1 00:00:20,490 --> 00:00:21,490 Harmonics and Switching Transients in Single Phase Transformers 2 00:00:21,490 --> 00:00:31,730 So far we have assumed that when you apply a sine wave voltage to a single phase transformer; 3 00:00:31,730 --> 00:00:38,170 the flux in the core is sinusoidal, and the current drawn by the transformer is also sinusoidal. 4 00:00:38,170 --> 00:00:42,460 However, in practice this is not the case; it is mainly due to the nonlinearity of the 5 00:00:42,460 --> 00:00:47,930 magnetic core used in a transformer, which is made up of ferromagnetic material. 6 00:00:47,930 --> 00:00:54,290 So, a practical transformer you will have a core on which the windings will be placed 7 00:00:54,290 --> 00:00:57,100 and on the other winding the load may be connected. The applied voltage we assumed to be sinusoidal, 8 00:00:57,100 --> 00:01:01,790 if we neglect the resistive drop in the winding, then the counter induced emf must completely 9 00:01:01,790 --> 00:01:08,780 cancel the applied voltage; that is the counter induced emf e should be equal to V i. 10 00:01:08,780 --> 00:02:24,560 Hence, the flux which is given by N 1 d phi d t equal to e will be equal to this will 11 00:02:24,560 --> 00:02:35,110 be sine wave. So, the flux in the core will be sine wave; and the flux will also lag the 12 00:02:35,110 --> 00:02:44,510 applied voltage by 90 degree. Now, had the magnetic material be in linear; the current 13 00:02:44,510 --> 00:02:50,110 drawn the magnetizing current drawn by the transformer; it also have been a sine wave. 14 00:02:50,110 --> 00:02:59,550 However, this is not the case in practical transformer. In a practical transformer the 15 00:02:59,550 --> 00:03:09,270 no load current, the magnetizing current and the flux are related by the B-H characteristic 16 00:03:09,270 --> 00:03:20,060 on the material; due to this non-linear B-H characteristic we will soon see that the no 17 00:03:20,060 --> 00:03:28,840 load current drawn by the transformer will not be a sine wave. But it will be distorted 18 00:03:28,840 --> 00:03:37,640 with different harmonic components. Let us examine this a little more carefully. 19 00:03:37,640 --> 00:03:52,770 Let us say this is the applied voltage waveform. The applied voltage waveform V i and the flux 20 00:03:52,770 --> 00:04:03,710 waveform which will lag the voltage by 90 degree; it will be somewhat like this the 21 00:04:03,710 --> 00:04:16,100 core flux. Now, let us try to find out what will be the current due to in order to establish 22 00:04:16,100 --> 00:04:26,880 this flux in 23 00:04:26,880 --> 00:04:33,880 the core; for that we will have to consider the B-H characteristic of the material. So, 24 00:04:33,880 --> 00:04:53,960 for the current let us see that at time t equal to 0 when applied voltage was 0 and 25 00:04:53,960 --> 00:05:17,439 increasing; the flux voltage was at its negative peak. This was the value of the flux and hence 26 00:05:17,439 --> 00:05:21,290 the corresponding current was this. So, current was at negative maximum. 27 00:05:21,290 --> 00:05:32,409 Let us say this corresponds to the negative maximum of the current. Now, as the voltage 28 00:05:32,409 --> 00:05:49,819 is increased now flux is increasing and while increasing in a B-H curve the flux follows 29 00:05:49,819 --> 00:06:18,069 this path. So, we see the current when the current become 0 the flux still has a negative 30 00:06:18,069 --> 00:06:37,090 component; that is the current will become 0 before the flux becomes 0. So, this is the 31 00:06:37,090 --> 00:06:42,879 point let us say where the flux will become 0; where some current still exist, some flux 32 00:06:42,879 --> 00:06:47,599 still exist. When the flux becomes 0 we find that the current has become already small 33 00:06:47,599 --> 00:06:53,919 positive. So, this corresponds to the point; again when the flux reaches its positive peak 34 00:06:53,919 --> 00:07:00,620 which coincides with the zero crossing of the voltage the current also reaches its positive 35 00:07:00,620 --> 00:07:08,490 peak. So, this is the corresponding point of the current. And since the B-H curve is 36 00:07:08,490 --> 00:07:18,319 symmetric around x axis; the positive peak