N-type Mosfet as switch

Hello

I am working on a project that uses a IRFZ44N mosfet as a switch to turn disconnect a load that is connected to a battery. I am using a LM393 comparator to turn of the voltage at the gate when the battery output becomes 16V. I have gotten it to work at 21V and up to 2A. The problem that I am having is that I need it to be able to run at 2.7A, but the second I go over 2, the mosfet begins to overheat and stops working. Everything else in the circuit seems to be working as intended. I have tried to use the mosfet in both the low and high side, but the issue recurs, I also used an IRF3205, which is supposed to be a high current mosfet but this one suffered the same fate. My setup has the gate connected to the LM393, which sends 5V as long as the battery output is greater than 16V, drain is connected to the load, and source to (-) on the battery. Would someone please be able to tell me why this might be happening and if here is a specific type of mosfet that would suit this application better. I saw I might be able to use a low-side switch but I had never heard of these devices before, so I appreciate any help with this.

IRFZ44NPBF Infineon Technologies | Discrete Semiconductor Products | DigiKey

The LM393 is an open-collector device, therefore largely unsuitable for driving FETs without use of a pull-up resistor of some sort. Subsequent to that, a 5V drive level is a bit weak for a device with a 4v threshold. Even so, either device named should be able to deal with a ~3A load with a Vgs of 5v.

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I have a 2.8K ohm pullup resistor which draws power from a 5V regulator. Should I be drawing that power straight from the source instead?

Hello @esteban.montes,

You may be interested in this article. It provides some of the background describing why MOSFETs are difficult to drive. In your case, it sounds like your MOSFET is entering a high channel resistance condition causing high I^2R losses. You may have better performance using one of the MOSFETS mentioned in the article.

Please let us know if you were successful.

Best wishes,

APDahlen

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@esteban.montes, I have to ask vs assume- do you have the mosfet bolted to a proper heatsink? - with some sort of thermal interface material (thermal paste, heat pad, etc)?

As @rick_1976 pointed out, this FET requires a more voltage to ‘fully turn on’ than one with a lower Vgs threshold, so it’s going to act far more like a resistor (generating a lot of heat) than a fully-on switch.


Assuming you are heatsinking it well, the STP60NF06L looks like a far better option for being driven with 5V…

Thank you, this has been very useful!

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Yes, I realized that only 3.5V were getting tough to the gate and so I fixed that. As for the heat sink, I am waiting on some thermal adhesive to arrive given that right now, i just have the heatsink taped to the mosfet.

Hello @APDahlen and @Brian_German ,
I read your article and implemented a voltage regulator to make sure the mosfet constantly gets 5V at the gate unless disconnected by a comparator. I tested it in the electric load at 21V and 1A, it shot up to over 200 degrees (f) almost instantly. Previously it was able to get up to 2A before heating up.do you know why this might be happening?

Hello @esteban.montes,

Is your design related to this simulation (similar but not exact parts used):

My concern is that you have a 5 VDC regulator in place of the 12 VDC regulator. This would make the MOSFET very hard to drive.

Looking forward to you response.

Sincerely,

Aaron

Not exactly, I am trying to make a low voltage disconnect for a M18 Milwaukee battery, the 5V regulator is only being used to power the comparator and to set the reference voltage at which to do the cutoff.
The mosfet drain is connected to the return from the load and source is connected to ground. The gate is connected to the comparator and a 10V regulator, so the comparator sends those 10V to the gate as long as the battery output is higher than the reference voltage on the comparator, which I have set to 16V. When the battery output goes below 16V, the comparator reroutes what comes from that 10V regulator to ground, which will make the tool get disconnected from the battery.

Hello @esteban.montes,

I think I understand, but could you please include a schematic.

Sincerely,

Aaron

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Hello @esteban.montes,

This may help.

  • common ground for all

  • If I recall, the comparator is an open drain/collector device. It needs to work against a resistor (10 k in this example). The comparator output current is typically in the 20 mA range. As drawn in your original schematic, the 10 V regulator likely destroyed the comparator.

  • The 1 k prevents MOSFET overcurrent when the MOSFET is turned off (value not critical).

  • R6 forms a undesirable voltage divider with the 10 k resistor. Recommend replacing it with a 15 VDC Zener diode.

Sincerely,

Aaron

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My apologies, I forgot to draw a resistor, I have a 1.5K where you drew the 10K. I will make the changes you have suggested and give it a try. Thank you

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Hi @esteban.montes ,
I’d add some positive feedback to the comparator circuit to increase the hysteresis. Depending on the impedance of your voltage source (battery), the shown circuit may become unstable and start oscillate at the trip point.
In addition, the shown circuit consumes some milliamps of current while in “low battery” state, draining the battery eventually. Maybe that is not a concern, but worth pondering.
Cheers, heke

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True @heke!

That instability could also cause the MOSFET to overheat.

Sincerely,

Aaron

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Hello @heke
thank you for the feedback. I am aware of the consumption of the battery while in low voltage so that is why I have it to cutoff at 16V instead of 15.
I am not entirely sure of what you mean about the positive feedback, this is my first time trying to make a circuit out of my classes.

Thank you.

Hi @esteban.montes ,

No worries.
A positive feedback is used to pull the comparator input(s) further away from the trip point, in order to prevent output oscillations. That pull is called hysteresis, i.e. the trip voltage depends on the output state of the comparator.
Take a look at here:

So, I’ve redrawn your circuit, adding AP’s suggestions, with some further modifications explained below the picture.

Batt_SW

The resistor R7 adds the positive feedback, Thus Turn ON trip point is 1.5V higher that Turn OFF. Added also a few capacitors and took the reference voltage from the 10V regulator, saving one chip. Note that in your original circuit the voltage divider for comparator + input "Vin" is not properly wired.

Alternatively your could use e.g. TPS26600 IC, which integrates all the functionality + adds reverse polarity protection and load short circuit protection.

cheers, heke

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Thank you @heke,

I was just getting started with this when I saw your note.

Nice to see only one regulator.

Sincerely,

Aaron

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