ADE5166_08 AD [Analog Devices], ADE5166_08 Datasheet - Page 53

• Manufacturer Part Number: ADE5166_08
• Description: Single-Phase Energy Measurement IC with 8052 MCU, RTC, and LCD Driver
• Manufacturer: AD [Analog Devices]

Voltage Channel RMS Calculation
Figure 58 shows details of the signal processing chain for the
rms calculation on the voltage channel. This voltage rms esti-
mation is done in the ADE5166/ADE5169 using the mean
absolute value calculation, as shown in Figure 58. 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 (Bit 2) in the MODE2 register (Address 0x0C).
With the specified full-scale ac analog input signal of 0.5 V, the
output from the LPF1 in Figure 58 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 ADE5166/ADE5169 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 ADE5166/ADE5169 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 8 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
t i
v
p
p
( )
( )
rms
) (
(
t
t
t
rms0
)
= V
=
=
=
=
is the rms measurement without offset correction.
VI
v
2
rms0
2
(
t
×
−
×
)
VI
+ 64 × VRMSOS
I
V
×
rms
sin(
t i
cos(
sin(
(
register.
)
ω
2
ω
t
ω
)
t
)
t
)
rms
register. The voltage rms
Rev. 0 | Page 53 of 148
(5)
(6)
(7)
(8)
The average power over an integral number of line cycles (n) is
given by the expression in Equation 9.
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 9, that is, VI. This
is the relationship used to calculate active power in the ADE5166/
ADE5169. 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 (see Figure 59).
Because LPF2 does not have an ideal brick wall frequency response
(see Figure 60), 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. 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
VI
–4
–8
=
0
1
nT
1
INSTANTANEOUS
POWER SIGNAL
∫
CURRENT
i(t) = √2 × i × sin(ωt)
0
nT
Figure 60. Frequency Response of LPF2
p
Figure 59. Active Power Calculation
) (
VOLTAGE
v(t) = √2 × v × sin(ωt)
t
dt
3
=
VI
FREQUENCY (Hz)
p(t) = v × i – v × i × cos(2ωt)
10
ADE5166/ADE5169
ACTIVE REAL POWER
SIGNAL = v × i
30
100
(9)