Chapter 9 Installation and Wiring
9-2
4

) Power consumption block diagram of PLC systems

5) Power consumption of each part

(1) Power consumption of a power supply part
Approximately 65% of the power supply module current is converted into power 35% of that 65% dissipated as heat,
i.e., 3.5/6.5 of the output power is actually used.
• Wpw = 3.5 / 6.5 {(I5V x 5) + (I24V x 24)} (W)
where, l5v: 5VDC circuit current consumption of each part
l24v: 24VDC circuit average current consumption of output part (with points simultaneously switched ON).
Not for 24VDC power supplied from external or power supply part that has no 24VDC output.
(2) Total 5VDC power consumption
The total power consumption of all modules is the power of the 5VDC output circuit of the power supply part.
• W5V = I5V × 5 (W)
(3) Average DC24V power consumption (with points simultaneously switched ON)
The total power consumption of all modules is the average power of the DC24V output circuit of the power supply part.
• W24V = I24V × 24 (W)
(4) Average power consumption by voltage drop of output part (with points simultaneously switched ON)
• Wout = Iout × Vdrop × output points × the rate of points switched on simultaneously (W)
I
out : output current (actual operating current) (A)
V
drop : voltage dropped across each output load (V)
(5) Average power consumption of input parts (with points simultaneously ON)
• W
in = lin ×E × input points × the rate of points switched on simultaneously (W)
I
in : input current (effective value for AC) (A)
E : input voltage (actual operating voltage) (V)
Main Unit
power
supply
part
A
C power
Supply
CPU part input part
output part
(transistor)
special
module
Expansion module
I5V
External
24VDC
power
Supply
I24V
5VDC line
24VDC line Load Load
Output Current.
(IOUT)×Vdrop Input Current
(IIN)×Vdrop
Input part
Output part
(Transistor)
Output
Current
(IOUT)
Input
Current
(IOUT)
Input
Output Current.
(
IOUT)×Vdrop
Input Current
(IIN)×Vdrop
Output
Current
(IOUT)
Input
Current
(IOUT)