RE: [CAD_CAM_EDM_DRO] Currect size relay for E-STOP

This is not correct, because you are using a "power" equation on a device
that doesn't use power.

The equations you cite are used for calculating the way max voltage and max
current are related for stuff that DOES dissipate power, like resistors,
etc.

An ideal relay has either infinite(contact open) or zero(contact closed)
resistance, so it will not use or dissipate ANY power. Even "Real" relays
have very low contact resistance, and so we can approximate their behavior
by assuming their resistance is zero.

This means that a relay either has "some" voltage and zero current across
it(when the contacts are open), or zero voltage and "some" current across
it(when the contacts are closed.

Relay current ratings are the maximum that the contacts can handle for
sustained current flow, AND still be able to "break" the current flow
without excessive arcing, which could cause the contacts to become welded
shut. They are rated as "Xamps at Y volts, because the contacts can only
reliably break that current if the voltage causing that current is within
the spec. For example, most relays have a much higher voltage spec for AC
than they do for DC, because with DC it's harder to extinguish the arc.

Relay contact ratings are the maximum that they can handle, without being
able to mathematically compute the other quantities using Ohm's law.

Sorry.

So in your example, the relay rated for 25A at 28VDC wikk NOT be a good
choice for use at a higher voltage, because even at rather low current
levels, you may experience arcing when you open the contacts, which can burn
the contacts, or even weld them together.

Andy Wander
Verrex Corporation


-----Original Message-----
From: Michael Gamber [mailto:mgamber@...]
Sent: Thursday, April 01, 2004 2:31 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: [CAD_CAM_EDM_DRO] Currect size relay for E-STOP


I intend to clamp a wire-wound resistor across the filter cap on my
68.7v DC PS.

I have 3 75v 2A servos running. Assume the purpose of this is to
absorb energy when the e-stop is pressed and the AC is cut to the PS.

If I use a relay rated at 25A @ 28v DC does this mean that it is
good for 10A @ 70v DC by using OHM's law?

P=E*I
28v * 25A = 700W

so

I=P/E

700W / 70v = 10A

Currect?

And is this enough, or do I need more?



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