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There are many heatsink calculation Q&A's on SE that I've looked through. One very reasonable answer tells me that I can calculate heatsink C/W as follows: $$\frac{T_j - T_a}{P_d} - R_{JC} = R_{SA}$$ where Rsa is the C/W that I need to achieve. In one LDMOS we are considering, the numbers substitute as follows: $$\frac{175 - 25}{272} - 0.55 = 0.00147$$

But 0.00147 C/W is impossibly small!

Another LDMOS we are considering shows the following C/W: $$\frac{225 - 25}{690} - 0.29 = -0.000145$$ Yes, that is a negative C/W.

What am I doing wrong here?

Daniele Tampieri
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KJ7LNW
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  • Why does the second one need to dissipate 690 Watts and the first one only 272 Watts? That doesn't make sense unless it is two different designs. Maybe the second one would work with 272 W. Or how are you calculating Pd that you are plugging in to your formula? – user57037 Mar 21 '21 at 03:49
  • Just to clarify, Pd in that equation is the power dissipated by the transistor. In the datasheet, I think the power they are listing is the RF TX power (not 100 percent sure because RF is not my field...). If that is correct, then the dissipation will be Pout / eff - Pout. – user57037 Mar 21 '21 at 03:58
  • The 690W-Pd device has a 76W TX power, and the 272W-Pd device has 320W of TX power. Strange indeed! – KJ7LNW Mar 21 '21 at 19:49
  • But what is the design TX power you are trying to achieve? – user57037 Mar 21 '21 at 19:59
  • @mkeith, 76W at 146MHz. – KJ7LNW Mar 21 '21 at 20:38

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Your dissipated power numbers are the "Total Device Dissipation @ TC = 25°C" from the datasheets.

This is not a power you can achieve in practice. It is like saying "the Bugatti Chiron can go at 500km/h". Yes, but you need a special road for that, which is not easy to find. To achieve "Total Device Dissipation @ TC = 25°C" you need a special heat sink that is not easy to find, since it can maintain the case at 25°C no matter what the power is. Since there is thermal resistance between the case and the sink, say 0.1°C/W, this means at 690W the sink must be is -69°C below the 25°C case. So bring on the liquid nitrogen.

Basically, "Total Device Dissipation @ TC = 25°C" is purely the result of a mathematical formula. Its use is to tell the designer "If your power goes near that, there is no way, you need another transistor." And your question shows that is true: if you use this power, you need a magical heat sink.

So your formula is correct, but the power you use is wrong. You must use the power that it will actually dissipate in your application (which is not your RF power).

bobflux
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  • Ah, so the OP probably read total device dissipation and thought it was how much the device actually dissipates. Did not understand that it is a limit (and really more of a theoretical limit...). So those transistors might work, but OP needs to state actual TX power intended for the design. – user57037 Mar 21 '21 at 17:46
  • @bobflux, I understand that I need to use the Pd for heat dissipation (as opposted to TX power), but the datasheet isn't very clear on the subject. Did you see anything in the PDFs that might hint at the truth here? – KJ7LNW Mar 21 '21 at 19:52
  • It's the "@ TC = 25°C" mention – bobflux Mar 21 '21 at 19:54
  • The @TC=25°C makes sense for their Rjc numbers, but any idea why Pd is 690 for a 76W device? I thought Pd was independent of TC... – KJ7LNW Mar 21 '21 at 20:40
  • 76W is the RF output power, not the dissipated power. Yes actual Pd is independent of Tc because Pd depends only on how the device is used in the circuit, but... maximum allowed Pd depends on Tc via RthJC and maximum junction temperature. – bobflux Mar 21 '21 at 20:52
  • So assuming I could sink the heat, this is saying that I could run this at 690W without damage, even though typical use is far below that? – KJ7LNW Mar 21 '21 at 22:22
  • It's just a model based on a simple equation that assumes the whole silicon die is at the same temperature, it does not account for hot spots, uneven thermal conduction or other real life "details". – bobflux Mar 21 '21 at 22:43
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The first one is telling you that it is not likely to be practical to remove so much heat from the device even with 25C ambient (and I question whether you can rely on 25C ambient... is this in an air-conditioned room?)

The second one tells you that even with Rjc of zero, you cannot dissipate enough heat for the device to survive. You would actually need to cool ambient down below 25C to get any margin.

Stepping back a bit, what both results are telling you is that this isn't going to work. You need to lower your power output or add more active devices in parallel to accommodate your output level. I assume this is an RF transmitter you are designing. But whatever it is, the transistor cannot dissipate this much power.

user57037
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  • Other than bringing the "ambient" temperature down, one can flow more ambient air through a heat sink. I believe that all gets absorbed into the Pd calculation of this model. – Abel Mar 21 '21 at 03:56
  • @Abel no. If the calculation shows a negative thermal resistance is required, then even with the case pinned at 25C, the device cannot be cooled below Tjmax. – user57037 Mar 21 '21 at 03:59
  • Tip: You can also use HTML entities `Ω`, `μ`, `°`, `×`, etc. as well as `...` and `...` in the posts (but they don't work in the comments). – Transistor Mar 21 '21 at 19:06
  • @mkeith, I could lower the power output, but the devices are spec'ed for these TX power levels and their graphs support that. I opened a ticket with NXP, maybe they can give an answer that makes sense. – KJ7LNW Mar 21 '21 at 19:51
  • @KJ7LNW you are still not getting it. The heatsinking is not based on TX power. The heatsinking is based on dissipated power of the transistor. The efficiencies are given in the datasheet, so you can calculate the dissipated power based on the TX power and the efficiency and the frequency. Then you can figure out how much heatsink you need. – user57037 Mar 21 '21 at 20:01
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    dissipation = Pin - Pout. Pout is TX power. Pin = Pout/eff. So dissipation = Pout/eff - Pout. That is how you calculate dissipation from datasheet. The efficiency depends on frequency and output power, so you need to use the right operating point. The Greek letter eta is the symbol for efficiency: η. – user57037 Mar 21 '21 at 20:08
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    @mkeith, I understand: 76÷0.68−76 = 35W and thats about what I would expect. But...why is Pd 690W? That doesn't seem to fit anywhere for this 76W FET, which is what I'm asking NXP. Do you know why Pd is 690? – KJ7LNW Mar 21 '21 at 20:35
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    By long standing tradition, MOSFET's report power dissipation based on the unrealistic condition of keeping the case cooled to 25C while the junction is at its maximum temperature. So basically the maximum dissipation is a kind of figure of merit calculated based on Tj(max), and Rjc when Tc is 25C. It may be useful when comparing with other transistors using the same rating scheme, but it doesn't tell you how much power you can really dissipate. I have little exposure to RF MOSFET's but what I say is true for switching FET's. Probably just the same tradition. – user57037 Mar 21 '21 at 20:43
  • For switching FET's, they are also starting to report more useful data such as Rja (junction to ambient) or Pdis based on Ta instead of Tc. – user57037 Mar 21 '21 at 20:45
  • @mkeith, great info, especially about tradition. If you roll the note that Pd should be "dissipation=Pout/efficiency-Pout" for heatsink calculation and the "fet rating tradition" comments into your answer then I think that will check the box. – KJ7LNW Mar 21 '21 at 22:30