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Biasing Transistor

Discussion in 'Electronic Basics' started by Jack// ani, Nov 13, 2004.

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  1. Jack// ani

    Jack// ani Guest

    Hi all,

    Placing a finger on the base pin of this darlington configuration
    glows the LED (in fact reaches very close to saturation)!

    I learned that you need a threshold voltage of about 0.7volts across
    the base-emitter junction in order to bring transistor into
    conduction.

    So does this imply that my finger having a potential of 0.7volts??

    Thanks



    VCC
    +
    |
    |
    .-.
    | |
    | |
    '-'
    |
    |
    V LED
    -
    |
    -------|
    | |
    |/ |
    o--------o| |
    |> |
    | |
    | |/
    -----|
    |>
    |
    |
    |
    |
    |
    |
    ===
    GND

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  2. More like the finger has a resistance of a few K's. For real saturation just
    moisten the digit.
     
  3. I think the OP was saying that ONLY one finger was used. He's acting kind of
    like a radio antenna, I suspect.

    Jon
     
  4. Jack// ani

    Jack// ani Guest

    Oh yeah, exactly.
     
  5. Ban

    Ban Guest

    ..
    In the first place you will need more than +1.2V, because the darlington has
    2 junctions in series.
    And your finger will have an AC-potential, which you can measure with an
    oscilloscope. This is mainly because of 60Hz mains field radiating all over
    the place. The transistor conducts only half of the time, so the intensity
    will be less than half of the maximum. Since the current gain of a
    darlington is very high (maybe 10000 or even more) you will need only a uA
    to bias the base into conduction.
     
  6. Jack// ani

    Jack// ani Guest

    whoops, I forget that fact.
    Oh yes, I can clearly see a 50Hz ac/noise. But why there exist only
    50Hz, what about my cell and cordless phones which emits MHz, KHz
    frequencies…

    Any help is highly appreciated
     
  7. Ban

    Ban Guest

    Really? You remind me of my youngest daughter, one question answered, two
    new ones are popping up immediately. :)
    Check out the input impedance of the darlington, we have seen already
    1.2V/1.2uA = 1Megohm DC-resistance and now calculate the input capacitance
    of that darlington. Hint: it is high because of the Miller-capacitance.
    And then you can calculate the 3dB frequency of that lowpass, and you will
    see it is low.
    And BTW, what kind of antenna is your finger? Must have some loss at 800MHz?
    You need a tuned circuit for RF to establish some power transfer.
     
  8. Jack// ani

    Jack// ani Guest

    Thanks
     
  9. Rich Grise

    Rich Grise Guest

    If it's that simple to make a touch sensor, then howcome all
    those touch sensor circuits are so complicated?

    Thanks,
    Rich
     
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