Power Consumption Of Each Part; Current Consumption Of Power Module - Honeywell ML200 Series Installation And Commissioning Manual

Power consumption of each part

1. Current consumption of power module

The current conversion efficiency of the power module is around 70 %. Radiation consumes 30%, and
the current consumption is 3/7 of the output power. The calculation is as follows:
W = 3/7 {(I5V X 5) + (I24V X 24)} (W)
pw
I (DC Current consumption of DC 5V circuit of each module (internal current
5V) consumption).
I (DC Average current consumption of DC 24V of output module (current
24V) consumption of simultaneous ON point).
If DC 24V is externally supplied or a power module without DC 24V is used, it is not applicable.
2. Sum of DC 5V circuit current consumption: The DC 5V output circuit current of the power
supply module is the sum of current consumption of each module.
W 5V = I5V X 5 (W)
3. DC 24V average current consumption (current consumption of simultaneous ON point): DC
24V output circuit average current of power module is the sum of current consumption of each
module.
W 24V = I24V X 24 (W)
4. Average current consumption by output voltage drop of output module (current consumption of
simultaneous ON point).
i. W
out
ii. I : Output current (current in actual use) (A)
out
iii. V
drop
5. Input average current consumption of input module (current consumption of simultaneous ON
point).
Chapter 2 - Calculating the power supply for components
= I X V X Output point X Simultaneous ON rate (W)
out drop
: Voltage drop of each output module (V)
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