For example, suppose a relay contact drives a load at 120VAC, 1/2 A. Since this example has
an AC power source, first calculate the peak values:
Now, find the values of R and C:
If the contact is switching a DC inductive load, add a diode across the load as near to load
coil as possible. When the load is energized, the diode is reverse-biased (high impedance).
When the load is turned off, energy stored in its coil is released in the form of a negative-
going voltage spike. At this moment, the diode is forward-biased (low impedance) and shunts
the energy to ground. This protects the relay contacts from the high voltage arc that would
occur as the contacts are opening.
For best results, follow these guidelines in using a noise suppression diode:
• DO NOT use this circuit with an AC power supply.
• Place the diode as close to the inductive field device as possible.
• Use a diode with a peak inverse voltage rating (PIV) at least 100 PIV, 3A forward current or
larger. Use a fast-recovery type (such as Schottky type). DO NOT use a small-signal diode
such as 1N914, 1N941, etc.
• Be sure the diode is in the circuit correctly before operation. If installed backwards, it short-
circuits the supply when the relay energizes.
I
peak
= I
rms
x 1.414, = 0.5 x 1.414 = 0.707 Amperes
V
peak
= V
rms
x 1.414 = 120 x 1.414 = 169.7 Volts
R (
) =
C (µ F) =
10
I
2
V
10 x I
x
, where x=
50
V
1 +
=
10
0.707
2
= 0.05 µ F, voltage rating
170 Volts
x=
50
169.7
1 +
= 1.29
R (
) =
169.7
10 x 0.707
1.29
= 26
, 1/2 W, ± 5%
Chapter 3: I/O Wiring and Specifications
3–12
DL205 Installation and I/O Manual, 2nd Edition
3
