My IRL540N is smoking
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I am using an IRL540N MOSFET to drive a 12 V pump (about 6 A) and a 12 V solenoid valve (about 1 A max). It’s smoking after just two uses. I need to know two things.
- Is it still alright? It’s still working and I have switched off power. When do I know it’s best to replace it?
- How do I prevent the smoking? It’s getting too hot. What’s a good heat sink for such applications? I’m surprised it got so hot as it’s winter and the ambient temperature is close to 12-13°C. If it’s in this condition in this weather, then when summer brings 45°C it will just go nuclear. I’m scared.
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- 2How are you driving it? How are you cooling it? – winny 1 hour ago
- 2Show a schematic. – Rohat Kılıç 1 hour ago
- I have a public IRL540N schematic for everybody to use: circuitlab.com/circuit/b45c4a7snvv9/irl540n_2023feb0103 – tlfong01 43 mins ago
- 1@tlfong01 that’s mighty nice, but might have nothing to do with DribbleNibble’s schematic, so not that helpful here. – Marcus Müller 39 mins ago
- There are many reasons to heat up the IRL540N. Perhaps (1) your input control signal is not right, (2) Yor IRL540N is fake IRF540N, (3) Not adequate heat sink, (4) No flyback diode, etc. – tlfong01 36 mins ago
- One suggestion is to use a high Vgs, so Ron is small and therefore less heating up. – tlfong01 27 mins ago
- 1@tlfong01 that’s good answer material, but you should probably put that in an answer, not comments. And, generally, when you say “one suggestion is”, you should also say under which conditions, and why – this is engineering, not “try this and see if it still burns”. “Rules of thumb” are cool, but only when they come with a set of rules that whoever applies them understands – otherwise, that’s a recipe for surprise disasters. – Marcus Müller 21 mins ago
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Look out also for temperature coefficient of Ron. If you get a MOSFET whose Ron increases with temperature and you don’t drive it hard enough on or give enough cooling, this can cause magic smoke very quickly. The Ron you see on the graph on the datasheet rises, the higher it goes the more power it tries to lose …
Basically Ron, tempco of Ron and thermal resistance form a complex system which needs to remain in a stable state. Being conservative about calculations and generous with heatsinking goes a long way to achieving this.
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answered 12 mins ago
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At 7A, and assuming you are supplying at least 5V Vgs (Which is where the on state resistance is specified), I make the power dissipation about 5 watts, so you will want a small heatsink to keep the thing happy, bolting it to the side of a metal case should suffic.
Note that a common trap for beginners is looking at Vgs(th) and thinking that is all the gate drive you need, when it is actually the point at which the thing just starts to conduct a tiny bit (250uA in this case), to get the mosfet hard on you need typically double whatever Vgs(th) is, details will be in the graphs.
If you try to switch this with 3.3V logic it will absolutely smoke.
There are much better mosfets for this that will run with no heatsink.
The front page of mosfet datasheets doesn’t usually lie exactly, in the same sense that a government minister addressing parliament doesn’t usually lie exactly…
They are however heavily influenced by marketing’s desire to publish the best possible numbers, and particularly for things like maximum drain current a smart designer will go to the graphs instead.
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answered 31 mins ago
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When do I know it’s best to replace it?
After it smoked, it’s necessary to replace it. See the “Absolute Maximum Ratings”, first page of the datasheet. You’ve clearly exceeded that maximum temperature.
Physically, failure modes of overheated semiconductor devices involved migration of dotants, degradation of insulation, degradation of thermal coupling, … So, things you don’t easily see. An IRL540N is between 1 and 2€ in single quantities – so definitely not worth risking it.
How do I prevent the smoking? It’s getting too hot.
It would seem that way, yes! So, you need to stop it from getting so hot.
What’s a good heat sink for such applications? I’m surprised it got so hot as it’s winter and the ambient temperature is close to 12-13°C.
You didn’t install it on a heatsink to begin with? I’m a bit surprised: The Junction-to-Ambient thermal resistance from the datasheet says 62 °C/W, so, with a loss of only 3 W, you’d be far into “beyond repair” territory.
Anyways, I’m not sure why you have that loss. The currents you’re switching are low, and at your (starting) temperature and this current, the on-resistance should be far below 1 Ω. In fact, looking at Fig 1. from the datasheet, at 7A your Drain-Source voltage would be < 0.3V for any reasonable Gate voltage, so you’d be converting 0.3 V · 7 A = 2.1 W into heat. Sure enough, that’s a lot, but according to the datasheet thermal resistance should heat up your case to little more than 120°C hotter than ambient – not smoking hot.
So, three things are possible: you’re using a gate control voltage that’s too low, the current is higher than you say, or you’re swithing the MOSFET on and off a lot – but these are motors, so I’m pretty sure you wouldn’t want to PWM them?
In any case, when you have something that produces ~ 2W of heat, yeah, a heatsink is desirable – but a cheap one should do. All heatsinks will come with some specified thermal resistance. That, added to the thermal resistance from the MOSFET datasheet from junction to cooling surface, will tell you how hot the chip gets when you generate some defined heat power.
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answered 25 mins ago
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