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Solution:
TIS: 6 * 5W = 30W
TMS: 3 * 10 * 5W = 150W
We need a power supply with a capacity of min. 180W.
The power supply 48V/10A can exactly produce 480W (branches
533W out of the net), but there must be a reasonable reserve of
20%. The internal power supply is sufficient in this case.
This example is a basic power calculation in order to estimate.
Larger installations need a more detailed calculation that takes into
account the cable length of the branches
Loss of voltage
The voltage of an interface unit may not be lower than 36V. To
calculate the loss of voltage one must take into account the number
of interface units, the devices connected to it and the length of the
cables (with its losses) connecting the interface units.
A program, PowerCalc, has been created to calculate this. The
program asks some details: the power supply, the number of
interface units, the connected devices, the branch and the distance
between the interface units. The program calculates the voltage at
each interface unit and the total cable length. If the voltage
becomes lower than 36V, a new branch must be created from the
central.
PS5500
Description
The power supply PS5500 is a 19" rack mount version. The front
panel is equipped with a rocker mains switch with built-in indicator
and two mains fuses.
The PS5500 is available in different versions, where the output
voltage is 48V and the output current is 10A or 20A.
The use of an additional power supply is needed when the internal
power supply of the CPU5500 is insufficient, or the need of more
than the available branches is needed (and therefore an extension
unit is needed).
Features & Controls
Front panel mounted mains fuses.
On/Off rocker switch with built-in indicator.
48 V DC LED indicator.
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