If I'm understanding correctly, the only pin of the FET connected to my generated negative signal should be the gate, which is perfectly acceptable

Unfortunately not. The 0V connections must be common.
You also have the MOSFET wired as a source follower, so the maximum voltage you can get at the source will be about 3V more than the gate voltage (say 8V from a 5V gate signal).

 

TI has several inverting charge pumps that output -5.00V for +5.00V in. If you want to roll your own I will post a circuit using two CMOS inverter ICs, two capacitors and one resistor that does the same. Here's one of the TI ICs:

LM27762 https://www.ti.com/product/LM27762

 

Descendants of Intel's ICL7660 still work and are low-cost.

https://www.ebay.com/itm/295917976434?itmmeta=01HYJ6PK89QAXEC8E9Y4F2SB1P&hash=item44e615fb72:g:MZYAAOSwJA5k9vkN&itmprp=enc:AQAJAAAA4FROaewgFMPmxCEycQlzq9e9xfzPqT10x9/yaw1Fpy/+j4g/2JlqNEO2OGkr1N2jsr2tvwmGDyeSl/D2smFmjOAtVSQ6rzjYEXO1Tgr7OWi8Y/aiC9IU/4LTGOUkokg6JFRp/ogrXWnOHOHUBdGRnRpl1qQaSw0SGWcyPZQJ6QUB3I8Y2JFz0eloyYeZQ/qpiCgEz2/zZiFUVLgFW8gbd/GgJg1wXx+AJ9p+QTN8ZYuwQ7YSYagI4UCzqUBh4O8qDsp3h4WSZoOTmujTTuzfs9F7Vkw0JPbCUvJA88/MT75q|tkp:Bk9SR6S02sb0Yw

 

Descendants of Intel's ICL7660 still works and is low-cost.

https://www.ebay.com/itm/295917976434?itmmeta=01HYJ6PK89QAXEC8E9Y4F2SB1P&hash=item44e615fb72:g:MZYAAOSwJA5k9vkN&itmprp=enc:AQAJAAAA4FROaewgFMPmxCEycQlzq9e9xfzPqT10x9/yaw1Fpy/+j4g/2JlqNEO2OGkr1N2jsr2tvwmGDyeSl/D2smFmjOAtVSQ6rzjYEXO1Tgr7OWi8Y/aiC9IU/4LTGOUkokg6JFRp/ogrXWnOHOHUBdGRnRpl1qQaSw0SGWcyPZQJ6QUB3I8Y2JFz0eloyYeZQ/qpiCgEz2/zZiFUVLgFW8gbd/GgJg1wXx+AJ9p+QTN8ZYuwQ7YSYagI4UCzqUBh4O8qDsp3h4WSZoOTmujTTuzfs9F7Vkw0JPbCUvJA88/MT75q|tkp:Bk9SR6S02sb0Yw

Forget about charge pumps and look at his proposed circuit in post #17.
There’s no common ground, and he is hoping to use the depletion MOSFET as a source follower switching 250V DC.
It needs a complete rethink.

 

Here’s a bit of a back-of-a-fag-packet drawing of what is required, and it achieves the normally-on function.

 

Forget about charge pumps and look at his proposed circuit in post #17.
There’s no common ground, and he is hoping to use the depletion MOSFET as a source follower switching 250V DC.
It needs a complete rethink.

Why is it a bad idea to use a MOSFET as a load switch in a source follower configuration?

 

Here’s a bit of a back-of-a-fag-packet drawing of what is required, and it achieves the normally-on function.
View attachment 322960

Thanks! This seems to be what I'm looking for!

 

Why is it a bad idea to use a MOSFET as a load switch in a source follower configuration?

Because the output voltage depends on the gate voltage.
With the gate at 0V, the source will be at3V.

 

Why is it a bad idea to use a MOSFET as a load switch in a source follower configuration?

In order for the voltage drop across the FET to be its absolute minimum possible value (lowest Rdson), the gate must be more positive than the source by several volts. For most power MOSFETs, 10 V is the recommended value. That is, higher than the circuit Vcc. There are gate driver chips that have a charge pump built-in for this.

ak

 

In order for the voltage drop across the FET to be its absolute minimum possible value (lowest Rdson), the gate must be more positive than the source by several volts. For most power MOSFETs, 10 V is the recommended value. That is, higher than the circuit Vcc. There are gate driver chips that have a charge pump built-in for this.

ak

Does this apply to depletion mode MOSFETs as well? Would I have a much lower Rdson at Vgs =10V vs Vgs =0V with a depletion mode MOSFET?

 

Does this apply to depletion mode MOSFETs as well? Would I have a much lower Rdson at Vgs =10V vs Vgs =0V with a depletion mode MOSFET?

Yes. Unlike JFETS, the gate voltage of a depletion MOSFET can be taken positive, and it may (depending on the device) reduce RdsOn.