ade7166 Analog Devices, Inc., ade7166 Datasheet - Page 58

• Manufacturer Part Number: ade7166
• Description: Single-phase Energy Measurement Ic With 8052 Mcu, Rtc, And Lcd Driver
• Manufacturer: Analog Devices, Inc.

ADE7566/ADE7569/ADE7166/ADE7169
Voltage Channel RMS Calculation
Figure 62 shows details of the signal processing chain for the
rms calculation on the voltage channel. The voltage channel
rms value is processed from the samples used in the voltage
channel waveform sampling mode and is stored in the unsigned
24-bit V
The update rate of the voltage channel rms measurement is
MCLK/5. To minimize noise in the reading of the register, the
V
crossing of the voltage input. This configuration is done by
setting the ZXRMS bit in the MODE2 register (0x0C).
With the specified full-scale ac analog input signal of 0.4 V, the
output from the LPF1 in Figure 62 swings between 0x28F5 and
0xD70B at 60 Hz (see the Voltage Channel ADC section). The
equivalent rms value of this full-scale ac signal is approximately
0d1,898,124 (0x1CF68C) in the V
measurement provided in the ADE7566/ADE7569/ADE7166/
ADE7169 is accurate to within ±0.5% for signal input between
full scale and full scale/20. The conversion from the register
value to volts must be done externally in the microprocessor
using a V/LSB constant.
Voltage Channel RMS Offset Compensation
The ADE7566/ADE7569/ADE7166/ADE7169 incorporate a
voltage channel rms offset compensation register (VRMSOS).
This is a 12-bit signed register that can be used to remove offset
in the voltage channel rms calculation. An offset can exist in the
rms calculation due to input noises and dc offset in the input
samples. One LSB of the voltage channel rms offset is equivalent
to 64 LSBs of the rms register. Assuming that the maximum
value from the voltage channel rms calculation is 0d1,898,124
with full-scale ac inputs, then 1 LSB of the voltage channel rms
offset represents 3.37% of measurement error at −60 dB down
of full scale.
where V
ACTIVE POWER CALCULATION
Active power is defined as the rate of energy flow from source
to load. It is the product of the voltage and current waveforms.
The resulting waveform is called the instantaneous power signal
and is equal to the rate of energy flow at every instant of time.
The unit of power is the watt or joules/second. Equation 10 gives an
expression for the instantaneous power signal in an ac system.
where:
v is the rms voltage.
i is the rms current.
rms
register can also be configured to update only with the zero
V
v
t i
p
p
( )
rms
( )
) (
) (
t
t
t
rms0
rms
= V
=
=
=
=
register.
is the rms measurement without offset correction.
VI
v
2
rms0
2
) (
t
−
×
×
VI
+ 64 × VRMSOS
I
×
V
sin(
t i
cos(
) (
sin(
ω
2
ω
t
ω
)
t
)
t
)
rms
register. The voltage rms
Rev. A | Page 58 of 144
(10)
(7)
(8)
(9)
The average power over an integral number of line cycles (n) is
given by the expression in Equation 11.
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 component of the
instantaneous power signal p(t) in Equation 11, that is, VI. This
is the relationship used to calculate active power in the ADE7566/
ADE7569/ADE7166/ADE7169. 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 in
Because LPF2 does not have an ideal brick wall frequency
response (see Figure 64
ripple due to the instantaneous power signal. This ripple is
sinusoidal and has a frequency equal to twice the line frequency.
Because of its sinusoidal nature, the ripple is removed when the
active power signal is integrated to calculate energy (see the
Active Energy Calculation section).
0xCCCCD
0x19999A
0x00000
P
–12
–16
–20
–24
–4
–8
VI
=
0
1
nT
1
INSTANTANEOUS
POWER SIGNAL
∫
CURRENT
i(t) = √2 × i × sin(ωt)
0
nT
Figure 64. Frequency Response of LPF2
p
Figure 63. Active Power Calculation
) (
VOLTAGE
v(t) = √2 × v × sin(ωt)
t
dt
3
), the active power signal has some
=
VI
FREQUENCY (Hz)
Figure 63.
p(t) = v × i – v × i × cos(2ωt)
10
ACTIVE REAL POWER
SIGNAL = v × i
30
1
00
(11)