of the current wave form and the negative 37 00:07:18,319 --> 00:07:26,940 peak of the current waveform will be equal; while returning though the flux takes a different 38 00:07:26,940 --> 00:07:35,509 path because of hysteresis. So, we see that when the current becomes 0 39 00:07:35,509 --> 00:07:48,930 the flux still has some positive value. Therefore, the corresponding point here will be let us 40 00:07:48,930 --> 00:08:00,240 say this. And when the flux becomes 0 the current has already become negative and here 41 00:08:00,240 --> 00:08:12,849 again it is negative peak. Therefore, the current waveform will look somewhat like this; 42 00:08:12,849 --> 00:08:31,080 which is not sinusoidal although it is symmetric around y axis. But it is not sinusoidal unlike 43 00:08:31,080 --> 00:08:40,900 the applied voltage waveform V m; V i or the flux waveform phi m. So, this current waveform 44 00:08:40,900 --> 00:08:50,920 is not sinusoidal; of course it can be divided any into Fourier components it will have. 45 00:08:50,920 --> 00:08:57,019 So, the current drawn i m will have a fundamental component i 1 m plus harmonic components. 46 00:08:57,019 --> 00:09:14,920 And because of its odd nature it 47 00:09:14,920 --> 00:09:48,630 will have only odd harmonics; that is where i 1 m equal to I 1 m sine omega t cos omega 48 00:09:48,630 --> 00:10:14,350 t plus sine omega plus phi and i n m equal to I n m sine n omega t plus n phi. 49 00:10:14,350 --> 00:10:34,620 So, this is the source of mostly the source of harmonics in a single phase transformer. 50 00:10:34,620 --> 00:10:44,360 And mostly in the magnetization current that this harmonic will be observable; there is 51 00:10:44,360 --> 00:10:50,890 one interesting thing to note here that the zero crossing of the flux waveform and the 52 00:10:50,890 --> 00:10:55,730 current waveform do not coincide. In fact, the zero crossing of the current waveform 53 00:10:55,730 --> 00:11:00,180 leads the zero crossing of the flux waveform by some angle; although the peak of the current 54 00:11:00,180 --> 00:11:04,640 waveform coincides with the peak of the flux waveform. If one decomposes the current waveform 55 00:11:04,640 --> 00:11:18,610 into its Fourier components at the fundamental we will find that the current waveform has 56 00:11:18,610 --> 00:12:10,540 a component which is large component; which is in phase with the flux waveform phi m. 57 00:12:10,540 --> 00:12:22,690 And a smaller component which is leads the phi m and in fact it is in phase with the 58 00:12:22,690 --> 00:12:30,200 supply voltage V i. So, the out of the phase that is component 59 00:12:30,200 --> 00:12:39,220 of i m which is in phase with phi m. And in quadrature with that is at 90 degree lagging 60 00:12:39,220 --> 00:12:44,860 V i is responsible for setting up the flux in the core. While the in phase component 61 00:12:44,860 --> 00:13:17,540 of I m that is the component which is in phase with the applied voltage V i supplies the 62 00:13:17,540 --> 00:13:19,810 hysteresis loss; of course this magnetization current waveform will also be modified by 63 00:13:19,810 --> 00:13:24,740 the eddy current component. However, this component is sinusoidal because the flux is 64 00:13:24,740 --> 00:13:42,230 sinusoidal which introduces a sinusoidal voltage in the core. And sinusoidal current in the 65 00:13:42,230 --> 00:13:49,920 core material which will be compensated by drawing sinusoidal current from the source 66 00:13:49,920 --> 00:14:00,420 which will be in again in phase with V i. So, the in phase component sinusoidal component 67 00:14:00,420 --> 00:14:12,160 of V i will increase that will make the waveform less distorted. 68 00:14:12,160 --> 00:14:25,449 Therefore, the no load current of a transformer will have a in phase component; which is due 69 00:14:25,449 --> 00:14:37,130 to in phase component which is in phase with the voltage which is due to the hysteresis 70 00:14:37,130 --> 00:14:47,649 and eddy current loss. And a component which is perpendicular which is lagging V i by 90 71 00:14:47,649 --> 00:14:59,160 degree and in phase with the flux waveform which establishes the flux. In addition to 72 00:14:59,160 --> 00:15:07,670 these two fundamental components there will be harmonics; odd harmonics that is third 73 00:15:07,670 --> 00:15:10,689 harmonic, fifth harmonic, seventh harmonic etcetera of the fundamental frequency; which 74 00:15:10,689 --> 00:15:15,569 are due to the saturation of the core material. The deeper into saturation the transformer 75 00:15:15,569 --> 00:15:21,779 walks, the more will be harmonics and the nature of the current waveform will become 76 00:15:21,779 --> 00:15:22,779 very peaky. 