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I've spent a few hours inside my car radio and I've found that when I short a hidden pin called AUX-ON to VCC, I can pipe music right into the it and it's treated like a regular AUX input. Great success.

Now, I have a switched jack that I'd like to use to short VCC to AUX-ON whenever I plug in a cable. The jack I'm using is Radio Shack 274-0246, the diagram is below. I can't figure out any way to do this without having the potential of grounding the audio to the vehicle ground (which could, I assume, distort quality) or without having the potential of sending 12v back through the audio cable into the phone. I know there MUST be a way to do this with a semiconductor or two, but I don't even know where to start. All I know is I need to pass current from pin 2 to 3 (or 4 to 5) and when that current stops (the jack mechanically separates the contacts when plugged in) then VCC can go to AUX.

enter image description here

EDIT: I forgot to mention, the unit is expecting balanced inputs, however.. I don't think I've ever seen an ipod or phone with balanced outputs. Should that be taken into consideration with this circuitry?

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  • \$\begingroup\$ Consumer equipment is off-topic. question should be closed. \$\endgroup\$ May 8 '12 at 9:10
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    \$\begingroup\$ This question is not about consumer equipment. It's about how I can sense the interruption of power and use that to trigger something else... I mentioned that it's in a radio just to provide some context, the radio has nothing to do with the question. \$\endgroup\$
    – kavisiegel
    May 8 '12 at 9:19
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    \$\begingroup\$ Why can't you just connect AUX-ON to VCC all of the time? \$\endgroup\$ May 8 '12 at 9:21
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    \$\begingroup\$ I am going to have to disagree with Leon, although this started as a consumer electronics question, the actual question here is "How can I short 2 contacts when I plug in a headphone jack" This is something I could see being a common question for many electronics projects. It is a little on the broad side, but I think it is acceptable. \$\endgroup\$
    – Kellenjb
    May 8 '12 at 11:25
  • \$\begingroup\$ @Cybergibbons - Because then the radio doesn't work. These switching jack diagrams were (and still occasionally are) very confusing to me, and I agree with Kellenjb that this is a good question. \$\endgroup\$ May 8 '12 at 13:43
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You don't even need a semiconductor.

Audio jack insertion detect

In this circuit, when no jack is connected, GPIO will be pulled down to (almost) ground R4. When you insert the jack, R4 becomes disconnected, and R2 weakly pulls GPIO up to AVDD.

In your case you'd have VCC instead of AVDD. And you'd have AUX-ON instead of GPIO.

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  • \$\begingroup\$ I've adapted your drawing to what I'm trying to do: i.imgur.com/CYTjp.png - is that set up correctly? And if so, how does that stop current from VCC to AUX at all - all that's between them is a resistor? Also, what does a capacitor do in this case? Thanks a ton! \$\endgroup\$
    – kavisiegel
    May 8 '12 at 17:07
  • \$\begingroup\$ @kavisiegel - The resistor won't stop the current, but it will reduce it to almost nothing. The capacitors will also block any DC voltage from the jack reaching the phone. \$\endgroup\$ May 8 '12 at 18:28
  • \$\begingroup\$ Did I mix up GND and AUX in my drawing? I just can't visualize how the current would flow here, heh. In a normal state, with nothing plugged in, I see the following circuit: (VCC) -> (J1 pin 11) -> (J1 pin 3) -> (R4) -> (GND) AND (VCC) -> (R2) -> (AUX) With the jack plugged in, VCC only goes to R2 and AUX, not ground. Am I misinterpreting? \$\endgroup\$
    – kavisiegel
    May 8 '12 at 18:55
  • \$\begingroup\$ @kavisiegel - GND shouldn't be connected to AUX when the jack is plugged in. You want AUX to be pulled up to VCC don't you? \$\endgroup\$ May 8 '12 at 20:06
  • \$\begingroup\$ That's correct, yes. If my adapted drawing looks correct, could you perhaps explain how it works a bit? Because I just can't grasp how VCC connects to AUX in any way except for through R2. It seems like all the switching is done on GND? (Side question - Is R3 actually doing anything?) \$\endgroup\$
    – kavisiegel
    May 8 '12 at 20:20

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