I am attempting to us a voltage shut off switch with some pre-existing hardware. I do not understand transistors well enough yet and am looking for clarification. I desire for V2 to be able to go to the output but also be able to shut it off.

Q1's part number is EMH1T2R and it has the two resistors internal. When I turn on M1, Q1's base only comes down about 100mV.

What am I missing here to be able to shut Q1 completely off?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Are you interested in using MOSFETs exclusively? Or would you be willing to consider BJTs where you currently use MOSFETs? If you would, can you also add some information about what's at the other end of \$M_2\$'s drain? You show \$R_6\$ as a kind of "pull-down." But this doesn't say much about what's "observing" that voltage and what current compliance it may require. Can you talk a little about what's being attached there (at the output?) \$\endgroup\$ – jonk Jun 7 at 5:27

The IRF530 FET has a high gate threshold, and will not reliably turn on with 3.3V gate to source. Use a logic-level FET that's specified for 3.3V gate-source, or use another EMH1T2R in place of M1.


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