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ADE7758
Voltage RMS Gain Adjust
The ADC gain in each phase of the voltage channel can be
adjusted for the rms calculation by using the voltage rms gain
registers (AVRMSGAIN, BVRMSGAIN, and CVRMSGAIN).
The gain of the voltage waveforms before LPF1 is adjusted by
writing twos complement, 12-bit words to the voltage rms gain
registers. Equation 11 shows how the gain adjustment is related
to the contents of the voltage gain register.
Content of VRMS Register ?
Data Sheet
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 (the low-pass filter) to obtain the
average active power information on each phase. Figure 65
shows this process. The active power of each phase accumulates
in the corresponding 16-bit watt-hour register (AWATTHR,
BWATTHR, or CWATTHR). The input to each active energy
register can be changed depending on the accumulation mode
setting (see Table 22).
Nominal RMS Values Without Gain ? ? ? 1 ?
VRMSGAIN ? (11)
?
?
?
2 12
0x19999A
INSTANTANEOUS
POWER SIGNAL
p(t) = VRMS × IRMS – VRMS × IRMS × cos(2 ω t)
For example, when 0x7FF is written to the voltage gain register,
the RMS value is scaled up by 50%.
0x7FF = 2047d
2047/2 12 = 0.5
Similarly, when 0x800, which equals –2047d (signed twos
complement), is written the ADC output is scaled by –50%.
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/sec. Equation 14 gives an expression for the instantaneous
power signal in an ac system.
v ( t ) = √2 × VRMS × sin( ωt ) (12)
ACTIVE REAL POWER
SIGNAL = VRMS × IRMS
VRMS × IRMS
0xCCCCD
0x00000
CURRENT
i(t) = 2 × IRMS × sin( ω t)
VOLTAGE
v(t) = 2 × VRMS × sin( ω t)
Figure 65. Active Power Calculation
Because LPF2 does not have an ideal brick wall frequency
response (see Figure 66), 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.
i ( t ) = √2 × IRMS × sin( ωt )
(13)
Because the ripple is sinusoidal in nature, it is removed when
where VRMS = rms voltage and IRMS = rms current.
p ( t ) = v ( t ) × i ( t )
p ( t ) = IRMS × VRMS ? IRMS × VRMS × cos (2ωt)
(14)
the active power signal is integrated over time to calculate the
energy.
0
The average power over an integral number of line cycles (n) is
given by the expression in Equation 15.
–4
? p ? t ? dt ? VRMS ? IRMS
p ?
where:
1 nT
nT 0
(15)
–8
–12
–16
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 component of the instantaneous power
signal p ( t ) in Equation 14, that is, VRMS × IRMS . This is the
–20
–24
relationship used to calculate the active power in the ADE7758
1
3
8 10
30
100
for each phase.
Rev. E | Page 30 of 72
FREQUENCY (Hz)
Figure 66. Frequency Response of the LPF Used
to Filter Instantaneous Power in Each Phase
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