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I have a 3.5mm audio jack with a switch build in. When you plug a jack in the switch opens. I would love to use this build in switch as a power on switch for a small device so when i plug in a head phone jack it closes a circuit and powers on the device. I don't know how to use the open circuit of the audio jack to close a power circuit.

Can anyone help me figure this out?

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  • \$\begingroup\$ Is the switch in the jack plug or the jack socket? \$\endgroup\$ – JIm Dearden Mar 2 '17 at 17:08
  • \$\begingroup\$ Which configuration of switched jack socket is it? There are many. Please post the switch type schematic. Also what are you plugging in? Is it really a two channel headphone jack or a single channel device. \$\endgroup\$ – Trevor_G Mar 2 '17 at 17:19
  • \$\begingroup\$ BTW.. your title is miss-leading... this has little to do a signal invertor. \$\endgroup\$ – Trevor_G Mar 2 '17 at 17:22
  • \$\begingroup\$ Its a simple mono 3.5mm audio jack with an extra part that is pushed apart breaking the circuit when a head phone jack is inserted. Maybe i used the wrong word but i wanted to invert the signal so low would be high and high would be low. \$\endgroup\$ – jardane Mar 2 '17 at 18:40
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I figured it out, i just needed to use an NPN transistor. enter image description here

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  • \$\begingroup\$ Your load will then not see GND but GND+Vce,sat. If that's good enough for your application then fine! If not, then you may want to consider a high side switch instead - implementable with a PNP for example. \$\endgroup\$ – Enric Blanco Mar 2 '17 at 23:03
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Just use a pull-up resistor to Vcc and a connection to GND and you're set. Use the ENABLE signal (active high) for triggering power-up as needed.

schematic

simulate this circuit – Schematic created using CircuitLab

Or you could also use the jack switch to turn ON a PNP BJT and switch a load (or a relay if you want to switch higher voltages):

schematic

simulate this circuit

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  • \$\begingroup\$ Maybe i am just being thick but how would the pull-up resistor work? Not sure what the Enable Pin means in this example. What would a full diagram look like if you were using a 3V battery the audio jack and the pull up resistor for example. \$\endgroup\$ – jardane Mar 2 '17 at 18:17
  • \$\begingroup\$ The above circuit is broken. I'll try to correct it later if the person who posted this answer doesn't get to it first. \$\endgroup\$ – Dwayne Reid Mar 2 '17 at 20:04
  • \$\begingroup\$ Im glad its not just me \$\endgroup\$ – jardane Mar 2 '17 at 20:06
  • \$\begingroup\$ Yes, there's an error in the second schematic. I can't correct it now, I'll do it when I'm in front of a PC. Thanks, @DwayneReid ! \$\endgroup\$ – Enric Blanco Mar 2 '17 at 20:12

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