Saw this in the specs of a momentary switch :
Note: AC Inductive Load PF = 0.6 - 0.7 DC Inductive Load L/R = 7 msec
I suppose I should be ashamed, but... Someone care to tell me what "PF" and "L/R" mean here ?
-- André Majorel "After 15 minutes, I wanted to marry her. After a half hour, I completely gave up the idea of snatching her purse." -- _Take the Money and Run_
Andre Majorel wrote in news: snipped-for-privacy@atc5.vermine.org:
PF might be power factor. Not sure about L/R.
Lostgallifreyan wrote in news:Xns9B7565A9E6395zoodlewurdle@216.196.109.145:
A bit of Gooling suggests that L/R is just inductance over resistance, in analysis of the inductive load the switch is meant to be used for.
That is exactly right, as consider a dc suppy to an inductive relay coil, the operating switch has to be able (reliably) to break the coil current. But being inductive, there is a much greater amount of arcing at the switch contacts. So a contact, or switch must be rated for its voltage, current carrying capacity, and its load breaking ability (especially when there is inductance in a DC circuit). And of course, the greater the L/R ratio, and higher voltages, then greater arcing will occur, to the extent that the contact nay not be able to break the arc at all if it does not have an adequate rating.
Peter
Yup.
Nice to see it so well explained. I've never seen L/R specified before but I'm not into heavy power where it's likely a more significant issue.
Graham
Eeyore wrote in news: snipped-for-privacy@hotmail.com:
Actually, you might. :) You're into audio amps, right? Toroidal transformers are a classic switched inductive load. The main reason it's not much of an issue is that it's cheap to overspecify a switch on that scale. Or more likely, easy to just do what others have done. Which is not easy if you're switching a new design of industrial motor...
I liked that explanation too. Helps to clarify it. Closest I knew of this before is that a coil of thin wire won't arc so bad because its own resistance is a strong factor in what happens when the circuit is broken. But large amplifiers have windings just heavy enough to be important to consider. Though a few tests of a real switch are as good a way to find out as calculations, maybe. If a 1A current switches with a crack like a 10A current I'll use a 10A switch.
Power Factor?
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