Calculating a Simple PCB Trace Width

[jack0987]

I am using a one ounce per square foot printed circuit board single sided. High frequency is not an issue.

At 5 volts DC and a max of 0.2 amps, I set about trying to calculate the trace width that would be needed.

Attempting to use this calculator
https://www.circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/
I came up with 1.28 mils.

Doing the math of 5 times 0.2 , I come up with 1 watt max for the trace.

I think a hair is thicker than 1.28 mils and this is going to carry a watt of power. The calculator has no entry line for volts. What if the voltage was 500. We now have 100 watts of power. Will the trace width still be 1.28 mils?

Calculating the wire size needed and then applying that to the trace width seems a far better way to do this?

Can anyone give me a method that makes sense to me?

 

[Pommie]

The voltage is irrelevant as it is not across the trace. Only the current matters as this is what the trace will be carrying. The smallest trace I use is 0.012" which, according to various pages, can carry over 0.5A.

Mike.

 

[ChrisP58]

The width of a PCB trace is only about current, not power. Voltage is important only to determine how far one trace is from another trace, pad or other conductive object.

As for power dissipated by the trace. If your 1.28 mil wide trace was 64 inches long, and you put 5 volts to the ends, your trace would dissipate about 1 watt. (1)

Realistically, a 1.28 mill trace width is to small to be practical with standard processes. Most board shops want 6-8 mils minimum.


Note 1. Assuming that the trace stayed at 25C.

 

[Cicero]

Yeah, it'll only dissipate 1W if the actual traces resistance alone was 25Ω...which it obviously isn't.

I try push my track widths as big as I can on a board, with a general rule of 7-10mils minimum.

 

[jack0987]

I am having the most awful time accepting that voltage does not matter.

On minimum trace width, I have etched boards with trace widths of 27 mils successfully. Being a hobbyist, I could probably cut that in half and be ok but I do use larger traces when possible.

How about distance between traces and pads. I usually use not less than about 3/10 inch roughly. Makes it easier to solder.

 

[KeepItSimpleStupid]

Resistance = pL/A where p=rho or a materials constant for copper called resistivity in units of ohm-cm.
L is the length and A is the cross-sectional area. Just as in wiring, you care about the voltage drop as first order problem.

You may be looking at power the wrong way: P=(V2-V1)/I V2-V1 is the drop across the trace.

Just as in wiring you care about the insulation resistance. That's like a nearest neighbor thing at a lower or higher potential. You may find cuts in the board to increase this. Teflon standoffs may be required to manage leakage paths.