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I have a project in which I have to "close a digital switch" using either a 5V or a 3.3V supply. The load will be at 12V, with a required current of around 200mA.

As a newbie I thought to use an N-channel MOSFET such as the FQP30N06L datasheet.

While the transistor will actually provide me more than enough current, according to the datasheet, either at 3.3V or 5V, I will be in the saturation region and not in the ohmic region.

I thought I was fine with that, but reading around I found out you should be in the ohmic region to use the MOSFET as a switch.

Why so? Also: What are the downsides of my configuration (i.e. switching in saturation region)?

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    \$\begingroup\$ No, you should be firmly in the ohmic region at 200 mA drain current and a Vgs of 5 volts. \$\endgroup\$
    – Andy aka
    Commented Jun 25, 2018 at 11:22
  • \$\begingroup\$ Are you saying that I am reading the datasheet wrong? It could be, thanks \$\endgroup\$
    – Moriambar
    Commented Jun 25, 2018 at 11:26
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    \$\begingroup\$ That and you need understanding how a MOSFET works if you want to understand saturation region (vs triode region). In saturation region the MOSFET will determine the current. A switch lets the load determine the current so a MOSFET in saturation is not a nice witch. To use a MOSFET as a switch, simply apply as much Vgs as you can/is allowed. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 25, 2018 at 11:32
  • \$\begingroup\$ Are you intending to use the MOSFET to switch the low side of the load (i.e. close the circuit to GND)? Or is the MOSFET intended to switch 12V to the high side of a load that is referenced to GND? If the latter you should be looking at a P-channel MOSFET. \$\endgroup\$
    – Michael Karas
    Commented Jun 25, 2018 at 11:38
  • \$\begingroup\$ I am going to switch the low side. \$\endgroup\$
    – Moriambar
    Commented Jun 25, 2018 at 11:40

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The ideal switch has infinite resistance when off and 0 resistance when on. When a MOSFET is being used as a switch, you want it to look as much like 0 resistance as possible when on. This happens when the part is being driven hard, which is in the ohmic region.

However, don't get hung up on labels. Look at the datasheet and see what maximum on-resistance is guaranteed for what minimum gate voltage. If you can supply that gate voltage and that on-resistance is acceptable, then go ahead and use that FET.

If you had provided a link to the datasheet of your FET, I could have commented on its suitability to your problem. My jellybean FET for such purposes is the IRLML2502.

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