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Connecting an 'open collector' type data pin to a DC output - result?

Discussion in 'Electronic Basics' started by Jukka Aho, Aug 13, 2003.

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  1. Jukka Aho

    Jukka Aho Guest

    Hi!

    Here's the problem:


    DEVICE #1

    .-------------.
    .--||---||-----| 2.5 mm | "Tip" = 'open collector' type
    < || || | stereo | data port (0...5V)
    `--||---||-----| plug |
    `-----. | "Ring" = DC output (5V)
    T R S | |
    `-+---+-' "Sleeve" = GND
    | |
    | |
    | |//
    | //
    | //
    | //|
    |// |
    // |
    // |
    //| |
    | |
    DEVICE #2 | |
    .-+---+-. "Tip" = 'open collector' type
    | | data port #1 (0...5V)
    .-----´ |
    .--||---||-----| 2.5 mm | "Ring" = 'open collector' type
    < || || | stereo | data port #2 (0...5V)
    `--||---||-----| plug |
    `-------------´ "Sleeve" = GND
    T R S


    I intend to connect the devices together (for some remote
    control / programming experiments.)

    Coincidentally, the connectors and signals are almost identical
    on both devices, but Device #2 has two 'open collector' type
    data ports (on "tip" and "ring") whereas Device #1 only has
    one (on "tip"), and designates the "ring" connection as
    a +5V DC output instead.

    The second data port is not needed for this application;
    I would just like to safely disregard it.

    Now, what I'd like to know is what would happen if I dared
    to connect these devices together using a simple, directly
    connected cable like the above? To recap, both ends have a
    common ground, but one has an 'open collector' type data port
    (0...5V) on the "ring" whereas the other end would connect
    a 5V DC supply there. This "ring" data port is not needed
    for anything, but would I fry it or short something with
    this kind of connection?

    * * *

    The devices in question are a LANC camcorder and a TI
    programmable/hackable calculator. The intended application
    is to use the calculator as a wired remote control device
    for the camcorder.

    Of course I could simply disconnect the "ring" wire to be
    on the safe side. However, I would much rather use the
    standard pre-made cable that came with the calculator if
    it is at all possible (since this way others could too
    easily use the same program without having to build a
    special cable for themselves.)

    * * *

    I would appreciate any ideas and insights. Electronics is
    not really one of my strong points, and I have a bit hard
    time figuring out how the circuit would behave in this case.

    --?
    znark
     
  2. Steve

    Steve Guest

    I believe OC port number 2, while "not needed for anything"
    would effectively be short circuited to +5V when it did
    operate. That is not good. While I do not know the equipment
    you are using, the potential current through the driver would
    most likely exceed 100ma, and would most likely be destructive.

    At a minimum, I think you need a series resistor in the "ring"
    line. Better not to connect that line through from end to end.
    If OC port number 2 could be guaranteed to never operate, then
    it could be connected end to end - but it may be an "accident
    waiting to happen"

    It could also be that you need a (seperate) pullup resistor from
    "tip" to "ring" (or more exactly the +5V) signal.
    10 kilohm is often good for this. Put a meter on the signal
    and see if there is a stable +5V when the OC driver is OFF.
    If so, no resistor rqd. If there is much less than 5V, add 10k.

    Hope your project goes well.

    Steve
    http://www.airborn.com.au/circuits/index.html
     
  3. Jukka Aho

    Jukka Aho Guest

    Thanks for your comments. I suspected something like the above.

    Actually, I could program the second device (a programmable
    calculator with two open collector type I/O ports) to always
    pull up the open collector data port #2 to +5V. This, to my
    understanding (am I correct?), would effectively prevent
    short-circuiting the +5V DC power source at the other end
    of the cable to the common ground, as long as everything
    else works as intended.

    However, even if I can control the pin in my _own_ code, I
    still cannot guarantee that it will _always_ stay at +5V.
    Someone might leave the cable connected and run some other
    program that deliberately pulls the same pin down to the
    ground level. (Moreover, the calculator is powered by
    batteries, and when they run out, there will not be any
    potential left between the OC pin #2 and ground, which
    - according to my limited understanding of the matters -
    would translate to a short-circuit situation again.)
    Yep, I think I will just have to resort to building another
    cable for this purpose, and leaving the DC output line
    disconnected. (Shame - it would have been nice if I could
    have used the standard link cable that comes standard with
    the calculator, as it is otherwise physically compatible.)

    Thank you for your help.
     
  4. Rich Grise

    Rich Grise Guest

    Can you assure that your equipment will _never_ try to output
    a low? If that's the case, (output transistor off all the time)
    then it essentially doesn't matter what the input is (as long
    as it's between 0 and 5).

    Hope This Helps!
    Rich
     
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