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AN3322 数据表(PDF) 23 Page - STMicroelectronics |
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AN3322 数据表(HTML) 23 Page - STMicroelectronics |
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23 / 38 page ![]() DocID18304 Rev 1 23/38 AN3322 Computational issues and compensation algorithms power computed from these samples is not the same for every period of mains. This causes the parasitic signal with the frequency FH. After some time (Tx) the coverage of the period is the same as it was at the beginning (in relation to the phase of the measured signal). This time Tx is the period of the frequency FH. Figure 12. Parasitic frequency FH in detail(1) 1. Y-axis: voltage, X-axis: cycles of 50 Hz. In this example, the sampling of 44 samples per period is synchronized with a frequency of 49.93785 Hz. During every period of the mains, the last sample is displaced by time Tdiff_sample. Equation 24 Tdiff_sample = 1/50Hz - 1/49.93785Hz = 0.00002489 s In every cycle of the mains, all the samples are shifted by Tdiff_sample. They again pass their initial position in the time Tshift. Equation 25 Tshift = (1/ 50 Hz) / Tdiff_sample = 803.5 cycles (of 50 Hz) Since the computed instant active power is a function with double the frequency of the input voltage and current (negative instant voltage and negative instant current again give instant positive power), the Tshift should be divided by two: Equation 26 1/FH = (803.5 / 2) * 0.020 s = 8.035 s That the specific sample again passes its initial position does not mean that it is exactly in the same position. This happens after even more cycles of mains. This causes the fluctuation of the instant power by additional disruption of signal with FL. The period of FL is, in this example, 17,500 cycles of mains that is equal to 350 s (obtained by a simulation). The jitter in instant power brings jitter to the accumulated energy value that brings the jitter to the LED pulse output generation. In order to obtain correct instant power, during every period of the mains, or over a short period of time, it is necessary to provide coherent sampling. |
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