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Thermal resistances should be additive. Your thermal resistance should be the Rjc + Rpaste + Rha. I suppose you might think of it being negative in the sense it's supposed to improve conduction into the heatsink, but that goes against what I have been taught in the past. The assumption for Rha is that the heat is already "inside" the heatsink and Rha would capture the resistance of the flow from within the heatsink to the ambient. The Rpaste probably omits the resistance from case to paste, and from paste to heatsink surface. I might be wrong, but that's just how I've been taught to interpret it.
If you have the thermal paste areaC/W (or areaK/W) then by balancing units, you have to multiply by the application area. Lets say we take your figure of 0.061cm^2C/W. If you have a TO-220 package (a fairly common one), the area of the contact is about the rectangular portion behind the chip area (excluding the tab hole to simplify calculations) - lets say 12.95mm x 8.5mm based on this drawing. This means the area is 1.107225cm^2, multiplied by the figure before gives us 0.067540725C/W as the result. In other words, given proper thermal coupling (i.e. paste application), the paste contribution should be negligible. Using the other figure - say 2.4W/mK, would probably need to know the thickness of the film.
Despite this, I've rarely ever bothered working out this because most of the time, the rule of thumb I often use is to add another 0.5 to 1C/W for the case-to-heatsink resistance contribution just to be pessimistic and lump in all of the unknown case-to-paste reistance, paste bulk resistance, paste-to-heatsink resistance. If you can't handle this amount of additional thermal resistance, you can't handle a 1-degree rise in ambient and that would be extremely borderline in terms of design.
Thanks Gough, that helps with the thermal paste. Was I right in my other calcs? That is, without a heatsink, I use Rja and with a heatsink, I use Rjc?
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In general, yes,
Rja is Junction to Ambient, so that's the figure you use assuming no case. Note that some datasheets will have some specific stipulations about Rja - e.g. assuming a certain leg length or soldered with a certain leg to a minimum amount of area of copper for example.
Rjc is Junction to Case, so that's the figure you're interested in if you are trying to work out the total thermal resistance of a system with a heatsink which is mounted to the case of the device. It's also useful to work out how far above Tcase the Tjunction is for for a given power dissipation so you can compensate and work out what Tjunction is, given a Tcase measurement and knowing the Pdissipated.
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Just for completeness of the thread, W/mK is Watts per metre Kelvin. The higher the number, the more efficient the heat transfer.
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There is term missing from datasheets, which is the thermal resistance from case to heatsink. The case-to-heatsink term is affected by how much contact area they have, which depends on how flat the surfaces are, which is not controlled by any single datasheet, so it isn't often listed, but you can look it up for various materials. This resistance term can be reduced by themal compound because although thermal compound has themal resistance it is lower than the thermal resistance of the tiny air gaps between case and heatsink.
The heatsink-to-air term is only accurate under specific conditions, because it is significantly affected by airflow and dust and humidity, and airflow can be caused by anything, including the hot air rising off the heatsink. And of course the ambient air temperature is constantly changing as the heat coming off the heatsink changes it. And then there are external heat sources to consider, like sunshine, that can add 1KW per square meter. It sounds like you know all this stuff, but it is mentioned just in case it is useful....
When you need to squeeze the last bit of cost and performance out of a system there are a a lot of factors to calculate, including how long the system needs to operate because even moderate temperatures will reduce lifespan. I generally do rough calculations without thermal paste for a temperature target well below maximums listed in datasheets and add thermal paste to provide extra margin.
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I'm pretty much aware that there's a lot that I don't have experience with yet, that's for sure, but intuition tells me that ambient isn't likely to stick at 25c as things warm up! I do suspect better understanding comes with time and experience but I have to start somewhere so if I'm on the right lines then I've got a good point to start from. From the calculations I've made, based on my understanding above, I've selected heatsinks that keep temperatures well under maximums (around 60% at worse case) as a tradeoff between cooling vs design size.
Interestingly, the one part that may give me some grief is a MOSFET but its data sheet does list Case-to-Heatsink which I've included in my calculation! Modelling with LTSpice shows that it may be dissipating up to 12W worse case, depending upon usage (power supply), so needs a good heatsink. This is the one part I'll need to carefully monitor when testing as that worse case calculation gives a temperature of 80% of maximum - typically, I think it will be a lot less.
