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ADE7762
i ( t ) = I O + 2 × ∑ I n × sin ( ( n ω t ) ( β n ) )
V× I
2
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS
ACTIVE POWER SIGNAL
∞
n = 1
where:
i ( t ) is the instantaneous current.
(4)
I O is the dc component.
0V
CURRENT
VOLTAGE
I n is the rms value of current harmonic n .
β n is the phase angle of the current harmonic.
Using Equation 3 and Equation 4, the active power, P, can be
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS
ACTIVE POWER SIGNAL
expressed in terms of its fundamental active power (P 1 ) and
harmonic active power (P H ).
P = P 1 + P H
V× I
2
× cos(60°)
where:
P H = ∑ V n × I n cos φ n
0V
VOLTAGE CURRENT
60°
Figure 13. DC Component of Instantaneous Power Signal
P 1 = V 1 × I 1 cos φ 1
φ 1 = α 1 ? β 1
∞
n = 1
(5)
(6)
NONSINUSOIDAL VOLTAGE AND CURRENT
The active power calculation method also holds true for
nonsinusoidal current and voltage waveforms. All voltage and
current waveforms in practical applications have some har-
monic content. Using the Fourier transform, instantaneous
voltage and current waveforms can be expressed in terms of
their harmonic content
φ n = α n ? β n
As can be seen from Equation 6, a harmonic active power
component is generated for every harmonic, provided that
harmonic is present in both the voltage and current waveforms.
The power factor calculation has been shown to be accurate in
the case of a pure sinusoid. Therefore, the harmonic active
power also correctly accounts for power factor because
v ( t ) = V O + 2 × ∑ V n × sin ( n ω t + α n )
∞
n = 0
where:
v ( t ) is the instantaneous voltage.
(3)
harmonics are made up of a series of pure sinusoids. A limiting
factor on harmonic measurement is the bandwidth. On the
ADE7762, the bandwidth of the active power measurement is
14 kHz with a master clock frequency of 10 MHz.
V O is the average value.
V n is the rms value of voltage harmonic n .
α n is the phase angle of the voltage harmonic.
Rev. 0 | Page 13 of 28
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