77 00:15:22,779 --> 00:15:31,330 However, when you since the magnetization current is non sinusoidal one would expect; 78 00:15:31,330 --> 00:15:39,050 if one assumes the winding resistance that the applied voltage is sine wave; the current 79 00:15:39,050 --> 00:15:48,629 drawn i 0 has harmonic. Then one should accept the induced voltage e also to have harmonics. 80 00:15:48,629 --> 00:16:01,990 Hence, even the flux waveform will possibly not be exactly a sine wave which may or may 81 00:16:01,990 --> 00:16:18,610 not increase the harmonics in i 0 depending on the nature of the B-H characteristics. 82 00:16:18,610 --> 00:16:45,250 However, since i 0 magnitude is very small and also the state resistance is small e and 83 00:16:45,250 --> 00:17:19,569 phi m can be for all practical purpose assume to be sine wave. So, in a transformer if the 84 00:17:19,569 --> 00:17:25,610 applied voltage is sinusoidal then the core flux is almost sinusoid. But the magnetization 85 00:17:25,610 --> 00:17:32,970 current drawn is distinctly non sinusoidal; it has fundamental components and odd harmonics. 86 00:17:32,970 --> 00:17:49,970 However, when we put a load on the transformer; I have already mentioned that there will be 87 00:17:49,970 --> 00:17:59,750 a reflection current i L dash flowing the transformer. Since, the core flux is a sine 88 00:17:59,750 --> 00:18:03,559 wave the induced voltage on the secondary side is also a sine wave. And if the load 89 00:18:03,559 --> 00:18:09,279 is a resistive or inductive a linear load; then the current I L will be a sine wave. 90 00:18:09,279 --> 00:18:12,970 And hence the, its reflection i L dash will also be a sine wave. 91 00:18:12,970 --> 00:18:28,169 Now, i 0 is much smaller in a normal transformer i 0 is much smaller compare to i L dash. So, 92 00:18:28,169 --> 00:18:34,960 when transformer is loaded I L will dash will dominate over i 0. Therefore, the waveform 93 00:18:34,960 --> 00:18:38,670 will look very much sinusoidal. And although there will be harmonics of the magnetization 94 00:18:38,670 --> 00:18:42,110 current present. But it will not be apparent since the magnetization current consistute 95 00:18:42,110 --> 00:18:45,549 only a very small fraction of the total output current of the transformer. 96 00:18:45,549 --> 00:18:50,110 So, a transformer single phase transformer at no load draws predominate draws a non sinusoidal 97 00:18:50,110 --> 00:19:00,750 current from the supply consisting of fundamental and odd harmonics. However, on load, the fundamental 98 00:19:00,750 --> 00:19:09,520 component also include the reflected load currents of the fundamental component increases 99 00:19:09,520 --> 00:19:17,370 but the harmonics do not increase. Therefore, on load transformer current appears to be 100 00:19:17,370 --> 00:19:29,799 more or less sine wave. At this point one should be careful about how one should connect 101 00:19:29,799 --> 00:19:49,529 a transformer for performing certain test. In this circuit we have assumed r 1 to 102 00:19:49,529 --> 00:20:07,940 be small. However suppose we want to perform a no load test on this transformer. And the 103 00:20:07,940 --> 00:20:14,169 available supply voltage is larger than the transformer voltage rating. Then one possibility 104 00:20:14,169 --> 00:20:21,210 could have been to connect the transformer like this. And then connecting a resistance 105 00:20:21,210 --> 00:20:28,340 of large value and connecting a transformer across the supply. 106 00:20:28,340 --> 00:20:58,960 Since, we are doing a no load test then the current drawn by the transformer will be non 107 00:20:58,960 --> 00:21:13,220 sinusoidal. And the voltage; actual voltage across the transformer will also be non sinusoidal 108 00:21:13,220 --> 00:21:30,149 this will make the flux non sinusoidal with predominantly third harmonic component. 