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PM4100APD 参数 Datasheet PDF下载

PM4100APD图片预览
型号: PM4100APD
PDF下载: 下载PDF文件 查看货源
内容描述: 电能表评估模块 [Energy Meter Evaluation Module]
分类和应用:
文件页数/大小: 14 页 / 645 K
品牌: SAMES [ SAMES ]
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PM4100APD
sames
The capacitor C1 is inserted to compensate for the phase shift
caused by the current transformer. The following equation
shows how to calculate the capacitor value for a phase shift of
0.18 degrees.
C = 1 / (2 x
p
x Mains frequency x R8 x tan(Phase shift angle))
C = 1 / (2 x
p
x 50 x 1M x tan(0.18 degrees))
C = 1.013µF
CTs with a low phase is recommended.
CURRENT SENSE INPUT
Resistors R2 and R3 (figure 3 and figure 4) define the current
level into the current sense inputs of the device. The
PM4100APD module can be configured to use any type of
shunt or CT, but to ensure proper current sensing it is advisable
to use a shunt or CT that will give a voltage drop of at least 20mV
at maximum supply current. The resistor values are calculated
for an input of 16µA
RMS
on the current sense inputs at rated
conditions. The design equation is as follows:
R2 = R3 = (I / 16µA) x R
SH
/ 2
L
SHUNT Sensing
I
L
= Line current
R
SH
= Shunt value
CON2
3
2
1
CURRENT
R2
R3
IIN
IIP
GND
REFERENCE VOLTAGE
The on chip reference current is determined from a biasing
resistor R4 connected between pin 3 of the device and V
SS
.
This is 47kW for the SA2102D and SA4102A, and 24kW for the
rest of the devices compatible with this module.
ANALOG GROUND (GND)
The GND pin of the device is connected to the analog ground
plane, which is halfway between V
DD
and V
SS
.
Figure 3: Current sensing using a shunt
SENSING METHOD (CT OR SHUNT)
Provision is made for the use of either a Current Transformer
(CT) or a Shunt on the application module. For use with a CT,
the terminating resistor (R1) and phase compensation
capacitor (C1), as well as jumper (J1) must be inserted. The
jumper references one side of the CT to ground for sensing.
When using a CT the current sensing resistors (R2 and R3)
must be inserted according to the equation stated below.
R2 = R3 = (I
L
/2500/16µA) x R
1
/2
I
L
= Line current
1:2500 = CT ratio
CT Sensing
CON2
3
2
1
CURRENT
GND
J1
CT
R1
CT ONLY
R2
R3
IIN
IIP
Figure 4: Current sensing using a CT
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