current measurementinductormosfet
So I have the below circuit simulation on LTSpice and trying to understand why whenever I set V3 Low or Hi I always get current through L1. M2 acts as an enable switch, so what I was expecting is that when V3 was high no current would flow, but that's not the case.
You can just use the PMOS, and use a pin switched from output low to high impedance (e.g. input) to switch it on/off:
The 10kΩ resistor holds the FET off when the ON-OFF signal is left floating (i.e. your pin is an input/high impedance)
Then when the pin is driven low, the gate is pulled to ground and the FET turns on. You need to make sure the P-ch MOSFET you use has a suitably low Vgs(th) (turn on voltage threshold - some can be quite high) so you want a "logic level" MOSFET.
Simulation:
Place the load between the 7.5 Volt supply and the MOSFET drain, and you will get switch-like behavior.
simulate this circuit – Schematic created using CircuitLab
In your current layout, the source is floating depending on current through the LED and resistor. Hence, Vgs is not 0 to 5 Volts as you assumed, but much lower, depending on where Source is floated to at a point in time.
What is needed is for the "switch" to conduct when given a positive signal from the Arduino, thus pulling its Drain node to ground (or close to it), thereby expressing the desired ~7 Volts across the LED + R3.
Best Answer
Sure. You are misusing M2. As a p-type it has a body diode with the anode at the source and cathode at the drain. So, whenever M1 is conducting, the body diode is forward biased, and current flows.
To do what you want, move M2 to the other side of the solenoid and reverse its orientation. That is, connect the source to your +12, and the drain to the solenoid.