How to Calculate the ampacity of a tarce on IMS board - Page 1
Forum section: electronics / projects

author: Electrons
author_rank: Contributor
timestamp: March 23, 2023, 12:21:15 pm
content: HiI am designing an IMS board and have not found any calculators for knowing how much current does a trace carry at 1oz / 2oz .PCB way specifies IMS with multiple conductivities from 1W/m.k up to 3W/m.kany idea ?

author: ajb
author_rank: Super Contributor
timestamp: March 23, 2023, 08:50:27 pm
content: How much current are you looking to handle? Hundreds of mA? Amps? Tens of amps?Hundredsof amps*?The current handling capacity of a track is determined (mostly) by how much power dissipation you're willing to accept. Any calculator that tells you a given track can handle X amps is really telling you how much current it will take to make that track dissipate a certain amount of power, and/or exhibit a certain amount of temperature rise. Temperature rise is dependent on the power dissipation and the thermal resistance, so there are a few assumptions backed into that number, in particular the thermal resistance between the track and the ambient environment, and the amount of temperature rise you can tolerate will depend on what 'ambient' means. Even on FR4, the 'ambient' may be elevated by the dissipation of other components of the circuit before you even account for copper dissipation.If you're designing with a metal core PCB, then presumably you have a lot of power to dissipate overall, and the 'ambient' is not going to be ~25C air, especially if the PCB is mounted to a heatsink. Generally a better approach would be to consider the amount of heat dissipation you can tolerate in the PCB tracks as part of your overall thermal budget along with the dissipation of any active circuitry and then feed this into your thermal design calculations. In general, power dissipated by tracks is probably negligible for most metal core PCB applications for any reasonable track width. Of course the allowable voltage drop along the track is a limiting factor on current handling as well, but that's unlikely to be THE limiting factor in most metal core PCB applications.Wider tracks will be somewhat more efficient on account of dissipating less power due to their lower resistance, and they may improve thermal coupling between the devices they connect to and the metal core by serving as sort of heatspreaders across the surface of the insulating layer (the benefit of this is moderated by the very high thermal conductivities and small layer thickness, though).Quote from: Electrons on March 23, 2023, 12:21:15 pmThat's not a very useful number on its own. Presumably they're referring to the bulk thermal conductivity of the insulation layer, but to get a thermalconductancevalue you can use directly in a temperature rise calculation you'd need to eliminate the distance term from the denominator. To do that, you'd multiply the conductivity by the area (cross section of the heat path) and divide that by the thickness (length of the heat path). Assuming all your dimensions are meters, that gives you(m * m * W) / (m * m * K), so the meters on top and bottom cancel out and you get W/K. So you'd need to know the thickness of the dielectric and then plug in the area of the track, then you can multiply by the dissipation of the track to get a temp rise vs the substrate.* A CM once told me a story about building some 'metal core' PCBs where the 'core' was a 1" thick aluminum bus bar for some crazy military battery charging system handling hundreds of amps. Track resistance was a significant factor there due to its effect on the current sharing of the devices on the PCB. Apparently on the prototype all of the power transistors popped one after the other all the way down the PCB until they rearranged the bus connections!