Trace Width

[POTL]

Hi
I want to clarify, the thickness of the tracks is 16 mils is suitable for effects?
9-18 volts, not more than 100 mA
Various online calculators say that you can make tracks and are thinner, but still I want to be certain that 16-mils is good choise

[287m]

Yes

[POTL]

Yes

100% you do not worry that the width in the power path will not be too thin?

[PRR]

You need 0.7mil for 100mA (unless there are complicating factors).

I don't think you can GET 0.7mil on commodity PCB fabrication.

Nothing wrong with using 20X the width you absolutely need.

Use the fabricator suggested minimum (any smaller will be trouble); or however big you like so you can eyeball your work.

[R.G.]

I commonly use 10, 12, 15 - and 25 if there is no need to make things very compact.

For your next step up, you can calculate the actual resistance of the trace. The resistance of any resistor is given by the equation
R = pL/A
where p = the basic resistance of a square of the material; in copper's case, this it 1.68 *10^-8 ohm-meters. L is the length of the conductor and A is the area, in meters and meters squared, respectively.

Since PCB traces are commonly measured in mils (i.e. 1/1000 of an inch) we can convert that to mils. One meter is 39.37 inches, which is 39,370 mils. So an ohm-meter is an ohm-39,370 mils, and 1.68 *10^-8 ohm-meters is 1.68*3.937*10^-8*10³ = 6.614*10^-5 Ohm-mil

So then you multiply this by the length of the trace (in mils, 1/1000 of an inch) divided by the area of the trace. One ounce copper is 1.4 mils thick. So the area of a trace is 1.4 mils times the width of the trace.

And a one-inch trace that is 10 mils wide is 6.614*10^-5* 1000mils / (1.4mil * 10mil) = 4.724*10^-3 or 4.7 milliohms.

A trace 1 mil wide and 1 inch long would be then 47milliohms. So a 16 mil wide trace 1 inch long is 47/16 = 2.94 milliohms. A 16 mil wide trace that's 1.5 inches long (1500 mils) would be 2.94milliohms * 1.5 = 4.41 milliohms.

All the math just tells you that for traces in the 10 mil to 25 mil range, you're down in the range of milliohms per inch, which is pretty small.

A 100ma current through an inch of 16 mil wide trace would generate 0.294 millivolts, or 294 micro-volts.

Yeah, ought to cause you no problems in a pedal.

That's what all that math stuff is for.

[Rob Strand]

Various online calculators say that you can make tracks and are thinner, but still I want to be certain that 16-mils is good choise

The calculators (assuming they are correct) give a minimum.   You can go thicker (and make manufacturing less problematic).

[bluebunny]

Various online calculators say that you can make tracks and are thinner, but still I want to be certain that 16-mils is good choise

The calculators (assuming they are correct) give a minimum.   You can go thicker (and make manufacturing less problematic).

+1

That's the trouble with these online calculators.  For example, you can indeed use a 390 ohm resistor in series with a red LED in a 9V circuit.  It won't burn out.  But it will burn your retina.    So not so useful after all...

It's really worthwhile learning a tiny bit of theory and doing a tiny bit of arithmetic, as R.G.'s reply shows.

[POTL]

thanks to everyone

[electrosonic]

My preference, 70 mil pads and 16 mil traces. The pads are large enough to be easy to solder and by the OSH Park design rules I can fit a trace between two IC pads.

Andrew.

[POTL]

My preference, 70 mil pads and 16 mil traces. The pads are large enough to be easy to solder and by the OSH Park design rules I can fit a trace between two IC pads.

Andrew.

funny, I also like to use 70 mils for pads