you can do is apply the d c voltage directly across the load for some time say ten millisecond 173 00:16:24,370 --> 00:16:32,190 and apply a minus v across the load for another ten millisecond time and go on repeating the 174 00:16:32,190 --> 00:16:37,890 cycle now this is a twenty millisecond or fifty hertz cycle you can expand this is a 175 00:16:37,890 --> 00:16:44,800 fourier series and you will find that this has a fundamental component like this has 176 00:16:44,800 --> 00:16:49,440 a fundamental component like this there are other harmonics which can be either filtered 177 00:16:49,440 --> 00:16:53,860 or ignored so this is you know its a starting point for d c to a c conversion its a very 178 00:16:53,860 --> 00:16:59,550 very simple example now what are we doing here essentially is we are connecting the 179 00:16:59,550 --> 00:17:04,699 load directly across the d c source for some duration of time we are connecting the load 180 00:17:04,699 --> 00:17:08,800 directly across the d c source for some time and what are we doing we are reverse connecting 181 00:17:08,800 --> 00:17:13,050 the load for certain duration of time you connect the load for some ten milliseconds 182 00:17:13,050 --> 00:17:17,199 reverse connect for ten milliseconds you connect it for ten milliseconds and reverse connect 183 00:17:17,199 --> 00:17:21,530 you go on doing this way so you get some plus v and minus v plus v and minus v across the 184 00:17:21,530 --> 00:17:26,380 load and a square waveform as v saw before so you will get a bipolar square wave now 185 00:17:26,380 --> 00:17:31,510 this square wave has a fifty hertz fundamental component though it also has harmonics now 186 00:17:31,510 --> 00:17:37,330 which can be filtered or ignored now so this is a starting point for d c to a c conversion 187 00:17:37,330 --> 00:17:43,740 now we can certainly do better than that some examples are here now let us say this is what 188 00:17:43,740 --> 00:17:49,770 we have so square wave now it has a fixed fundamental voltage oh that is determined 189 00:17:49,770 --> 00:17:55,190 by this v whatever is this v now if you want to change the fundamental voltage of the square 190 00:17:55,190 --> 00:17:59,240 wave the only way to do this is to change v that is like operating an inverter with 191 00:17:59,240 --> 00:18:03,960 a variable d c voltage which is not what we do commonly these days 192 00:18:03,960 --> 00:18:12,280 so we try and operate it with a fixed d c voltage now so this amplitude cannot be modified 193 00:18:12,280 --> 00:18:17,190 in fact the amplitude cant be any higher than this but you can reduce this amplitude how 194 00:18:17,190 --> 00:18:24,150 can you reduce this amplitude lets say you apply this v this v and for a short duration 195 00:18:24,150 --> 00:18:31,160 of time you can have a notch like this you can have a notch like this similarly the negative 196 00:18:31,160 --> 00:18:35,980 half cycle you can apply minus v as shown here and for a short duration of time you 197 00:18:35,980 --> 00:18:43,780 can apply plus v like this you can go on with the cycle this green waveform is similar to 198 00:18:43,780 --> 00:18:54,419 the original waveform which was you know i mean the black one except that it has a notch 199 00:18:54,419 --> 00:18:58,340 now what happens it is minus v whenever you have this notch here 200 00:18:58,340 --> 00:19:03,280 so the fundamental voltage is kind of reduced here you get a different value of fundamental 201 00:19:03,280 --> 00:19:09,360 voltage you get a different value of fundamental voltage you can by varying this notch with 202 00:19:09,360 --> 00:19:15,630 for example you can vary the fundamental voltage now this notch what we are doing is we are 203 00:19:15,630 --> 00:19:21,160 applying plus v and we are applying minus v so if you look at you know the interval 204 00:19:21,160 --> 00:19:26,510 during this notch this period you know the interval pertaining to this notch what you 205 00:19:26,510 --> 00:19:32,120 need is actually a sinusoidal wave which is as shown here this is the desired one what 206 00:19:32,120 --> 00:19:37,880 you are actually applying is this minus v which you are you know like this there is 207 00:19:37,880 --> 00:19:43,940 a substantial error between what is applied and what is needed what is needed is a sign 208 00:19:43,940 --> 00:19:49,330 wave and what is which is almost close to its positive peak and what you want is you 209 00:19:49,330 --> 00:19:53,340 know what you manage to apply is minus v so there is a very big difference between 210 00:19:53,340 --> 00:19:58,420 what is needed and what is applied during this interval of time so it is better to apply 211 00:19:58,420 --> 00:20:05,430 a slightly different kind of a waveform so let us say you apply same plus v here theres 212 00:20:05,430 --> 00:20:12,110 a notch as shown by the red colored waveform there is a notch but during this notch the 213 00:20:12,110 --> 00:20:18,700 voltage applied is only zero it is not equal to minus v the same way you go around you 214 00:20:18,700 --> 00:20:24,520 apply a notch so what have we done now we are applying a notch during which the voltage 215 00:20:24,520 --> 00:20:29,980 is equal to zero or not equal to minus v this way we may be able to produce same amplitude 216 00:20:29,980 --> 00:20:34,940 but the instantaneous error between the desired waveform namely the sine wave and the actual 217 00:20:34,940 --> 00:20:40,380 applied waveform namely this pulsed waveform you can see that it is not very different 218 00:20:40,380 --> 00:20:45,170 it is much less than what it was before so what we normally want to do is we want 219 00:20:45,170 --> 00:20:50,200 to apply either plus v or zero during the so called positive half cycle of the waveform 220 00:20:50,200 --> 00:20:54,940 and we want to apply minus v or zero during the so called negative half cycle of the a 221 00:20:54,940 --> 00:21:00,370 c waveform that we decided to so it need not be one notch it can be many notches or it 222 00:21:00,370 --> 00:21:03,809 can be seen as several pulses as indicated in this waveform here you can see that there 223 00:21:03,809 --> 00:21:08,040 are several pulses now and these determine you know there is some amount of fundamental 224 00:21:08,040 --> 00:21:12,140 voltage and the width of these pulses can be controlled if you control the width of 225 00:21:12,140 --> 00:21:16,020 these pulses that i have indicated here you know you will be able to carry on you will 226 00:21:16,020 --> 00:21:22,970 be able to adjust your amplitude of ah the sine wave or the fundamental ah component 227 00:21:22,970 --> 00:21:28,720 of this waveform now you can also do better by modifying the 228 00:21:28,720 --> 00:21:35,289 widths i mean modulating the widths in a particular fashion here so by doing all this you will 229 00:21:35,289 --> 00:21:41,360 not only be able to control the amplitude you will also be able to reduce the harmonics 230 00:21:41,360 --> 00:21:46,080 substantially this is what we will be trying to do in a more systematic fashion and at 231 00:21:46,080 --> 00:21:50,360 greater depth in the later lectures to do now for now what we are looking at is what 232 00:21:50,360 --> 00:21:55,460 is it that we expect the power converter to do what kind of converter is required for 233 00:21:55,460 --> 00:21:59,710 d c to a c inversion that is our question so that is what we are looking at so this 234 00:21:59,710 --> 00:22:06,110 converter should be capable of taking an input v d c input v and producing an output which 235 00:22:06,110 --> 00:22:13,990 is either equal to plus v or equal to minus v or equal to zero that is what we want 236 00:22:13,990 --> 00:22:19,771 so the switching network should be capable of connecting the load directly across the 237 00:22:19,771 --> 00:22:25,799 d c source that is you are producing an output equal to plus v if you call v d c as the input 238 00:22:25,799 --> 00:22:30,780 voltage d c input voltage it is it should be capable of applying plus v d c between 239 00:22:30,780 --> 00:22:36,870 the two output terminals again this switching network should be capable of reverse connecting 240 00:22:36,870 --> 00:22:40,880 the load across the d c source that is it should be able to produce apply an output 241 00:22:40,880 --> 00:22:46,049 voltage of minus v d c also it should be able to short the load which means the output voltage 242 00:22:46,049 --> 00:22:50,991 is zero it should be capable of doing these three in the case of d c to d c converter 243 00:22:50,991 --> 00:22:56,230 for example in the case of a buck converter we saw that you know we wanted the network 244 00:22:56,230 --> 00:23:03,220 to be able to connect the load across the source and we also saw that you know the network 245 00:23:03,220 --> 00:23:09,520 was required to short the load but this is different this is now different 246 00:23:09,520 --> 00:23:14,400 in the case of d c to a c conversion what we want is we also want a negative voltage 247 00:23:14,400 --> 00:23:18,690 to be applied we also want the load to be reverse connected across the source whenever 248 00:23:18,690 --> 00:23:22,970 it is necessary now so these are the requirements on the switching network so how should the 249 00:23:22,970 --> 00:23:29,730 switching network look like here we have something like this this is load let us presume that 250 00:23:29,730 --> 00:23:34,480 the load is inductive that s how many loads are its r l or it can be an active load where 251 00:23:34,480 --> 00:23:38,799 some resistance inductance and there is also certain is here now that is the kind of load 252 00:23:38,799 --> 00:23:42,960 we are looking at and there is certain amount of inductance in this load now 253 00:23:42,960 --> 00:23:51,330 so what are we going to do here so this load what you want is you want to apply either 254 00:23:51,330 --> 00:23:57,470 plus v or minus v or zero across this one simple way to do that is to have a single 255 00:23:57,470 --> 00:24:02,169 pole double throw switch like this where the pole is connected to a and the two throws 256 00:24:02,169 --> 00:24:12,790 are connected across plus and minus so a can be either positive or negative similarly you 257 00:24:12,790 --> 00:24:20,110 have b so now this b this is another single pole double throw switch the pole of that 258 00:24:20,110 --> 00:24:25,429 switch is connected to the other load terminal b and the two throws are connected across 259 00:24:25,429 --> 00:24:33,110 the power supply the d c supply so b can also be positive or negative it can be either connected 260 00:24:33,110 --> 00:24:39,070 to the positive bus or negative bus and so you have all these combinations a can be positive 261 00:24:39,070 --> 00:24:45,740 b can be negative which would mean you can have plus v d c applied across the load a 262 00:24:45,740 --> 00:24:51,270 can be negative b can be positive which means you can apply minus v d c across the load 263 00:24:51,270 --> 00:24:56,610 how about zero a can be positive and b also can be positive that will make it zero and 264 00:24:56,610 --> 00:25:01,390 also a connected to negative and b connected to negative that will also give you zero so 265 00:25:01,390 --> 00:25:08,860 you can short the load in two different manners by connecting both the poles through the top 266 00:25:08,860 --> 00:25:13,500 throws or both the poles through their to their bottom throws so there is an option 267 00:25:13,500 --> 00:25:19,750 that is available here as far as you know if you look at this option what matters is 268 00:25:19,750 --> 00:25:24,520 the state of the switch is different but the output voltage is the same so we will use 269 00:25:24,520 --> 00:25:30,741 this in kind of coming up with certain modulation methods later on so now lets say that we want 270 00:25:30,741 --> 00:25:34,210 to do this now this is the kind of switching network that we want and this switching network 271 00:25:34,210 --> 00:25:37,980 satisfies the requirement that we are talking about what is the requirement it should be 272 00:25:37,980 --> 00:25:42,980 capable of applying plus v d c minus v d c or zero across the load which it is capable 273 00:25:42,980 --> 00:25:47,470 of doing so so it is doing fine now the next point is to you know issue is how 274 00:25:47,470 --> 00:25:51,670 do we realize the single pole double throw switch that is our next question now how do 275 00:25:51,670 --> 00:26:04,260 we realize that lets just say we have its clearer part of it we need to find out what 276 00:26:04,260 --> 00:26:09,750 are the conduction blocking requirements now this is load what you apply here is this is 277 00:26:09,750 --> 00:26:15,059 an a c load you know what is produces on the a c side of the inverter it is an a c load 278 00:26:15,059 --> 00:26:20,570 so the current here is really bidirectional what you really have is a bidirectional current 279 00:26:20,570 --> 00:26:25,240 the during the positive half cycle the current flows in one direction and in the negative 280 00:26:25,240 --> 00:26:32,650 half cycle it flows in the other direction so what you want to be what you want to do 281 00:26:32,650 --> 00:26:41,990 is to connect a to the positive throw or the negative throw as you wish independent or 282 00:26:41,990 --> 00:26:47,640 irrespective of the direction of current the current may be flowing from left to right 283 00:26:47,640 --> 00:26:53,539 or right to left you want to connect a to plus that is what let us say you have done 284 00:26:53,539 --> 00:27:03,950 now so between a and lets call this as this is pole p and lets call this as t one between 285 