I plan on running about 15A through a trace on a PCB for about 5 seconds. Does the trace need as large as a trace calculator would say?
In other words for 15A how big of a trace on a 2oz copper board would i need?
I plan on running about 15A through a trace on a PCB for about 5 seconds. Does the trace need as large as a trace calculator would say?
In other words for 15A how big of a trace on a 2oz copper board would i need?
The real question is how hot do you want your copper to get, the copper heats near instantaneously as the resistance and current in the copper determine the heat in watts that is dissipated. This heat is usually dissipated into air which is usually considered to be 25C.
To estimate the temp rise, a calculator can be used of 10C above ambient will require a trace roughly 250mil wide.
20C will need ~160mil
30C will need ~130mil
40C will need ~110mil
These calculations are a steady state (after the copper has heated and reaches thermal equilibrium), the heat then spreads out to other parts of the board, especially if there are adjacent layers of copper. I would worry about the temperature rise as a worst case. 5 seconds is more like DC for thermal so at some point the wire will most likely reach a steady state condition. PWM with a duty cycle can be treated differently as the average power drops.
The second question is will resistance in the milliohms be a problem for your design? Probably not but something to think about.
Thermal TimeConstant of 1micron cube of copper is 9.6 nanoSeconds. Thermal TimeConstant of 10micron cube of copper is 100X slower, or 960 nanoSeconds. Thermal TimeConstant of 100micron cube of copper is another 100X slower, or 96 microseconds. Thickness of standard CU foil (1 ounce/foot^2) is 35 microns, close enough to 31.6 microns so we can with confidence say the timeconstant of 35micron cube of CU foil is 9.6 microSeconds.
To unload heat through FR-4 is tough, what with that ThermalResistance being 200X higher than CU. So you only path is along the CU trace. And that Thermal Resistance is 70 degree Centrigrade per watt per square of foil. A trace 10 mils wide and 1 inch long has 100 squares, thus Rthermal is 70*100 = 7,000 degree Centigrade per watt.
The temperature rise of 30 deg may be acceptable, but I suspect the loss in power to the motor from voltage drop along the tracks.
I would consider adding a busbar which comes in strips with thruhole pins or equiv.. There are many ways to do this.
If a thermal calculator indicates heat rise is excessive then perhaps excessive voltage drop and crosstalk may occur on switching reactive currents from ESL of tracks and/or ESR of capacitive load or V=LdI/dt of reactance.