amp rating vs. inductive amp rating

ganutenator · Jan 9, 2021

What are the odds that a 2.0 amp relay output card will be able to handle the inductive load rating of aprox. 0.625A. (24VAC s.v. coil 15 watts)?

Is there a formula for converting amp ratings to inductive load?

lfe · Jan 9, 2021

In the manuals for that coil it should indicate the power factor, a number less than 1. Then:

Apparent current = Active current / power factor

Apparent current is what really circulates through the cable.
Active current is in your case 0.625A

Jobbe9000 · Jan 9, 2021

Check the manual of your relay card.

My guess is that it can handle the load, but there might be a derating in the number of switching cycles that it can do before worn out.

Not sure what you mean by amp rating to inductive load.. most AC equipment load is noted in VA when the reactive load is large enough to actually matter.

With that and the power factor, you can find the resistant load and the reactive load using cos sine calculations.

James Mcquade · Jan 9, 2021

it will handle the load, but not for very long. the contacts will weld together.
the surge current for that load (.625 amps) is typically 6-10 times that or 3.75 to 6.25 amps. the bigger issue is the breaking of the current, this is where the opening contacts try to arc (weld) together. even if you use a rc network in the terminals, i believe you will have issues from personal experience. would a triac output card work?
hope this helps,
james

Steve Bailey · Jan 9, 2021

I concur with James' observations. Solid state output modules generally have higher inrush ratings than contact output modules. But not all PLC models offer 24 VAC output modules. If yours does, a 2A, 24VAC output module would be a better choice than a contact output. If there aren't enough 24 VAC solenoids to justify a module strictly for them, consider an interposing device that would be easier to replace when it fails.
In any event you will improve your odds by installing suppression.

ganutenator · Jan 9, 2021

thanks all.
I've always used relays for this reason and others.
Only reason I asked is, this time space is an issue.
Totally remember now that the inrush current is the dealio w/ the s.v.
Triac's ring a bell.
Now to see if Wago offers triac output cards.

kallileo · Jan 10, 2021

As James Mcquade said the problem with inductive load is the wear on the contacts when they open.
Normally 15W should be handled easily by a 2Amps relay output but it depends on the number of operation per hour so you must to be careful.
I would definitely add a varistor on the load even if you decide to use a solid state output card.
Usually solenoid valves with AC coils are provided with voltage spike suppression built in to the connector.

Like the following which has both LED and varistor built in.
https://www.mandmcontrols.com/danfo...s-power-led-plus-varistor-indicator-connector

lfe · Jan 10, 2021

Both AC and DC coils accumulate energy that is released on disconnection with a voltage spike that can produce a spark in the relay contacts. This is how engine spark plugs work.

As other said for me the best for AC coils is to put a varistor on the coil terminals so the spike does not flow through the cables which by induction could affect other circuits

JesperMP · Jan 11, 2021

'Inductive' may mean many things.
Small solenoid - then a varistor or RC-snubber could be enough.
Large solenoid or a small spindle motor - then a proper contactor is the solution.
Many actuations - then an SSR is the solution.
If space is a concern, then consider a mini contactor like this one:
https://www.se.com/ww/en/product-range/714-tesys-sk/

amp rating vs. inductive amp rating

ganutenator

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