00:27:03,950 --> 00:27:10,190 p and t one what do we want we want a switch which can conduct in either direction the 286 00:27:10,190 --> 00:27:15,800 direction of conduction will depend upon the load current so we want a switch which can 287 00:27:15,800 --> 00:27:20,289 conduct in either direction when the pole p is connected to throw t one like this what 288 00:27:20,289 --> 00:27:24,870 happens between p and t two obviously the switch between p and t two is off and it is 289 00:27:24,870 --> 00:27:30,170 blocking certain potential ah what is the polarity of that potential since a has been 290 00:27:30,170 --> 00:27:32,070 or the pole p has been connected to the to throw 291 00:27:32,070 --> 00:27:36,080 so what you have is the this is connected to the positive d c bus voltage so this is 292 00:27:36,080 --> 00:27:41,740 point is positive and the throw t two is connected to the negative of the d c bus and the entire 293 00:27:41,740 --> 00:27:46,330 d c bus voltage gets separate across that what you get between p and t two is the entire 294 00:27:46,330 --> 00:27:52,049 d c bus voltage and p is positive with respect to t two that is what we find here this is 295 00:27:52,049 --> 00:27:57,549 the kind of conduction blocking requirements we want now now let us look at this side here 296 00:27:57,549 --> 00:28:02,281 let us say this b or this pole is connected to the bottom throw so here also you need 297 00:28:02,281 --> 00:28:08,159 this switch to conduct in both the directions depending upon whatever is the load now since 298 00:28:08,159 --> 00:28:13,649 the pole is connected to the bottom throw here the pole is at a negative potential and 299 00:28:13,649 --> 00:28:19,460 the throw is at positive potential and this is blocking a potential v d c as shown here 300 00:28:19,460 --> 00:28:24,260 now so both the top switch that is between pole 301 00:28:24,260 --> 00:28:29,549 and the top throw or between the pole and bottom throw the requirement is just the same 302 00:28:29,549 --> 00:28:34,460 that is the switch should be able to conduct in both the direction and it should be able 303 00:28:34,460 --> 00:28:41,950 to block a potential of a particular polarity the top pole to throw you know if you consider 304 00:28:41,950 --> 00:28:46,990 the top pole to throw the throw voltage will be positive with respect to the pole if you 305 00:28:46,990 --> 00:28:51,299 connect the bottom you know connect consider p and t two then it has to block a potential 306 00:28:51,299 --> 00:28:54,909 where p will be positive with respect to t two this is the requirement that you want 307 00:28:54,909 --> 00:29:00,830 now so you want to come up with a particular set of electronic switches now what you do 308 00:29:00,830 --> 00:29:07,590 is you can have a switch like this so you can have a bidirectional conduction here you 309 00:29:07,590 --> 00:29:12,360 can have a bidirectional conduction any any direction the load current might be flowing 310 00:29:12,360 --> 00:29:15,850 now so if the load current is in one direction 311 00:29:15,850 --> 00:29:22,940 the transistor will conduct if the load current is in the opposite direction this can conduct 312 00:29:22,940 --> 00:29:29,580 and let us say this switch is also on correspondingly depending on the load current direction the 313 00:29:29,580 --> 00:29:34,860 transistor will conduct or the diode will conduct and when you have such a state what 314 00:29:34,860 --> 00:29:41,730 happens a the load terminal a or the pole of this single pole double throw switch this 315 00:29:41,730 --> 00:29:47,380 leg is a single pole double throw switch this is pole to throw this is pole p this is throw 316 00:29:47,380 --> 00:29:54,700 t one this is throw t two so you have a bidirectional conduction between p and t one and p the potential 317 00:29:54,700 --> 00:29:59,591 at p is now equal to the positive you know its connected to the positive ah terminal 318 00:29:59,591 --> 00:30:03,640 of the d c source now so what happens to the switch between p and 319 00:30:03,640 --> 00:30:11,240 t two it blocks it blocks a potential like this so essentially the transistor is blocking 320 00:30:11,240 --> 00:30:16,730 a potential where its collector is positive with respect to emitter similarly what happens 321 00:30:16,730 --> 00:30:21,520 here this is been conducting so therefore b is at negative potential and this is at 322 00:30:21,520 --> 00:30:28,880 positive potential this is blocking a potential like this this is in one of the states let 323 00:30:28,880 --> 00:30:34,270 us say when the output voltage is equal to plus v d c since this is connected here the 324 00:30:34,270 --> 00:30:38,690 load is like this a is connected to the positive bus b is connected to the negative bus and 325 00:30:38,690 --> 00:30:49,250 we have plus v d c applied here now let us look at a scenario where minus v d c is applied 326 00:30:49,250 --> 00:30:56,029 i am indicating all this in green colour the load current once again is a bidirectional 327 00:30:56,029 --> 00:31:00,399 current it can be either flowing in one direction or the other direction 328 00:31:00,399 --> 00:31:08,799 so now which switches are on the top switch is on and this bottom switch is on now so 329 00:31:08,799 --> 00:31:15,149 you have the top being on so plus and this side is minus this is the potential that is 330 00:31:15,149 --> 00:31:19,740 applied now when the top is on this has to conduct and this has to conduct depending 331 00:31:19,740 --> 00:31:23,390 upon you know what is the whatever is the direction of current flow now if the direction 332 00:31:23,390 --> 00:31:26,440 of current flow is like this then the transistor conducts if the direction of current flow 333 00:31:26,440 --> 00:31:31,769 is the other way the diode conducts similarly here also you know it depends on the direction 334 00:31:31,769 --> 00:31:34,960 of current the transistor conducts in this direction the diode conducts in the opposite 335 00:31:34,960 --> 00:31:41,490 direction these are all bidirectional current switches switches with bidirectional conduction 336 00:31:41,490 --> 00:31:44,120 capability now what happens to the other two switches 337 00:31:44,120 --> 00:31:50,830 which are blocking a potential so this is connected to the positive bus this is connected 338 00:31:50,830 --> 00:31:54,880 to the negative bus so this transistor is blocking a potential positive with respect 339 00:31:54,880 --> 00:31:57,950 to negative and here again it is blocking a potential which is positive with respect 340 00:31:57,950 --> 00:32:04,860 to negative now so whatever we wanted here we are able to realize what we want here is 341 00:32:04,860 --> 00:32:10,779 between a pole p and lets say throw t one we wanted a switch which has bidirectional 342 00:32:10,779 --> 00:32:17,070 