# Hard temperatures rises for a MOSFET

I'm a mechanical engineer and a high school teacher that is about to start mechatronics and electronics.

I've built this simple circuit that has the purpose to switching on and off a 12 V / 25 W lamp by means of a MOSFET and an optocoupler:

All works fine, but after a few minutes the temperature of the MOSFET rises to over 90 °C.

Recently I've added a heat sink to the MOSFET, but with poor results. On the IRF730's datasheet I can read that it can handle more than 7 A. Why is the temperature rise so hard?

• Why are you keeping the gate voltage down to 6.22V? At what gate voltage (Vgs) in Rds(on) guaranteed? Lower Vgs gives higher Rds, wasting more power. Even at Vgs=10V it can be 1 ohm, dissipating a lot of power, needing a fairly big heatsink. There are better MOSFETs...
– user16324
Commented Feb 27, 2021 at 16:08
• It is great that you are branching out. Congratulations on putting together a circuit that at least works somewhat! Commented Feb 27, 2021 at 18:49
• What do you mean by "hard temperature rise"? Very high temperature rise? Very quick rise (high rate of change of temperature)? That the temperature rise is hard to get rid of? Or something else? Commented Feb 28, 2021 at 15:31

According to datasheet:

But you're not driving the gate with 10V, rather it's close to 6-7V so RdsON will be higher.

Note this is a guaranteed maximum value. Actual value may be lower. As Andy says in his answer, if the FET follows the "typical" characteristics graph, its RdsON will probably be around 0.7 ohms, but notice the text in the corner of the "typical" characteristics graph that says "20µs pulse, Tc=25°C". And according to fig.4, RdsON increases with temperature, but according to fig.3, threshold voltage decreases with increasing temperature. There is a lot of dispersion in the threshold voltage value, so it's impossible to tell exactly how much except by measuring the FET, which isn't that interesting. And you might get that special snowflake, stubborn FET with a threshold voltage at the upper end of the guaranteed range, that really needs the extra volts on the gate.

It's a lot less headache to use the guaranteed RdsON spec.

The simulator screenshot you show says the FET will dissipate 3W, that could be pretty realistic. In any case, without a heat sink, it will fry.

Solution:

• Get a FET with lower RdsON, if you just want to switch a lamp you don't need a fast FET, so basically anything with RdsON lower than say 50 mOhms. No need for a 400V like IRF730 to switch 12V, you can use a 30V or 40V FET.

• Drive the gate with the full 12V by removing R2 and D4, also you could set R5 to a higher value to make sure the LED doesn't use most of the available optocoupler output current.

Note the opto isn't really necessary, unless you need the isolation.

Extra info:

RdsON is a compromise with speed and rated voltage. Higher rated voltage FETs will usually have higher RdsON. If you need the voltage rating, then there's no choice, but if you're not using the actual voltage the part is capable of, you still get the high RdsON.

Lower RdsON means a bigger FET which will be slower. So if you care about speed (say, in a switching converter) then it's worth it to spend time to find the right compromise. But to switch a lamp, you don't care about speed...

This FET is meant for fast high voltage switching, at the cost of higher RdsON.

So, set search criteria to RdsON < some value that means it will stay cool without heat sink, and hit "sort by price" button. Since you have 12V to drive the gate, no need to spend time looking for FETs that work with lower voltage gate drive too.

• For a specific recommendation, the common and cheap IRF540 will more than suffice. It's my go-to MOSFET when I just need a switch and don't need anything particularly fancy. Commented Feb 27, 2021 at 16:42
• Is there a simple explanation for the high on-resistance of the IRF730? e.g. the channel has to be physically long enough to not break down with 400V across it, therefore even when conducting the best it can, there's still significant resistance? And/or the gate insulation has to be thicker (lower capacitance), giving it less control over the channel? That might help to intuitively understand why most high-voltage MOSFETs would have this resistance problem (assuming that's true), and thus be a poor choice for this application even if driven with a higher gate voltage. Commented Feb 28, 2021 at 3:16
• @PeterCordes your intuitive explanation is good enough and pretty much not limited to MOSFETS. Anything (including a mechanical switch) rated for higher voltage will have higher Ron if other factors are kept similar. Commented Feb 28, 2021 at 7:39
• With the higher Ciss then lower output dV/dt and L/R ratio makes the diode redundant. Commented Feb 28, 2021 at 13:26

I've builded this simple circuit that has the purpose to swiching on ad off a 12 V / 25 W lamp

That's a load current of about 2 amps (25 watts divided by 12 volts).

