Heat Dissipation of a PCB - What Am I Missing? - Page 1
Forum section: electronics / projects

author: Paul Bryson
author_rank: Contributor
timestamp: March 26, 2018, 10:40:54 pm
content: I have been looking at thermal resistance and heat dissipation in a PCB - where the PCB itself is the primary heat sink. My admittedly simplified analysis indicates that the thickness of the copper (1 oz vs 2 oz) has nearly zero effect on the total heat dissipation. However, repeatedly in app notes I read recommendations to use 2 oz copper for high power dissipation. For example, from TI, "AN-2020 Thermal Design By Insight, Not Hindsight",(http://www.ti.com/lit/an/snva419c/snva419c.pdf), claims a 25% improvement in thermal resistance of a 58 cm² board by changing from 1 oz to 2 oz copper.The thermal resistance from the PCB to still air is two or more orders of magnitude greater than the thermal resistance of the copper - how can thicker copper matter? Am I missing some path or mechanism by which the copper thickness has more of an effect than I think?My Thinking:The thermal resistance to still air for 1 cm² is given by various sources as between 600 - 1000 °C/W. Which would be 300 - 500 °C/W, if both sides of the PCB are considered.Assume a series of concentric annular rings of copper foil surrounding a heat source, each with area = 1 cm². In regards to total thermal resistance, ?, each one of these rings is essentially in parallel. Or more precisely, each additional ring is in parallel with the combination of all of the rings inside it - the resistance of the copper is added in series. Say the first ring has an inner radius of 0.25 cm. I calculate that the thermal resistance across this annular ring (1 oz copper) is 10 °C/W; and each subsequent 1 cm² ring has an ever decreasing resistance (10,3,2,1,...). The resistance to air for each additional 1 cm² ring is constant. Even for the first 1 cm², the difference in total ? between 1 oz and 2 oz is less than 1%. For a larger board the difference is nearly zero.I built a spreadsheet that treated the rings as lumped components, with the following results:for 1 cm²: ? (1 oz) = 527 °C/W; ? (2 oz) = 522 °C/Wfor 10 cm²: ? (1 oz) = 51.9 °C/W; ? (2 oz) = 51.8 °C/Wfor 50 cm²: ? (1 oz) = 10.35 °C/W; ? (2 oz) = 10.35 °C/W

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: March 26, 2018, 11:21:08 pm
content: What matters is the temperature of individual devices. I'd think 2 oz traces leading away from a device would be way more effective than 1 oz. The board itself has a given area and orientation, so total power dissipated will be about the same regardless. You need to consider a real world case. For another example, consider a power resistor connected to an electrolytic filter cap, not an uncommon situation. If the trace is heavy and thick, the cap will run at a high internal temperature and fail years earlier. A lightweight trace, or longer, will increase the cap life substantially. In either case the power resistor dissipates the same amount. It's just a question of where.

author: Paul Bryson
author_rank: Contributor
timestamp: March 26, 2018, 11:39:35 pm
content: At first sight you would think that thicker copper would make a difference; but, think of it this way:You have a 1 ohm resistor (copper) in series with a 100 ohm resistor (PCB to air). If you change the 1 ohm resistor to 2 ohms the total resistance just doesn't change very much.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 27, 2018, 07:22:16 am
content: There is a property of lateral heat spreading, which depends upon the conductivity of the board relative to the conductivity of air.Suppose you have a slug of, well, some device that's dissipating heat, a D2PAK's copper tab, say. That tab, being solid copper, can be reasonably assumed as a constant-temperature boundary condition, and a power source. That is, it's dissipating heat into the board, and all points of the board that touch it are held at the same temperature.Within a short distance of the slug, there is not much board area, and little power dissipation. This area heats up, and power continues to spread, laterally.As the heat continues to spread, the total area increases, and gradually, it cools down. This slows the spread of heat, so that 90% of the heat is dissipated within a given area, even if the board is infinite.I once solved this equation, finding a solution in the form of the Bessel function. Namely, the zeroth order, where the value of the function corresponds to temperature (as a function of radius) and the first zero can be taken as the lateral spreading distance. (I'm probably remembering this very poorly, because a zero is certainly not part of an asymptotic function, which should be the correct result. Note that the "radius" is not the geometric value, but a dimensionless product taking into account board and air conductivities.)Anyway, there is theoretical basis for the simple case (linear thermal conductivity), and you can run the diff eq if you like.The practical result is this: PCB heat spreading radius is approximately as many inches as half the total ounces.So, for a D2PAK on 2oz two-sided PCB (with vias and heat spreading on both sides), expect about a 1" radius, or ~6.3 in^2 (pi, then x2 sides). The corresponding convection constant, in mixed units, is about 150 K.in^2/W (what a horrible unit, I know), so expect about 20 K/W, or a maximum dissipation of 5W at 100C temp rise (about the most you'd ever want to push, for a commercial grade product)....I'm playing with a board right now that's a combined total of 14 ounces. I can dissipate 10W in the middle of it, at ~40C peak temp rise (that's the temp on top of the components). Downside: it's almost impossible to solder!Tim