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ADE5569 bảng dữ liệu(PDF) 59 Page - Analog Devices

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Giải thích chi tiết về linh kiện  Single-Phase Energy Measurement IC with 8052 MCU, RTC, and LCD Driver
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ADE5569 bảng dữ liệu(HTML) 59 Page - Analog Devices

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ADE5166/ADE5169/ADE5566/ADE5569
Rev. B | Page 59 of 156
Note that the active power is equal to the dc component of the
instantaneous power signal, P(t), in Equation 9, that is, VI. This
is the relationship used to calculate active power in the ADE5166/
ADE5169/ADE5566/ADE5569. The instantaneous power signal,
P(t), is generated by multiplying the current and voltage signals.
The dc component of the instantaneous power signal is then
extracted by LPF2 (low-pass filter) to obtain the active power
information (see Figure 64).
INSTANTANEOUS
POWER SIGNAL
P(t) = V × I – V × I × cos(2ωt)
ACTIVE REAL POWER
SIGNAL = V × I
0x19999A
VI
0xCCCCD
0x00000
CURRENT
I(t) = 2 × I × sin(ωt)
VOLTAGE
V(t) = 2 × V × sin(ωt)
Figure 64. Active Power Calculation
Because LPF2 does not have an ideal brick wall frequency response
(see Figure 65), the active power signal has some ripple due to
the instantaneous power signal. This ripple is sinusoidal and has
a frequency equal to 2× the line frequency. Because of its sinu-
soidal nature, the ripple is removed when the active power signal is
integrated to calculate energy (see the Active Energy Calculation
section).
FREQUENCY (Hz)
–24
1
–20
3
10
30
100
–12
–16
–8
–4
0
Figure 65. Frequency Response of LPF2



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