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ADE5166 数据表(PDF) 51 Page - Analog Devices

部件名 ADE5166
功能描述  Single-Phase Energy Measurement IC with 8052 MCU, RTC, and LCD Driver
PDF  148 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

ADE5166 数据表(HTML) 51 Page - Analog Devices

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Preliminary Technical Data
ADE5166/ADE5169/ADE5566/ADE5569
Rev. PrB | Page 51 of 148
PHASE COMPENSATION
The ADE5166/ADE5169/ADE5566/ADE5569 must work with
transducers that can have inherent phase errors. For example,
a phase error of 0.1° to 0.3° is not uncommon for a current
transformer (CT). These phase errors can vary from part to part,
and they must be corrected to perform accurate power calcula-
tions. The errors associated with phase mismatch are particularly
noticeable at low power factors. The ADE5166/ADE5169/
ADE5566/ADE5569 provide a means of digitally calibrating
these small phase errors. The part allows a small time delay or
time advance to be introduced into the signal processing chain
to compensate for small phase errors. Because the compensation
is in time, this technique should only be used for small phase
errors in the range of 0.1° to 0.5°. Correcting large phase errors
using a time shift technique can introduce significant phase
errors at higher harmonics.
The phase calibration register (PHCAL[7:0]) is a twos complement,
signed, single-byte register that has values ranging from 0x82
(−126d) to 0x68 (+104d).
The PHCAL register is centered at 0x40, meaning that writing
0x40 to the register gives 0 delay. By changing this register, the
time delay in the voltage channel signal path can change from
−231.93 μs to +48.83 μs (MCLK = 4.096 MHz). One LSB is
equivalent to a 1.22 μs (4.096 MHz/5) time delay or advance. A
line frequency of 60 Hz gives a phase resolution of 0.026° at the
fundamental (that is, 360° × 1.22 μs × 60 Hz).
Figure 37 illustrates how the phase compensation is used to
remove a 0.1° phase lead in the current channel due to the
external transducer. To cancel the lead (0.1°) in the current
channel, a phase lead must also be introduced into the voltage
channel. The resolution of the phase adjustment allows the
introduction of a phase lead in increments of 0.026°. The phase
lead is achieved by introducing a time advance into the voltage
channel. A time advance of 4.88 μs is made by writing −4 (0x3C)
to the time delay block, thus reducing the amount of time delay
by 4.88 μs, or equivalently, a phase lead of approximately 0.1° at a
line frequency of 60 Hz (0x3C represents −4 because the register is
centered with 0 at 0x40).
1
1
0
1
0
0
1
70
PGA1
IP/IPA
IN
I
ADC 1
HPF
24
PGA2
VP
VN
V
ADC 2
24
LPF2
V
I
60Hz
0.1°
I
V
CHANNEL 2 DELAY
REDUCED BY 4.48µs
(0.1°LEAD AT 60Hz)
0x0B IN PHCAL[7:0]
PHCAL[7:0]
–231.93µs TO +48.83µs
60Hz
1
1
DELAY BLOCK
1.22µs/LSB
Figure 37. Phase Calibration
RMS CALCULATION
The root mean square (rms) value of a continuous signal V(t) is
defined as
×
=
T
rms
dt
t
V
T
V
0
2
)
(
1
(1)
For time sampling signals, rms calculation involves squaring the
signal, taking the average, and obtaining the square root. The
ADE5166/ADE5169/ADE5566/ADE5569 implement this method
by serially squaring the input, averaging them, and then taking
the square root of the average. The averaging part of this signal
processing is done by implementing a low-pass filter (LPF3 in
Figure 38, Figure 40, and Figure 41). This LPF has a −3 dB cutoff
frequency of 2 Hz when MCLK = 4.096 MHz.
()
)
sin(
2
t
V
t
V
ω
×
=
(2)
where V is the rms voltage.
()
t
V
V
t
V
ω
=
2
cos
)
(
2
2
2
(3)
When this signal goes through LPF3, the cos(2ωt) term is attenu-
ated and only the dc term Vrms2 (shown as V2 in Figure 38) goes
through.
V
LPF3
INPUT
V2 (t) = V2 –V2 cos(2ωt)
V2 (t) = V2
V(t) = √2 × V sin(ωt)
Figure 38. RMS Signal Processing
The Irms signal can be read from the waveform register by setting
the WAVMODE register (0x0D) and setting the WFSM bit in
the Interrupt Enable 3 SFR (MIRQENH, 0xDB). Like the current
and voltage channels waveform sampling modes, the waveform
data is available at sample rates of 25.6 kSPS, 12.8 kSPS, 6.4 kSPS,
or 3.2 kSPS.
It is important to note that when the current input is larger than
40% of full scale, the Irms waveform sample register does not
represent the true processed rms value. The rms value processed
with this level of input is larger than the 24-bit read by the wave-
form register, making the value read truncated on the high end.



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