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Buzzer voltage drop

Discussion in 'Electronic Basics' started by pata, Oct 17, 2004.

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  1. pata

    pata Guest

    I have an electronic buzzer (ICC pn: BS2316L-06) I am going to put it into
    a circuit with some LED's and other components. It draws 25ma and is rated
    at 6V, how do I know what the voltage drop across the buzzer is? Is it 6V?
    I need to know this so I can calculate what size resistor to put in the
    circuit.

    Thanks
    Pat
     
  2. You can probably assume that if you provide it with 6 volts it will
    pass about 25 ma.
     
  3. pata

    pata Guest

    I understand that but how does that tell me what the voltage drop across it
    is?
     
  4. CFoley1064

    CFoley1064 Guest

    Subject: Buzzer voltage drop
    The 25mA listedfor the electromechanical buzzer is an average -- sometimes a
    lot more, sometimes less. It will work with a series resistor, but not too
    well.

    Let's assume you're running off a 13.8V supply. In order to get it to run
    well, you might want to use a zener like this (view in fixed font or M$
    Notepad):


    13.8V
    +
    |
    1N4737A |
    Vz=7.5V/-/
    ^
    |
    |
    |
    | .---|
    '---| |
    .---| |
    | '---|
    |
    |
    ___ |/
    o-|___|-o-|2N3904
    Control 10K | |>
    Signal .-. |
    10K| | |
    | | |
    '-' |
    | |
    === ===
    GND GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Good luck
    Chris
     
  5. pata

    pata Guest

    Thanks, I understand that better. I am an ME and I struggled in our 1
    required EE class!
     
  6. The voltage drop across it is the voltage you supply to it. If you
    have 12 volts available and want 6 volts to appear across the buzzer
    while it draws 25 ma, assume the buzzer looks like 240 ohms and put
    another 240 in series with it. They (the buzzer and the resistor
    should split the 12 volts into two approximately 6 volt drops.
     
  7. Of course, I left a gotcha in there to see if you would ask the next
    question. The buzzer draws some pulsed waveform that averages 25 ma
    when you apply 6 volts across it. There may be large instantaneous
    current variations. You may need to add a capacitor in parallel with
    the buzzer to soak up all those variations and hold something close to
    the average drop across the buzzer to make ir work correctly.

    I would guess that 10 microfarads or so might be enough.

    Capacitors are sort of like fly wheels (if voltage is like rotational
    speed and current is like torque) in this application. Internal
    engines put out an average torque at some speed but without a
    flywheel, the speed varies all over the place each turn. :)
     
  8. pata

    pata Guest

    OK, thanks. Let's forget about the cap for now so I don't have an overload.
    Let's say my simple circuit consists of a 12V source, an LED that requires
    25ma and has a Vf of 3.2v, and the buzzer. To figure out the correct
    resistor the calculation would be 12-(3.2+6) /.025 = 112 ohms. Is this
    correct?

    Also, theoretically if the my source voltage was 9v would I not need a
    resistor at all?

    Don't leave me any gotchas! I need it explained to me like I was 5.

    Patrick
     
  9. That is he general idea.
    Leds have very large current swings for very small voltage changes, so
    they are not modeled by ohm's law, very well.

    Normally you need some resistance in series with LEDs to make sure the
    current is controlled. But the buzzer may perform that resistor
    function, approximately. But it would worry me.
     
  10. Rich Grise

    Rich Grise Guest

    No! No! No! Don't get the newbie started off all on the wrong foot.
    A capacitor is a _spring._ An _inductor_ is a flywheel. "current"
    means "movement", after all, right? :eek:)

    Cheers!
    Rich
     
  11. Rich Grise

    Rich Grise Guest

    John Popelish explained fairly well about your series circuit, and
    you're spot-on, so far. :)

    You can kind of think of an electric current something like the flow
    of water in pipes. Pressure is voltage, current is flow rate, and
    resistance is friction.

    Well, an LED, as John Popelish said, doesn't act like an ordinary
    resistance. In a resistance, current, or movement, increases
    proportionally to pressure, or voltage, and inversely to friction,
    or resistance. An LED is like a greased chute - very little
    change in pressure causes a great change in flow, unless something
    else restricts that flow, in this case an "ohmic" resistor.

    And a capacitor is like a split tank with a rubber baffle, and
    an inductor is like a positive displacement turbine/pump coupled
    to a flywheel. :)

    Try googling on basic electronics tutorials and stuff.

    Hope This Helps!
    Rich
     
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