Of course the hot spot is the hottest temperature! That deserves a big DUH!
Maybe the thermal
There's that bar graph thingie on the side.
but the resolution is
6 degs C gradient from hot spot to nearby copper, with 1 watt dissipated on the opposite side.
It's useful because all of my designs work.
Well, it's a bit under 1 watt, since the resistors on top are dissipating 1 watt. So the theta of that hot spot is at least 6 K/W, probably more. James and I estimated that his via field was about 4 K/W. So the spreading resistance is more than the via resistance.
But in reality
It demonstrates typical thermal spreading resistance in a copper plane. That sounds like useful info to me.
Do you do much electronic thermal design? Show us some.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
I would add that surface contact resistance can be a pain. And dominate the total thermal resistance. A standard hex 1/4 inch stand off (nickel plated brass) 4-40 hole, screwed into a copper plate has a junction resistance of ~5 (Thermal) ohms thermal ohms=deg K/ watt. (or worse if not tightened "properly") (that's with no goop in the gap.)
Re: electrical-thermal modleling Charge = Joule and so, Capacitance (Farad) = Q/V = Joule/deg. K (the heat capacity)
The 1 gram of Aluminum is a nice number, but obscures the correspondence.
(Oh and everything has a heat capacity of about 3 J/cm^3... heat capacity scales with volume not with the mass (density) of stuff.)
There is the big problem. You don't listen! Max temperature meaning max of the camera range. Saturation. No distinction about temperature within the hot spot.
You mean the display? Do you set the range or does the camera?
And I believe over 10 degrees between the two sides. But what is the range of temperature *within* the hot spot. Your camera shows it as isothermal, but that can be due to saturation. You have to use the device properly to get useful info.
You said this was a test board. So that point is moot.
Which actually means they are in the same ballpark and so even a moderate change in either one will impact the overall result.
So your work can't stand on its own? You make claims that are *not* substantiated by the info you post and in fact so far everything you have provided shows the vias under the pad *will* have a useful impact on the result.
The FLIR can measure up to 250C or some such. In the mode that I used it, it automatically scales the temperature bar to the min and max temps in the field of view. It's not saturated. Why would any thermal imager saturate at 38C?
The heat source on the top has dimensions greater than the PCB thickness, and that creates a hot spot on the bottom that is pretty much isothermal near its center, sort of an inverted parabola or something, kinda flat on top. That's all logical.
In that mode, it does. Nothing's saturated. I guess I could zoom in on the spot, to see its temperature profile in more detail... if I can find the board.
The Flir ain't saturated. The spot is pretty isothermal, in that the lateral gradient is low near the center. The real point is that the copper is hot, and half an inch away it's down near ambient. It doesn't conduct heat very well.
I do calculations and experiments and learn. Then I design stuff that works. What do you have against that?
So, you don't do thermal design. Thought as much. Do you design electronics at all?
Here's a TO220 Caddock resistor bolted to a big sheet of 0.062 thick aluminum.
formatting link
Note that even 1/16 thick aluminum has substantial thermal spreading resistance. Just an inch away, the temperature rise has dropped in half. That has implications to sinking things like voltage regulators to a chassis.
1.4 mil (1 oz) copper is way worse. So the trick is to maximize the area that the heat is initially dumped into. The tighter the via field, the higher theta.
I'm surprised that I have to explain stuff this simple.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
2D thermal conduction is like that, though. In equilibrium, for a uniform insulated plate, the temperature flux through any annular ring is constant, so the temperature gradient goes as -1/r and the temperature profile goes like -log r.
As you say, if the copper slug in the package is wider than the board is thick, vias round the edge (but still just under the slug) do more than those in the middle.
In fact, if the slug is perfectly isothermal, the board is well insulated, and the aggregate thermal resistance of the edge vias is very low, the region inside the via ring will be at exactly the same temperature as the slug, i.e. the inside vias will do nothing whatsoever. Reality isn't quite this neat, of course, but it's clearly true that given the same number of vias of a given size, it's better to distribute them around the edges of the slug rather than put them in the centre.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Mitigated a bit by convection. This stuff is complicated!
Right! Move them outside the slug a bit.
Reality isn't quite this neat, of course, but it's clearly
One fix for James' problem would be to glue a metal disk to the bottom of the board, to effectively increase the diameter of the via array.
I was recently doing some studies to bring down the temperature of an FPGA (the ring oscillator experiment and such, wound up being a 16 page report). Gluing a pin-fin heat sink to the top of the FPGA dropped the silicon temp by 4K. The fins didn't do any good; the benefit was from the thermal spreading of the base of the sink. A flat aluminum square did just as well.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
Which is unrealistic, of course. A perhaps more useful (but likely equally unrealistic* :) ) formulation, where all points on the surface have constant thermal resistance to an infinite sink, yields a dependency something like J_0(r). Which emphasizes the notion of a lateral spreading distance: the temperature is highest in the center, near the peak of the Bessel function, and dropping off to zero away from there. More material thickness or conductivity gets you more distance and thus more total dissipation; more distance without any meat behind it results in wasted space.
*Problems concerning heat dissipation and convection being notoriously anti-analytic, of course. That said, power laws are usually a good fit, and who knows, maybe a T^(3/2) or T^2 power is easier to integrate than that silly Bessel function.
"Kinda" flat? It would be spherical through the relatively insulating PCB and then relatively higher conductivity of the copper would spread it making it look more even.
The point of the device is that it isn't very good at showing what the temperature difference is in what you call a hot spot. But that is mostly irrelevant since this is not a board with vias which completely change the picture.
You have already given numbers that disprove your case. I believe you
these two numbers are close enough that a change in either one significantly impacts the sum.
I don't appreciate your faulty logic.
Now that you provided real data you don't seem to want to discuss it.
We are discussing the claims you made which I say are not supported by your data. In fact, now that you have provided hard data your claims are clearly wrong. Do you wish to discuss your numbers?
My first choice would be right round the edge of the slug, some inside and some outside, and then stitching the top pour to the heatsink layer for some reasonable but poorly specified additional radius to reduce the net thermal resistance.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Can't be an oscillatory Bessel function, because the heat equation is first order. I_0 and K_0, possibly, if your cows are cylinders instead of spheres. ;)
Which emphasizes the notion of a lateral spreading
You just need a copy of Gradshteyn & Ryzhik. ;) Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
That's my qualitative take. Qualitatively, the goal is to launch as much heat into the bottom pour, over as wide an area as possible.
The area under the tab is going to be, by definition, the hot spot, since it's where the heat originates, and it's so small compared to the total dissipating surface. So, spreading that wider is better.
I was quasi-tempted to LT-Spice it for fun, but wound up helping roof a house instead!
You do seem to collect groupies of that sort. Must be your air guitar skills. ;)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Ok, I take it you just can't figure out how to justify your poor logic. That's all I was trying to do, point out that you could do a better job of understanding the issues instead of declaring the interior vias to be of no value without supporting evidence.
He supported it fine. Yes, of course in the real world "everything depends on everything else", but most of the contributions are insignificant - which is what makes design at all possible. "Understanding the issues" includes the ability to see where these contributions are going to be insignificant, so that attention can be concentrated where it is useful. Otherwise every time you consider heat you will end up needing some monstrous 3D FEA thermal modelling program just to place a DPAK. And even when you modelled it, you still won't know what to do, you won't *understand* anything.
I actually use one of those to read SED from my phone when I'm stuck someplace. It's dramatically better than GGroups.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
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