Rockwell Automation 1747-ACNR15 ControlNet Adapter Module User Manual - Manual (Page 97)

Publication 1747-UM003A-EN-P
Understanding Your SLC 500/1746 Control System B-5
Worksheet for Selecting a 1746 Power Supply
Make copies of this worksheet as needed. For a detailed list of device load
currents, refer to the SLC 500 price sheet, product instruction sheet, or
appropriate product data.
Consider future system expansion when selecting a power supply.
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slot _______
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Peripheral
Device
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slot _______
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slot _______
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slot _______
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Peripheral
Device
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Procedure
Required Power Supply for this Chassis: 1746-
1. For each slot of the chassis that contains a module, list the slot number, the catalog number of the module, and its 5V and 24V
maximum currents.
Chassis Number: _______ Chassis Number: _______
Catalog
Number Maximum Currents
5V 24V
2. Add the power supply loading currents of
all the system devices ( at 5V and 24V).
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3. Compare the Total Current required for the chassis with the Internal Current Capacity of the power supplies.
To select the proper power supply for your chassis, make sure that the power supply loading current for the chassis is less than the
internal current capacity for the power supply, for both 5V and 24V loads.
Internal Current Capacity
5V
24V
Catalog Number 1746-P1 2.0A | 0.46A
Catalog Number 1746-P2 5.0A | 0.96A
Catalog Number 1746-P3 3.6A | 0.87A
Catalog Number 1746-P4 10.0A | 2.88A (70W maximum)
Required Power Supply for this Chassis: 1746-
Catalog
Number Maximum Currents
5V 24V
Total Power
Total Current:
When using the 1746-P4 power supply, use the formula below to calculate
total power consumption of all the system devices (at 5V and 24V). Note that
the 1746-P4 total power supply loading currents cannot exceed 70 Watts. If
you are not using a 1746-P4 power supply, proceed to step 3.
Total Current:
Total current
at 5V Total current
at 24V Total Power Total current
at 5V Total current
at 24V User Current
at 24V
The user current at 24V listed below is for example only . The current
required depends on the application.
User Current
at 24V
24V) = W + 24V) x 5V) x x 5V) + x24V) + ( + x x 24V) = W
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