Tristated output to two common Base transistor issue


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  1. #1
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    Hi,

    There are ony two 22 ohm resistors for the 20 LEDs.


    Best regards,

    Luciano
    Last edited by Luciano; - 6th September 2007 at 23:55.

  2. #2
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    Red face Mea culpa!

    BrianT advised me by private message that I had oversimplified my earlier analysis, and he is right. I looked only at the voltage divider created by the resisitors built into the transistors and neglected to consider the diode drop across the base-emitter junctions of the transistors. The circuit, redrawn, looks like this:

    <IMG SRC="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1980&stc=1&d=118913663 8">

    With the diode junctions in place (a nominal 0.7 VDC drop across each), two voltages change: At A, 4.3 volts; at C, 0.7 volt. At B, it remains 2.5 volts. Both transistors are still "turned on" if there is no voltage present (a high-impedance state) at B.
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    Last edited by RussMartin; - 7th September 2007 at 05:33.
    Russ
    N0EVC, xWB6ONT, xWN6ONT

    "Easy to use" is easy to say.

  3. #3
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    Hi Roger,

    After testing my circuit on a breadboard I have found out
    that the circuit works, but the current available for the LEDs
    is too low for your application where the forward current of
    the LEDs is 50mA.

    Best regards,

    Luciano

  4. #4
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    Unhappy to bad...

    So,

    Seems I have no other choice but to buy a 20 Outputs PIC.

    Thank you all anyway for all your propositions and the time you spent on my problem.

    BTW Luciano, I made the circuit too and yes, available current is far to low...
    Last edited by flotulopex; - 7th September 2007 at 20:16. Reason: added
    Roger

  5. #5
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    Question

    Luciano: After testing my circuit on a breadboard I have found out
    that the circuit works, but the current available for the LEDs
    is too low for your application where the forward current of
    the LEDs is 50mA.
    Roger: I made the circuit too and yes, available current is far to low...
    Is it possible that the series base resistor (2.2K) is too large? The "brute force" method to assure that the switching transistor saturates is to give the base about ten times as much current as it needs to turn on.

    Remember, the base current controls the collector current: Not enough base current means not enough collector current. Try 820 ohms, then on down (680, 560, and so on) until it works.

    ---------------

    I did a little looking on line. Is this the kind of thing you are doing?

    http://www.customdynamics.com/knight..._light_bar.htm

    and the cool video:

    http://www.customdynamics.com/videos...derz_video.htm
    Last edited by RussMartin; - 7th September 2007 at 23:22.
    Russ
    N0EVC, xWB6ONT, xWN6ONT

    "Easy to use" is easy to say.

  6. #6
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    Hi Russ,

    Saturation is not possible, here are the results:




    Best regards,

    Luciano

  7. #7
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    I was just wondering if something like that could work.
    <img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1986&stc=1&d=118926122 9">
    But after I made some trials, it is still not doing the trick
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    Roger

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