# Mosfet on positive side of wire

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#### donperry

##### New Member
Most examples I've seen place the mosfet in line with the negative side of the wire pair (when used as a switch).
So the circuit is completed when the FET switches on and completes the negative to ground.

What I want to accomplish is to have the mostfet complete the connection of a wire to a positive side, not negative.
I hope i'm not to muffled in my explanation.

I basically have a remote wire i need to switch on to a 12V, and using GND is not an option.

#### ronsimpson

##### Well-Known Member
Use a P-mosfet.
OR
Using a N-mosfet you will need to have the gate at ground for off and then move the gate to 24 volts for the mosfet to be on. (the gate to source voltage should be 8 to 15 volts)

Use the P mosfet it is easy. (you can connect a G-S resistor to turn off the mosfet) This will keep the G-S voltage at zero for off. The gate will be at 12V from ground. To turn on the mosfet you will pull the gate to ground.
(this is a mirror image of how the N-mosfet works)

Here is an internet example where a transistor is added so a computer can drive a P-mosfet.
Power from +12V goes to Vout when the gate is near ground.

#### crutschow

##### Well-Known Member
To avoid possible confusion, the arrows in that diagram indicate the direction of electron flow (negative to positive).
Conventional current flow is in the opposite direction.

#### donperry

##### New Member
So the reason for the transistor in that diagram is to drive a regular fet from a MCU, If i am using a logic level P - Fet would it be still needed?

#### Colin

##### Active Member
Your "logic Level" would have to be 0v (low) and 12v (high) if you don't use a transistor.

#### MikeMl

##### Well-Known Member
So the reason for the transistor in that diagram is to drive a regular fet from a MCU, If i am using a logic level P - Fet would it be still needed?
Yes, a logic-level P-Fet with its Source tied to +12v begins turning on as the Gate voltage is lowered from +12V to ~10V. The Vgs at which it turns on hard is ~-4V (Gate is more negative than the Source, so the Gate voltage would be ~8V).

The output pin on your micro swings from 0V to either 3.3V or 5V, so the NPN is needed as a 'level-shifter".

#### crutschow

##### Well-Known Member
To turn off the P-MOSFET its gate voltage has to be essentially equal to the source voltage (Vgs = 0).
Since the source is at +12V, that means the gate drive voltage has to also go to +12V.
And since your micro output can't go to that high a voltage you need a level shifter transistor to do that, as MikeML noted.
It has nothing to do with whether the MOSFET is logic-level or not.

#### donperry

##### New Member
Understood

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