Hi all,

I have here a personal project involving switching Canadian mains power (nominally 120V @ 60hZ, probably operating on a 15A circuit).

I have some relays which are all marked with both "5A/250VAC" and "5A/30VDC". I want to control the power supply to a laser printer with a relay (solid state or not, I don't care; I just thought I'd see if the ones I have here would work).

I am thinking that the 5A current rating on the contacts holds regardless of the voltage. So while the thing could supply a maximum 1250W load at 250VAC, it could manage at best a 600W load at 120VAC. Is this correct?

Also, with electromechanical relays, should I derate the figures a bit for safety? Say, down to 300 or 400W max. for a laser printer, to account for startup current in the motors and fuser etc?

The particular printer in question is in my sleeping sister-in-law's room so I cannot check its wattage rating until tomorrow.


Torben

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I don't think I am qualified to answer (Considering the importance) But that is the way I was taught. Power is the issue...

Always derate. Margin, Margin, Margin...

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That was the first thing I thought but then I figured that I want confirmation. Taking the power as the max rating would seem to indicate that the relay is rated to conduct 10.4A at 120V. If I assume that the contacts are limited to 5A regardless of voltage then at least I err on the side of not starting a fire. That's why I want to get a solid answer on this.

I'm a big fan of overbuilding. I don't like rebuilding.


Torben

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Yes and No.

The 5A rating will hold for AC voltage up to 250V.

However, 5A rating holds for DC voltage up to 30V DC only. i.e. one should not use it to switch a DC load of 5A at 35V DC.

Hope it is clear.

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L.Chung

 

Thanks for the reply! Yes, I understand that part. I'm just wondering about how to safely convert the amperage ratings for different AC voltages.

I mean, you say the 5A rating will hold up to 250VAC--but does that mean that I should still not exceed 5A at 120VAC, or am I OK up to 10.4A at 120VAC? I believe it's the former but I would like confirmation.


Torben

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Curiosity was framed. Ignorance killed the cat.

 

Your caution is advisable. I would wait for several responses to get a good sample then take the average mean of all input...

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I think I'd take a heavily weighted average.


Torben

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Curiosity was framed. Ignorance killed the cat.

 

The maximum AC rating is probably determined by the physical size of the contact area and connection wire.

The maximum DC rating is probably determined by the contact separation distance, speed of separation and whether there is any help to extinguish the arc from external means like using permanent magnet. In most cases, this will be less than the AC current rating.

So, one cannot switch 50A at 10V AC using a 5A 110V rated relay. The contact will simply melt.

A 5A rated relay can only used for 5A maximum from 0V to rated AC voltage.

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L.Chung

 

I think if we stick to AC voltage, then Torben's thinking is on the right track. I also think for this case, we should discount DC as it does not seem apply to Torbens application.
That is not to say DC specs are not important, but if I understand correctly, Torben is concerned with AC.

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hi Torben

All you'll ever need to know about relays is here:

http://pewa.panasonic.com/pcsd/tech_info/pdf/rti.pdf

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Eric
"Good enough is Perfect"

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Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

No, don't convert! As L.Chung states:

I think this is correct, and seems to be in accordance with the excellent pdf Eric posted. (Thanx! BTW)

It seems that higher voltages need smaller conductors to carry the same current, so it is logical that the contacts can still switch at 250V, the same current they switch at 120V.

If you need to switch more current, then change relays. The pdf points out that the power factor must be taken into consideration, so what kind of load is a switching supply? I think all of the motors, solenoids, etc. in a printer are run from the printer's psu, which will take care of their particular start up loads...

I like brick sh*t houses; build once, run forever.

 

OK, great! I mean, it would be even greater if my relays here could handle this printer, but no go. The ratings labels on the back are in Chinese (I think) but the ratings themselves state 110-127V, 6A, 600W print. I think I'll order some 8 or 10 A relays instead.

Thanks for all the replies (and for the awesome PDF, Eric!)


Torben

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Curiosity was framed. Ignorance killed the cat.

 

L.Chung posted the simplest, most accurate response possible. It's a 5A relay.

 

Cool--that's what I got from what he said as well. My relays are not up to the task.

Luckily, I remembered this afternoon that I have the igniter circuit board from an RV furnace which includes an Omron G8P-1A4-CF 9VDC relay, which at 15A/125VAC believe should be more than adequate for switching power to a 600W 6A laser printer, despite the fact that from what I gather it's nowhere near being a nice easy resistive load.

I intend to fit an RC network across the output lines to the load and a diode across the coil. The PDF that Eric posted gives the following hint for sizing the RC network components:

. . .so it seems like 100Ω and 4.7µF would be good starting points, but I think I may need to change them since the load isn't purely resistive.


Torben

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Curiosity was framed. Ignorance killed the cat.