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PM2102DPD Datasheet(PDF) 6 Page - Sames

Part # PM2102DPD
Description  Energy Meter Evaluation Module
PDF  16 Pages
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Manufacturer  SAMES [Sames]
Direct Link  http://www.sames.co.za
Logo SAMES - Sames

PM2102DPD Datasheet(HTML) 6 Page - Sames

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PM2102DPD
http://www.sames.co.za
PRELIMINARY
SETUP EXAMPLES
The following examples can be used for the setup
conditions indicated.
Example 1
Example 2
Rated voltage:
220V
Rated current:
80A
Sensing element: SHUNT (80A, 50mV, 625µ )
Pulse constant:
800 imp/kWh
Motor constant:
100 imp/kWh
Calculate R6 and R7:
R6 = R7 = (IL / 16µA) x RSH / 2
= 80A / 16µA x 625µ
/ 2.
= 1.5625 k
A standard value of 1K6
is chosen.
The pulse constant is derived from the equation below:
Thus using table 3, we see that R0 = 1, R1 = 1, and R2 = 0.
Rated voltage:
220V
Rated current:
20A
Sensing element: SHUNT (80A, 50mV, 625µ )
Pulse constant:
3200 imp/kWh
Motor constant:
100 imp/kWh
Calculate R6 and R7:
R6 = R7 = (IL / 16µA) x RSH / 2
= 20A / 16µA x 625µ
/ 2.
= 390.625
A standard value of 390
is chosen.
W
W
W
W
W
W
W
W
LED p/kWh =
1160 p/sec x 1 / DF_LED) x [3600 /
((Vnom x Imax) / 1000)]
=
1160 p/sec x 1 / DF_LED) x [3600 /
(220 x 80) / 1000]
800 =
1160 p/sec x (1 / DF_LED ) x 204.54
1 / DF_LED =
0.0033717
DF_LED =
296
MOTOR p/kWh = LED p/kWh / DF_MO
100 = 800 / DF_MO
DF_MO = 8
The pulse constant is derived from the equation below:
MOTOR p/kWh = LED p/kWh / DF_MO
100 = 320 / DF_MO
DF_MO = 32
Thus using table 3, we see that R0 = 1, R1 = 0, and R2 = 0.
Rated voltage:
220V
Rated current:
40A
Sensing element: CT (80A max rated, 1:2500 low phase shift
internal impedance of 50 )
Pulse constant:
1600 imp/kWh
Motor constant:
100 imp/kWh
When a CT is to be used as the sensing element, the
terminating or burden resistor (R16) must also be calculated
and inserted.
R16 is dependent on the type of CT used and can normally be
found on a CT datasheet specified as "burden resistor, Rb".
The terminating or burden resistor for the TZ76V CT is 5.4
yielding 86.4mV at 40A rated.
Calculate R6 and R7:
R6 = R7 = (IL / 2500 / 16µA) x Rterminating / 2
= 40 / 2500 / 16µA x 5.4 / 2
= 2700
CT ratio = 2500
A standard value of 2K7
is chosen.
LED p/kWh =
60p/sec x (1 / DF_LED) x [3600/((Vnom x
Imax)/1000)]
=
1160 p/sec x (1 / DF_LED) x [3600 / (220 x
20) / 1000]
3200 =
1160 p/sec x (1 / DF_LED ) x 818.18
1 / DF_LED =
0.003371
DF_LED
=
296
IL= Line current
Example 3
W
W
W
W
Please note that there should ideally be a voltage drop of
86.4mV but not less than a
minimum of 20mV across the
burden resistor at I
.
ma
x



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