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How does this work?

Discussion in 'General Electronics Discussion' started by TheMaster, Aug 15, 2016.

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  1. TheMaster

    TheMaster

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    Jul 20, 2016
    I've seen multiple threads on various websites asking this question, but none some to fully answer it.
    How do those electric pen/bubble gum shocker pranks work?
    On the inside they look like this:
    [​IMG]
    Four small button cells, attached to an unknown white component, attached to a high ratio step-up transformer.
    Somebody on some forum tried to explain it with this circuit:
    [​IMG]
    This cannot be right as far as I know, as given those values the capacitor would discharge in a millisecond.

    How do these toys succeed in giving an estimated ~300V ~2ma, continuous shock using 4 small high internal resistance cells?
    And no, I don't want to build one, I simply want to understand how it works.
     
  2. davenn

    davenn Moderator

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    1,875
    Sep 5, 2009
    Google L/C oscillator circuit to see how the inductor (8Ω winding) of L1 interacts with the capacitor in parallel with it. Then come back and tell us if you figured it out :)
     
  3. TheMaster

    TheMaster

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    Jul 20, 2016
    Well I considered something of the sort. Does the inductor continuously recharge the capacitor?
    In most examples I see of lc circuits on the internet, the battery is connected with a two way switch.
    In this example the battery is always connected when the device is on.
    Wouldn't the transformer dampen the voltage until the oscillations quickly stop?
    No, I can't say that I figured it out...
     
  4. Sunnysky

    Sunnysky

    470
    118
    Jul 15, 2016
    it jolts on each quick press then release repeated and uses 3.0V lithium cells.
    It charges the inductor with current then oscillates when released.
    An air coil prevents saturation and resistance limits the input current.
     
  5. TheMaster

    TheMaster

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    Jul 20, 2016
    I have the exact pen in the picture. It definitely jolts on the press, not on the release.
    And it seems to jolts continuously. It tried it for at least 3 seconds until I took it apart.
    The transformer is the black shrink wrapped component. It is not air core, both the primary and secondary are tightly wound around a ferrite column. It seems so very simple but i still don't get how it works
     
  6. hevans1944

    hevans1944 Hop - AC8NS

    4,495
    2,092
    Jun 21, 2012
    Do you actually have the pen shown in your image? Really? This image and the schematic were copied from this website, but without the explanation of the circuit operation which appeared below the schematic.

    In a nutshell: 22 μF capacitor is charged from battery through 5.6 kΩ resistor... time constant 0.1232 seconds, so about 0.616 seconds to recharge after switch is opened. Closing the switch, and keeping it closed, dumps capacitor charge into primary inductance, after which the energy originally stored in the capacitor sloshes back and forth between capacitor and inductance of primary. Some energy is lost with each cycle of oscillation, so the output is a damped high-voltage sinusoidal waveform. How long this lasts depends on the loss mechanisms, but it could conceivably be several seconds. Release the switch (open it) and about a half-second later it is ready to go again. Adult children's toy.

    I personally prefer the old-fashioned, wind-up, Joy Buzzer-in-the-palm-of-your-hand novelty toy. No electricity, so safe for use with my heart pacemaker/defibrillator implant. No batteries to purchase either. If you have a firm handshake, the "fun" continues until the mechanism runs out of spring power. Buy one here! And at other places on the Internet.

    [​IMG]
     
  7. TheMaster

    TheMaster

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    Jul 20, 2016
    I do have the exact type of pen as seen in the photo, it doesn't mean I took the photo. They cost $2 on amazon.
    When I run it in the simulator with those values the oscillations last for several hundredths of a second and then stop. Then again I don't have the exact L value, but I tried an assortment of values and was not able to get anything which demonstrates the pen's performance accurately.
     
  8. hevans1944

    hevans1944 Hop - AC8NS

    4,495
    2,092
    Jun 21, 2012
    What is the purpose of the simulation? Is it to verify the posted example circuit works as advertised? Clearly that circuit will not produce a sustained output!

    Have you measured the inductance and resistance of the primary and secondary windings on the ferrite-core transformer? Have you located and measured the capacitance of the alleged 22 μF capacitor? Have you determined by actual measurement whether the high-voltage oscillations on the secondary are damped or sustained? You need all this information before you can even consider simulating the actual device and creating a model of the circuit.

    Maybe there is more to the particjular pen you have than meets the eye, What is that white blob? You need to really carefully disassemble it and identify all the components and draw up your own schematic, from which you can then begin a simulation model. Heck, this contraption might even contain a semiconductor wired up as an oscillator! In fact, it has to have an active semiconductor oscillator drawing power from the battery if it produces sustained oscillations when the switch is closed! There are circuits on the web that show a UJT (unijunction transistor) serving this purpose. If you really do experience an indefinitely sustained output when the switch is closed, there has to be an active semiconductor oscillator present to supply power to make up for losses. Why semiconductor? 'Cause I don't see no stinkin' vacuum tubes.

    If you don't try to simulate the transformer, and just simulate the inductance of the transformer primary, you should be able to create a damped sine wave from the example circuit that lasts for as long as you please with a sufficiently large capacitance. The losses that cause damping will be mainly those in the charging resistor, since it is always effectively in parallel with the capacitor. Since you are just simulating the real operation, you can eliminate the charging resistor and battery entirely from the simulation and set an initial voltage on the capacitor at the start of the simulation. Then the losses will be whatever you specify for the inductor winding resistance, unless your capacitor model is lossy.

    Before creating a model for simulation it helps to gain as much information about the real circuit as possible. Measure the oscillation frequency and see how it compares with the both the simulation oscillation frequency and the frequency you get if you plug in measured values for the capacitance and the inductance... assuming you can even find something in the pen that looks like a capacitor. If the high-voltage oscillations are sustained and not damped, the example circuit is definitely wrong. There has to be an active oscillator in there somewhere that draws power from the battery and makes up for losses in an LC resonant circuit. The capacitance part of LC may not even be a real component... just parasitic capacitance of the windings!

    Interesting problem, explaining how a $2 item can produce a sustained shocking experience. Please tell us what you discover.
     
  9. TheMaster

    TheMaster

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    0
    Jul 20, 2016
    Yes, I suspect that circuit diagram does not describe this device. It really, REALLY does seem to shock indefinitely, I held it down for at least 5 seconds or so before I disassembled it (It was rather painful and unpleasant). It doesn't seem as if the jolt becomes increasingly weaker for the duration of the zap.
    I have no idea what that white component is. I know very little about the makings of electronic components, but I cut it open with pliers and on the inside it is composed of some white powdery material. If there is some sort of capacitor in there, it can't be very large, as the components you see in the photo are physically very small.
    If I remember correctly, a + and - exit the battery compartment and attach to two ports on the white component. The two transformer input legs also attach to two ports on the component, and then the user is shocked by the transformer output and the batteries -.

    Unfortunately I cannot at the moment continue to test it because its been disassembled, but I think it is safe to say that no explanation readily found on the internet explains how it works correctly, and I have not found other circuits that can use such small batteries to deliver such a current at a high voltage through a big load.
     
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