# My mosfet is heating, can't understand why

Hi guys,

I've created a circuit in order to activate a load (a little refrigerator) via arduino.
The circuit it's very simple :

Arduino digital pin 2, ground, 5V pins are connected to this mosfet module : http://tinkerkit.com/en/Modules/T010020 that is using this mosfet : http://pdf1.alldatasheet.com/datasheet-pdf/view/227583/IRF/IRFS3806PBF.html
The arduino will provide a Gate power of 5 V, that should be necessary to grant a passage of 10 Amps (i will need no more than 4,2 tough)

To the module it's connected also a 15 v DC power supply , that will power the load.
The load is a little refrigerator : 12 V DC required Voltage, 50 Watt of power. (yes i know that it's less than the power supply, but I tried anyway to use that power supply )

The circuit works, the problem is that the mosfet it's overheating.

What do you think it's the problem? I think it could be the fact that im using 15 v instead or 12 v, or maybe could be an inrush of current in switching on the load ?

It could just be that is how hot it gets when switching the load. Using a higher voltage will give you more current, do you know how much? Then take that current, square it and multiply by the value of the on resistance of the FET. What do you get? It could be you need a heat sink.

Here's my calculations :

load : 12 V, 50 W -> resistance = 2,88 ohm (looks strange?)
current at 12 V = 4.16 Amp
at 15 V = 5.20 Amp

Mosfet has a 15 mOhm resistance so, Power in the mosfet at 15 V = 0,4 Watt (datasheet say maximum it can dissipate = 70 W).

Tomorrow (it's night in my country now) if I find time I open the refrigerator and try to see if more things are written inside.

(as usual, thanks for your time)

That TinkerKit page says that module uses an IRF520. Is your module different?

edit: That would make a BIG difference. The mosfet you have on the link above has a Rds = 15.8 mOhm. The IRF520 has a Rds = .27 Ohm = 270 mOhm

Uhm you right they told it's the 520... but on mine module's fet it's clearly written :
"IRFS3806
P208J
5G 7I"

Have you measured the voltage across the mosfet when it is on? Maybe it is not on all the way?

I would never try to use 5 volt gate drive for a standard power mosfet. Either change to a logic level power mosfet or add a needed gate driver circuit to provide a full +10vdc of gate voltage.

I... think that a piece of 2 mm aluminum 100 mm on a side and drilled such that the device is mounted in the center, you might make it a channel (U shape but flat in the center for the Mosfet and a good Mica washer with a copious dollop of elephant snot on either side would work well. The transistor and the PCB (Heat-Sink) in the drawing is WAY too small. I don't remember the supply voltage but if it is under 15 - 18V... I again would recommend a "High Side" switch, a PNP transistor emitter from the +HV supply a 1K resistor from the PNP base to the to the + HV supply, a 1K resistor from the PNP collector to the gate of the moset and a 1K resistor from the base of the PNP to the collector of the NPN and 22K resistor from the collector of the NPN to ground a 1K resistor from the base of the NPN to the Arduino output... and set the pin hh to turm on the mosfet. I do hope that is understandable, IT is the esiest way to go but it requires a bread-board of some kind, Or you could remove the Fet from the PCB and use it to wire the new mosfet on the heat-sink which may well be too big But it is better to have too much... Much Better than having too little and if it is too much, they'res a lot of room for other power devices... Too. and there are very inexpensive "High Side" driver IC's that can be had from the 'bay or from Farnel or Arrow... You'll have no problems with either solution and I am partial to the discrete because I have some "Perfect" PCB's for that and other small jobs AND I have All the parts, the driver Ic's are of limited use a handful of small parts have many uses.

Doc

It is already on a circuit board with a "signal amp", according to the webpage. It is Arduino pin compatible.
http://tinkerkit.com/en/Modules/T010020

The question is whether it is working right or not. If the voltage across the mosfet is more than 100mv, then I would check the gate-source voltage to insure the "signal amp" is working. I'll guess that is the other IC on the circuit board.

Or strip the mosfet from the board and use it to connect to a larger Mosfet or a larger heat-sink (Preferred)... Been there and done that... Just once, My employer was watching.. With a fire extinguisher in his hands as a "Joke"...

Doc

So, today i've tried to power the mosfet with a 9V battery. I've taken some measure :
(dV_x stands for difference of voltage at the sides of x)
dV_supply = 14.8 V

It was not heating at the beginning, i let him go some minute and after that he started heating (much less than last time tough).
I guess the problem was an insufficient gate voltage...It's strange that the datasheet of the mosfet i think it's in (reading his name) says it should work.

It's strange that the datasheet of the mosfet i think it's in (reading his name) says it should work.

What makes you say that?
The data sheet you posted on the first post quite clearly says it requires 10V to fully turn it on, in the Rds(on) line.

You are making the classic beginner mistake of thinking the Vgs(th) threshold of 4.0V is only the point where the FET starts to turn on, not when it is on fully.

What makes you say that?
The data sheet you posted on the first post quite clearly says it requires 10V to fully turn it on, in the Rds(on) line.

You are making the classic beginner mistake of thinking the Vgs(th) threshold of 4.0V is only the point where the FET starts to turn on, not when it is on fully.

In the RD(on) line the datasheet say you need 10 volt on the gate to let 25 Amps pass through.
I just need 4-5 Amps, so I've looked for fig 1 pag 3, in which one you may see that 5 Volt allows 5 amps passing in without entering in the saturation area, therefore having the little resistance you expect from a conductor.
I did not look the gate threshold voltage, because I've learned (thanks to you ) that doesn't mean the fet it's fully on.

I would never try to use 5 volt gate drive for a standard power mosfet.

That will depend on the current you are trying to switch, among others.

I just need 4-5 Amps, so I've looked for fig 1 pag 3, in which one you may see that 5 Volt allows 5 amps passing in without entering in the saturation area, therefore having the little resistance you expect from a conductor.

No that is wrong.

In that mode your FET is not fully on. It is only with a fully on FET you get minimum heating. The fact that it says you can get 40A at 10V is besides the point, you need that FET to be on, the current will be limited by your load. Yes 5V will allow 5A to flow but what will the resistance be at that gate voltage. It will be high and that is why your FET is overheating.
The 40A at 10V is also mythical in that at 40A you will be burning off more heat than the device can stand. Such things are just spec talk.
So to stop your FET from overheating give the gate 10V, then it will run the coolest it can. However even at that you might still need a heat sink.

dhenry:

I would never try to use 5 volt gate drive for a standard power mosfet.

That will depend on the current you are trying to switch, among others.

Not true for me. When I want a transistor to act like a simple on/off switch, I want to be able to force it on as fully as possible (minimum Ron) to make the device dissapation (heat) as small as possible.

I want to be able to force it on as fully as possible (minimum Ron) to make the device dissapation (heat) as small as possible.

True for you.

But it doesn't mean you will always need to drive the switching device to the maximum allowable Vbe or Vgs - my only point here.

But it doesn't mean you will always need to drive the switching device to the maximum allowable Vbe or Vgs

Yes it does if you want to do a proper design and minimise component heating. Every rise of 10 degrees C half's the life of a device.
Mind you if you want to do a crap design then you are in good company on the net.

Yes it does if you want to do a proper design and minimise component heating.

Why do you always want to minimize component heating?

dhenry:

Yes it does if you want to do a proper design and minimise component heating.

Why do you always want to minimize component heating?

I think he state his reason, to maximize device long term reliablity. Unless the application is to heat something up, heat is very much an enemy of the circuit designer.