On by default switch

[kuon]

I have a power supply which has an inhibit pin that must either be left floating for the power supply to be on, or pulled down for the power supply to be off.

I want the power supply to be off by default when power is applied and be turned on when a control line is pulled high.

So basically, when the control line is low, the circuit must conduct, and when the control line is high it must be open circuit and let it float.

I would usually use a transistor with a pull up resistor, but the documentation of my power supply clearly state that the inhibit pin must not be pulled up.

Do you have recommendation on how to implement this?

[oPossum]

Normally closed (Form B) optocoupler

For example... https://www.mouser.com/datasheet/2/240/CPC1106N-1548424.pdf

[kuon]

Thanks, didn't think of the optocoupler approach. Will do perfectly.

[NiHaoMike]

So basically, when the control line is low, the circuit must conduct, and when the control line is high it must be open circuit and let it float.

I would usually use a transistor with a pull up resistor, but the documentation of my power supply clearly state that the inhibit pin must not be pulled up.

Transistor without a pullup resistor.

[David Hess]

A p-channel JFET or p-channel depletion mode MOSFET if you can find one will do that.

[magic]

To turn a p-channel JFET off its gate needs to be driven higher than voltage on the inhibit input which is unknown. For an n-channel JFET, negative voltage would be required.
A normally-on OptoMOS doesn't seem like such a bad idea in these circumstances.

[Jeroen3]

Perhaps you can also use a resistor. The current source on the pin is probably finite, so you could pull it down.
Then you only have to put enough voltage on the chip to turn it on again. With a slight loss in the resistor.