GE 90-30 PLC Series Installation And Hardware Manual page 391

F
solenoid draws 1.0 Amp. The cylinder advances and retracts once every 60 seconds that the
machine is cycling. It takes 6 seconds to advance and 6 seconds to retract.
Since the cylinder takes equal time to advance and retract, both solenoids are on for equal lengths
of time: 6 seconds out of every 60 seconds, which is 10% of the time. Therefore, since both
solenoids have equal current draws and on-times, our single calculation can be applied to both
outputs.
Use the formula Average Power Dissipation = Voltage Drop x Current Draw (in Amps) x Percent
(expressed as a decimal) of on-time:
Then multiply this result by 2 since we have two identical solenoids:
Also in this example, the other 14 output points on this 16-point module operate pilot lights on an
operator's panel. Each pilot light requires .05 Amps of current. Seven of the pilot lights are on
100% of the time and seven are on an estimated 40%.
For the 7 lights that are on 100% of the time:
Then multiply this value by 7:
For the 7 lights that are on 40% of the time:
Then multiply this value by 7:
Adding up the individual calculations, we get:
Step 4: Input Calculations for Discrete Input Modules
A Discrete Input Module requires two calculations, one for the module's signal-level circuits,
which was already done in Step 1, and one for the input circuits. Note that the power dissipated by
the input circuits comes from a separate power source, so are not included in the figure used to
calculate PLC power supply dissipation in Step 2. We will assume that all input circuit power
delivered to these modules is eventually dissipated as heat. The procedure is:
F-4 Series 90-30 PLC Installation and Hardware Manual – August 2002
1.5 x 1.0 x 0.10 = 0.15 watts per solenoid
0.15 watts x 2 Solenoids = 0.30 watts total for the two solenoids
1.5 x .05 x 1.00 = 0.075 watts per light
0.075 watts x 7 lights = 0.525watts total dissipation for the first 7 lights
1.5 x .05 x 0.40 = .03 watts per light
0.03 watts x 7 lights = 0.21 watts total dissipation for the other 7 lights
0.30 + 0.525 + 0.21 = 1.035 watts for the module's total output calculation
Find the value for the Input Current in the "Specifications" table for your input module in
the Series 90-30 I/O Module Specifications Manual, GFK-0898.
Multiply the input voltage times the current value times the estimated percent of on-time
to arrive at average power dissipation for that input.
GFK-0356Q