### Author Topic: Trace width - amps only?  (Read 2683 times)

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#### TomS_

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##### Trace width - amps only?
« on: July 13, 2016, 10:23:07 pm »
Hi everyone.

I would like to understand why only amps are considered when calculating the width of traces for carrying power.

Heres a tool I found:

http://circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/

I would have thought that voltage would also play a part, since watts = volts * amps, and surely higher wattages require fatter traces? For example, 5V at 10A is very different to 12V at 10A? But none of the tools I have come across so far seem to take voltage in to consideration and Im a bit stumped as to why.

Is anyone able to help shed any light on this?

Thanks!
« Last Edit: July 13, 2016, 10:25:59 pm by TomS_ »

#### Dubbie

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##### Re: Trace width - amps only?
« Reply #1 on: July 13, 2016, 10:30:39 pm »
Voltage only plays a part when it comes to insulation against arcing.

The same cross section wire can carry more power (watts) the higher you make the voltage.

Thats why here in NZ, our 240V mains cables can carry twice as much power as the primitive  120V US mains voltage. (given the same size cables)

Ours are more likely to kill you if you touch the wrong thing however!

This is one reason why cross country power transmission lines run at ludicrously high voltages up to 1/2 million volts sometimes.

There are a few other wrinkles but that is the gist of it.

R
« Last Edit: July 13, 2016, 10:42:03 pm by Dubbie »

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#### The Soulman

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##### Re: Trace width - amps only?
« Reply #2 on: July 13, 2016, 10:40:15 pm »
Nah, 10 amps=10 amps, if your trace has a resistance of 0,1 ohms you will lose 1 volt and will dissipate 10 watts in heat, regardless if you drop from 5 to 4 volt or from 12 to 11 volt or even from 100 to 99 volt.

And what dubbie says....

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#### Kilrah

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##### Re: Trace width - amps only?
« Reply #3 on: July 13, 2016, 10:44:13 pm »
P = U * I
U = R * I

-> P = R * I^2.

Shows you that voltage is not relevant to power dissipation, and that's becasue it's the voltage across the element you're looking at - i.e. is only a result of the current flowing, it has nothing to do with the supply.
« Last Edit: July 14, 2016, 07:23:56 am by Kilrah »

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#### alsetalokin4017

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##### Re: Trace width - amps only?
« Reply #4 on: July 13, 2016, 11:25:07 pm »
P = U * I
U = I / R

-> P = I^2/R.

Er... want to try that again?

« Last Edit: July 13, 2016, 11:28:14 pm by alsetalokin4017 »
The easiest person to fool is yourself. -- Richard Feynman

#### alsetalokin4017

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##### Re: Trace width - amps only?
« Reply #5 on: July 13, 2016, 11:47:36 pm »
Hi everyone.

I would like to understand why only amps are considered when calculating the width of traces for carrying power.

Heres a tool I found:

http://circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/

I would have thought that voltage would also play a part, since watts = volts * amps, and surely higher wattages require fatter traces? For example, 5V at 10A is very different to 12V at 10A? But none of the tools I have come across so far seem to take voltage in to consideration and Im a bit stumped as to why.

Is anyone able to help shed any light on this?

Thanks!

Let us be very clear and explicit in an attempt to answer your question correctly. Yes, Watts = Volts x Amps, but in the case you are concerned with the Volts value is the _voltage drop_ across the circuit element you are working with, that is, the PCB trace.  Since V=IR by Ohm's Law (not I/R as the previous poster has it) the voltage drop is equal to the current times the resistance of the trace. The resistance of the trace is obviously related to the thickness of the trace. The power dissipated _in the trace_ is found by P=I2R where R is the resistance of the trace.

The rest is elementary algebra.

V=IR (Ohm's law)
so
I=V/R and R=V/I   (rearranging terms algebraically)

and since Power = V x I, substituting algebraically we obtain

P = (IR) x I    that is    P = I2R
and equivalently
P = V x (V/R)    that is   P = V2/R

-- always recalling that the V value of concern here is the Voltage Drop across the circuit element (the trace) when the applied current is flowing through it.
The easiest person to fool is yourself. -- Richard Feynman

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#### Kilrah

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##### Re: Trace width - amps only?
« Reply #6 on: July 14, 2016, 07:24:44 am »
Er... want to try that again?
Ouch. Seems I'm getting too old to be posting at 2am after a couple of beers

#### TomS_

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##### Re: Trace width - amps only?
« Reply #7 on: July 14, 2016, 10:11:08 am »
Thanks everyone for your replies. Amazing resources in this forum!

I must be mixing things up when I think of power in watts in this respect. I guess that explains why none of these tools consider voltage in the calculation.

#### Brumby

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##### Re: Trace width - amps only?
« Reply #8 on: July 14, 2016, 12:00:46 pm »
I think the water analogy works well here.

The width (and thickness) of the trace is equivalent to the inside of a water pipe.  The greater the cross sectional area, the greater the current that can flow - whether it be water or electrons.

Voltage is equivalent to pressure - the higher it is, the more you need to contain it.  With water, that means thicker pipe walls and with electricity it means thicker insulation.  Since air is an insulator, then "more of it" means wider gaps between traces, isolation slots and the like.

Once you have this fundamental understanding, the rest is straightforward ... and the math is easy.

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#### Zbig

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##### Re: Trace width - amps only?
« Reply #9 on: July 14, 2016, 12:58:58 pm »
I must be mixing things up when I think of power in watts in this respect. I guess that explains why none of these tools consider voltage in the calculation.

But the power is expressed in Watts, so nothing wrong with your understanding here. What I feel might be the case with your understanding is the common misconception many beginners fall under and that would be, despite knowing the Ohm's law in general, still thinking about the voltage and the current as somehow "decoupled".

Let's attack this from a different angle: if the resistance of the given length of trace is constant (and we ignore the temperature effects now for the sake of simplicity), both voltage and current are strictly dependent. You change one, the other follows. So, let's assume you have a trace whose resistance measures 2 Ohms end to end. If you apply a voltage difference of 5V across it, the current of 2.5A will flow. To put it different way, you NEED 5V to "force" 2.5A current through 2 Ohm resistance so there's no point stating both. Continuing, in this case it is a mistake to argue that "5A@5V is more power than 2.5A@5V" or that "2.5A@10V is more power than 2.5A@5V". Your resistor is 2 Ohm so you simply cannot assume any arbitrary voltage and any arbitrary current. If you measure 5A of current, that means there's 10V across your trace's ends. If you measure 3V across your trace, that means there's 1.5A flowing through it. There is no magical box that lets you dial both of them independently through a given constant resistance.

I think at least some of this misunderstanding results from the way DC power adapter manufacturers specify their parameters. Far to many times have I heard that "you cannot use this power supply with this device: it provides 5V, 2A and the device needs 5V, 1A so it will burn out". Once more, not the case. How you should read the DC power bricks' specs is: both of them are constant 5V voltage sources so both of them will be good for my device that needs 5V at 500 mA. Where they differ is, the beefier one should be able to safely provide a current of up to 2A once loaded with an equivalent resistance low enough for the 2 amps of current to be able to flow at 5 volts and that would be 2.5 Ohms.
« Last Edit: July 14, 2016, 01:00:51 pm by Zbig »

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