5
\$\begingroup\$

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.

  1. Is it still alright? It's still working and I have switched off power. When do I know it's best to replace it?

  2. 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.

Below is my schematic.

enter image description here

\$\endgroup\$
21
  • 2
    \$\begingroup\$ How are you driving it? How are you cooling it? \$\endgroup\$
    – winny
    Commented Feb 1, 2023 at 9:03
  • 3
    \$\begingroup\$ Show a schematic. \$\endgroup\$ Commented Feb 1, 2023 at 9:33
  • 5
    \$\begingroup\$ @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. \$\endgroup\$ Commented Feb 1, 2023 at 10:41
  • 2
    \$\begingroup\$ 1kΩ between gate and driving signal (Arduino) is a bit much. I'd recommend something in the range 10Ω and 100Ω. 1kΩ could slow down the rising edge too much, which is another source of inefficiencies. \$\endgroup\$
    – Velvet
    Commented Feb 1, 2023 at 11:58
  • 2
    \$\begingroup\$ @MarcusMüller Yes, the resistor drawn in the schematic is a pull-down. I'm referring to the resistor described in the text above the image: "only difference being that I have connected a 1K between drive signal and gate basis a suggestion I received elsewhere". It would be less confusing if OP would update the schematics. \$\endgroup\$
    – Velvet
    Commented Feb 1, 2023 at 12:12

3 Answers 3

6
\$\begingroup\$

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 dopants, 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.

\$\endgroup\$
5
  • \$\begingroup\$ The current could at most be 10A, but I'm not imagining it being that much. I am controlling the gate by Arduino Nano with 1K between gate and drive for a reason I don't really understand but was told its good practice. I have updated my question, please read it. I am indeed switching the MOSFET on and off very frequently but not as frequently as it would eventually work, I'm expecting it to on/off every 15 seconds for 8-10 hours. I think I'll get a big heat sink. So just to be sure, no flashy resolution required? Just a regular aluminum heatsink would do? \$\endgroup\$ Commented Feb 1, 2023 at 11:48
  • \$\begingroup\$ ah every 15s is not "often"; I was thinking of 100 times per second. You're fine on the switching losses – the amount of time your MOSFET spends between "completely off" and "completely on" is negligible. \$\endgroup\$ Commented Feb 1, 2023 at 11:56
  • 1
    \$\begingroup\$ In your picture, the resistor is not in a place that makes very much sense to me \$\endgroup\$ Commented Feb 1, 2023 at 11:57
  • \$\begingroup\$ @user253751 yep, thanks \$\endgroup\$ Commented Feb 1, 2023 at 17:35
  • \$\begingroup\$ Ghe 1k resistor is a bad idea in this case. In some contexts it helps damp gate ringing. Here it reduces drive voltage by about 0% - and you need all you can get. || The FET is margional for this task. It has and Rdson (on resistance) in the 0.5 to 1v range in your application. Extreme hotness is to be expected. You has get FETs with Rdson of a few 10's of milliOhms. \$\endgroup\$
    – Russell McMahon
    Commented Feb 2, 2023 at 9:57
12
\$\begingroup\$

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.

\$\endgroup\$
9
  • \$\begingroup\$ I'm using the digital output from an Arduino Nano (I'm assuming its 5V) to drive it as I had thought being a logic level MOSFET it would be sufficient. So you are saying a 3.3V logic MOSFET would be a better choice with a Nano? Can you suggested some examples of those better mosfets you mentioned? \$\endgroup\$ Commented Feb 1, 2023 at 11:31
  • 2
    \$\begingroup\$ Well the right thing to do is to add a low side gate driver so you can hit the gate with 10V or so, but absent something like that, a IRLB8748PbF manages 6.8mR at 4.5V Vgs, so about 330mW rising to half a watt when it heats up, rather better then 3-5W of the old IRL540 (Note the 30V VDSS limit). The even more impressive MCAC130N04 manages just 2.5mR under the same conditions, reducing the heat load to just 122mW. Both are in stock at Digikey. Generally the really good stuff is SMT these days. \$\endgroup\$
    – Dan Mills
    Commented Feb 1, 2023 at 12:13
  • \$\begingroup\$ @DribbleNibble There's the Arduino Nano, which runs at 5 V, and the Arduino Nano 33 (BLE/IoT), which runs at 3.3 V (and has additional bluetooth or "Internet of Things" features, though that's not important here). If you have the plain Nano, it'll output 5 V, but if you have the Nano 33 it'll only output 3.3 V. \$\endgroup\$
    – Hearth
    Commented Feb 1, 2023 at 14:11
  • 1
    \$\begingroup\$ Ah, I see -- Look at the Id vs Vds graph, find the operating point on the Vgs and Vds calculate the power and Rds_on, etc. \$\endgroup\$
    – Dave X
    Commented Feb 2, 2023 at 4:26
  • 1
    \$\begingroup\$ At 12V either will be fine, but I didn't know your voltage. Do note that spikes and surges have to fit within that 30V limit, so you do need a clamp diode. \$\endgroup\$
    – Dan Mills
    Commented Feb 2, 2023 at 10:39
6
\$\begingroup\$

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.

This app note gives a decent overview. In the past when I did this, I always did some prototyping, making sure to try a few different devices and measure temperature rise on the device body at maximum current (plus some error margin) - I feel that real world tests go a long way to confirming whatever paper calculations or simulations you came up with, to make sure you didn't miss anything important. This is not stuff you want to make an error with (especially in equipment which might be left running unattended, which is most equipment).

Also don't forget to be pessimistic about ambient temperature - especially if you are designing gear which might will run in a rack, with many devices all conspiring to push "local" ambient temperature up above what you expect.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.