I am designing a circuit where I have a microcontroller driving a four-channel logic level N MOSFET module. The simplest way for my circuit to work is to have the microcontroller run "HIGH" on the MOSFET gate for extended period of time (1 hour on for every 10 seconds off). Assuming the heat dissipation is within specifications, is it generally "acceptable" to have a MOSFET on this long? Or would an experienced circuit designer find a way to accomplish the same thing by inverting the logic?

For reference, I am designing an LED neon sign. One of the MOSFETs is meant to "flicker" the entire light. Turning the MOSFET off briefly cuts the power supply. Turning it back on reestablishes the power. The other MOSFETs turn on/off different parts of the light. I only have N channel MOSFETs, I am trying to avoid breadboards/resistors, etc. I believe I am able to dissipate enough heat to avoid a heat sink.

Here is a circuit below. The 100 ohm resistors are just placeholders. The bottom three 3.3 V are the microcontroller I/O pins. The top 12 V is the LED power supply. I will add pull-down resistors to keep the I/O pins and MOSFET gates low.

Enter image description here


3 Answers 3


No, it is not necessarily bad practice to have a MOSFET always on. They are frequently used as power switches.

Since you only have N-channel MOSFETs, you probably want to do it something like this:


simulate this circuit – Schematic created using CircuitLab

This is called low-side switching. You have done that properly with your two elements, but you need to move your main switch to the low side as well. This way your signal inputs are ground referenced. If the sources of your MOSFETs are not at ground you won't be able to activate the gate properly, because how 'on' a MOSFET is depends on the gate to source voltage.

SIG1 turns the whole thing on and off, while SIG2 and SIG3 control the individual segments.

You also will want pull-downs on the gates so they turn off reliably.

Make sure you use logic-level MOSFETs that will be fully turned on by 3.3V (don't use the threshold voltage for this).

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    \$\begingroup\$ Thank you for this. It looks like its important to always keep the source pin of the mosfet grounded, that way, when you apply 3.3V at the gate, you are actually getting the full 3.3V difference between the source and gate. If you see in my circuit, the mosfet circled green by Andy, after the source pin there is a resistor then goes to ground. Does this not pull it low? I do see that in my original schematic I have the load downstream of the mosfet switch. I remember I read you are supposed to have the mosfet switch downstream of the load, no doubt due to the issue you are describing. \$\endgroup\$ Commented Jan 5, 2021 at 16:06
  • \$\begingroup\$ @domerwannabe Current passing over that resistor will cause a voltage drop. This means the source of that MOSFET will not, in fact, be at ground, but at whatever voltage that resistor is dropping. If you have a resistor (or load) it needs to be at the drain. \$\endgroup\$ Commented Jan 5, 2021 at 16:08
  • \$\begingroup\$ Okay, I will rearrange the circuit to ground the source pin. I do understand how currently the voltage prior to that resistor will be at the value of whatever the resistor is dropping. I suppose in my mind, how its setup is identical to having a pulldown resistor. I am trying to understand why a pulldown off the I/O pin turns the mosfet gate and I/O low, but does not put the mosfet source low. \$\endgroup\$ Commented Jan 5, 2021 at 16:24
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    \$\begingroup\$ I am supposing that it has to do with the fact that (incidental) pulldown is wired in parallel, and not in series. \$\endgroup\$ Commented Jan 5, 2021 at 16:27
  • \$\begingroup\$ Yes, that is a great way to look at it (parallel vs series). The gate pull down resistors are to drain the charge from the gate's capacitance. When you first turn the signal off to the gate, the voltage stays high while it is draining over that resistor (for the same reason mentioned above), but as the capacitance drains the voltage goes down, eventually no more current flows, and at that point the voltage across the resistor is now zero. \$\endgroup\$ Commented Jan 5, 2021 at 16:34

Here's the problem: -

enter image description here

You need a PNP BJT or a P channel MOSFET to adequately switch on or off the 12 volt rail. And, whatever transistor you choose, you'll need a ground referenced transistor to activate the top-side transistor because you can't use a 3.3 volt logic signal on a transistor connected to 12 volts.

Apart from that you need gate pull-down resistors of circa 100 kΩ to 1 MΩ.

  • \$\begingroup\$ Thanks Andy, Yes I will add a pulldown. Can you explain what you mean by a source follower? \$\endgroup\$ Commented Jan 5, 2021 at 15:17
  • \$\begingroup\$ Source follower (or common drain). \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2021 at 15:19

Assuming the heat dissipation is within spec, is it generally "acceptable" to have a mosfet on this long?

There is no reason not to have a MOSFET on forever, provided it's within heat limits.


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