Yea the MCU that's providing this signal is an STM32 board, so the signal really gets low. I ordered some 2N7000's to try using that as the switch to the system, but I'm looking in to working a stopping RCD snubber into my circuit, I'll let you guys know what I find. I'll be checking up on this thread over the weekend, and thanks again for the input everyone.

I do not like inductive loads

 

You say the load is inductive but you also say the power goes to the converter after the switch - is the inductive load powered by the converter or by the switch?

Can you use an integrated load switch? :
https://www.fairchildsemi.com/datasheets/FD/FDC6324L.pdf

 

You say the load is inductive but you also say the power goes to the converter after the switch - is the inductive load powered by the converter or by the switch?

Can you use an integrated load switch? :
https://www.fairchildsemi.com/datasheets/FD/FDC6324L.pdf

The load is powered by the converter, but the converter cannot be turned off to disable driving the load because I'm doing this project on a prototyping board where access to its enable pin isn't available, it's hardwired ON (or high), so this buck converter will keep trying to pull in current to run the modem it's powering unless I cut off the power running to it. I'm designing this switch as a quick solution until we get more resources available to make our own PCB to which I can avoid this problem altogether

And I'm gonna try and pick one of those up, although the internals of the kit you linked me to look exactly like the circuit I'm about to set up so I'll wait and see if replacing the BJT with an N-channel MOSFET works...

Thanks for the link though! I might use that package in a future application!

 

If you start with the BJT off, power flows through the inductor and D/S on the MOSFET. Once you activate the BJT, you begin pinching off the MOSFET and you will see current through the inductor fall, while current throught he 100K Ohm resistor climbs.

At this point, electrons are now flowing around the inductor and the MOSFET, through the BJT & resistive paths to the 12V source.

If you again turn the BJT off, you will see current through the 100K resistor fall, current through the inductor (and MOSFET) climb, returning to the original state.

I didn't see what your inductor was, so I simply assumed 1nH for my calcs and I assumed a 260-Ohm resistor for R1, since you'd specified a range.

The BJT needs a pull-down resistor on it, or it may never fully turn off, because your inductive aspect begins powering the BJT through the BE junction. Unless you have a pull-down (like 2K-Ohm) on the base of the BJT, once the leakage current drops below the knee, it won't discharge much further, and may quasi float near the knee because it takes so long to get there (in electronic time).

Poorly designed circuit.

 

When you pinch off the fet, it causes current to flow in such a way that your resistor paths become pullups, turning the fet back on.

I thought we were talking about 2N7000 - you don't pinch those off, you enhance them.