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ADE7752/ADE7752A
i ( t ) = I O + 2 × ∑ V n I × sin ( n ω t β n )
This is the correct real power calculation.
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS
REAL POWER SIGNAL
∞
n = 0
(2)
where:
V × I
2
0V
CURRENT
i ( t ) is the instantaneous current.
I O is the dc component.
I n is the rms value of current harmonic n .
β n is the phase angle of the current harmonic.
VOLTAGE
Using Equations 1 and 2, the real power, P , can be expressed in
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS REAL
POWER SIGNAL
terms of its fundamental real power ( P 1 ) and harmonic real
power ( P H ).
V × I
2
× cos(60 ° )
0V
P = P 1 + P H
where:
VOLTAGE
60 °
CURRENT
P 1 = V 1 × I 1 cos φ 1
φ 1 = α 1 ? β 1
(3)
P H = ∑ V n × I n cos φ n
Figure 16. DC Component of Instantaneous Power Signal
Conveys Real Power Information PF < 1
∞
n = 1
(4)
NONSINUSOIDAL VOLTAGE AND CURRENT
The real power calculation method also holds true for nonsin-
usoidal current and voltage waveforms. All voltage and current
waveforms in practical applications have some harmonic
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 4, a harmonic real power compo-
nent 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 real power must also
correctly account for power factor since it is made up of a series
∑ V n × sin ( n ω t + α n )
v ( t ) = V o + 2 ×
where:
∞
n = 0
(1)
of pure sinusoids.
Note that the input bandwidth of the analog inputs is 14 kHz
with a master clock frequency of 10 MHz.
v ( t ) is the instantaneous voltage.
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. C | Page 13 of 24
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