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I am trying to figure out how to properly calculate the minimum contact surface area based on the maximum required power dissipation. Let me explain..

I am creating a custom semi-circular metal tab that makes an electrical connection to a metal strip. The metal tab will be connected to a system that dissipates 7 watts. Assuming that I know that I need to support a maximum power current draw of ~.55A @ 12 V (this is where the ~7W power dissipation comes from), how can I best calculate the size of the contact surface area that the two metal contacts need to share in order to support the specified current dissipation. Please see the very crude drawing below. I am trying to determine the contact surface area highlighted in the blue circle.

enter image description here

Here are the givens...

  1. I have control over the choice of metal for both the contact and strip

  2. I can decide the thickness and size of both contacts.

  3. A voltage drop of no more than .5 Volts may be tolerated

  4. A temperature rise of no more than 10 C may be tolerated

  5. The load has a current draw of ~.55 A @ 12 V (i.e. ~7W)

I am not looking for anyone to do the work for me, I am just looking for some help with the proper equations to use. Thanks.

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  • \$\begingroup\$ How much current do you need? Contacts are specified by current not power. Or is the voltage negotiable, if you have a switch-mode PSU? Second, please clarify if there is any heat transfer from the 7 W box to the rails, or just power? \$\endgroup\$
    – tomnexus
    Commented Aug 25, 2015 at 5:01
  • \$\begingroup\$ It is all about the point contact resistance. Your contacts won't be dissipating 7 watts I hope, if that is your load. Even carbon brushes would conduct to your load with virtually no heating. Just about any common conductor would be fine, even the fingers in an edge connector would be overkill. But you want something plated to prevent oxidation and increasingly poor contact, and something that doesn't lose its spring over time... again, fingers from an edge connector should be fine. \$\endgroup\$
    – R Drast
    Commented Aug 25, 2015 at 11:45
  • \$\begingroup\$ @tomnexus the load is operating at 12V, which means a current draw of .58 A. And no, it is just the power connection. I can see how I might have phrased the question poorly. I am really more concerned with being able to handle the current draw. \$\endgroup\$
    – Willis
    Commented Aug 25, 2015 at 14:41
  • \$\begingroup\$ Thanks. Last question is how much it needs to move, if at all, and how fast? For static contacts a simple wire spring on a thin brass strip would work, think of the battery terminal in a flashlight. For continuous motion you need a spring-loaded carbon brush, more like a motor commutator. Temperature rise will be easy, reliability over time and use will be harder, low cost manufacture hardest. \$\endgroup\$
    – tomnexus
    Commented Aug 25, 2015 at 17:14
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    \$\begingroup\$ If you want to design your contact on a purely theoretical bases you will most probabely have to do finite element analysis. If an empiric solution is also possible get a bunch of switches with proper spec and analyse them in detail \$\endgroup\$
    – christoph
    Commented Aug 28, 2015 at 19:53

1 Answer 1

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You need to calculate contact resistances to find an answer to your question.

Contact resistance splits in two parts. Resistance coming from the small slot:

$$ \Lambda \approx\frac{3{,}7}{E^{*} \cdot \rho \cdot l} \cdot F_N $$

with Lambda being the conductivity in Siemens, E* the effective E-modulus of your materials, rho the specific resistance of the metal, and Fn the normal Force exhibited from the contact. l is the root mean square of surface height (aka roughness).

The effective E-modulus is approximated by

$$ E^* = \frac{E}{2\cdot(1-\nu^2)} $$

with nu beeing the Poisson's ratio.

You see, in this formula the area of contact doesn't show up. It is much more important to reach a sufficient contact force.

The other part is the resistance of surface layers, like oxides. If you resort to gold or palladium surfaces you can neglect it. Otherwise it is difficult or even impossible to calculate. If you have at least minimal movement between contact surfaces, oxide layers may be broken up. Other ways of getting rid of oxides is reaching the wetting current. AFAIK wetting currents are in the single mA range for many contacts. So this is perhaps no problem for you.

To get an idea how force and resistance are connected I recommend viewing a bunch of datasheets from manufactures of spring loaded contacts. Kitagawa has some with both values, and PTR gives maximum currents, too.

Sample contact data (PTR)

Kitagawa grounding contacts

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