Olin Lathrop wrote:
> Track width requirement doesn't come from power, it comes from current.

Well, that makes sense... Ohm's Law in action.

> It seems most tables on the net show the temperature rise as a function of
> current and width, but not the resistance or voltage drop or some other way
> to compute the same thing.  Advanced Circuits has (did about a year ago at
> least) a calculator that does give you all these things.  If I remember
> right, their web site is www.4pcb.com.

Yep - Tips and Tools ==> trace width calculator, which links to 
circuitcalculator.com.

For 10A, 1oz copper, and a rise of 10C, that's coming up with 283 mils. 
That seems fairly sane... but may make it difficult to wire up the 
terminal block connectors I was intending to use:

   Input voltage is 240V AC RMS, or 1.414*240=339.41V peak. Round that 
to 340 to make the math easier.

   Per Art of Electronics page 841, "a good rule is 5 volts per mil". 
Let's run with that. 340/5=68mils minimum clearance (i.e. the distance 
from the centre of track1 to the centre of track2, minus half the width 
of track1, minus half the width of track2). Rounding this up to 70mil 
clearance...

   Working backwards, to keep a 70mil clearance with a pin pitch of 200 
mils, that means each track will be about 130 mils wide, at most. With 
1oz copper and 10C temperature rise, that's ~5.75A at most...

So it looks like I'll be using one pair of terminal blocks for each L 
contact, one pair for each N. I get the feeling a bag of 25 connectors 
might not go as far as I'd originally thought.

Oh well, at least I have a reasonable idea what I'm doing now... :)

Thanks,
-- 
Phil.
piclist@philpem.me.uk
http://www.philpem.me.uk/
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