conduction capability and therefore you have a and a anti parallel diode which has the 343 00:32:17,070 --> 00:32:22,149 bidirectional current capability then you want let us say between p and t two when when 344 00:32:22,149 --> 00:32:26,350 p and t one are connected t one p and t two should be able to block a potential with this 345 00:32:26,350 --> 00:32:30,120 polarity you see that you can block it with this polarity 346 00:32:30,120 --> 00:32:33,830 similarly if pole is connected to throw t two you needed a bidirectional current carrying 347 00:32:33,830 --> 00:32:41,399 capability which you have here and you wanted the top switch to block a potential with polarity 348 00:32:41,399 --> 00:32:47,179 t one being positive with respect to p which it does now so what we do here is now from 349 00:32:47,179 --> 00:32:52,640 a single single pole double throw switch we moved on to two single pole double throw switches 350 00:32:52,640 --> 00:32:57,370 because now you see the load has two terminals now it has an a c load it can conduct in both 351 00:32:57,370 --> 00:33:01,340 the directions one terminal is connected to one of the poles the other terminal is connected 352 00:33:01,340 --> 00:33:04,590 to another pole and the load is assumed to be inductive as i mentioned a little earlier 353 00:33:04,590 --> 00:33:10,649 now and the two throws you know you they are connected across the d c bus you see this 354 00:33:10,649 --> 00:33:15,480 throw t one here and the throw t one here also or connected to the positive terminal 355 00:33:15,480 --> 00:33:20,590 of the d c bus similarly throw t two and throw t two here or both connected to the negative 356 00:33:20,590 --> 00:33:25,809 terminal here so the rule that any inductance or you know 357 00:33:25,809 --> 00:33:30,740 current source is connected in series with pole p and the capacitance or voltage source 358 00:33:30,740 --> 00:33:35,029 is connected across throw t two is what you are doing now this is a switching network 359 00:33:35,029 --> 00:33:40,940 which can apply either plus v d c minus v d c or zero across the load and it satisfies 360 00:33:40,940 --> 00:33:45,570 these conditions you know that pole is connected to the inductive inductance or current source 361 00:33:45,570 --> 00:33:49,940 and throws are connected across this so that is how we have come to this now and the next 362 00:33:49,940 --> 00:33:54,010 stage of this is realizing the i mean electronic realization of the single pole double throw 363 00:33:54,010 --> 00:33:58,240 switches which is what we did just a while back now so let me just clear off all this 364 00:33:58,240 --> 00:34:14,069 let me just clear up so that we see a clear picture of the inverter 365 00:34:14,069 --> 00:34:21,770 so now you have a voltage source inverter which is been realized using some transistors 366 00:34:21,770 --> 00:34:26,409 i mean or i g b t let us say with the anti parallel diodes this is what you have now 367 00:34:26,409 --> 00:34:30,349 these such devices are commonly available and voltage source inverter is very very popular 368 00:34:30,349 --> 00:34:36,339 these days so we look at how do you do further control etcetera as we go about now what is 369 00:34:36,339 --> 00:34:43,820 clear is you have these two either the top is on or the bottom is on this is a single 370 00:34:43,820 --> 00:34:49,320 pole double throw switch so if the top is on the bottom has to be half and vice versa 371 00:34:49,320 --> 00:34:58,230 so if the top transistor in a is gated like this the bottom has to be gated like this 372 00:34:58,230 --> 00:35:07,390 complementary to that these two are complementary to one another so this is how they are gatedif 373 00:35:07,390 --> 00:35:14,030 the top is on then the potential here is equal to plus if the bottom is on the potential 374 00:35:14,030 --> 00:35:20,460 at a is equal to is equal i mean a gets connected to the negative bus of this the negative terminal 375 00:35:20,460 --> 00:35:25,599 of this d c supply alright so similarly when you have b this b can be connected either 376 00:35:25,599 --> 00:35:29,610 to the positive supply or the negative supply here also these two transistors are switched 377 00:35:29,610 --> 00:35:34,070 in a complementary fashion the top transistor on the bottom transistor are switched in a 378 00:35:34,070 --> 00:35:38,089 complementary fashion if the top transistor is switched like this the bottom transistor 379 00:35:38,089 --> 00:35:45,270 is switched like this so you know there is also something called dead time that is the 380 00:35:45,270 --> 00:35:49,740 outgoing transistor is turned off slightly ahead of the turning on of the getting on 381 00:35:49,740 --> 00:35:53,200 of the incoming transistor which is called dead time we will ignore it for now we will 382 00:35:53,200 --> 00:35:58,430 deal with it a much later in this course now so you but of now let us say that the transistors 383 00:35:58,430 --> 00:36:02,760 are you know they have instantaneous turn on turn off they they dont actually but we 384 00:36:02,760 --> 00:36:07,359 will ignore the switching transitions for the time being we will deal with them as ideal 385 00:36:07,359 --> 00:36:12,850 devices so ideal devices do not drop any voltage when they are when they are on do not conduct 386 00:36:12,850 --> 00:36:17,099 any current when they are off and their switching transitions are instantaneous so these are 387 00:36:17,099 --> 00:36:21,940 the attributes of ideal switches and lets regard all these switches as ideal now so 388 00:36:21,940 --> 00:36:27,420 you will switch them like this and whenever a you know the top and lets call this as a 389 00:36:27,420 --> 00:36:33,780 plus and lets call this as a minus lets call this as b plus lets call this as b minus so 390 00:36:33,780 --> 00:36:39,730 whenever you have a plus and b minus on [nose] you have a particular potential that is v 391 00:36:39,730 --> 00:36:46,430 d c d whenever you have a minus and b plus or on you have the output potential is equal 392 00:36:46,430 --> 00:36:55,630 to minus v d c and whenever a plus and b plus are on or a minus and b minus are on the load 393 00:36:55,630 --> 00:36:59,849 is simply shorted what happens is a kind of free wheeling just the stored energy in the 394 00:36:59,849 --> 00:37:04,020 load continues to flow through that there is no transfer of power between the d c source 395 00:37:04,020 --> 00:37:09,030 and load during that interval of time so this is how you look at this is a single 396 00:37:09,030 --> 00:37:15,580 phase inverter now what we want to see is also can we inward d c into three phase a 397 00:37:15,580 --> 00:37:22,940 c because more often you need three phase a c what should we do now now the load does 398 00:37:22,940 --> 00:37:26,670 not have just two terminals a