Hello, sorry if termonolgy is incorrect, I need to get a simple on/off +12v signal from a feed wire, unfortunately the signal wire switches between +5v and +12v when it is off or on, it never switches off to 0v.

I need something that will sit idle, open circuit with a 5v supply, but will close the circuit/switch when it sees 10v or above

I had thought of using a 10v or 12v voltage regulator like a LM12, will these sit idle if voltage is below the regulated rating, or would it perform erratically if ran on too low a voltage

space is very tight, I am very restricted on the size,

I would really appreciate some help with this or any ideas I could try

It will be powering some LED lights, current will be below 1A

many thanks

GT390 (Rob)

 

As far as I know a regulator will just sag below its normal output voltage if fed with too low an input, or at least a 5v regulator I once worked with did.

My knowledge is limited but you could perhaps use a comparator set to go high on 10v feeding a MOSFET or something like that.

 

Do the 12v and 5v supplies share a common ground? Could you use the +5v as the ground to an optoisolator and feed your signal wire to the other pin through an appropriate resistor? Then, use the output of the optoisolator to switch your LEDs...

 

How much current are you going to draw from the 12V supply?

 

Use a comparitor with a reference voltage of 10 volts. Anything above will output 10 volts, anything below will output zero volts.

Connect the output of the comparitor to a nice beefy Darlington transistor that can handle your one amp load.

Good luck.

 

Use a comparitor with a reference voltage of 10 volts. Anything above will output 10 volts, anything below will output zero volts.

Connect the output of the comparitor to a nice beefy Darlington transistor that can handle your one amp load.

Good luck.

A 10V reference is hard to come by if your only source of power is the one you're trying to switch. There are ways to use a series zener to accomplish this.
A PNP Darlington will drop about 1.4V Vce when it's on. It will also not truly disconnect from the input until the output is below about 5.5V, unless the Darlington has no internal base-emitter resistors. The reverse current path is through the CB junction. An NPN Darlington has similar problems.
Attached is a circuit that does completely disconnect from the input when the input is 5V. Of course, there is no way to anticipate the input, so when it goes back to 5V, it will drag the output down to about 10V before it cuts off.
A relay can be used in place of the MOSFETs. The turn-off delay will be milliseconds instead of microseconds, but the isolation is basically perfect.