jfetmicrocontrollermosfetswitching
I have a JFET that requires negative voltage to the base in order to close and 0V to be open. I have the following power rails: 5V , -12V & GND (0V).
What would be the simplest way to switch the JFET's base between 0V & -12V using a GPIO pin from the MCU?
Why not use a relay - it can be driven from the 5V logic, possibly with a transistor to help it and the relay contacts would just connect in parallel with the load. What you haven't said is how fast you need to do the load bypass switching (is there any PWM involved) and what the maximum load current is - this may drive you down the MOSFET route.
Here's the relay I'd consider if it is just to act as a dumb load on/shunt switch: -
BTW - in your 2nd circuit there is no gate over-voltage protection when the power supply is greater than 20V (most MOSFETs can deal with maybe 15 to 20V on their gate but not 50V).
Best Answer
You could tag a voltage divider off an PNP like this
simulate this circuit – Schematic created using CircuitLab
Note Vcc shown above is expected to be 5V. If your logic output is at a lesser voltage you would need to recalculate R1 appropriately.
Or if ground needs to really be ground, the circuit below may be better. Note the control signal inversion.
simulate this circuit
In both cases, you need to ensure Vhi of your output port is high enough to fully turn off the PNP transistor. If not, either use an open-collector output with a pull-up on the base, or drop the emitter voltage a little with a diode or two.