ADE7761
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS
ACTIVE POWER SIGNAL
The low frequency output of the ADE7761 is generated by
accumulating this active power information. This low frequency
inherently means a long accumulation time between output
pulses. The output frequency is, therefore, proportional to the
average active power. This average active power information can
in turn be accumulated (for example, by a counter) to generate
active energy information. Because of its high output frequency
and therefore shorter integration time, the CF output is propor-
tional to the instantaneous active power. This is useful for
system calibration purposes that would take place under steady
load conditions.
V × I
2
0V
CURRENT
VOLTAGE
INSTANTANEOUS
POWER SIGNAL
INSTANTANEOUS
ACTIVE POWER SIGNAL
DIGITAL-TO-
FREQUENCY
V × I
2
F1
F2
× cos(60°)
0V
CH1
CH2
ADC
ADC
HPF
MULTIPLIER
DIGITAL-TO-
FREQUENCY
LPF
VOLTAGE
CURRENT
CF
60°
Figure 21. Active Power Calculation over PF
INSTANTANEOUS
POWER SIGNAL –p(t)
INSTANTANEOUS
ACTIVE POWER SIGNAL
Nonsinusoidal Voltage and Current
V × I
p(t) = i(t).v(t)
WHERE:
The active power calculation method also holds true for
nonsinusoidal 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:
V × I
2
v(t) = V × cos(ϖt)
i(t) = I × cos(ϖt)
V × I
p(t) =
{1 + cos (2ϖt)}
2
TIME
Figure 20. Signal Processing Block Diagram
v(t) =V + 2 × ∞ V ×sin(hωt + α )
(1)
Power Factor Considerations
∑
O
h
h
The method used to extract the active power information from
the instantaneous power signal (by low-pass filtering) is still
valid even when the voltage and current signals are not in
phase. Figure 21 displays the unity power factor condition and
a displacement power factor (DPF = 0.5), that is, current signal
lagging the voltage by 60°. If one assumes that the voltage and
current waveforms are sinusoidal, the active power component
of the instantaneous power signal (dc term) is given by
h ≠0
where:
v(t) is the instantaneous voltage.
VO is the average value.
Vh is the rms value of voltage harmonic h.
αh is the phase angle of the voltage harmonic.
∞
i(t) = IO + 2 × I ×sin(hωt +β )
(2)
(V × I/2) × cos(60°)
∑
h
h
h ≠0
This is the correct active power calculation.
where:
i(t) is the instantaneous current.
IO is the dc component.
Ih is the rms value of current harmonic h.
βh is the phase angle of the current harmonic.
Rev. A | Page 16 of 28
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