I've been struggling to put together a working circuit that uses 3.3 V to switch a 12 V relay (that then switches 240 V). Having tried a BJT solution (which didn't work), I then thought I'd try a MOSFET based approach. My reasoning is thus:

  • My 3.3 V power supply (powered by 12 V) can supply a limited amount of current, so don't want to expend much on driving the base of a BJT. Would prefer to rely solely on voltage and no/low current driving a MOSFET gate.
  • Similarly, driving the relay coil, prefer to use the 12 V rather than 3.3 V current.

I breadboarded my latest solution and it worked, i.e. a signal from my PIC pin was able to drive the 12 V through the MOSFET and turn on the relay. On that basis, I went ahead and got some PCBs manufactured. Testing the PCB circuit however, I find the solution is not working.

Measurements on the circuit show the following:

Signal Pin OFF Drain=12.2 V, Gate=0 V, Source=0 V

Signal Pin ON Drain=12.2 V, Gate=3.2 V, Source=2.6 V

Measuring current between Source and pin-1 of relay shows 0 A regardless of state of signal pin. FWIW, the resistance across the relay coil is 500 Ω.

I'm using a 2312 MOSFET which has the following description

The 2312 uses advanced trench technology to provide excellent RDS(ON), low gate charge and operation with gate voltages as low as 2.5 V.

Here is the circuit schematic. MOSFET and relay

Why isn't the MOSFET turning on?

Looking forward to finding out what's wrong with this circuit (btw, should also mention I have tried something similar in a low-side switching mode...also didn't work).

NB: Having read the response comments, I believe I must have wired the MOSFET on the low side when I breadboarded it, and then mistakenly wired it on the high side for the PCB prototype. Looks like I'll have to run off some more PCBs with the MOSFET on the low side.

I have now reconfigured the relevant part of my circuit to look like this.


Cheers, Mike


2 Answers 2


The relay coil needs to be connected between 12V and the drain. The source should be tied to ground.

Remember mosfets are controlled by the voltage between g and s. Not g and gnd. In your circuit once the source starts to conduct it rises therefore reducing Vgs until the fet starts turning itself off.

By tying s to GND you can make a solid 3.3V or 0V Vgs causing the fet to conduct between source and drain or not.

  • \$\begingroup\$ Thankyou for the explanation....I now see where I was going wrong. I was mistakenly referring to the voltage between gate and GND. \$\endgroup\$
    – NomadAU
    Sep 29, 2023 at 3:41

Invert the whole project. Keep Q1 Source on ground.
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