Go Down

### Topic: A general electronics question... (Read 1 time)previous topic - next topic

#### ReCreate

##### Dec 11, 2009, 03:27 am
I have this relay that is rated at .5 amps at 125 volts, How would that scale down to, say, 5 or 12 volts? How many amps?

#### PaulS

#1
##### Dec 11, 2009, 03:31 am
Rated at 0.5 amps @ 125 volts on the input side or the output side?

volts * amps = volts * amps

So, 0.5 amps at 125 volts would equal 5 amps at 12.5 volts or 12.5 amps at 5 volts.

#### ReCreate

#2
##### Dec 11, 2009, 03:32 am
In the output side. And thanks.

#### retrolefty

#3
##### Dec 11, 2009, 04:47 amLast Edit: Dec 11, 2009, 04:55 am by retrolefty Reason: 1
Quote
In the output side. And thanks

Trust me a relay with a contact rating of .5 amps will not switch a 12.5 amp circuit, no matter what the switching voltage is. They will quickly burn up.

When one talks of a relays output, it means the contact ratings. It's the maximum current and maximum voltage that the contacts are rated to switch on and off. Note that the contact current ratings can be different depending on if it switching AC or DC voltages. The current rating can not be scaled just because one lowers the voltage switched. A one amp contact rating should not be forced to flow more current just because the voltage is lowered. The manufacture will mark on the case or in their datasheet what the maximum current ratings are and at what voltage, and if AC or DC current maximums are different. Switching DC is harder on contacts then AC as the zero crossings in AC helps shorten the arcing time.

When one talks about a relays input, it means it's coil voltage and current requirement and if designed for AC or DC voltage.

A relays specifications are fixed and there is no real modification one can do to make it work with either input or ouput voltages it wasn't designed for.

Lefty

#### AWOL

#4
##### Dec 11, 2009, 08:58 amLast Edit: Dec 11, 2009, 10:14 am by AWOL Reason: 1
What Lefty said - imagining that your contacts can handle higher current just because the voltage is lower is a dangerous assumption, and one that will probably end up with your contacts welded shut or even simply burnt out.
(I2R and all that stuff)

#### Grumpy_Mike

#5
##### Dec 11, 2009, 04:01 pm
Just for the beginners reading this:-

if you contacts are rated at 0.5 amps that does not mean you have to switch 0.5 amps, it just means the contacts will handle current up to 0.5 Amps. Any more and you will damage the contacts.

#### elwing

#6
##### Dec 11, 2009, 04:31 pmLast Edit: Dec 11, 2009, 04:32 pm by elwing Reason: 1
hum, just to add some more details:

P = U*I everyone knows that

but your relay does not disipate 125*05=62,5Watts...

or if you prefer an other absurdity what about R = U/I?

that would give you a 62,5Ohms... whenever it's for the closed or openned state it sounds quite bad either...

it's obviously something else...

0,5amps means that when the contact is closed you can draw up to 0,5amps on your relay, while 125V means that when the contact is opened you can have up to 125V on the relay.
simple as that...

so, when commuting a 1Volt source you won't be able to draw more than 0,5amps without damaging your relay.

#### retrolefty

#7
##### Dec 11, 2009, 04:38 pm
Quote
hum, just to add some more details:

You got it!

Or to put it in other words:

Relay contacts are not resistors, they do not follow ohm's law calculation. Rather their ratings are mechanical safety type maximum limits that should not be exceeded. Welded or burned up contacts are not an unusual failure mode for relays when load problems or short circuits develop.

Lefty

#### elwing

#8
##### Dec 11, 2009, 04:48 pm
Quote
You got it!

Or to put it in other words:

Relay contacts are not resistors, they do not follow ohm's law calculation. Rather their ratings are mechanical safety type maximum limits that should not be exceeded. Welded or burned up

wrong, they are fully passive (the contact part I meant) they indeed follow the Ohm law.

the problem with theses two values (125V and 0,5A) is that they are not mesured at the same time, but with the relay in the two extreme condition, that is closed or openned, the resistance of the contact should be really low when the contact are closed, the voltage drop when closed should be under 1V (through it depend of the relay caracteristics). leading in a much more acceptable power dissipation of ~1V * 0,5A = 0,5Watts...

#### westfw

#9
##### Dec 11, 2009, 05:33 pm
Actually, I  think the current rating of a relay has more to do with the transitions between on and off than the current that can be carried in the "on" state; sometimes you can "cheat" if there won't be current trying to flow during the actual on/off events.  (but the conclusion is the same as retro/etc.)

Simple mechanical switches have similar ratings and similar issues.  As some people discover the first time they try to discharge a HV cap through such a switch...

#### ReCreate

#10
##### Dec 11, 2009, 07:06 pm
Woah woah woah people. I get it now.  OK OK... >_<

Go Up