Rockwell Automation 1747-L40 SLC 500 Fixed Hardware Style - Manual (Page 38)

Chapter 1
Selecting Your Hardware Components
1–19
Example
Increasing the load current by 100mA 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:
R (Ohms)
V (Volts)
I (Amps)
Resistor value (Ohms) = Applied voltage (Volts) / Desired current (Amps)
= 24 / 0.1
P (Watts) = I 2 (Amps) x R (Ohms)
Actual Power (Watts) = (Desired Current) 2 x Resistor Value
Resistor size = 2 x Actual power (Watts)
= 2 x 2.4
Round resistor size to 5 Watts.
= 2.4 (Watts)
=
24V = your applied voltage
Need 100mA of load current to reduce the transient to < 2.5 ms. (taken from graph on
previous page)
= 240 (Ohms)
= (0.1) 2 x 240
= 4.8 (Watts)
You need a resistor rated for 240 Ohms at 5 Watts to increase the load current
by 100mA; thus decreasing the transient time from approximately 7 ms to
less than 2.5 ms.