single phase load has only two terminals which we called 399 00:37:26,670 --> 00:37:30,790 as a and b earlier now lets say there are three terminals and lets follow the notation 400 00:37:30,790 --> 00:37:36,339 r y b now we have three terminals now now the load is presumed to be inductive now so 401 00:37:36,339 --> 00:37:43,540 every terminals of the load should be connected to a pole so you cant have just two poles 402 00:37:43,540 --> 00:37:49,640 you must have three poles or you must have three numbers of single pole double throw 403 00:37:49,640 --> 00:37:53,630 switches instead of two numbers that you had before now this is what you mean by adding 404 00:37:53,630 --> 00:37:58,119 an additional leg earlier the voltage source inverter had only two legs the single phase 405 00:37:58,119 --> 00:38:00,750 voltage source inverter because the load has only two terminals 406 00:38:00,750 --> 00:38:07,040 now you have three so you you have an additional leg added in here now the configuration is 407 00:38:07,040 --> 00:38:11,820 essentially the same now the three phase load is connected you now to these points or which 408 00:38:11,820 --> 00:38:18,910 are indicated as r y n b here now so the midpoint of the d c is a convenient point for considering 409 00:38:18,910 --> 00:38:22,720 all our measurement this point may not be available but analytically we want to express 410 00:38:22,720 --> 00:38:28,121 this potential or you know potential at this pole r or at this pole y or at this pole b 411 00:38:28,121 --> 00:38:33,590 with respect to some point of reference and this o or the d c bus midpoint or the d c 412 00:38:33,590 --> 00:38:39,320 bus neutral as it is variously called is is a convenient reference so we will say we are 413 00:38:39,320 --> 00:38:44,920 o for example which means the voltage at this pole r measured with respect to o this can 414 00:38:44,920 --> 00:38:48,310 be either plus b d c by two or minus b d c by two 415 00:38:48,310 --> 00:38:53,730 if r is connected to the top throw v r o equals plus v d c by two if it is connected to the 416 00:38:53,730 --> 00:38:58,410 bottom throw v r o is equal to minus v d c by two this is what we have in all the three 417 00:38:58,410 --> 00:39:04,760 legs fine so you know you have now a three face inverter which is you know which has 418 00:39:04,760 --> 00:39:09,170 been represented in terms of single pole double throw switches how do you electronically realize 419 00:39:09,170 --> 00:39:14,650 its pretty much the same as in previous cases now so this current r you know it can be in 420 00:39:14,650 --> 00:39:18,910 either direction if you look at r the current may be flowing either in this direction or 421 00:39:18,910 --> 00:39:24,310 it can be flowing in the opposite direction so if r is connected to let us say the positive 422 00:39:24,310 --> 00:39:30,480 throw this has to be capable of conducting in both the directions similarly if it is 423 00:39:30,480 --> 00:39:34,490 connected to the bottom throw it should once again be capable but let us say r is connected 424 00:39:34,490 --> 00:39:38,210 to the positive throw what happens to the bottom throw mean what happens between the 425 00:39:38,210 --> 00:39:43,910 pole and the bottom throw there is a switch which is open which is off and this switch 426 00:39:43,910 --> 00:39:49,230 is blocking a voltage and what is it blocking r is connected to the positive terminal of 427 00:39:49,230 --> 00:39:54,680 the d c source so it is plus and throw is connected to the bottom of d c so it is minus 428 00:39:54,680 --> 00:40:00,290 it is blocking a potential equal to v d c the d c bus voltage v d c 429 00:40:00,290 --> 00:40:09,010 if you look at the other state of this single pole double throw switch lets show that in 430 00:40:09,010 --> 00:40:16,680 red colour where you know the pole is connected to the bottom throw then what you need is 431 00:40:16,680 --> 00:40:22,339 the bottom switch should be able to conduct in both the direction and how about the top 432 00:40:22,339 --> 00:40:30,200 switch the top switch is now off and this is positive this is negative and it is blocking 433 00:40:30,200 --> 00:40:35,650 a potential of p d c this is what you have in the top switch now so what you need is 434 00:40:35,650 --> 00:40:42,000 once again you need a switch between the pole and throw which can conduct in both the directions 435 00:40:42,000 --> 00:40:46,940 and which can block a potential of only one polarity which needs to block a potential 436 00:40:46,940 --> 00:40:49,030 of only one polarity as we have indicated here now 437 00:40:49,030 --> 00:40:53,790 so its just same as the same thing that we used earlier you have a top transistor with 438 00:40:53,790 --> 00:40:57,079 an anti parallel diode and a bottom transistor with an anti parallel diode this will serve 439 00:40:57,079 --> 00:41:03,730 the purpose now so this for example can conduct in both the directions if you the bottom transistor 440 00:41:03,730 --> 00:41:07,670 is on this can conduct either in this the transistor will conduct in this direction 441 00:41:07,670 --> 00:41:10,860 or the diode will conduct in this direction depending on the direction of your load current 442 00:41:10,860 --> 00:41:16,829 here whether it flows this way or that way and the top transistor if it is off it will 443 00:41:16,829 --> 00:41:22,180 be blocking the d c bus voltage with such polarity as indicated here the same thing 444 00:41:22,180 --> 00:41:26,930 applies to the other next two so here also you know what you find is basically 445 00:41:26,930 --> 00:41:31,600 the top on the bottom switches are switched in a complementary fashion so that every leg 446 00:41:31,600 --> 00:41:37,990 is a single pole double throw switch and you have three numbers of single pole double throw 447 00:41:37,990 --> 00:41:43,040 switches here so the poles are once again you know connected to the load terminals which 448 00:41:43,040 --> 00:41:48,950 are inductive load being inductive and the throws are connected across the power d c 449 00:41:48,950 --> 00:41:55,140 supply so now this is a three phase voltage source inverter lets quickly take a look at 450 00:41:55,140 --> 00:41:59,579 three phase current source inverter what are we trying to do here we want to produce a 451 00:41:59,579 --> 00:42:04,150 c from d c but this d c is what you have is a d c voltage available and you want to produce 452 00:42:04,150 --> 00:42:06,329 an you know you get an a c voltage output at that 453 00:42:06,329 --> 00:42:10,920 now what let us say what you have is not a d c voltage source but a d c current source 454 00:42:10,920 --> 00:42:18,329 what do you do its actually fairly simple so you look at this you look at a single phase 455 00:42:18,329 --> 00:42:23,540 voltage source inverter so what do you