109 00:21:30,149 --> 00:21:46,440 The flux waveform, instead of looking sinusoidal may look something like this; this is the 110 00:21:46,440 --> 00:22:26,809 sine wave component of the flux, this is third harmonic component of the flux. So, the resultant 111 00:22:26,809 --> 00:22:42,679 flux waveform near show predominantly peaky nature; additional resistance is connected 112 00:22:42,679 --> 00:22:51,020 here; in which case because of saturation this will be significantly higher than the 113 00:22:51,020 --> 00:23:06,669 fundamental magnitude of the flux. So, the current drawn by the circuit may be 114 00:23:06,669 --> 00:23:16,890 significantly large. Therefore, it is not desirable to make a circuit connection like 115 00:23:16,890 --> 00:23:25,990 this; hence for doing no load test under such situation it will be preferable instead of 116 00:23:25,990 --> 00:23:32,419 using a resistance to drop the voltage. One should use an auto transformer, variac, in 117 00:23:32,419 --> 00:23:40,240 order to reduce the voltage to the desired level. Because the induced voltage in the 118 00:23:40,240 --> 00:23:47,990 auto transformer will be almost sinusoidal. So, far in this figure we have assumed that 119 00:23:47,990 --> 00:24:09,580 the transformer is operating in steady state; that is the given a sine wave applied voltage 120 00:24:09,580 --> 00:24:41,100 the flux waveform is also a sine wave. However, when the voltage is first switched on in the 121 00:24:41,100 --> 00:24:50,950 transformer it under goes a transient before the flux becomes settles down to its final 122 00:24:50,950 --> 00:25:04,779 sine wave value. During this period the transformer can experience very larger current; let us 123 00:25:04,779 --> 00:25:07,190 see how? 124 00:25:07,190 --> 00:25:22,480 Again, let this is the transformer on no load. If we assume neglect the transformer winding 125 00:25:22,480 --> 00:25:40,141 resistance; then we can write e equal to N d phi m by d t equal to v i equal to V sine 126 00:25:40,141 --> 00:26:00,650 omega t or phi m equal to V by N minus V by omega N cosine of omega t plus and constant 127 00:26:00,650 --> 00:26:07,750 of integration. At time t equal to 0 one would expect that core the de energized. So, phi 128 00:26:07,750 --> 00:26:33,590 m at time t equal to 0 may be assumed to be 0. So, at time equal to 0 or C equal 129 00:26:33,590 --> 00:26:43,280 to V divided by N omega or therefore, phi m assuming at t equal to 0; the voltage was 130 00:26:43,280 --> 00:27:06,630 0 phi m will be V by N omega 1 minus cosine of omega t. Let us plot this. 131 00:27:06,630 --> 00:27:22,260 So, this is the applied voltage sine omega t applied voltage v i; we have switched on 132 00:27:22,260 --> 00:28:29,700 at t equal to 0 when the voltage was 0. Then the flux waveform phi m is given by of the 133 00:28:29,700 --> 00:28:54,080 form phi m equal to phi max into 1 minus cos omega t. At t equal to 0 it was 0; at omega 134 00:28:54,080 --> 00:29:32,559 t equal to pi by 2 it will become phi max. And omega t equal to pi it will become 2 phi 135 00:29:32,559 --> 00:30:27,530 max. So, the flux waveform will cosine wave which is shifted by phi max. 136 00:30:27,530 --> 00:31:11,350 Now, let us see what will be 137 00:31:11,350 --> 00:31:35,840 the consequences in term of currents? As we have already mentioned a practical transformer 138 00:31:35,840 --> 00:31:39,169 core will exhibit a predominate saturation characteristics along with hysteresis. Let 139 00:31:39,169 --> 00:32:09,850 us say transformer is usually designed to operate at a flux density of phi max; maximum 140 00:32:09,850 --> 00:32:19,920 flux in flux phi max. However, we see that because of switching initially the flux raises 141 00:32:19,920 --> 00:32:31,059 to almost double the value; due to non-linearity of saturation the current will not be double 142 00:32:31,059 --> 00:32:40,980 it will be very large; incases it can be 50 to 100 times the normal no load peak current. 