Philip Pemberton wrote:

> Olin Lathrop wrote:
>> 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... 

These calculations are for a long track. You can go smaller on a short
piece of track. It will of course become hotter at the smaller piece,
but the larger copper around it will be able to dissipate some of the
additional energy. Possibly the connector helps, too.

Remember that the clearance is a voltage thing and needs to be there
always. But the width of the track is a question of dissipating the
energy, and as long as you can dissipate the energy, you can have
smaller and wider parts in the track.

Gerhard
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