| 数据搜索系统,热门电子元器件搜索 |
|
ADE7756ARS 数据表(PDF) 24 Page - Analog Devices |
|
|
|||||||||||||||||||||||||||||
ADE7756ARS 数据表(HTML) 24 Page - Analog Devices |
|
24 / 32 page ![]() REV. 0 ADE7756 –24– reduce the ripple. Also, averaging the output frequency by using a longer gate time for the counter will achieve the same results. t E(t) VIt VI 4 fl(1 + 2 fl/8.9Hz SIN(4 fl t) Figure 33. Output Frequency Ripple ENERGY CALIBRATION By using the on-chip zero-crossing detection on Channel 2 the energy calibration can be greatly simplified and the time required to calibrate the meter can be significantly reduced. To use the zero-cross detection the ADE7756 is placed in calibration mode by setting Bit 7 (CMODE) in the Mode register. In Calibration Mode the ADE7756 accumulates the Active Power signal in the Active Energy register for an integral number of half cycles, as shown in Figure 34. The number of half-line cycles is specified in the SAGCYC register. The ADE7756 can accumulate Active Power for up to 255 half-cycles. Because the Active Power is integrated on an integral number of line cycles, the sinusoidal component is reduced to zero. This eliminates any ripple in the energy calculation. Energy is calculated more accurately and in a shorter time because integration period can be shortened. At the end of an energy calibration cycle the SAG flag in the Inter- rupt Status register is set; this will cause the SAG output to go active low. If the SAG enable bit in the Interrupt Enable register is enabled, the IRQ output will also go active low. Thus the IRQ line can be used to signal the end of a calibration also. Another calibration cycle will start as long as the CMODE bit in the Mode register is set. Note that the result of the first calibration is invalid and must be ignored. The result of all subsequent calibration cycles is correct. From Equations 5 and 11, E t VIdt VI fl Hz tdt nT nT () – /. cos ( ) 00 12 8 9 2 ∫ + ∫ ω (13) where n is an integer and T is the line cycle period. Since the sinusoidal component is integrated over an integer number of line cycles, its value is always zero. Therefore: E t VIdt nT () = ∫ 0 (14) E t VInt () = (15) APOS [11:0] 23 0 AENERGY[39:0} ACTIVE POWER SIGNAL – P LPF2 WAVEFORM [23:0] 39 11 0 SAGCYC[7:0] CCCDh CALIBRATION CONTROL ZERO CROSS DETECT LPF1 0 00h CHANNEL 2 ADC FROM MULTIPLIER Figure 34. Energy Calculation in Calibration Mode CALIBRATING THE ENERGY METER Calculating the Average Active Power When calibrating the ADE7756, the first step is to calibrate the frequency on CF to some required meter constant, e.g., 3200 imp/kWh. In order to determine the output frequency on CF, the average value of the Active Power signal (output of LPF2) must first be determined. One convenient way to do this is to use the calibra- tion mode. When the CMODE (Bit 7) bit in the Mode register is set to a Logic 1, energy is accumulated over an integer num- ber of half-line cycles as described in the last section. Since the line frequency is fixed at, say, 60 Hz, and the number of half-cycles of integration is specified, the total integration time is given as: 1 260 × × Hz number of half cycles For 255 half-cycles this would give a total integration time of 2.125 seconds. This would mean the energy register was updated 2.125/1.1175 µs (4/CLKIN) times. The average output value of LPF2 is given as: Contents of AENERGY at the end Number of times AENERGY was updated [: ] [: ] 39 0 39 0 Or equivalently, in terms of contents of various ADE7756 regis- ters and CLKIN and line frequencies (fl): Average word LPF AENERGY fl SAGCYC CLKIN () [: ] [: ] 2 39 0 8 70 = ×× × (16) where fl is the line frequency. Calibrating the Frequency at CF Once the average Active Power signal is calculated it can be used to determine the frequency at CF before calibration. When the frequency before calibration is known, the Calibration Frequency Divider register (CFDIV) and the Active Power Gain register (APGAIN) can be adjusted to produce the required frequency on CF. In this example, a meter constant of 3200 imp/kWh is chosen as an appropriate constant. This means that under a steady load of 1 kW, the output frequency on CF would be, Frequency CF imp kWh Hz () / . = × == 3200 60 60 3200 3600 0 8888 min sec |
|
链接网址 |
| ALLDATASHEET是否为您带来帮助? [ DONATE ] |
关于 Alldatasheet | 广告服务 | 联系我们 | 隐私政策 | 数据表链接 | 链接交换 | 制造商名单 All Rights Reserved©Alldatasheet.com |
| Russian : Alldatasheetru.com | Korean : Alldatasheet.co.kr | Spanish : Alldatasheet.es | French : Alldatasheet.fr | Italian : Alldatasheetit.com Portuguese : Alldatasheetpt.com | Polish : Alldatasheet.pl | Vietnamese : Alldatasheet.vn Indian : Alldatasheet.in | Mexican : Alldatasheet.com.mx | British : Alldatasheet.co.uk | New Zealand : Alldatasheet.co.nz |
|
Family Site : ic2ic.com |
icmetro.com |