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Mail Archives: geda-user/2020/01/15/12:50:31

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From: "Nick Bowler (nbowler AT draconx DOT ca) [via geda-user AT delorie DOT com]" <geda-user AT delorie DOT com>
Date: Wed, 15 Jan 2020 12:28:02 -0500
Message-ID: <CADyTPEzksBzCOezSWzHxukpmXqOyJavk1kDWBe_dbUM-RKW2RQ@mail.gmail.com>
Subject: Re: [geda-user] Maximum current for thermals?
To: geda-user AT delorie DOT com
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On 1/15/20, Richard Rasker (rasker AT linetec DOT nl) [via
geda-user AT delorie DOT com] <geda-user AT delorie DOT com> wrote:
> I have a hopefully simple question: I'm putting a 16A terminal block in
> a ground plane, with Ø 3 mm solder pads and a 1.3 mm hole (0.85 mm
> annulus width), with 0.2 mm clearance. Copper thickness is 0.035 mm
> (~1.5 mil)
>
> Connection with a thermal gives 4 connections with a width of 0.2 mm
> each, same as the clearance. This seems a bit skimpy, so I usually make
> these high-current connections without thermals.
>
> Still, I wonder what the approximate current capacity of these thermals
> is? I know from experience that a few amps is no problem at all, but I'm
> pretty certain that they'll fail upwards of 25A or so. If possible, I
> prefer thermals for easier soldering, so a more accurate estimate of the
> limits would be appreciated.

Resistivity of copper ρ is about 1.7 × 10^-8 ohm*m at 20°C, with
temperature coefficient α about 3.9 × 10^-3 / °C.

Divide ρ by the trace thickness (0.035mm) and width (0.2mm) to get:

  ( 2.4 * (1 + α(T - 20°C) ) mOhm/mm

where T is the trace temperature.

Let's say we don't want things hotter than 50°C so we'll plug that
number in and our 0.2mm long thermals to get each thermal having
a resistance of about 0.5 mOhm.  There are 4 of them so if each
carries 1/4 of the current then:

  25A: each thermal will dissipate 20mW or 80mW altogether.
  50A: each thermal will dissipate 80mW or 0.3W altogether.
  100A: each thermal will dissipate 0.3W or 1.2W altogether.

How hot the thermals will get is a system design question of
how well you are heatsinking that generated heat away.  They are
presumably connected to a large copper plane which will help.

Cheers,
  Nick

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