Current and PCB Traces

The current carrying capacity is going to depend on the trace width, plating thickness, and allowable temperature rise. As the trace width and plating thickness go up the resistance goes down and the temperature rise drops with it. The resistance will be a function of the (volume) conductivity of copper.

Here is a link to a website that I found with a quick google search:

I don't know how accurate the results are but if you can find a couple of calculators that agree or find the formulas for copper conductivity and temperature rise you could verify the results.

Your PCB can certainly handle it but you do need to worry about trace width

Dan

huh?

is this suppose to be a link? Cause its not working...

Well, Since 20awg wire can only do about 1 amp(thats what it says on the insulation of the 20awg wire I have) then it seems that a trace of 0.040in just falls short(assuming a depth of about 0.005in(a guess but it doesn't seem that deep on my boards)?

Jon

ok.

Thanks, Jon

It really isn't that complicated.

The thickness on the boards is based upon the plating thickness in oz (per square foot I think). You choose the thickness of the plating you desire, .5oz, 1oz, 2oz, or 3oz and knowing this value you can then determine the necessary width of the trace to allow X amount of current at Y temperature rise. If your current draw is operating at a high frequency you will need to account for skin effect which will decrease the effective thickness of your copper.

There are a lot of variables there. For example, is it an internal or external trace. How much of a temperature rise do you want on the trace. How thick is the trace.

For example, a typical temperature rise of 10 degrees Centigrade and at 2 amps: EXTERNAL TRACE:

INTERNAL TRACE:

Brian

ok... atleast I have some general ideas now.

Thanks, Jon

You're confusing the current rating of the wire's *insulation* with the thermal characteristics of the bare copper on a PCB. Insulation melts at a much lower temperature than copper[1].

A given cross section of copper will have a certain temperature increase for a given current. More current, more heat. More copper, less heat. You choose a copper size based on how much heat you can tolerate. For example, in a 70F room with a 110F limit on heat, you design for a 40F increase in heat due to resistive losses in the copper.

Wiring is different - it's rated according to how hot it can get before the *insulation* fails and you get a short.

Get a copy of the old, free PCBTEMP (used to be available from UltraCAD, but I believe it's no longer available at their site). It's a handy tool for such things.

Cheers

PeteS

http://66.102.7.104/search?q=cache:sN5Zsx5RCxAJ:

Ok, I think I understand. Basicaly the in both cases heat will be generated but in the case of wire the insulation has a lower melting point than on a PCB(which is due to the type of material and the size?)?

Thanks, Jon

Yes.

Technically, the insulation has a maximum safe temperature, and your PCB has a maximum safe temperature, with "safe" defined accordingly for each. Your project could have a different and unique definition of "safe" for example. Enclosure temp, thermal cooling of other components, heatsinks, etc.

But in general, you can't use the current ratings of wires as a guide to the current ratings of traces.

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It's a message ID - copy it to your clipboard, migrate to news:alt.binaries.schematics.electronic and paste the message ID in the appropriate "find message ID" place.

Good Luck! Rich