How much current can a PCB handle?

Try

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-Mike

Try looking at this on-line calculator:

I suggest using a safety factor of at least THREE times above the expected peak, and of course use fuses, circuit breakers, or thermal fuses/cutouts..

You may want to go to an even lower resistance if the board is particularly long, to minimize voltage drops and heating.

And of course there's always dual-sided boards or ones with thicker copper. I recall looking at a street-light controller PCB (with Tubes!), which had particularly heavy copper as it had to carry many amps.

I successfully did a 30A switching power supply on a small (2.5 inch by

1.2 inch) board. That had 2 oz copper for each plane, however.

A larger backplane I did had to distribute 70A, and I used 2 x 2 oz planes for that (because of the distances involved and the fact there were thousands of holes for the plug-in board connectors ;)

In both cases I used the PCBTemp program which yu can still occasionally find floating around.

Cheers

PeteS

I posted some tips a while back on this topic here:

Regards Dave :)

I've seen these calculators before but I thought they were primarily for circuit traces (tracks) instead of just copper areas (pours). Also, they max out at 35A and 0.4" (the ones I've seen). At any rate, is the "copper area" they produce as part of the results the same as the copper area that would be needed for a copper pour/area? i.e. if a particular result on these calculators says "4000 mils copper area" does that mean a 2" x 2" copper pour would produce the results it says? I always thought the copper area result they gave was some sort of interim result that was needed for the calculator to find the remainder of the results. I never considered that it might mean the actual copper area (answer to my question).

I've found the pcbtemp software but I'm not sure how to handle copper pours/areas or planes with it? As far as I can tell, I must enter a track width? What do I enter for a width so that I get the proper results? It seems that I could enter just (the bigger) of two dimensions of a rectangular copper pour/area/plane? Length doesn't affect the current, it is only used for voltage drop. It seems that this calculator has a flaw? For example, if I enter a width of, say, 1 inch (1000 mil) and a length of 0.1" (100 mil), 10 deg C rise, it says I can handle 35 amps with 1mV drop. If I enter 100 mil wide, 1 inch long, 10 deg C rise, it says 7.5 amps, 20mV drop. Quite a big difference for the exact same area of copper...

This just makes me not trust it 100%, because I'm not sure how to use it. Any pointers for using these calculators/PCBtemp software for copper pours/areas/traces?

Here's a table of suggested values for a 10C temperature rise. It's very conservative. There's a little formula for other temp rises. I'd be quite happy with 30C for non-continuous duty.

I've designed many high power audio amplifiers btw and the kind of currents you mention are quite routine in that application. 2 oz ( 70 um ) copper is a good idea btw.

Graham

Current capacity depends on cross-section area, not surface area. In your first example, you have a cross-section area of about 1500 mil^2, and your second is 150 mil^2. That's a ten times difference. The rest of the math is beyond my experience, but I'm not surprized that the program gives you different numbers. I'd be suspicious if it didn't.

For example, a trace 1 mil wide and 100 inches long will not have the same current capacity as a 100 mil trace 1 inch long.

The areas is actually important for radiating and conducting the heat away though.

Graham

Take a look at PC motherboards. Some processors draw upwards of

75A, two of 'em on a board...

Also think about the connectors.

--
  Keith

Are you saying it depends on where the connections are, within the trace?

The way I described what is puzzling me, the 0.1" wide by 1" trace VS the 1" wide by 0.1" long trace, they would look exactly the same on a board (copper amount). If you are looking at it one way, and rotate the board 90 degrees, it looks like the second way I describe. The same exact copper pour, yet the calculator gives two VERY different results depending on which one you consider to the be "width" and which you consider to be the "length" (basically, which way you are orienting the board when viewing it, in the case I described). I assume this either means there is a bug in the calculators, or that the placement of your via/hole/pin within the trace matters greatly.

I'll try to draw it quick with some text below.

================================

• 100mil wide *

================================

Where the * is a via/hole/pin. This would be the "1 inch long by

100mil wide" version (above) that I described, and the "100 mil long by 1 inch wide" version (below). The below version is a little impractical, but you can see how it is the same copper area, just a different via/hole/pin placement which changes the wording of width and length. Is this the important fact that I'm missing? That via/hole/pin placement matters?

================================ - * 100 mil long * ================================ -

See how the copper is the same amount? I can definitely see how the via placement would/could affect the current capacity (I guess), but I am asking if it does? This is an especially important question when I am trying to figure out the current capacity of a particular power or ground plane, or copper pour/area. If the via locations do in fact matter as much as (or more than) the size of the copper which they are connected to, than this will greatly affect how much current a particular pour/area/plane can handle, right? i.e. If I have an entire plane (say 3" x 3" board) dedicated to power, if I place a via in the top center and bottom center, it is much different current handling capacity than if I place them both in the center, 0.125" apart? Am I all screwed up? Am I thinking logically? I'm confusing even myself now :(

In message , dated Tue, 15 Aug 2006, Andrew writes

I think when you get the message, you will kick yourself.

Think about wire instead of copper foil. I have this copper wire which is 0.5 inches square and four inches long. It has a resistance from one end to the other (4 inches) of X ohms.

Now I have this copper billet, which is 4 inches wide by 0.5 inches thick and 0.5 inches long. It has a resistance from one end to the other (0.5 inches) of Y ohms. Y is not equal to X!

Yes, unless the board conductor is a geometrically-square sheet, the resistance does depend on where on the edges you connect to it. Even with a square, it does vary a bit if you have only point contacts.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Yeah, hence the "beyond my experience" comment. I know it matters, I just don't know how. I figured *one* obvious discrepancy was enough to discourage the "they should be the same" theory.

If you put three 1k resistors in series, does it have the same total current capacity as three 1k resistors in parallel? It's a similar question to what you ask.

So where the connections are matter *alot*. I get it now, I think.

The "from one end to the other" phrase you list is the key component. The ends meaning where the connections are made. You could take the same copper billet and place the connections (ends) the "long" way and it would be the same as the copper wire, right? If so, I understand completely.

I just didn't realize how important the closeness of the connections was for current capability, when talking about an entire copper area or plane.

Would this be correct to say as well: That the resistance is determined by the closeness of the connections. The copper that expands outward (width) is for heat sinking.. ?

If so, is there a point where the width stops affecting thte resistance, at least significantly enough to consider?

(Sorry if these are ridiculous questions, but I've never considered how resistance is affected by shapes and where connections are, etc)

In message , dated Tue, 15 Aug 2006, Andrew writes

Yes. There is a formula, which usually uses Greek letters, but I spare you that.

resistance = resistivity x length/area

resistivity is something you find from tables of material properties. It's measured in funny units - ohm.metres. Don't bother why, until your studies have progressed a bit.

Maybe. It doesn't work well as a heat sink if it's expanded a lot.

Yes, but it's difficult to quantify.

Don't worry, but you may be happier posting to sci.electroncis.basics, where the constitution says that there are NO stupid questions. There may, however, be stupid answers, such a 'use a PIC'. (;-)

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Especially if you're trying to answer the question in $subject :-)

Q: How much current can a PCB handle? A: Use a PIC.

John, surely you have IEC/EN 60065 or 60950 to hand ? Isn't there a formula in there too ? Or is it just track spacing and I'm imagining it ?

Graham

In message , dated Tue, 15 Aug

2006, Eeyore writes

Spacing; nothing about current capacity of traces and sheets.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK