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SA2002HPA 参数 Datasheet PDF下载

SA2002HPA图片预览
型号: SA2002HPA
PDF下载: 下载PDF文件 查看货源
内容描述: 单相双向功率/电能计量IC,具有瞬时脉冲输出 [Single Phase Bi-directional Power / Energy Metering IC with Instantaneous Pulse Output]
分类和应用: 脉冲
文件页数/大小: 12 页 / 149 K
品牌: SAMES [ SAMES ]
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SA2002H
TYPICAL APPLICATION
In figure 11, the components required for stand alone power
metering application, is shown. The application uses a shunt
resistor for the mains current sensing. The meter is designed
for 220V/40A I
MAX
operation. The most important external
components for the SA2002H integrated circuit are the current
sense resistors, the voltage sense resistors as well as the bias
setting resistor.
sames
VOLTAGE DIVIDER
The voltage divider is calculated for a voltage drop of 14V +
5%(14.7V). Equations for the voltage divider in figure 9 are:
RA = R1 + R2 + R3
RB = R12 || (R11+P1)
Combining the two equations gives:
(RA + RB) / 220V = RB / 14.7V
A 5k trimpot will be used in the voltage channel for meter
calibration. The center position on the pot is used in the
calculations. P1 = 2.5kW and values for resistors R11 = 22kW
and R12 =1MW is chosen.
Substituting the values will result in:
RB = 23.91kW
RA = RB x (230V/14.7V - 1) = 333kW so the resistor values of
R1, R2 and R3 are chosen to be 110kW.
BIAS RESISTOR
R13 defines all on-chip and reference currents. With
R13=24kW, optimum conditions are set. Device calibration is
done on the voltage input of the device.
SHUNT RESISTOR
The voltage drop across the shunt resistor at rated current
should be at least 20mV. A shunt resistor with a value of 625µW
is chosen. The voltage drop across the shunt resistor is 25mV
at rated conditions (Imax). The power dissipation in the current
sensor is:
P=(40A)² x 625µW = 1W.
CURRENT SENSE RESISTORS
The resistors R6 and R7 define the current level into the
current sense inputs of the device. The resistor values are
selected for an input current of 16µA on the current inputs of
the SA2002H at rated conditions. According to equation
described in the Current Sense inputs section:
R6 = R7 = ( I
L
/ 16µA ) x RSH / 2
= 40A / 16µA x 625µW / 2
= 781.2W
A resistor with value of 820W is chosen, the 5% deviation from
the calculated value will be compensated for when calculating
resistor values for the voltage path.
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