What prevents load switch MOSFETs from always encountering thermal runaway?  - Page 1
Forum section: electronics / general

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: October 11, 2023, 11:09:33 pm
content: I have a MOSFET as a load switch. It passes 10A to a load. The Rds(ON) is 30 milliOhm. So the dissipation is 3W. Suppose the thermal resistance is 10C/W. Then the junction of the FET has a temp rise of 30C. Suppose we started at room temp (25C), so now the junction temp is 55C.The Rds(ON) of a FET increases with increasing junction temp, and the datasheets usually give a graph characterizing this change. Let's say for our FET the Rds(ON) is 1.5 times the original nominal 30 milliOhm. So now the Rds(ON) is 45 milliOhm. But the load is still drawing 10A so now the FET is dissipating 4.5W. So now the junction has risen 45C above 25C, so it's at 70C.So now we're at a junction temp of 70C, which means the Rds(ON) has increased to perhaps 1.6 times the nominal 30 milliOhms. Now we have even more dissipation which further raises the junction temp. You see where I'm going with this. It seems like there's no end. The junction temp keeps increasing, which keeps raising the Rds(on), which keeps increasing the dissipation ad infinitum, i.e. thermal runaway.Ok...so obviously in the real world, FETsdon'tencounter thermal runaway all the time like this. So that means I'm thinking about this wrong somehow. Set me straight. How am I thinking about this wrong?

author: thm_w
author_rank: Super Contributor
timestamp: October 11, 2023, 11:43:06 pm
content: The FET chosen is not suitable for the task then. You'd need one with lower Rds on, or to decrease the thermal resistance in some way (additional heatsinking, fan, etc.). Almost never want to design around the 25C spec.https://techweb.rohm.com/product/power-device/si/si-basic/6882/

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 12, 2023, 03:13:31 am
content: Indeed, the tortoise never reaches the line; it always halves its distance to the finish, and in the next instant, half again!Fortunately, reality is more than willing to disregard the foolishness of Zeno, and feedback systems converge smoothly to their limiting values, within whatever degree of accuracy we are willing to wait for. Or, y'know, diverge and oscillate or blow up, as the case may be, but in any case a real result is always had.If you have a function for Rds(on) vs. T, and T vs. Pd, you can calculate the steady-state point based on the lack of change, i.e. after substituting one into the other, d/dT = 0.Tim

author: Kim Christensen
author_rank: Super Contributor
timestamp: October 12, 2023, 03:34:15 am
content: Quote from: Jim from Chicago on October 11, 2023, 11:09:33 pmThe hotter something gets relative to ambient, the faster heat energy will be transferred to ambient. So as long as the heat transfer rate increases faster than the RDS(on) & power dissipation rate, then the FET will eventually stop getting hotter.

author: Siwastaja
author_rank: Super Contributor
timestamp: October 12, 2023, 12:22:25 pm
content: Simple: power dissipation capability increases roughly linearly with increasing temperature, yet Rds_on does not rise even close to linearly. In other words, when you double the temperature difference, you could double the power dissipation, but the dissipation did not double because of the rising Rds_on, maybe increased by just 10-20%.You can indeed do the calculation iteratively like you showed, 3-4 steps should converge close to final value. If it seems you have thermal runaway exceeding maximum die temperature, and you did calculate correctly, you have just too small of a MOSFET or too little cooling for it.

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: October 12, 2023, 04:59:22 pm
content: Thanks everyone. I see now that if I keep iterating the calculation, eventually I get a very small temperature rise that is not significant enough for me to distinguish on the graph of Rds(on) vs junction temp. Is there a more elegant way to perform this analysis besides manually iterating several times? Maybe some software? I have been literally and metaphorically burned in the past by picking a FET which started to reflow solder because it got too hot under max load.

author: Siwastaja
author_rank: Super Contributor
timestamp: October 12, 2023, 05:09:55 pm
content: Remember the datasheet curves are not exact science either, so such "inelegant" iterative calculation is IMHO enough, that's what I do. Thermal modelling can get pretty complex, so we tend to use simpler napkin/Excel calculations anyway. In this regard, iterating the Rds_on is one of the smallest problems. Finally, nothing beats testing the prototype. You can monitor at least the case/tab temperature, and given even 30-40% accurate Rth_case-heatsink model, you can calculate Tj pretty close from the Tc, Ta, P measurements.

author: DavidAlfa
author_rank: Super Contributor
timestamp: October 12, 2023, 05:11:37 pm
content: If you can't source a better fet or whatever, you can simply put more fets in parallel.

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: October 13, 2023, 01:42:28 am
content: Ok so if I were to do the analysis with this FET:https://www.vishay.com/docs/75583/sqm100p10-19l.pdfIn our application it needs to handle a max of 10A. When it's on our application Vgs is around -14V. So I'm gonna take the 19 milliOhm value from the table. For our products we do a worst-case analysis of 70C ambient air. So we have 10A^2 times .019ohm = 1.9W. For thermal resistance let's use 25C/W (I tested the original part we were using which has the same package and I came up with this padded value).So we have a rise of 1.9 * 25 = 47.5C above ambient, so Tj is 117.5C. So it looks like Rds(on) would be about 1.75 times the nominal according to the normalized graph.. 019 * 1.75 = .03325 ohm. So we have 3.325W dissipation which equates to temp rise of 83.1C... 70 + 83.1C = 153.1C.. At that junction temp we have Rds(on) about twice the nominal, or .038 ohm. So we have 3.8W dissipation, which equates to a temp rise of 95C. 70C plus 95C = 165C. At 165C Rds(on) is about 2.2 times nominal or .0418 ohm. 4.18W dissipation. 104.5C plus 70C = 174.5C.. Now we're at the max Tj of the part and Rds(on) is about 2.3 times nominal or .0437ohm.Is this a sound analysis? Would you agree this FET is not suitable for this application?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 13, 2023, 02:06:15 am
content: Yes, that sounds reasonable. Note also Rds(on) max at temp is specified. You can take that as a worst-case starting point, and if temperature falls on iterating, it's fine.If you can improve RthJA, or reduce max ambient or I_D, that will do, or otherwise pick a type with lower Rds(on).If you're limited on space, or cost, note that NMOS perform significantly better than PMOS (about 2.5x in terms of Rds(on) * Ciss). It may be worthwhile to craft a charge pump gate drive circuit, or use an IC with such.Tim

author: f4eru
author_rank: Super Contributor
timestamp: October 13, 2023, 07:40:17 am
content: typically, you also want (or need) an overload protection.Wiring short circuits, and other overload conditions should not destroy your driver.