microcontrollermosfet-driverrelayswitch-mode-power-supply
The above is part of my project design. 'BRAKE_CTL_3' is output pin from uController. It sends 0 or 1 to drive Q4 to turn it on or off. 'BRAKE_LSD_3' connect to one end of Relay. Other end of relay is connected to +12V. I would expect my design to turn relay on or off. However, relay is always turned on(closed). When I measured the pin output at 'BRAKE_LSD_3', the voltage switch from 0 to 3.3V instead of 0V to 12V(ideally). I dont know whats wrong with my design? Can anyone help?
One big question is whether this is an AC load or not. The solid state relay you are using is meant to switch AC loads only and WILL NOT shut off until the zero crossing occurs.
If you are using it for a DC load, then it'll stay on once switched on, and will not turn off until the current goes to 0 (usually at zero crossing with an AC power supply).
If it is an AC load, then you may need to design your snubber more carefully to match your load and power supply. You haven't given us enough information to help you with that part, though.
If you are trying to switch a DC load, then you need a different type of solid state relay.
You cannot increase the voltage by connecting GPIO pins in series, as they all are referenced to a common ground.
To drive your relay you should do something like:
simulate this circuit – Schematic created using CircuitLab
Best Answer
Your MOSFET is connected upside down.
The Source pin shound be connected to "ground"/0V and the Drain pin should be connected to your relay/"BRAKE_LSD_3".
Remember that most MOSFETs have a "body diode" in parallel with the Drain and Source.
If you look at the symbol of your MOSFET you'll see a little arrow in the middle - that gives you a hint as to which way the diode will allow current to flow through the MOSFET no matter whether its turned on or off.