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| 型號： | ADE7169ACPF16-RL |
| 廠商： | ANALOG DEVICES INC |
| 元件分類： | 模擬信號調(diào)理 |
| 英文描述： | Single-Phase Energy Measurement IC with 8052 MCU, RTC and LCD driver |
| 中文描述： | ANALOG CIRCUIT, QCC64 |
| 封裝： | 9 X 9MM, MO-220VMMD, LFCSP-64 |
| 文件頁數(shù)： | 56/140頁 |
| 文件大?。?/td> | 1359K |
| 代理商： | ADE7169ACPF16-RL |
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ADE7169F16
Preliminary Technical Data
Rev. PrD | Page 56 of 140
LINECYC
VALUE
CYCEND IRQ
LWATTHR REGISTER
Figure 43. Energy Accumulation when LINECYC changed
From Equations 13 and 18,
E
(
t
) =
∫
0
∫
0
+
nT
nT
f
VI
dt
VI
2
cos
9
1
(2
π
ft
)
dt
(20)
where:
n
is an integer.
T
is the line cycle period.
Since the sinusoidal component is integrated over an integer
number of line cycles, its value is always 0. Therefore,
E
=
∫
nT
VIdt
0
+ 0
(21)
E
(
t
) =
VInT
(22)
WDIV[7:0]
WATTOS[15:0]
WGAIN[11:0]
LPF1
+
+
LWATTHR [23:0]
ACCUMULATE ACTIVE
ENERGY IN INTERNAL
REGISTER AND UPDATE
THE LWATTHR REGISTER
AT THE END OF LINCYC
HALF LINE CYCLES
OUTPUT
FROM
LPF2
FROM VOLTAGE
CHANNEL
ADC
23
0
LINCYC [15:0]
48
0
%
ZERO CROSS
DETECTION
CALIBRATION
CONTROL
TO
DIGITALTO FREQUENCY
CONVERTER
Figure 44. Line Cycle Active Energy Accumulation
Note that in this mode, the 16-bit LINCYC register can hold a
maximum value of 65,535. In other words, the line energy
accumulation mode can be used to accumulate active energy for
a maximum duration over 65,535 half line cycles. At 60 Hz line
frequency, it translates to a total duration of 65,535/120 Hz =
546 seconds.
REACTIVE POWER CALCULATION
Reactive power is defined as the product of the voltage and
current waveforms when one of these signals is phase-shifted by
90°. The resulting waveform is called the instantaneous reactive
power signal. Equation 25 gives an expression for the instanta-
neous reactive power signal in an ac system when the phase of
the current channel is shifted by +90°.
v
(
t
) =
)
sin(
2
θ
+
ω
t
V
(23)
i
(
t
) =
)
sin(
2
t
I
ω
π
2
+
ω
=
′
sin
2
)
(
t
I
t
i
(24)
where:
θ is the phase difference between the voltage and current
channel.
V
is the rms voltage.
I
is the rms current.
q
(
t
) =
v
(
t
) ×
i
’(
t
)
(25)
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