ADE7933 datasheet(57/125 Pages) AD

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Data Sheet

ADE7978/ADE7933/ADE7932/ADE7923

Rev. D | Page 57 of 125

ACTIVE POWER CALCULATION

The ADE7978 computes the total active power on every phase.

The calculation of total active power includes all fundamental

and harmonic components of the voltages and currents. The

ADE7978 also computes the fundamental active power, that is,

the power determined only by the fundamental components of

the voltages and currents.

TOTAL ACTIVE POWER CALCULATION

Electrical power is defined as the rate of energy flow from source

to load and is given by the product of the voltage and current

waveforms. The resulting waveform is called the instantaneous

power signal, and it is equal to the rate of energy flow at every

instant of time. The unit of power is the watt or joules/sec. If an

ac system is supplied by a voltage, v(t), and consumes the current,

i(t), and the voltage and current contain harmonics, then

(

)

∑

∞

sin

)

(

(25)

(

)

∑

∞

sin

)

(

where:

Vk, Ik are the rms voltage and current, respectively, of each

harmonic.

φk, γk are the phase delays of each harmonic.

The total active power is equal to the dc component of the

instantaneous power signal, that is,

∑

∞

k I

cos(φk − γk)

This equation represents the total active power calculated in the

ADE7978 for each phase.

The equation for fundamental active power is

FP = V1I1 cos(φ1 − y1)

(26)

Figure 79 shows how the ADE7978 computes the total active

power on each phase. The ADE7978 first multiplies the current

and voltage signals in each phase. It then extracts the dc component

of the instantaneous power signal in each phase (A, B, and C)

using the LPF2 low-pass filter.

INSTANTANEOUS

PHASE A

ACTIVE POWER

CURRENT SIGNAL

FROM HPF

VOLTAGE SIGNAL

FROM HPF

LPFSEL BIT

CONFIG[5]

LPF2

APGAIN

AWATTOS

AWATT

Figure 79. Total Active Power Datapath

If the phase currents and voltages contain only the fundamental

component, are in phase (that is, φ1 = γ1 = 0), and correspond to

full-scale ADC inputs, then multiplying them results in an instan-

taneous power signal that has a dc component, V1 × I1, and a

sinusoidal component, V1 × I1 × cos(2ωt). Figure 80 shows the

corresponding waveforms.

INSTANTANEOUS

POWER SIGNAL

INSTANTANEOUS

ACTIVE POWER

SIGNAL: V rms × I rms

p(t) = V rms

× I rms – V rms × I rms × cos(2ωt)

53,982,544

V rms × I rms =

26,991,271

i(t) =

√2 × I rms × sin(ωt)

v(t) =

√2 × V rms × sin(ωt)

Figure 80. Active Power Calculation

Because LPF2 does not have an ideal brick wall frequency response,

the active power signal has some ripple due to the instantaneous

power signal. This ripple is sinusoidal and has a frequency equal

to twice the line frequency. Because the ripple is sinusoidal in

nature, it is removed when the active power signal is integrated

over time to calculate the energy.

Bit 5 (LPFSEL) of the CONFIG register (Address 0xE618) selects

the LPF2 strength. When LPFSEL is cleared to 0 (the default value),

the settling time is 650 ms, and the ripple attenuation is 65 dB.

When LPFSEL is set to 1, the settling time is 1300 ms, and the

ripple attenuation is 128 dB. Figure 81 shows the frequency

response of LPF2 when LPFSEL is cleared to 0. Figure 82 shows

the frequency response of LPF2 when LPFSEL is set to 1.

–5

–10

–15

–20

–25

0.1

FREQUENCY (Hz)

Figure 81. Frequency Response of the LPF Used to Filter Instantaneous Power

in Each Phase: LPFSEL Bit of CONFIG Register Set to 0 (Default)

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