参数资料
| 型号： | EVAL-ADE7759EBZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 20/36页 |
| 文件大小： | 0K |
| 描述： | BOARD EVALUATION FOR ADE7759 |
| 标准包装： | 1 |
| 主要目的： | 电源管理，电度表/功率表 |
| 已用 IC / 零件： | ADE7759 |
| 已供物品： | 板 |
| 相关产品： | ADE7759ARSZRL-ND - IC ENERGY METERING 1PHASE 20SSOP ADE7759ARSZ-ND - IC ENERGY METERING 1PHASE 20SSOP |
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ADE7759
V1P
HPF
0.4
V1
PGA1
ADC 1
20
0.3
V1N
V2P
20
LPF2
0.2
0.1
V2
PGA2
ADC 2
1
DELAY BLOCK
1.12 s/LSB
CHANNEL 2 DELAY
REDUCED BY 4.48 s
(0.1 LEAD AT 60Hz)
0.0
V2N
FCH IN PHCAL [7:0]
V1
V2
0.1
7
0
1 1 1 1 1 1 0 0
PHCAL [7:0]
–110 s TO +103 s
V2
V1
–0.1
–0.2
–0.3
60Hz
–0.4
54
56
58
60 62
64
66
60Hz
Figure 28. Phase Calibration
FREQUENCY – Hz
Figure 31. Combined Gain Response of the HPF and Phase
Compensation (Deviation of Gain in % from Gain at 60 Hz)
0.30
0.25
0.20
0.15
0.10
0.05
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 wave-
forms. 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/second. Equa-
tion 3 gives an expression for the instantaneous power signal in
an ac system.
0.00
–0.05
v ( t ) = 2 V ( ω t )
i ( t ) = 2 I sin( ω t )
(1)
(2)
–0.10
100
200
300
400 500 600 700
800
900
1000
where:
FREQUENCY – Hz
Figure 29. Combined Phase Response of the HPF and
V = rms voltage
I = rms current
Phase Compensation (100 Hz to 1 kHz)
p ( t ) = v ( t ) × i ( t )
p ( t ) = VI – VI cos( 2 ω t )
(3)
0.30
The average power over an integral number of line cycles (n) is
given by the expression in Equation 4.
∫ p ( t ) dt = VI
0.25
0.20
P =
1 nT
nT 0
(4)
0.15
0.10
0.05
0.00
–0.05
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 compo-
nent of the instantaneous power signal p(t) in Equation 3, i.e.,
VI . This is the relationship used to calculate active power in the
ADE7759. 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
–0.10
40
45
50
55 60
65
70
is illustrated in Figure 32. Since LPF2 does not have an ideal
“brick wall” frequency response (see Figure 33), the active power
FREQUENCY – Hz
Figure 30. Combined Phase Response of the HPF and
Phase Compensation (40 Hz to 70 Hz)
signal will have some ripple due to the instantaneous power
signal. This ripple is sinusoidal and has a frequency equal to twice
the line frequency. Since the ripple is sinusoidal in nature, it will
be removed when the active power signal is integrated to calcu-
late energy—see Energy Calculation section.
–20 –
REV. A
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| EVAL-ADE7816EBZ | 功能描述:BOARD EVALUATION FOR ADE7816 RoHS:是 类别:编程器，开发系统 >> 评估演示板和套件 系列:- 标准包装:1 系列:PSoC® 主要目的:电源管理，热管理 嵌入式:- 已用 IC / 零件:- 主要属性:- 次要属性:- 已供物品:板，CD，电源 |
| EVAL-ADE7854EBZ | 制造商:Analog Devices 功能描述:EVALUATION BOARDS - Boxed Product (Development Kits) |
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