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Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>


I started with wider traces, but got scared of cross-over between
traces, narrowed them to 125 mils, put them on both sides of the board,
and increased weight to 3 oz copper(both sides).


>Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

>Hi Mark,

>These things can be somewhat difficult to calculate.  I didnt spend
>any time figuring out the scale factor on your trace widths and I
>assume the board is using 2 oz copper (being an outer layer).  Im
>thinking that what you have done is probably reasonable.  I believe
>there are two issues with pc layout - heat generated in the trace
>(and you need to know the Irms to figure this out), and metal
>migration (peak current is the issue here and resulting current
>densities).  I just dont have any numbers.  Im wondering if you ever
>turn the board if there is any reason to not fatten up the traces as
>much as possible.

>Gerry R.

>Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>
>Good question.  Maybe you can help me here.  My trace calculations are
>based on pure DC and I do not know how to correct for the high
>frequency.  What I did is come close to the cross-sectional area of the

>CDE942 leads so on that basis of that I should be able to push 500
>Now here is what I tried with math.  See if you think I'm on the right
>Based on simple DC trace calculations I have:
>24 amps give a 10 Deg C rise.
>38 amps a 38 Deg C rise.
>The above values are based on dual(top and bottom) 125 mil traces in 3
>oz copper.
>Now at 460 BPS I have 2130 uS "off-time" (cap charging time) and 43 uS
>"on-time" for ring-up and ring-down with my RSG(Note - Terry's testing
>suggests the "on-time" might dramatically increase).  So maybe we can
>compute the average current during the whole break cycle to consider
>the thermal effect.
>If we have 500 amps during the "on-time" and near zero during the
>"off-time" then the average current is 43/2173*500 = 10 amps avg
>Subsequent to my board design, Terry found the "on-time" might jump to
>150uS.  If we use that measure then 150/2173*500 = 35 amps avg current.
>Critique of this calc would be much appreciated.
> >Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
> >Hi Mark,
> >How much current do you estimate the traces (high current path) on
> >your layout can take??
>Gerry R.