Please remember to use mulitple layers of copper with a ring of vias around
your current carrying pin AND the return pin (s). Most people forget that
there is .0007-.0012" of connection on a single layer. SO if your power
plane is on a single layer, put a circle of plane (copper) around several
layers encompassing as many vias that are appropriate for your application.
You also need to realize that the thermal spokes are for removing the
through hole pin and hand soldering. If your application gets thrown away
or never repaired you could have a direct connection. The reflow oven
heats the entire assembly and the thermal spokes are not necessary for
assembly.
Karl
From: David Baldwin <[log in to unmask]>
Reply-To: "(Designers Council Forum)" <[log in to unmask]>,
David Baldwin <[log in to unmask]>
To: [log in to unmask]
Subject: Re: [DC] Current Capacity of Thermal Reliefs
Date: Tue, 8 May 2007 15:52:55 -0700
Bill:
Thanks for the feedback.
PCB Libraries has a lot of good documentation (which is to IPC Standards) on
the thermal size, as well as a through hole calculator to calculate it for
you. It's all good stuff.
The formula for calculating the via circumference is 2 x pi (3.14159) x r}.
The main issue is when I need to handle 30 amps, and I have to have a
thermal relief, how do I know it can handle the current required? I can add
up the spokes and say I have a 200 mil (oops, 5.08mm) wide trace, but since
it only is 0.010" (.25mm) long, then there should be a factor which states
it can handle more current due to the extremely short distance. I did
exactly what you said, but how can I be sure? I'm taking my best educated
experience and I believe everything is fine, but a tool to confirm things
would be nice.
Thanks,
Dave
At 03:39 PM 5/8/2007, Brooks,Bill wrote:
>Dave,
>
>You will find the formulas for isothermal relief for through hole lands in
>the IPC-2222 section 9.1.2 Thermal Relief in Conductor Planes.
>
>Basically it tells you to use 60% of the land size as the 'thermal width'
>and then divide it up equally by however many ties you require. The gap
>between the pad and the plane can typically be near the same width as the
>web connection width unless you are dealing with some voltage spacing to
>another conductor which is a different matter all together.
>
>Regarding the current... if you have four .025 in. wide ties to a land, you
>would add them together to be equivalent to a .100 in. trace. Figure out
>what the current carrying capacity is for a .100 in. trace of the copper at
>the thickness you are using and you will be very close.
>
>Also the copper in the hole, if that is your application, can be calculated
>by converting the hole into a trace... just imagine splitting the barrel of
>the hole longwise and flattening it out... you then have a sheet of copper
>equivalent to the width and length of the rectangle by the thickness of the
>plating in the hole... It's not really rocket science... but you do have to
>use Pi in the calculation. :)
>
>I don't know of a calculator for that task but is sounds like a great
>project. Any ambitious programmers out there want to take it on?
>
>Best regards,
>
>Bill Brooks
>PCB Design Engineer, C.I.D.+
>Tel: (760)597-1500 Fax: (760)597-1510
>Datron World Communications, Inc.
>Vista, California
>
>
>-----Original Message-----
>From: David Baldwin [mailto:[log in to unmask]]
>Sent: Tuesday, May 08, 2007 2:35 PM
>To: [log in to unmask]
>Subject: [DC] Current Capacity of Thermal Reliefs
>
>Is there a calculator for the current carrying capacity of a
>component thermal relief?
>
>Thanks,
>
>Dave
>
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