have here you have a load here and you have the 456 00:42:23,540 --> 00:42:28,230 d c supply connected here you have the load and the d c supply connected here you just 457 00:42:28,230 --> 00:42:37,210 interchange these two you just interchange these two you get a single phase current source 458 00:42:37,210 --> 00:42:44,650 inverter you have a d c current source i d c the current source has two terminals and 459 00:42:44,650 --> 00:42:48,880 you know you should provide a path for this current to flow through so both these terminals 460 00:42:48,880 --> 00:42:54,620 are connected to poles this is connected to pole of one of the switch and this is connected 461 00:42:54,620 --> 00:42:59,230 to another pole of another switch now so this pole this is also a pole this is also 462 00:42:59,230 --> 00:43:04,770 a pole now so these poles you know they are either connected to the top throw t one or 463 00:43:04,770 --> 00:43:08,950 bottom connect or bottom throw t two at any given instant now and t one and t two are 464 00:43:08,950 --> 00:43:18,589 connected together to one of the load terminal sorry similarly the bottom two throws of the 465 00:43:18,589 --> 00:43:22,920 two s p d t switches are connected together and brought out as a single load terminal 466 00:43:22,920 --> 00:43:29,220 now so what you now have is the load terminals across the throws you have the load connected 467 00:43:29,220 --> 00:43:33,640 and the load is now a voltage stiff load i mean it is capacitive load you have a capacitance 468 00:43:33,640 --> 00:43:38,670 c a fairly big capacitance c connected across that this is an a c capacitance now because 469 00:43:38,670 --> 00:43:41,839 it actually has an a c voltage across this now 470 00:43:41,839 --> 00:43:47,730 so you know what do you want here you basically you have a current source and this current 471 00:43:47,730 --> 00:43:51,579 source you are providing a path for this current source to flow through now what you want here 472 00:43:51,579 --> 00:43:56,650 at this point is a bidirectional current let us say you want current i to flow like this 473 00:43:56,650 --> 00:44:03,110 so if a you know if a is connected to the top throw it will flow like this if b is connected 474 00:44:03,110 --> 00:44:07,079 to the bottom throw it will flow like this and current will flow back this will be the 475 00:44:07,079 --> 00:44:12,300 direction alternatively let us say you want the current to flow in an opposite direction 476 00:44:12,300 --> 00:44:17,150 let us say you want the current to flow in this direction so if you want the current 477 00:44:17,150 --> 00:44:22,670 to flow in this direction it flows here so what you do is it flows from t two to p then 478 00:44:22,670 --> 00:44:29,280 this current source on the next path is p to t one and back here so you what you do 479 00:44:29,280 --> 00:44:34,550 is you are injecting alternating current into this load circuit and this load as a capacitance 480 00:44:34,550 --> 00:44:39,119 connected to it in parallel to smoothen out the voltage waveform that you will get as 481 00:44:39,119 --> 00:44:42,730 we discussed here very very i mean similar to what we discussed while we were discussing 482 00:44:42,730 --> 00:44:47,060 a voltage boost converter or a current buck converter now 483 00:44:47,060 --> 00:44:51,050 so this is what you do now you you manage to pass an a c current through this now so 484 00:44:51,050 --> 00:44:55,440 this a c current so the voltage across this is also actually an a c voltage now so next 485 00:44:55,440 --> 00:45:00,870 step that you will have is to realize this switch now one thing that we observe is the 486 00:45:00,870 --> 00:45:05,290 load current is you know this is not the load current this is the source current it is it 487 00:45:05,290 --> 00:45:10,329 has a fixed direction so if you look at a between any pole and throw what you need is 488 00:45:10,329 --> 00:45:15,380 now a uni directional current conduction you dont need a bidirectional current flow as 489 00:45:15,380 --> 00:45:19,690 you needed in a voltage source converter it is now a unidirectional source now but how 490 00:45:19,690 --> 00:45:27,650 about the voltage blocking so let us say that this pole is connected to the top throw and 491 00:45:27,650 --> 00:45:33,700 the current is flowing as indicated here so what happens pole is connected here and 492 00:45:33,700 --> 00:45:38,160 what happens between this pole p and throw t two the load voltage appears across that 493 00:45:38,160 --> 00:45:44,010 the load voltage it can be of either polarity so the load voltage can be this positive and 494 00:45:44,010 --> 00:45:48,190 throw two negative or it can be pole positive and the throw two i mean pole negative and 495 00:45:48,190 --> 00:45:53,140 throw two positive it can be of either polarity p may be positive with respect to t two or 496 00:45:53,140 --> 00:45:57,900 p may be negative with respect to t two it depends on what the load voltage is at that 497 00:45:57,900 --> 00:46:05,579 instant so what you need is between p and t two a switch which in the in its off condition 498 00:46:05,579 --> 00:46:10,420 can block a voltage of either polarity so now what you once again need is a two quadrant 499 00:46:10,420 --> 00:46:14,869 switch but in a voltage source what we needed was a two quadrant switch in terms of the 500 00:46:14,869 --> 00:46:17,640 current conduction now what you need is a two quadrant operation 501 00:46:17,640 --> 00:46:21,770 in terms of the voltage blocking you want a switch which can conduct in only one direction 502 00:46:21,770 --> 00:46:27,690 but it should be capable of blocking voltages of either polarity so what do we do now lets 503 00:46:27,690 --> 00:46:34,520 say this is a simple option a thyristor is actually a natural option a thyristor can 504 00:46:34,520 --> 00:46:42,300 actually conduct in only one direction and it can block you know a potential in the off 505 00:46:42,300 --> 00:46:47,730 state and i mean if it is a reverse bias it will block a potential as indicated here if 506 00:46:47,730 --> 00:46:52,920 it is forward biased it can still block a potential as long as it is not fired on so 507 00:46:52,920 --> 00:46:57,110 a thyristor is a two quadrant switch as we discussed in the very first lecture which 508 00:46:57,110 --> 00:47:00,990 is capable of conducting in one direction but blocking voltage of either capability 509 00:47:00,990 --> 00:47:05,740 it is a natural option for such a current source inverter but should you want to use 510 00:47:05,740 --> 00:47:10,890 a transistor or an i g b t what you can do is you can connect a transistor with a diode 511 00:47:10,890 --> 00:47:14,160 in the so called anti series fashion as we discussed before 512 00:47:14,160 --> 00:47:19,290 so both of them conduct actually