Rockwell Automation 1747-L5xx SLC 500 Modular Hardware Style User Manual - Manual (Page 53)

Publication 1747-UM011G-EN-P - June 2008
Selecting Your Hardware Components 53
EXAMPLE
Increasing the load current by 100 mA decreases the transient
time from approximately 7 ms to less than 2.5 ms. To calculate
the size of the resistor added in parallel to increase the current,
use the following information:
24V = your applied voltage
Need 100 mA of load current to reduce the transient to <2.5 ms.
(taken from graph).
R (W) = V (Volts)/I (Amps)
Resistor value (Ohms) = Applied voltage (Volts)/Desired current
(Amps) = 24/0.1 = 240 W
P (Watts) = 1
2
(Amps) x R (W)
Actual Power (Watts) = (Desired Current)
2
x Resistor Value =
(0.1)2 x 240 = 2.4 (Watts)
Resistor size = 2 x Actual power (Watts) = 4.8 W =
approximately 5 W
Use a resistor rated for 240 W at 5 W to decrease the transient
time from approximately 7 ms to less than 2.5 ms.