FET's are normally connected between Ground and Load, therefore the load is alway connected to the positive rail....like in the green block of picture
Is it possible to connect it between the load and positive rail like in the red block. The reason I ask is since the circuit will be used on a vehicle in conjunction with existing circuitry and would rather not have 12V running to the load all the time..... If that makes sense.
You have selected an N-channel mosfet that usually connects to 0V.
If you use it as a source-follower like in your second sketch then its gate must be at +22V to +24V for it to fully turn on. Its gate must be 10V higher than its source voltage to fully turn on.
Use a P-channel Mosfet instead. Its source goes to +12V and its drain connects to the load which has its other wire at ground. When the gate is at the positive supply voltage then the P-channel Mosfet is turned off. When its gate is at 0V to +2V then the P-channel Mosfet is fully turned on.
Either way, the 12V is NOT powering the load all the time. When the FET is turned off, as in your first drawing, there is no current, and thus no power.
Sorry, but when you refer to high side / low side.... Does high side mean when the FET sit between the load and +, and low side when the FET is between the Load and -?
R1 feeding 12V from +12V rain, causing T3 to open, when MCU goes high, T1 (N-ch) closes causing R1 and R3 to become a voltage devider, and with the right values, turning T3 (P-ch) on and become closed?
R1 feeding 12V from +12V rain, causing T3 to open, when MCU goes high, T1 (N-ch) closes causing R1 and R3 to become a voltage devider, and with the right values, turning T3 (P-ch) on and become closed?
Yes that should work quite nicely. A 2N7000 series would be good for the first FET.
Bob
PS: I can see gates open and close, just like doors. However your understanding of "open" is the same as an "open" circuit; no conduction. This conflicts with another logical interpretation, that when a flood gate is open, current can pass through. It is best to describe the FET switch as ON or OFF. A positive pulse from the MCU to the gate of T1 turns on T1, causing a negative pulse on the voltage divider to the gate of T3. This turns on T3.