The IRF730 MOSFET just isn't very suitable for a 2 amp load: -

Whether the gate control voltage is 6 volts or 10 volts you will get about 2.5 watts to 3 watts dissipated in the MOSFET and that needs a heatsink.

Or, go for a MOSFET with much lower on resistance. You should be able to find one that when controlling 2 amps, the drain-source voltage should be less than 50 mV (0.1 watt dissipated).

• @M4Biz that's fine but allow me to point out that the gate drive voltage for this application using this MOSFET need not be as high as 10 volts and, the on-resistance will be typically something like 0.7 ohms and maybe 1 ohm maximum. I'm pointing this out because the answer you accepted appears to be saying something a little different. The devil is in the details when it comes to electronics. Commented Feb 28, 2021 at 10:39
• I would rather err on the side of caution. The datasheet guarantees a max RdsON value, but the curve is only "typical". Also when it gets hot, RdsON will be higher, but threshold voltage goes down, so that could cancel out, or not, depending on how the stars align. Commented Feb 28, 2021 at 11:06

On the IRF730's datasheet I can read that it can handle more than 7 A

Never trust the current ratings given on a MOSFET datasheet. They are obtained under artificial lab conditions you will never achieve in practice and are essentially useless. You can use them as a metric to non-rigorously compare the relative current capability of MOSFETs, but that is about all you can use them for.

At the very least calculate the energy dissipated due to conduction losses with P=I^2R and use the package thermal resistance in the datasheet (whose value is PCB and heatsink dependent so think critically when reading the value and the conditions it is given under) to find the ambient temperature rise. ANd when you do this, take note of the conditions that the RDSon is obtained under.

Also, if you are pushing the part and expect it to get warm or hot, you can't use the advertised "front-page" RDson or even the one listed in the data table which are at room temperature. You must use the graph that shows RDSon vs temperature further down in the datasheet (usually near the bottom). I usually use the worst case value which ranges between 1.5x to 2x the values listed in the table which are at room temperature.

For high frequency switching, you must also account for the energy dissipated in the actual transition between ON/OFF and OFF/ON, but that is a whole other can of worms. This is not your usage case, however.

• On power MOSFETs especially, I've seen FETs rated for over a thousand amps in TO-247 packages, and I feel like at a thousand amps the package itself would melt. I don't know where they get these numbers from but it's not something operating in our universe. Commented Feb 27, 2021 at 20:08
• @Hearth Some MOSFETs have two front-page current ratings (neither of which are to be trusted): the "actual" current rating and the package limited current rating. I don't remember how I know this, but I have the impression they directly access and actively cool the die to keep it at a constant temperature or something like that to get those numbers. So I'm imagining a just test setup with pipes and pumps all around the die. Commented Feb 27, 2021 at 20:10
• I'm not convinced those numbers don't just come from simulations, personally. Commented Feb 27, 2021 at 20:11
• One of the numbers is based purely on Tj max, Rds and Theta jc. This is traditionally how they did it. They calculated a thermal limit based on the assumption that the CASE temperature is maintained at 25C. Recently in a question here, someone pointed out that this number had a disclaimer in the datasheet because this was higher than the leadframe could withstand without fusing. So at best it is a kind of figure of merit rather than actual design data. Commented Feb 28, 2021 at 8:57
• @mkeith That's kind of like listing the payload of cargo plane as how much weight it would take before the fuselage lost structural integrity, under the assumption that the wings could produce infinite lift. heh Commented Feb 28, 2021 at 19:19

There are a lot of improvements that could be made to this design.

