trace width current consumption question

Logical in so far as the calculation is apparently correct. But also useless because 1 um trace width is no option for PCB. Suggest opposite method, put in minimal trace width available in your technology, e.g. 0.15 mm and check voltage drop and temperature rise.

my trace is 50 mil much larger( so i assume it can handle 18mA.
What temperature rise to put?
Thanks.

Can't the tool calculate temperature rise for 18 mA and your trace width?

Hi,

It's no rocket science to use Ohm's law and to calculate power dissipation.

And if 0.001mm (0.04 mil) results in 25°C temperature rise...
It also should be quite obvious when using amire than 1000 times wide trace the temperature rise will be ignorable small.
Just from simple math (P = I x I x R) one can exoect a 1000 times lower power dissipation.
Additionally a 1000 times thicker trace will spread the heat even better.

So isn't it obvious that the expectable temperature rise is less than 25°C / 1000?
So less than meaningless 0.025°C ...

It really does not need to be a physician or big skills in mathematics...

And again: every simple PCB layout tutorial will tell.

If the tool tells you, you need at least 0.04 mil (0.001mm)
... every width wider than this will be fine
... and you use maybe 10 mil as your srandard signal trace width --> it will be fine
.... or you use maybe 50 mil as your standard power trace width --> it will be more than fine

A picture:
If a tiny string of a spider web is rigid enough to stand the weight of a fly .... will a 1000 times thicker string also be strong enough?

Klaus