in the same direction and the transistor if it is off 513 00:47:19,290 --> 00:47:23,579 you know in the off state the transistor can block a potential with this polarity and the 514 00:47:23,579 --> 00:47:28,580 diode can block with another polarity so together the combination of these two switches can 515 00:47:28,580 --> 00:47:37,170 block a potential of either polarity as you know what a thyristor can do so you this is 516 00:47:37,170 --> 00:47:41,010 what you need now and you now you get that this is a current source inverter so if you 517 00:47:41,010 --> 00:47:45,470 have this conduction here you will also have if the let us say these two switches are on 518 00:47:45,470 --> 00:47:52,640 it will flow like this and the current kind of flows through this around this now so if 519 00:47:52,640 --> 00:48:01,589 it again you know in in in this situation lets just erase this part 520 00:48:01,589 --> 00:48:07,000 lets say the current is flowing like this so what happens here it depends on you know 521 00:48:07,000 --> 00:48:13,130 this switch is off and it has to block a potential and it has to block a potential you know which 522 00:48:13,130 --> 00:48:18,589 is actually really bipolar in nature it can be either positive or negative 523 00:48:18,589 --> 00:48:23,400 so this transistor is capable of blocking with this polarity and this diode is capable 524 00:48:23,400 --> 00:48:31,710 of blocking this polarity so the two of them together can block any potential of either 525 00:48:31,710 --> 00:48:36,780 polarity so you get a two quadrant switch here the same situation here also this can 526 00:48:36,780 --> 00:48:42,410 block a potential of this polarity and this can block a potential of this polarity together 527 00:48:42,410 --> 00:48:46,410 you know the series connection of these two switches can block a potential of either polarity 528 00:48:46,410 --> 00:48:52,000 so now what you have is this switching network can allow a current equal to i d c to flow 529 00:48:52,000 --> 00:48:57,540 here it is not just i d c it could be minus i d c so it can be plus or minus i d c that 530 00:48:57,540 --> 00:49:05,420 flows here and this current can also be zero if both these transistors are on let us say 531 00:49:05,420 --> 00:49:10,990 if both these transistors are on then the current gets a path to flow around like this 532 00:49:10,990 --> 00:49:17,740 alternatively let us say lets choose some other colour if these two transistors are 533 00:49:17,740 --> 00:49:21,970 on then the current gets to flow like this it does not flow through the load 534 00:49:21,970 --> 00:49:25,930 so the load side current will be equal to zero so like a voltage source inverter is 535 00:49:25,930 --> 00:49:30,310 capable of applying plus v d c minus v d c or zero this is capable of injecting plus 536 00:49:30,310 --> 00:49:36,839 i d c minus i d c or zero through the load this is a current source inverter now so how 537 00:49:36,839 --> 00:49:40,329 can you have a three phase current source inverter is one question before that lets 538 00:49:40,329 --> 00:49:44,579 just have this redrawn you know because this i have drawn it with some orientation here 539 00:49:44,579 --> 00:49:48,559 i am just drawing it with a different kind of an orientation now so the current source 540 00:49:48,559 --> 00:49:52,440 is shown separately as a source and the load is shown connected here now and you see that 541 00:49:52,440 --> 00:49:55,910 you know the current can actually flow in one of the directions the current can flow 542 00:49:55,910 --> 00:50:02,589 and if you will change it the direction the current can actually flow in this direction 543 00:50:02,589 --> 00:50:11,640 back alright so you know it is also possible for the current to be you know both these 544 00:50:11,640 --> 00:50:17,359 switches the top and what is indicated is top and bottom can be turned on so that the 545 00:50:17,359 --> 00:50:22,820 current source is just shorted and there is no transfer of power from between the from 546 00:50:22,820 --> 00:50:27,000 the current source or between the current source and the load now 547 00:50:27,000 --> 00:50:31,660 so let us say you want to do your three phase inverter now how do you get a three phase 548 00:50:31,660 --> 00:50:38,180 inverter is a question now so once again you know we are looking at single phase current 549 00:50:38,180 --> 00:50:43,640 source inverter and what we need now is a three phase current source inverter so what 550 00:50:43,640 --> 00:50:47,390 we need is a three phase current source inverter what is the difference between a single phase 551 00:50:47,390 --> 00:50:50,680 current source inverter and a three phase current source inverter is the difference 552 00:50:50,680 --> 00:50:57,400 in the d c source no is the same d c source i d c that s going to be used in both whether 553 00:50:57,400 --> 00:51:03,970 it is a single phase c s i or a three phase c s i so current source inverter c s i actually 554 00:51:03,970 --> 00:51:08,170 an abbreviation for that so whether it is a single phase or three phase its the same 555 00:51:08,170 --> 00:51:11,559 thing now so its the same current source so it has two 556 00:51:11,559 --> 00:51:17,200 terminals and these two terminals are connected to two poles now so you still need only two 557 00:51:17,200 --> 00:51:22,960 switches but what you now have is instead of a single phase load you have a three phase 558 00:51:22,960 --> 00:51:29,599 load a single phase load has two terminals which are connected to the two throws whereas 559 00:51:29,599 --> 00:51:36,850 what you now need what you now have is a three phase load and you need three throws so what 560 00:51:36,850 --> 00:51:42,150 we have is you know we need an additional throw here lets call this t three we also 561 00:51:42,150 --> 00:51:46,890 need an additional throw here in this switch lets call this t three 562 00:51:46,890 --> 00:51:52,280 so just as t one throws t one of the two are connected together and throws t two of the 563 00:51:52,280 --> 00:51:56,570 two are connected together the throws t three of the two should also be connected together 564 00:51:56,570 --> 00:52:01,849 it will be it will get a little clumpsy if i draw on this but anyway lets just draw this 565 00:52:01,849 --> 00:52:10,109 and see how it looks right so the two throw have been connected together and this becomes 566 00:52:10,109 --> 00:52:16,109 the third load terminal so you have one terminal two and three you have three different load 567 00:52:16,109 --> 00:52:23,720 terminals available now and you have capacitances across every two of them you have a capacitance 568 00:52:23,720 --> 00:52:30,990 c that is connected across very one of the every two of those terminals so these terminals 569 00:52:30,990 --> 00:52:38,380 can be