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Occasionally I am faced with a problem where I am faced with a device that needs two of it's inputs to be shorted together for something to activate (Think smoke machine or gate motor).

This is the circuit I have been using:

schematic

simulate this circuit – Schematic created using CircuitLab

The idea is that the resistance of the mosfet goes down when a signal is applied which simulates the short circuit. My concern is that this feels very unconventional.

Is this an acceptable method of doing this? are there more professional methods?

note: it isn't as simple as sending a '0' signal. That often isn't how these systems work.

thanks.

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  • \$\begingroup\$ You should really only do that if you have a common ground. You would be wiser to use a relay for the isolation benefits. \$\endgroup\$
    – Trevor_G
    Commented Feb 27, 2018 at 14:38
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    \$\begingroup\$ output2 is grounded. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 27, 2018 at 14:39
  • \$\begingroup\$ or an opto-isolator \$\endgroup\$ Commented Feb 27, 2018 at 14:39
  • \$\begingroup\$ Anyway, it resembles the open-drain output. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 27, 2018 at 14:47
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    \$\begingroup\$ @Trevor_G Ya, i figured that out after several hours of confusion \$\endgroup\$
    – Makoto
    Commented Feb 27, 2018 at 14:47

1 Answer 1

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The answer is, it depends.

If the gate drive is floating this will work fine, though note it has a polarity due to the intrinsic diode in the MOSFET.

schematic

simulate this circuit – Schematic created using CircuitLab

This however.. will not.

schematic

simulate this circuit

If you need to do multiples, you would need a common ground everywhere.

schematic

simulate this circuit

A better method would be to use a relay or better still an opto-coupled MOSFET.

enter image description here

The latter is also bi-polar due to the back-to-back MOSFETS which block the diodes.

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