Datasheet

ADE7763 Data Sheet
Rev. C | Page 26 of 56
)sin(ω2)( tVtv  (7)
)sin(ω2)( tIti  (8)
where:
V is the rms voltage.
I is the rms current.
)()()( titvtp 
)2cos()( tVIVItp  (9)
The average power over an integral number of line cycles (n) is
given by the expression in Equation 10.


nT
VIdttp
nT
P
0
)(
1
(10)
where:
T is the line cycle period.
P is the active or real power.
Note that the active power is equal to the dc component of the
instantaneous power signal p(t) in Equation 8, i.e., VI. This is
the relationship used to calculate active power in the ADE7763.
The instantaneous power signal p(t) is generated by multiplying
the current and voltage signals. The dc component of the instan-
taneous power signal is then extracted by LPF2 (low-pass filter)
to obtain the active power information. This process is illustrated
in Figure 53.
INSTANTANEOUS
POWER SIGNAL
p(t) = v
 i-v  i  cos(2  t)
ACTIVE REAL POWER
SIGNAL = v
 i
0x19 999A
VI
0xC CCCD
0x0 0000
04481-A-054
CURRENT
i(t) = 2
 i  sin(  t)
VOLTAGE
v(t) = 2
 v  sin(  t)
Figure 53. Active Power Calculation
Because LPF2 does not have an ideal “brick wall” frequency
response (see Figure 54), 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 to calculate energy—see the
Energy Calculation section.
FREQUENCY (Hz)
–24
1
dB
–20
3 10 30 100
–12
–16
–8
–4
0
04481-A-055
Figure 54. Frequency Response of LPF2