called as r y n b if you wish now the loads can be connected across these 570 00:52:38,380 --> 00:52:45,990 capacitances so thus you get a single phase i mean i am sorry you get a three phase current 571 00:52:45,990 --> 00:52:52,240 source inverter it is still a d c source but you have three phase ah load and you have 572 00:52:52,240 --> 00:52:58,321 three capacitors connected to this now so for this for a three phase c s i [noise[ what 573 00:52:58,321 --> 00:53:11,520 do you need you need a single pole triple throw switch you need single pole triple throw 574 00:53:11,520 --> 00:53:19,970 switches how many numbers two numbers so why do you need two numbers because you have a 575 00:53:19,970 --> 00:53:26,760 current source with two terminals why do you need two throws because your load is now a 576 00:53:26,760 --> 00:53:33,820 three terminal load so you need three so for a single phase c s i we just had you 577 00:53:33,820 --> 00:53:40,300 know two single pole double throw switches for a three phase c s i we have two single 578 00:53:40,300 --> 00:53:49,430 pole triple throw switches as we have shown here now you have two legs here and you have 579 00:53:49,430 --> 00:53:57,080 the single phase load connected here what you can do is you can have another leg connected 580 00:53:57,080 --> 00:54:08,190 like this you have one more leg connected like this and this becomes an additional load 581 00:54:08,190 --> 00:54:13,030 terminal you can have capacitance here and similarly between these two terminals you 582 00:54:13,030 --> 00:54:20,250 can have capacitance connected and load also connected here so you can change your single 583 00:54:20,250 --> 00:54:27,280 phase inverter into a three phase inverter let me erase this load parts 584 00:54:27,280 --> 00:54:32,450 so these are essentially the three terminals this would be called your terminal lets say 585 00:54:32,450 --> 00:54:39,530 or this would be become your terminal y this would be become your terminal b so this is 586 00:54:39,530 --> 00:54:43,880 you know a three phase current source inverter as i mentioned here thyristor is a natural 587 00:54:43,880 --> 00:54:48,309 choice for doing it so let us just summarize so what we would have seen is we have we have 588 00:54:48,309 --> 00:54:51,500 seen how you know voltage source inverter looks like and how a current source inverter 589 00:54:51,500 --> 00:54:56,089 looks like we have tried to derive this starting from our understanding of buck converter now 590 00:54:56,089 --> 00:54:59,780 so some points are made here how are these switches are connected in a voltage source 591 00:54:59,780 --> 00:55:06,809 inverter for some observations we are summarizing there are two single pole double throw switches 592 00:55:06,809 --> 00:55:11,680 in a single phase inverter in a three phase inverter you have three single pole double 593 00:55:11,680 --> 00:55:16,020 throw switches so as i mentioned here there are as many single pole double throw switches 594 00:55:16,020 --> 00:55:21,190 or as many poles as there are load terminals in a single phase load you have two terminals 595 00:55:21,190 --> 00:55:25,099 therefore you need you know the load the load is inductive so you need two poles or two 596 00:55:25,099 --> 00:55:29,170 single pole double throw switches in a three phase inverter you you have a three terminal 597 00:55:29,170 --> 00:55:33,840 load so three poles or three single pole double throw switches now so each load terminals 598 00:55:33,840 --> 00:55:38,369 is connected to a pole as indicated here and how about the throws the throws are connected 599 00:55:38,369 --> 00:55:42,170 across the d c bus the d c bus should never get shorted and therefore you always have 600 00:55:42,170 --> 00:55:46,520 the throws connected across the d c bus and what kind of switches you use you either use 601 00:55:46,520 --> 00:55:52,559 mosfet or i g b t with anti parallel diode so these devices are you know this ah an igbt 602 00:55:52,559 --> 00:55:56,579 with an anti parallel diode is capable of conducting in either direction and blocking 603 00:55:56,579 --> 00:56:00,540 a potential of only one polarity the collector being positive with respect to the emitter 604 00:56:00,540 --> 00:56:04,690 now on the other hand if you look at a current source inverter you what you need is you still 605 00:56:04,690 --> 00:56:09,520 need only two numbers of switches you know its a single phase inverter requires two single 606 00:56:09,520 --> 00:56:13,990 pole double throw switches if you go for a three phase inverter you still have only two 607 00:56:13,990 --> 00:56:20,299 but you need triple throw switches because what happens you know the current source are 608 00:56:20,299 --> 00:56:24,360 connected you know each source terminal is connected to a pole and the throws are connected 609 00:56:24,360 --> 00:56:29,200 across the load the load has three terminals and you have three throws here 610 00:56:29,200 --> 00:56:33,319 so you and in this case what you need is in a current source inverter you require a switch 611 00:56:33,319 --> 00:56:37,430 which can conduct in one direction but should be capable of blocking voltages of either 612 00:56:37,430 --> 00:56:42,849 polarity and a thyristor is a natural option for such requirement now or such an application 613 00:56:42,849 --> 00:56:47,349 now you can also alternatively use a transistor which has mosfet or i g b t with a series 614 00:56:47,349 --> 00:56:53,411 diode to realize this to this effect now so this is what we have seen now so what we need 615 00:56:53,411 --> 00:56:59,880 is now like we saw d c to d c conversion where we had to step down or step up a buck converter 616 00:56:59,880 --> 00:57:05,130 and boost converter today what we saw was we had d c to a c conversion 617 00:57:05,130 --> 00:57:09,060 the d c source may be available as a voltage source or current source and you know if it 618 00:57:09,060 --> 00:57:12,910 is available as a voltage source what we want is the load voltage should have plus v or 619 00:57:12,910 --> 00:57:18,530 minus v applied across it or a zero similarly current source should be capable of injecting 620 00:57:18,530 --> 00:57:23,800 plus sign minus sign or zero through that so we we just saw a few possibilities and 621 00:57:23,800 --> 00:57:28,310 you know this kind of finishes the lecture for today and in a subsequent lecture we will 622 00:57:28,310 --> 00:57:32,530 go to a slightly higher complex in a more complex ah d c to a c converter we will take 623 00:57:32,530 --> 00:57:39,410 up a multi level converter in the next lecture and thanks for your interest and your patience 624 00:57:39,410 --> 00:57:44,530 i hope this was interesting to you and hoping to see you again in the next lecture bye 625 00:57:44,530 --> 00:57:44,760 thank you