Philip Pemberton wrote:
> 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. 
>   
Do you really need the rise that low?
> 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. 
If you are only dealing with a pair of terminals the pads don't have to 
go in the middle of the tracks.
> With 
> 1oz copper and 10C temperature rise, that's ~5.75A at most...
>   
but if you allow a 20C temperature rise it's presumably about double 
that. Can your really not afford a 20C rise?

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