I've got a relay rated 1 A at 125 VAC. I need to switch around 6.5 A at 6 VDC. So is my math right:

(125 * 0.636) = 79.5 VDC
79.5 / 6 = 13.25

So the relay can handle 13.25 A @ 6 VDC?

Dan East

 

That's totally wrong.

Where did you get that formula from it's total rubbish!

Current is current, voltage is voltage, theres no calculation to convert one to the other!

Normally the DC current rating for a relay is lower than the AC (arcing forms more easilly with DC), so your 1A relay won't switch 6.5A DC. Buy a bigger relay rated to at least 6.5A.

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Huh? I thought wattage was wattage and voltage was voltage, but amperage was specific to voltage.

So 1 A at 125 VAC is 125 Watts. 6 A at 6 VCD is only 36 Watts. So a relay should be able to handle higher amperage at a lower voltage, right? I was also assuming that there was a conversion from VAC to VCD requiring the RMS since VAC is a sine wave.

Dan East

 

Okay, after I think about it more, I see what you're talking about. Amperage is the amount of electrons that flow (independant of how energetic they are). So you're saying that the points in a relay can only handle a certain flow of electrons, regardess of their energy level. Okay, that makes sense. I was thinking it was power based, which is why I was converting the amperage.

Dan East

 

The size (diameter) of a wire is determined by the current - not the voltage. The voltage determines the insulation requirements.

In a relay or any other contactor you have current carrying conductors that bring the power to the contacts - then the contacts. In a closed state the current most likely governs the contact characteristics - as well as the size of conductors that bring the current to the contacts. Current and voltage govern what happens in the moments after opening and before closing.

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stevez

 

Exactly, the 1A AC relay probably won't even switch 1A DC - he needs to buy a properly specced relay.

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Note also that a relay used to operate an AC motor (possibly other loads) should be horsepower rated. What that means is that the voltage rating, current rating and HP rating of the relay must meet or exceed your requirements.

Now, for some applications many of us just use what we have and hope for the best. That isn't done without some understanding that the consequences aren't distasterous or life threatening. What we've tossed around here is what might be considered good design practice. Good design practice ought to yeild a safe, reliable, predictable system. In your application you could overheat something by running too much current. If it's on your bench as part of an experiment it might not be a big deal - if you install it somewhere and the overheat condition causes a fire - that's something different.

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stevez

 

Your voltage x current= power calculations were for the load, not for the switch.

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