ADE5169 datasheet(59/156 Pages) AD

59 / 156 page

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

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

–20

100

–12

–16

–8

–4

Figure 65. Frequency Response of LPF2

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