143 00:32:40,980 --> 00:32:53,240 The current waveform as we see at we neglect the hysteresis at t equal to 0 it is 0; phi 144 00:32:53,240 --> 00:33:29,520 m reaches it at omega equal to pi by 2 reaches rated value. And then at omega equal to pi 145 00:33:29,520 --> 00:33:39,309 it reaches a very high value and this waveform repeats. 146 00:33:39,309 --> 00:33:48,230 So, if we look at a large scale; time scale we will find right after switching the current 147 00:33:48,230 --> 00:34:04,340 raises to very high value. And is almost entirely positive it does not become negative initially. 148 00:34:04,340 --> 00:34:15,380 Then slowly because of different losses in the core; the flux slowly gradually reduces 149 00:34:15,380 --> 00:34:39,159 and settles down to its final value; accordingly current also gradually reduces. So, this first 150 00:34:39,159 --> 00:34:51,440 peak can be as I said 50 to 100 times the normal steady state peak magnetization current; 151 00:34:51,440 --> 00:35:05,720 which means it can be 5 to 10 times the full load current 152 00:35:05,720 --> 00:35:21,369 of the transformer for first 1 or 2 cycles. This is called the in-rush phenomenon in a 153 00:35:21,369 --> 00:35:26,130 single phase transformer; this occurs due to the depending on the wrong switching instant 154 00:35:26,130 --> 00:35:59,280 of the voltage. It is interesting to note the instead of switching 155 00:35:59,280 --> 00:36:18,490 the waveform; the voltage at zero crossing had we switched it at the peak value; that 156 00:36:18,490 --> 00:36:40,260 is if the at t equal to 0 if the switched it at the peak value. 157 00:36:40,260 --> 00:37:01,800 Then, V i would have been given by instead of V sine omega it would have been equal to 158 00:37:01,800 --> 00:37:25,270 V cosine omega t that is equal to V sine omega t plus pi by 2. In which 159 00:37:25,270 --> 00:37:38,290 case that it is if we would have switched on at the positive peak, then the flux waveform 160 00:37:38,290 --> 00:37:55,050 would have been phi m would have been and if at t equal to 0 phi m was 0. Then c also 161 00:37:55,050 --> 00:38:03,329 would have been then phi m waveform would have been equal to V by N omega sine omega 162 00:38:03,329 --> 00:38:23,990 t equal to phi max sine omega t. And there would not have been any in-rush phenomenon. 163 00:38:23,990 --> 00:38:48,000 So, depending on when the transformer is switched on it is possible 164 00:38:48,000 --> 00:39:16,420 that there will be an magnetic in-rush in a single phase transformer; due to which the 165 00:39:16,420 --> 00:39:48,670 no load current of the transformer can raise to dangerously high level for initial few 166 00:39:48,670 --> 00:40:12,950 cycles. A protection system should be designed to handle this. Because it is difficult to 167 00:40:12,950 --> 00:40:23,329 precisely determine exactly at which instant a transformer will switched on. The situation 168 00:40:23,329 --> 00:41:03,400 will become worse since, a transformer at t equal to 0 the flux in the core may not 169 00:41:03,400 --> 00:42:22,859 be 0. Because of previous excitation of the transformer the 170 00:42:22,859 --> 00:44:54,660 there may be some flux remaining in the core this is 171 00:44:54,660 --> 00:46:02,780 called retentivity. So, if 172 00:46:02,780 --> 00:46:26,740 you have some remnant flux 173 00:46:26,740 --> 00:47:16,609 in the transformer, then the peak 174 00:47:16,609 --> 00:51:37,119 flux density may increase even more than twice the steady state peak flux; in 175 00:51:37,119 --> 00:52:12,840 which case the peak of the in-rush current will also increase accordingly; on load though 176 00:52:12,840 --> 00:52:47,930 this effect becomes somewhat less. Because the 177 00:52:47,930 --> 00:53:07,329 flux decays fast and the peak value of the 178 00:53:07,329 --> 00:53:12,319 current reduces faster. Therefore, in a single phase transformer, a switching transient during 179 00:53:12,319 --> 00:53:31,770 turn on may cause magnetic in-rush; due to which the no load current can increase to 180 00:53:31,770 --> 00:54:09,339 very high level this current slowly decays because 181 00:54:09,339 --> 00:54:57,630 the losses in the transformer. And finally settles downs to 182 00:54:57,630 --> 00:55:20,949 the normal magnetization current.