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ADE7756ARS 数据表(PDF) 24 Page - Analog Devices

部件名 ADE7756ARS
功能描述  Active Energy Metering IC with Serial Interface
PDF  32 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

ADE7756ARS 数据表(HTML) 24 Page - Analog Devices

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REV. 0
ADE7756
–24–
reduce the ripple. Also, averaging the output frequency by using
a longer gate time for the counter will achieve the same results.
t
E(t)
VIt
VI
4
fl(1 + 2
fl/8.9Hz
SIN(4
fl
t)
Figure 33. Output Frequency Ripple
ENERGY CALIBRATION
By using the on-chip zero-crossing detection on Channel 2 the
energy calibration can be greatly simplified and the time required
to calibrate the meter can be significantly reduced. To use the
zero-cross detection the ADE7756 is placed in calibration mode
by setting Bit 7 (CMODE) in the Mode register. In Calibration
Mode the ADE7756 accumulates the Active Power signal in the
Active Energy register for an integral number of half cycles, as
shown in Figure 34. The number of half-line cycles is specified
in the SAGCYC register. The ADE7756 can accumulate Active
Power for up to 255 half-cycles. Because the Active Power is
integrated on an integral number of line cycles, the sinusoidal
component is reduced to zero. This eliminates any ripple in the
energy calculation. Energy is calculated more accurately and in
a shorter time because integration period can be shortened. At
the end of an energy calibration cycle the SAG flag in the Inter-
rupt Status register is set; this will cause the
SAG output to go
active low. If the SAG enable bit in the Interrupt Enable register
is enabled, the
IRQ output will also go active low. Thus the IRQ
line can be used to signal the end of a calibration also. Another
calibration cycle will start as long as the CMODE bit in the
Mode register is set. Note that the result of the first calibration
is invalid and must be ignored. The result of all subsequent
calibration cycles is correct.
From Equations 5 and 11,
E t
VIdt
VI
fl
Hz
tdt
nT
nT
()
/.
cos (
)
00
12
8 9
2
+


ω
(13)
where n is an integer and T is the line cycle period.
Since the sinusoidal component is integrated over an integer
number of line cycles, its value is always zero.
Therefore:
E t
VIdt
nT
()
= ∫
0
(14)
E t
VInt
()
=
(15)
APOS [11:0]
23
0
AENERGY[39:0}
ACTIVE POWER
SIGNAL – P
LPF2
WAVEFORM [23:0]
39
11
0
SAGCYC[7:0]
CCCDh
CALIBRATION
CONTROL
ZERO CROSS
DETECT
LPF1
0
00h
CHANNEL 2
ADC
FROM
MULTIPLIER
Figure 34. Energy Calculation in Calibration Mode
CALIBRATING THE ENERGY METER
Calculating the Average Active Power
When calibrating the ADE7756, the first step is to calibrate
the frequency on CF to some required meter constant, e.g.,
3200 imp/kWh.
In order to determine the output frequency on CF, the average
value of the Active Power signal (output of LPF2) must first be
determined. One convenient way to do this is to use the calibra-
tion mode. When the CMODE (Bit 7) bit in the Mode register
is set to a Logic 1, energy is accumulated over an integer num-
ber of half-line cycles as described in the last section.
Since the line frequency is fixed at, say, 60 Hz, and the number
of half-cycles of integration is specified, the total integration
time is given as:
1
260
×
×
Hz
number of half cycles
For 255 half-cycles this would give a total integration time of
2.125 seconds. This would mean the energy register was updated
2.125/1.1175
µs (4/CLKIN) times. The average output value of
LPF2 is given as:
Contents of AENERGY
at the end
Number of times AENERGY
was updated
[: ]
[: ]
39 0
39 0
Or equivalently, in terms of contents of various ADE7756 regis-
ters and CLKIN and line frequencies (fl):
Average word LPF
AENERGY
fl
SAGCYC
CLKIN
()
[: ]
[: ]
2
39 0
8
70
=
××
×
(16)
where fl is the line frequency.
Calibrating the Frequency at CF
Once the average Active Power signal is calculated it can be used
to determine the frequency at CF before calibration. When the
frequency before calibration is known, the Calibration Frequency
Divider register (CFDIV) and the Active Power Gain register
(APGAIN) can be adjusted to produce the required frequency
on CF. In this example, a meter constant of 3200 imp/kWh is
chosen as an appropriate constant. This means that under a
steady load of 1 kW, the output frequency on CF would be,
Frequency CF
imp kWh
Hz
()
/
.
=
×
==
3200
60
60
3200
3600
0 8888
min
sec



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