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How to install a patch point?

Discussion in 'General Electronics Discussion' started by Yoa01, Aug 6, 2012.

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  1. Yoa01


    Jun 18, 2012
    I know, that title has horrible grammar. I'm sure you know what I'm talking about, though: I want to know how to install a patch point by a potentiomenter. I've tried Googling this and apparently it's a very well-kept secret among modders and such. What I want to do is this:
    At first, it would just be the pot's output signal to the wire on the far right. But, using a switchable jack (I have a couple and drew one here), I want it to be so that when the jack is not used, the pot reacts normally, but when it is being used the output is that only of the jack.

    This has been done on literally millions of things, yet I can't seem to figure it out. I've tried this design and a couple variations, but it seems I'm off somewhere. Can anyone help?
  2. shrtrnd


    Jan 15, 2010
    I don't know your application.
    If you're using the phono jack just like any device that ouputs to a speaker or a headphone
    type jack, they're designed for that purpose. (To work only the speaker or the headphone
    when you plug in the phono plug).
    Is your question about how to wire-up the headphone jack, or is there something else
    you're asking?
  3. Yoa01


    Jun 18, 2012
    Right, sorry, should have specified. I don't know how familiar you are with analogue synthesizers, but some models are/were "semi-modular,' meaning that you had the normal pots and such affecting the sound, but you could also use a patch cable and use whatever signal was going through that to replace the pot. It could also use the pot as sort of a guide, whereas whatever signal was going through the cable/jack would use the pot's setting and stay within that range, which was more common.
  4. gorgon


    Jun 6, 2011
    What is the impedance for the (patch) cable? Is it 600ohm line impedance or something else? Does the attenuator need to match impedance?

    TOK ;)
  5. Yoa01


    Jun 18, 2012
    And here is where I again feel like a total noob. After a bit of Googling and guesswork, we'll say 600, sure, but it seems a bit high. I'm using a 6" George L's instrument patch cable, if that helps at all.

    And what attenuator, the pot? If so, then I would assume so.

    I'll be honest, I'm not entirely the most electrically savvy person. Sorry!
  6. Yoa01


    Jun 18, 2012
    Ok, I also asked for help from Kevin Lightner (of, and he replied with exactly what I was thinking:

    "If you want a jack to over-ride a pot, the output from the pot (the center terminal) should be hooked to the normal on the jack. That is, when there's no plug inserted, the normal shorts to the tip of the jack and the pot sends its signal to where you want it. When a plug is inserted, it removes this normalled connection and instead the signal on the plug hits the tip and substitutes whatever the cable is carrying."

    My apologies, I always seem to have troubles explaining stuff. Thank you, though!
  7. Yoa01


    Jun 18, 2012
    Sorry to bring this topic back up, but I've done some research into the subject and have come up with three possible ways to do it, along with how to make a CV attenuator. Only problem is, I'm not sure if any would work.

    From what I've seen, most of these are using pots as rheostats. The CV inputs are mono jacks, signal outputs would be replacing the signal the pot was outputting (wiper position). Here are my conclusions:
    Let me explain:

    A: This was used in a mixer, with the setup like the grey input section here:
    B: This, I think, would make the CV input signal run through the pot and replace the wiper value with the CV value. At least, that's what it's described to do.
    C: I think this would work as a 5v attenuator.
    D: This one is the setup I referred to in the last post, using a switched jack, which seems the most logical.

    What I want to know is, would any of these (aside from C, which would probably work)take a CV signal, say, +3v, and use that to replace what the wiper is set to (say, +1v)? Say the use is to control filter cutoff.

    Thanks! And ask if you need clarification. I'll try.
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