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I know that IRFZ44N is not a logic level Mosfet, but I need it to control only 12v 1.2A (14.4W). Can I control it by applying only 5V on gate from an atmega microcontroller? I checked the datasheet and I also tested on breadboard by simply appling 5V from USB, and it seems to work fine, it remains cold all the time.

schematic

simulate this circuit – Schematic created using CircuitLab

I am new to electronics. Sorry for probably stupid question.

I tried to find the question on internet, but all the time people does not recommend to control a non-logic level mosfet directly from an microcontroller. Maybe they need to control high current circuits, but in my case it is just a short led strip.

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    \$\begingroup\$ For industrial design it is not the best way to work on the edge. For home project it is OK. If it works, use it. \$\endgroup\$
    – user263983
    Commented Nov 17, 2021 at 13:30
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    \$\begingroup\$ You need to put the pull down resistor on the MCU 5V pin, on the gate is acts as a voltage divider. Granted it is not a lot but eac millivolt means a lot to the MOSFET at these low drive voltages. The MOSFET is not logic level rated and at 5V will not be fully enhanced. Best chose another MOSFET. It will work but expect it to get warm. \$\endgroup\$
    – Gil
    Commented Nov 17, 2021 at 18:48

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It's a borderline scenario. See this extract from the Vishay data-sheet that you linked: -

enter image description here

I say "borderline" because the graph above is typical i.e. it's an average scenario and won't cover the extremities of how maybe a hundred devices (or more) might behave.

With 5 volts between gate and source, the IRFZ44N will typically drop about 0.1 volts when handling a drain current of 1 or 2 amps. It might rise to (say) 300 mV in some extremes and, it might be as lows as 30 mV in other extremes.

If the MOSFET drops 300 mV whilst conducting 1.2 amps, the power dissipation would be 0.36 watts and, due to self-heating, the MOSFET's conduction resistance may rise a bit causing the worst case volt drop of 300 mV and ambient temperatures to be more like 0.5 volts. Now, it's dissipating 600 mW. However, if your external load is defining the current flow, then you don't need to worry about thermal runaway.

Can you circuit handle that? Is it too much? Is there too much heat and losses?

Only you can say.

Check also the differences between other supplier offerings and the Vishay data-sheet that you linked. Make sure you use the latest data-sheet when doing this.

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From the datasheet:

enter image description here

At room temperature, with drain current \$I_D = 1.2A\$, and gate-source voltage \$V_{GS} = 5V\$, you can expect to see less than 0.1V between drain and source.

I imagine that's acceptable in this application, so for drain currents less than a couple of amps, this transistor can be used with 5V logic signals.

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  • \$\begingroup\$ A logic-level Mosfet is needed so use one instead of picking through hundreds of 10V Mosfets to find one that works with only 5V. \$\endgroup\$
    – Audioguru
    Commented Nov 17, 2021 at 19:34
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Let's have a look at the datasheet:

enter image description here

5V at the Gate is the second curve from the bottom of the graph. It seems to be OK for it to conduct enough.

But, looking at another graph:

enter image description here

5V at the gate is below the crossing of the two curves. It means that you are in a region where could be a thermal runaway : when the temperature rises, more current will flow and the temperature will rise more...

This could be risky for the chip if internally it is made of smaller parallel mosfets. At the end, only one of them will have to conduct all the current. And it may be too much for one single little mosfet.

source: On Semi - Thermal Stability of MOSFETs

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