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DC RC circuit with buzzer

Discussion in 'General Electronics Discussion' started by Darrell, Apr 20, 2016.

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  1. Darrell


    Apr 20, 2016
    IMG_20160419_182154757-1024x1820.jpg Hi there. I am trying to understand why this circuit from Frank Mims radio shack electronics lab manual works with the capacitor but not without it. Any help would be greatly appreciated.

  2. ramussons


    Jun 10, 2014
    Buzzer requires 12 mA to make a sound. The series 1K resistor will not permit that current. The capacitor charges to a voltage when the Buzzer starts sounding discharging the capacitor and finally stops.
    The cycle repeats.

    Replace the the Resistor in steps to 470 Ohms, then 330, 220... till the Buzzer starts sounding. It will be a flat tone. With the capacitor, it will be interrupted giving that "chirping" sound.
  3. Darrell


    Apr 20, 2016
    Thank you for answering. I've experimented around a bit and it seems that when you are just using a resistor, you do need to choose its value to get somewhere in the neighborhood of 12mA or so for good volume; but when you throw in the capacitor, that doesn't seem to be the case. For example, with a 1000 ohm resistor combined with a 470microfaradt capacitor, you get around a 6.3-6.5mA current at 9V and it works... Without the capacitor you get about half that and it doesn't work so well. The voltage drops shift too (which is to be expected). We go from 3.16v across the resistor and 5.8 across the buzzer with no capacitor to 6.51 across the resistor and 2.42 across the buzzer. It's almost as if the buzzer has a resistance of its own that varies depending on your voltage, R1 value, and whether or not you are running a capacitor parallel to it. That doesn't really make any sense and I'm pretty sure you are right about the charging and discharging - but it doesn't seem they are discrete events - except when you first close the circuit and then when you open it again... If you let the buzzer ring for an extended period of time, it doesn't seem that the capacitor cycles between charging and discharging as discrete events. It seems like they would be happening at the same time, eventually coming to an equilibrium of sorts... I guess I'm looking for some insight into the math behind all this if that makes sense, but I can't quite get at it.... Thanks again.
  4. duke37


    Jan 9, 2011
    I imagine that the buzzer takes a blob of current at each buzz. The capacitor is there to supply this current and is recharged between each buzz. If there is no capacitor, then the voltage drops too low and there is no sound and the average current will be low. The buzzer does have an internal resistance and also an internal inductance, the values will depend on the circuit.
    What you measure will depend on the type of meter. An analog meter will give an average voltage over time from which you can get average current.
  5. Darrell


    Apr 20, 2016
    Thank you for your help. So what I understand is that, because the meter gives you an average voltage/current/whatever you are measuring rather than an instantaneous value, the voltage/current that causes the buzzer to go off could actually be significantly higher than what the meter reads - i.e., the capacitor charges to the voltage which causes the buzzer to go off and discharges through the buzzer, but spends most of the time well below that voltage.
    From what you've told me, would it be fair to guess that the buzzer tends to be resistant to a source of constant voltage/current, kind of like a capacitor is resistant to a constant DC voltage/current, and that it's when you add in the capacitor that discharges when the buzzer sounds that you get a varying voltage/current that leads to a decrease in resistance from the buzzer and allows it to go off at much lower voltage/current values than you would need otherwise?
    Thanks again for your help.
  6. Alec_t


    Jul 7, 2015
    If you have access to an oscilloscope you could see what the voltage across the capacitor is doing.
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