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I'm new to electronics.

I'm trying to find out how this works out of curiosity.

http://www.amazon.com/gp/aw/d/B0050UNNQS

Does anyone have a circuit diagram for the circuit? Is the current safe for such a prank?

Internal picture: enter image description here


EDIT: Merged question from Trever Thompson

Help here would be greatly appreciated: This question has been asked several times on this site and elsewhere, and it seems as if the posters became satisfied with inaccurate answers.

How on earth do the electric shock prank pens/bubble gum work?
I have no intention of shocking anybody or recreating this circuit, I am simply baffled by how it operates. I have verified that all purported explanations readily available online are inaccurate or incomplete.

The above image shows the only two components in the circuit. I do not know for a fact what the white component is, but I have to assume it is a high value capacitor.

enter image description here

The above image is a literal layout of the circuit. The 50k load is the user's hand. Both switches are closed upon pressing the button. The switch is before the capacitor. This is based on closely observing the circuit and is 100% accurate unless someone can prove otherwise. I tested this with varying capacitance and inductance values and was unable to induce a significant current in the load for more than dozens of miliseconds. When testing I moved the virtual switches from the ground to the positive wire. When I move the switch after the capacitor I can have a long lasting current through the load, but the voltage/current are very low.

Edit: It is beginning to appear as if the white piece (Shown as cap above) is an active component/IC. There may not even be a capacitor in this circuit.

What I know:

  1. The device behaves as follows: When the pen is held and the button is pressed a mildly painful electric shock is felt. After about 3 seconds, the pain gradually decreases until it is not longer felt. Upon releasing and pressing again the full shock is felt once again, even if only released for the slightest moment. The button and grip are the electrodes.
  2. It uses 4 1.5V button cells in series (with high internal resistance)
  3. The shrink wrapped component is a tightly wound step up transformer. The secondary uses extremely thin wire. It apparently is an autotransformer, as it has three leads and is wrapped around one ferrite column.

What I think I know:

  1. I measured the frequency on the primary with a primitive oscilloscope. It was equal to a couple hundred hz, which is reasonable considering the pain of the shock.
  2. If the frequency is ~200hz, the capacitor (if the white component is a cap) would have to be over 100uF non polarized. This may be possible as I have discovered high capacitance low voltage smd caps. The primary inductance would have to be several mH.
  3. The turn ratio would (I believe) have to be at least 1:90. This surprised me because I read that most autotransformrs have a low ratio. I then encountered higher ratio custom ones (which is more reasonable if varying wire gauges are used).

What I want to know:

  1. What the white component is. Is it merely a capacitor? Does it serve other functions?
  2. How on earth does this circuit work. It seems so simple and then again I can't explain or emulate its behavior accurately in a simulator...

Help will be greatly appreciated, as this has kept me occupied for a long time.

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    \$\begingroup\$ You know, I could never find any solid schematics for these type of shock pranks. \$\endgroup\$ – Passerby Jun 13 '13 at 21:10
  • \$\begingroup\$ Also, normally you would want to wait for a day or so before accepting an answer, so more people can reply. \$\endgroup\$ – Passerby Jun 13 '13 at 21:11
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    \$\begingroup\$ "The above image is a literal layout of the circuit." I doubt that. I don't see a part labelled "unknown white blob". \$\endgroup\$ – WhatRoughBeast Sep 13 '16 at 13:33
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    \$\begingroup\$ Harsha Alva, how did you conclude that Jlm Dearden's answer matches the actual circuit of the "chewing gum"? \$\endgroup\$ – Daniel Tork Aug 18 '18 at 16:33
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Here's a little circuit that might fit the bill

enter image description here

The battery charges the capacitor (C1) through a resistor (R1). When the switch is operated the charge from the capacitor is dumped into the primary of the transformer. This produces a damped oscillation (LC circuit). The transformer secondary has a lot more turns than the primary and so the VOLTAGE part of the short burst of alternating current in the primary will be magnified by the turns ratio. The current part will be diminished. The total energy delivered to the 'shock' can be estimated from the energy store in the capacitor. (0.5 * C *V^2). For the values given = 0.5 * 22 * 10^-6 * 9^2 Joules or 891 micro joules - pretty harmless. The transformer is a small audio output type but others can be used such as the trigger transformer for a flash tube (hacked from a cheap camera) or you could even wind you own on a ferrite ring.

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4 variables.

