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PIC speark driver

Discussion in 'Electronic Basics' started by Michael, Apr 15, 2007.

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  1. Michael

    Michael Guest

    I'm generating a 1K Hz "tone" using an output pin on a PIC
    microcontroller. I'd like to increase the volume. I can (and have
    tried) using additional pins to source more current, but I'm hoping to
    use a very low pin count MC for the final version. Any ways to maximize
    the current without additional components? Otherwise, any simple
    circuits to accomplish function (e.g. single transistor).

    Currently PIC output pin is directly connected to 8 ohm speaker via 10uF
    cap.

    Thanks
     
  2. You can effectively double the voltage by using one pin on each speaker
    lead. You drive one pin high and drive the opposite pin low at the same
    time.
     
  3. Michael

    Michael Guest

    I'm generating a 1K Hz "tone" using an output pin on a PIC
    Doesn't this just yield the difference between a high pin output
    (~Vdd-0.7) and a low pin output (~0)? It would only yield double if a
    low pin output was the negative of a high. I tried it and the output
    level was lower than what I'm currently getting (probably from the ~.7V
    loss since it's currently connected between Vdd and a low pin output).
     
  4. john jardine

    john jardine Guest

    Would've thought you'll get 5V driving speaker current in one direction,
    then 5V driving speaker current in the other direction. Speaker cone
    movement 'll be full out, back to neutral, then full in. I.e twice the
    original distance.
    (and 4 times the power!)
     
  5. The peak current with a 5 volt possible swing is half the
    swing divided by the resistance. So in this case
    2.5V/8ohms=.31amperes. Obviously your PIC cannot deliver .3
    amps in either direction, so almost all of the 2.5 volt
    swing is dropped across the output stage of the PIC. Even 5
    outputs in parallel are limited by the total port current
    spec for the chip and will make it hot. There are two
    simple ways to get more volume to your speaker. One way is
    to use a transformer to get a less terrible impedance match
    between the output and the speaker.
    Something like this:
    http://www.tamuracorp.com/clientuploads/pdfs/engineeringdocs/MET-35.pdf
    at about $10 each. If you look through the surplus
    suppliers you may find one for a dollar or two.

    The other approach is to add a current booster stage to the
    output. The simplest would be an NPN and PNP transistor
    connected as a complementary emitter follower. This can
    boost the available peak current by something like 20 to 50
    times higher, and will keep the output current at the PIC
    within specs.

    This requires just the two transistors (e.g. 2N3904 and
    2N3906 or 2N4401 and 2N4403). Both bases connect to the PIC
    output, both emitters connect to the coupling capacitor to
    the speaker. The NPN collector connects to the 5 volt
    supply, and the PNP collector connects to ground.
     
  6. You may get some nasty (current) spikes while switching. I'd advise a small
    resistor between the collector of the NPN and the Vcc and a decoupling
    capacitor near the transistor circuit.

    petrus bitbyter
     
  7. I agree that a bypass capacitor from collector of the NPN to
    the ground return point for the speaker is a good idea. But
    there is so little extra voltage available, that I would
    probably try a ferrite bead on a lead between the collector
    of the NPN and the 5 volt supply, to retain as much voltage
    as possible for the transistor driver. If the supply cannot
    handle the current drawn by this driver, then it can't
    handle this volume from an 8 ohm speaker.
     
  8. Michael

    Michael Guest

    I'm generating a 1K Hz "tone" using an output pin on a PIC
    I see your point, but I tried it again and the results were the same.
    My understanding of analog electronics is limited, so maybe someone can
    explain why what you stated doesn't result (or what I might be doing wrong).
     
  9. My guess (I'm a hobbyist, nothing more) is that this is due to the
    driver impedance. Figure about 120 ohms source and 80 ohms sink. When
    you were driving just one pin, you only got one of those inserted.
    With two pins, you've always got both reducing current.

    Jon
     
  10. feebo

    feebo Guest

    your speaker will pull a lot more power that the PIC can source (or
    sink) directly... put a transistor in the way (same as you would drive
    a relay)... much better.
     
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