参数资料
型号: ADE7754ARZRL
厂商: Analog Devices Inc
文件页数: 17/44页
文件大小: 0K
描述: IC ENERGY METERING 3PHASE 24SOIC
标准包装: 1
输入阻抗: 370 千欧
测量误差: 0.1%
电压 - 高输入/输出: 2.4V
电压 - 低输入/输出: 0.8V
电流 - 电源: 7mA
电源电压: 4.75 V ~ 5.25 V
测量仪表类型: 3 相
工作温度: -40°C ~ 85°C
安装类型: 表面贴装
封装/外壳: 24-SOIC(0.295",7.50mm 宽)
供应商设备封装: 24-SOIC W
包装: 剪切带 (CT)
配用: EVAL-ADE7754EBZ-ND - BOARD EVALAUTION FOR ADE7754
其它名称: ADE7754ARZRLCT

ADE7754
If the VGAIN registers are used for apparent power calibration
(WATMOD bits in VAMODE register = 1 or 2), the voltage
rms values are changed by voltage gain register value as described
in the expression
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 compo-
nent of the instantaneous power signal p ( t ) in Equation 5 (i.e.,
VI ). This is the relationship used to calculate active power in the
? ?
? ?
Voltage rms register Phase A = ? rms × ? 1 +
? ? ?
?
?
AVGAIN ? ? ?
? ?
2 12
ADE7754 for each phase. The instantaneous power signal p ( t )
is generated by multiplying the current and voltage signals in
each phase. The dc component of the instantaneous power signal
in each phase (A, B, and C) is then extracted by LPF2 (low-pass
For example, when 7FFh is written to the voltage gain register,
the ADC output is scaled up by +50%. 7FFh = 2047d, 2047/
2 12 = 0.5. Similarly, 800h = –2047d (signed twos complement)
and ADC output is scaled by –50%. These two examples are
illustrated in Figure 21.
Voltage RMS Offset Compensation
filter) to obtain the active power information on each phase. This
process is illustrated in Figure 22. In a polyphase system, the total
electrical power is simply the sum of the real power in all active
phases. The solutions available to process the total active power
are discussed in the following section.
The ADE7754 incorporates a voltage rms offset compensation
INSTANTANEOUS
POWER SIGNAL
p(t) = V
I–V
I cos(2 t)
for each phase (AVRMSOS, BVRMSOS, and CVRMSOS).
These are 12-bit twos complement signed registers that can be
used to remove offsets in the voltage rms calculations. An offset
may exist in the rms calculation due to input noises and offsets
in the input samples. The offset calibration allows the contents
of the V RMS registers to be maintained at zero when no voltage
is applied.
1A36E2Eh
V. I.
D1B717h
ACTIVE REAL POWER
SIGNAL = V I
n LSB of the voltage rms offset are equivalent to 64
n LSB of
the voltage rms register. Assuming that the maximum value from
the voltage rms calculation is 1,898,124 decimal with full-scale
ac inputs, then 1 LSB of the voltage rms offset represents 0.07%
of measurement error at –26 dB below full scale.
V rms = V rms 0 + VRMSOS × 64
where V rmso is the rms measurement without offset correction.
The voltage rms offset compensation should be done by testing
the rms results at two non-zero input levels. One measurement
can be done close to full scale and the other at approximately
full scale/10. The voltage offset compensation can then be derived
from these measurements. See the Calibration of a 3-Phase
Meter Based on the ADE7754 Application Note AN-624.
ACTIVE POWER CALCULATION
Electrical power is defined as the rate of energy flow from source
00000h
CURRENT
i(t) = 2I sin( t)
VOLTAGE
v(t) = 2V sin( t)
Figure 22. Active Power Calculation
Since LPF2 does not have an ideal brick wall frequency
response (see Figure 23), 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.
Since the ripple is sinusoidal in nature, it is removed when the
active power signal is integrated to calculate the energy. See the
Energy Calculation section.
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/sec. Equa-
tion 5 gives an expression for the instantaneous power signal in
an ac system.
0
–4
–8
8Hz
v ( t ) = 2 V sin( ω t )
i ( t ) = 2 I sin( ω t )
where V = rms voltage and I = rms current.
p ( t ) = v ( t ) × i ( t )
p ( t ) = VI ? VI cos( 2 ω t )
(3)
(4)
(5)
–12
–16
–20
–24
The average power over an integral number of line cycles (n) is
given by the expression in Equation 6.
1
3
10
FREQUENCY (Hz)
30
100
P =
1
nT
nT
0
p ( t ) dt = VI
(6)
Figure 23. Frequency Response of the LPF Used
to Filter Instantaneous Power in Each Phase
REV. 0
–17 –
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