Go Down

Topic: My mosfet is heating, can't understand why (Read 11772 times) previous topic - next topic

retrolefty


Quote
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.

dhenry

Quote
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.

Grumpy_Mike

Quote
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.

dhenry

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


Why do you always want to minimize component heating?

retrolefty


Quote
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.


dhenry

Quote
Arduino digital pin 2, ground, 5V pins are connected to this mosfet module : http://tinkerkit.com/en/Modules/T010020


This module is actually quite interesting. Its description says:

Quote
Module Description: This module features an IRF520 power MOSFET transistor, a kick-back diode, a standard TinkerKit 3pin connector,a signal amplifier, a green LED that signals that the module is correctly powered and one yellow LED whose brightness depends on the input signal received by the module.


It is not sure what they meant by "signal amplifier". Most times, "amplifier" is an analog concept and doesn't apply to digital devices (which are either on or off). I would argue that if they indeed implemented an analog "amplifier" here, they would have been dead wrong.

If, on the other hand, they meant a "gate driver" by "signal amplifier", and the design is correct - that the gate driver would have driven the gate to sufficient levels, you are simply switching too much current in this case. ie, the mosfet has been turned on hard and it is just too much dissipation causing a failure.

It would help if you can read off the markings on the "signal amplifier".

Grumpy_Mike

Quote
Most times, "amplifier" is an analog concept

Not too sure where you learnt electronics but an amplifier is something that makes signals bigger, nothing to do with digital or analogue.

Quote
you are simply switching too much current in this case

No I don't think you get this at all. You don't say a mechanical switch is switching too much current, it is a switch with the smallest possible resistance. The same goes for a FET, you want the current to be determined by the load not the FET, especially not the paramaters of a device. Any current control needs to be a function of the design.

retrolefty

Well even a simple digital logic gate requires internal amplification, but it would be for current amplification not voltage, so to allow for a larger 'fan out' signal strength. A analog transconduction amplifier can be designed to provide significant current amplification while maintaning a 1:1 input to output voltage.

dhenry

I found a datasheet for an earlier version of this thing. If those guys got it right, it was designed to work with a 5v logic signal and this "signal amplifier" is a gate driver. The datasheet i saw indicated a puny irf520, so it isnt meant to handle lots of current.

Also, make sure you are using this to switch a DC source / load.

SurferTim

#24
Oct 01, 2012, 03:14 am Last Edit: Oct 01, 2012, 03:17 am by SurferTim Reason: 1
I presume the voltages you posted were supply voltage and voltage across the load. I can get the Vds from those.
14.8 - 14.3 = 0.5v
That is much higher than it should be at full on. That will generate some heat, even at 5 amps.

Did you measure Vgs while the mosfet was on?
add: that is the voltage from the gate pin to the source pin (ground) on the mosfet.

dhenry

The driver is a low-side gate driver.

Pin 1/2 (ch a) not used. pin 3 is the input signal (non-inverting), pin 4 ground. pin 5 output, pin 6 vdd, pin 8 gnd.

From the marking, it looks like FAN3229. Ch a not used.

The orange led is driven by the input.

If so, input drive level isnt the issue.

SurferTim

@dhenry: Thanks for that info. I did not find the schematic. Is VDD on that driver the load supply? I'm beginning to wonder about that tho, because the device you mentioned has a maximum supply voltage rating of 18 volts, and the module says it is good to 24 volts. It should be ok for 15 volts tho.

Here is the deal. There are two devices. A driver and a mosfet. It appears one is not working correctly. If the VGS is 10 volts or more, the mosfet is malfunctioning. If the VGS is lower than about 8 volts (maybe around 5?), the driver is malfunctioning or improperly designed.

Doesn't that sound logical?

vams

I could not find any schematic of the module, so if you could link the one you find i will appreciate, thanks.
I'd like to remind you that, notwithstanding what's written in the tinker kit site, the transistor in my module it's a IRFS803 (http://www.alldatasheet.com/datasheet-pdf/pdf/227583/IRF/IRFS3806PBF.html).

I'm quite sure now that the problem was an insufficient gate voltage.
Quote
Did you measure Vgs while the mosfet was on?
add: that is the voltage from the gate pin to the source pin (ground) on the mosfet.

it was too difficult to measure the gate-source voltage, the circuit it's in a bad position and with only two hands i can't measure it :)

Quote
I presume the voltages you posted were supply voltage and voltage across the load. I can get the Vds from those.
14.8 - 14.3 = 0.5v
That is much higher than it should be at full on. That will generate some heat, even at 5 amps.


yeah it's too high, that makes 0,1 ohm of resistance, while it should be 0,015 if fully on. That was with 9 volte at gate, today I'll try 10 V (don't know how) and we will see.
The fact is that now it looks quite useless driving a 15 volt with 10 volt...if i want to control it with arduino i'll have to do multiple fets switch? looks boring :(

Thanks to everyone.

dhenry

vdd is on the mosfet supply side. So the mosfet's gate is driven to its supply voltage - assuming Vdd meets uvlo, to protect the mosfet.

The schematic should be very simple, and you can figure it by looking at the pcb. It appears to be a 2-layer pcb in this case.

fungus


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 ?


Power is the voltage drop across a device multiplied by the current passing through it.

MOSFETS only get hot when they're not completely switched on. When they're completely on or off their resistance is so low/high that the power dissipated by the MOSFET is negligible.

They can get warm when they're switching at high speeds because they spend some time in the intermediate zone but I assume you're not doing that.

So...measure at the voltage on the gate pin then grab the datasheet for the MOSFET.
Advanced Arduino

Go Up