One other factor that often comes into play is if any parts may come into contact with people. Anything over 55C is considered too hot to touch (by me, but choose your own people-safe temperature).
I think I may have gone off track when I said above that the MOSFET may be the part to give me some grief - I think I was doing the thermal calculation wholly incorrectly (looking at the voltage drop Drain to Source * current). I think I should have been doing it as follows, could someone confirm?
IRF3205 - datasheet
Rds(on) = 8mOhm MAX
Tj = 175c
Pdiss = 200W @ 25c / linear derating 1.3 w/c
Rja = 62 c/w
Rjc = 0.75 c/w
Rcs = 0.5 c/w (case to heatsink!)
So given a current of 3A, then P = 3x3x0.008 = 0.072w
With no heatsink, junction temperature = 0.072 * 62 = 4.464c + ambient.
Ambient would have to rise to around 88c to cause a problem 175 -((88c-25c) * 1.3) = 93.1c max junction temperature vs 88+4.64 = 92.64 actual junction temperature.
Either I'm still off track, or I don't need a heatsink.
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That sounds about right - when fully on the FET won't need to dissipate much heat.
However the biggest heat source can be when the FET is turning on or off - the slew rate and the frequency can cause the power during switching to be very significant because the resistance in the FET is going from a very high value to a low value or vice versa.
So the heat will usually depend on how often the FET is switched on or off and how quickly the switching takes place. The gate capacitance can slow down switching so there is often an emphasis on driving the gate with a low impedance driver.
Calculating this seems to be fairly complicated but from the (limited) research I've done, then...
Switching frequency, F, of 200kHz
Rise time, Tr: 101ns
Fall time, Tf: 65ns
Power = (Tr*I*V + Tf*I*V) * F * 0.5 = (0.000000101*3*24 + 0.000000065*3*24) * 200000 * 0.5 = 1.2W [this formula I'm not sure on TBH, but the answer seems not unreasonable]
Temp with no heatsink would be 1.2*62 = 74.4 + ambient.
Ambient would have to rise to around 56c to cause a problem 175 -((56c-25c) * 1.3) = 134.7c max junction temperature vs 56+74.4 = 130.4c actual junction temperature.
56c might be conceivable depending upon the temperature rise of other components, ventilation in a case etc... so a small heatsink would be useful.
I like the way you are digging into all the calculations.
Are the rise and fall times from the FET datasheet (which may not include driver impedance) or from the driver signal (which may not include gate capacitance) or from simulation?
This is a generic question about determining need for and selection of a heatsink for a part - am I on the right lines?
From the data sheet for a part:
Tj - Junction Temperature: 125c
Rja - Junction-to-ambient: 40c/w
Rjc - Junction-to-case: 3 c/w
For the part, if I measure:
That would give me power dissipated of (2.93 - 1.04) * 1.53 = 2.8917W
With ambient temperature set at 25c, then:
- No heatsink, junction temperature will be: Pdiss * Rja + ambient = 2.8917 * 40 + 25 = 140.668c [heatsink definitely needed]
- With heatsink, sink-to-ambient resistance Rha 26 c/w, junction temperature will be: Pdiss * (Rjc + Rha) + ambient = 2.8917 * (3 + 26) + 25 = 108.8593c [way too close for comfort]
- With heatsink, sink-to-ambient resistance Rha 16 c/w, junction temperature will be: Pdiss * (Rjc + Rha) + ambient = 2.8917 * (3 + 16) + 25 = 79.9423c [much better - could improve with a bigger heatsink]
I haven't taken into account thermal paste between the part and the heatsink. I'm assuming that the thermal resistance of the paste is taken off the thermal resistance of the heatsink to improve it's efficiency?
- I'm confused about how to do this as the ratings I've seen are xW/mK (e.g. 2.4W/mK) or areaC/W (e.g. 0.061cm²C/W). However something like: Pdiss * (Rjc + Rha - Rpaste) + ambient.
I've tried to distill this from reading a variety of sources so I'm just checking that I'm on the right track.