Easy PCB trace width calculation

Question

I've tried this question and this and several other possible duplicates of this question, but I still can't wrap my head around this.

I'm using OSHPark for PCB production. They support a minimum of 6mil width tracing. I'm using a two-layer design, so they use FR4 170Tg/290Td for PCBs. They are 1 ounce copper on both sides.

My PCB as an integrated LiPol charger and a maximum consumption of 1A at 3.3V. It can be charged through a mini USB connector, so I'll assume a maximum of 1A at 5V of power being drawn. I don't mind my circuit to go 30º C above room temperature. If I follow the graph below, then my interpretation is that I start at 5V on top left and go to 30º which puts me roughly on the 26AWG equivalent. I then go down to 1oz and then left, getting ~15mils of width.

Does this make sense?

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UPDATE: I can't even read graphs. Since it says current in A, I start at 1.0, go right, down, left, and get around ~4mils. Is it right now? Is this completely independent of the voltage? Should I take the voltage into consideration for calculating the trace clearing?

Answer 1

Since it says current in A, it seems I should start at 1.0, go right, down, left, and get around 0.04 [inches].

Yes, basically just follow "Example #1". But you said you have 1 oz copper instead of 1/2 oz copper, so that should put you at about 20 mils trace width.

Is this completely independent of the voltage?

Yes

Should I take the voltage into consideration for calculating the trace clearing?

Yes, but at 5 V you won't have any particular clearance requirement. When your working voltage gets to 50 V, start worrying about clearancescreepage and personal safety.