EVAL-ADE7759E View Datasheet(PDF) - Analog Devices
Part Name
Description
MFG CO.
'EVAL-ADE7759E' PDF : 32 Pages View PDF
ADE7759
V1P
V1
PGA1
V1N
V2P
V2
PGA2
V2N
V2
V1
ADC 1
HPF
20
LPF2
20
ADC 2
0.1؇
1
DELAY BLOCK
1.12s/LSB
7
0
11111100
PHCAL [7:0]
–110s TO +103s
CHANNEL 2 DELAY
REDUCED BY 4.48s
(0.1؇ LEAD AT 60Hz)
FCH IN PHCAL [7:0]
V2
V1
60Hz
60Hz
Figure 28. Phase Calibration
0.30
0.25
0.20
0.15
0.10
0.05
0.00
–0.05
–0.10
100 200 300 400 500 600 700 800 900 1000
FREQUENCY – Hz
Figure 29. Combined Phase Response of the HPF and
Phase Compensation (10 Hz to 1 kHz)
0.30
0.25
0.20
0.15
0.10
0.05
0.00
–0.05
–0.10
40
45
50
55
60
65
70
FREQUENCY – Hz
Figure 30. Combined Phase Response of the HPF and
Phase Compensation (40 Hz to 70 Hz)
0.4
0.3
0.2
0.1
0.0
–0.1
–0.2
–0.3
–0.4
54
56
58
60
62
64
66
FREQUENCY – Hz
Figure 31. Combined Gain Response of the HPF and Phase
Compensation (Deviation of Gain in % from Gain at 60 Hz)
ACTIVE POWER CALCULATION
Electrical power is defined as the rate of energy flow from
source to load. It 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/second
Equation 3 gives an expression for the instantaneous power signal
in an ac system.
v(t) = 2 V(ωt)
(1)
i(t) = 2 I sin(ωt)
(2)
where
V = rms voltage,
I = rms current.
p(t) = v(t) × i(t)
p(t) = VI – VI cos(2ωt)
(3)
The average power over an integral number of line cycles (n) is
given by the expression in Equation 4.
P
=
1
nT
nT
∫
0
p (t ) dt
= VI
(4)
where T is the line cycle period. P is referred to as the Active or
Real Power. Note that the active power is equal to the dc com-
ponent of the instantaneous power signal p(t) in Equation 3, i.e.,
VI. This is the relationship used to calculate active power in the
ADE7759. The instantaneous power signal p(t) is generated by
multiplying the current and voltage signals. The dc component of
the instantaneous power signal is then extracted by LPF2
(Low-Pass Filter) to obtain the active power information. This
process is illustrated graphically in Figure 32. Since LPF2 does not
have an ideal “brick wall” frequency response (see Figure 33),
the Active Power signal will have some ripple due to the instanta-
neous power signal. This ripple is sinusoidal and has a frequency
equal to twice the line frequency. Since the ripple is sinusoidal
in nature it will be removed when the Active Power signal is
integrated to calculate Energy—see Energy Calculation section.
–20–
REV. 0
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