Fundamentals that need to be taught

1. Vgs gate voltage for low resistance needs to be at least 250% time the worst-case threshold Vt ( Vgs(th) ) for classic FET types from 2 to 4V. This means 10V min. for Vgs The subthreshold types <=1V only need >= 200% times the threshold voltage.

2. The FET you chose for this design is the wrong type. It for high voltage (400V) and not high current (1 Ohm) ( there are about 50k different FETs to choose from. Why this?)

3. Since the lamp temp rise is over 2000'C you want the driver temp rise to be <= 20'C or <=1% of the load. a 25W lamp 12V * 2.08A while the resistance hot ~ 6 Ohms and cold ~0.6 Ohm due to the temperature coefficient of tungsten (Tempco. or PTC) going to > 2500'K from cold to hot is ~ 10x**

4. Thus you want a FET with RdsOn <= 60 mOhm and Vgs >=2.5x Vt which will dissipate <=1/4 Watt for a 25 Watt bulb. (<1%)

5. the last detail is how you chose to drive the LED which reduces the voltage available to your Vgs drive.

6. There are better optoisolators and the net current gain worst case is << 1 due to optical loss but this can still work if improved when not in saturation mode (Vce>=0.7) where hFE ( and thus CTR) may drop to 10% of max CTR but as emitter-followers , as you have done but without the collector Zener drop and Rc. Use worst-case values.

7. There are better indicator LEDs e.g. 5mm > 10,000 mcd at 20mA only need 1 mA as an indicator thus drive the LED with current limiting R and gate Vgs with the same emitter voltage and collector to Vcc. So expect <1V drop for Vce and confirm with datasheet and simulation like Falstad’s which I use often here.

8. This means you must start over in your design and choose 6mA input and <2 mA output but supply the current to the LED and the same voltage to Vgs.

Otherwise the rest looks ok.

But you could protect the LED from incorrect reversal which will damage the IR LED with a series diode on the input.

The simulation indicates Rbulb 300'K= 288 mΩ, 2850'K = 5.475 Ω
LED in = 7.7mA , LED out = 2.2 mA , Vgs= 11.8V

I've used this Electrical forum for the first time and I'm very impressed for the real high quality of all replays. I'm a teacher, as I said, and I've very appreciated all explanations supported by diagrams and datasheets. But let me go to my results (for now only simultated on Multisim).

1. Setting Vds to 10 V, changes very little because the Pd is about 2,80 W instead of 3 W.
2. Replacing the mosfet with one with RdsON of about 50 mOhms (like IRFZ34N) is the solution. As you see in the attached multisim simulation, the Pd is now only 140 mW and, I think, the Mosfet will be much less hot (I think it will be almostcold).

I've also a suggestion for forum's administrators. I've much appreciated that there is a service that provide grammatical correction on the post published by users that have not English as native speaker. Nerverthless I see this correction only after I've published the post. The question is: it would be possible to see this correction prior to publish the post. Thanks to all for your amazing support.

• It's a bone of contention for me that anyone with sufficient reputation can make an amendment and screw things up. Any amendment does need peer approval though so it's very infrequent that this happens. You can always "roll-back" an amendment in case something gets lost along the way. BTW, it's not a forum; it's a Q and A site. Commented Feb 28, 2021 at 10:43
• Grammarly App or similar provides user feedback when writing. Commented Feb 28, 2021 at 13:20
• The meta information in the beginning belongs in comments (or could be deleted). Very few readers interested in the result will have any interest in the meta information. Commented Feb 28, 2021 at 15:34
• Your suggestion at the end is a very good one, but it doesn't belong in an answer. It can probably be tolerated in comments (I can not recommend posting on the meta site as it most likely will be summarily downvoted (I could be wrong)). Commented Feb 28, 2021 at 15:40
• Hi Peter. Thanks for your tip. I'm enchanted by the real high quality of this Q/A site. Thanks to all. Commented Feb 28, 2021 at 15:51