  • Vbat
  • Primary inductance
  • Turns ratio step-up tapped auto-transformer
  • C
  • R Skin resistance 1M ( tiny contact)

Simulation

enter image description here

CLose switch and Vout = Vbat * turns ratio Open switch and LC^2/2pi = 1/resonant f Positive HV is V=L di/dt

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  • \$\begingroup\$ In your simulation the battery voltage and internal resistance are unreasonable. The voltage in the actual device is 6V and the internal resistance (according to the datasheet) is at least 50ohms. Also, the skin resistance when gripping the pen is much lower than 1M. The step up ratio can then again be larger. Do you claim that the shocking action occurs when the switch is open? The switch opens (as for as I know) when you release the button, and are no longer in contact... I don't understand that \$\endgroup\$ – Trever Thompson Sep 13 '16 at 7:16
  • \$\begingroup\$ I agree on VBat & ESR. I wasn't sure but thought the voltage was between tip and hand so I used 1M from a dry finger to get more voltage on release. I also assumed release of switch occurred before release of tip. , I can simulate 100K which will reduce amplitude and duration \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 14 '16 at 4:28
  • \$\begingroup\$ So how do you consolidate the internal resistance? I have those small batteries at home and three in series gave about 40 ohms. Also, I am beginning to suspect strongly that the white piece is an active component. If it incorporates timer and a transistor which opens and closes the circuit everything may start making more sense. Your thoughts? \$\endgroup\$ – Trever Thompson Sep 14 '16 at 8:13
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The white object looks like a chip-on-board circuit (blob of epoxy over unpackaged integrated circuit). The basic scheme could be a spark coil (flyback type voltage boost, similar to a three-terminal automobile spark coil) or a vibrator-style inverter (like the old Model T spark generator).

The observed frequency indicates that an oscillator is in use, and one expects that a NE555 or similar astable is in the white component. If this is the case, it ought to have three wires, and/or some other connection than parallel to the coil primary winding.

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    \$\begingroup\$ I doubt very highly that there can be a chip there. It is certainly not a 555 as that would require programming with more passive components. Then again, perhaps there is some mechanism which opens the switch in there... The previous answer seems more in line but is still incomplete. \$\endgroup\$ – Trever Thompson Sep 13 '16 at 7:23
  • \$\begingroup\$ @TreverThompson Why do you doubt that? For these extreme-mass-market toys, you typically have a custom ASIC doing exactly everything necessary to operate, apart from things that has to be physically large (coils, capacitors). They aren't big. \$\endgroup\$ – pipe Sep 13 '16 at 8:09
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    \$\begingroup\$ Again I don't rule it out entirely. For you to make a claim that there is an active component in there you have to support it with evidence. Exactly what function would such a component perform? Secondly, I cut the blob open with pliers only to find a white grainy substance. The only thing I could begin to imagine is that there may be a switching component in there which re-opens the circuit after pressing the button, as that might support @TonyStewart 's answer. \$\endgroup\$ – Trever Thompson Sep 13 '16 at 8:22
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    \$\begingroup\$ +1 I have no doubt this is correct. The COB is a simple oscillator (probably high frequency divided down) and an output driver. \$\endgroup\$ – Spehro Pefhany Sep 13 '16 at 20:54
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    \$\begingroup\$ You're right , about 4 year ago I had one of them, but I lost my patience to work with that tiny and brittle chip and broke it. There must be some kind of ASIC, like in some calling bell sound generators, they packs pretty nice song inside such white epoxy chips , no external componcomponents. \$\endgroup\$ – Arnab Sep 16 '16 at 13:05
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When the switch is switched on, the primary inductance causes the current to build up in a ramp. When the switch is turned off the energy stored starts the resonant circuit on the primary side thereby inducing a high voltage in the secondary. The principle is the same as in the old fashioned ignition systems in cars

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My buddy had one of these for years. It's not "electronic" at all. It's a thin wire bent into a spring which snaps onto the unsuspecting victim's fingernail in a "mousetrap" fashion. Since it happens so quickly and isn't expected, it feels like a shock.

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  • \$\begingroup\$ There are multiple versions. This one does in fact deliver an electric shock. \$\endgroup\$ – Matt Clark Jun 13 '13 at 15:35
  • \$\begingroup\$ @jay, the image I posted had a coil, 4 button cells and something that looked like a piezoelectric element. \$\endgroup\$ – Harsha Alva Jun 13 '13 at 15:37
  • \$\begingroup\$ Ah, didn't see the image until after I posted! Agreed, it's most likely an induction coil. Pass some current through the loop, turn it off, and BAAM! Generate a nice high-voltage low-current shock! \$\endgroup\$ – Jay Greco Jun 14 '13 at 7:19
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Most likely this gadget has a piezoelectric element in it, which generates a large voltage at extremely tiny current when the device is activated. You feel a shock from it but it's not generally harmful. Similar technology is used in gas barbeque igniters.

EDIT: Based on the picture, I'm not so sure any more. The watch batteries could be used to source a voltage multiplier circuit of some form to generate the shock potential. The heat-shrunk component could be an inductor to which a DC current is applied then interrupted to generate the high voltage, or could be a capacitor.

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  • \$\begingroup\$ Here's an image.. Please login with donotbug and bugmenot as username and password respectively. www.electro-tech-online.com/attachments/electronic-projects-design-ideas-reviews/58386-shock-gum-circuit-shocking.jpg Thanks @madmanguruman \$\endgroup\$ – Harsha Alva Jun 13 '13 at 15:32
  • \$\begingroup\$ It's just a simple induction coil. I built a similar device myself once (back in high school) into a cigar box, powered by a D cell. I handed it to a friend, who naturally opened it to see what was inside. He was NOT happy! \$\endgroup\$ – Dave Tweed Jun 13 '13 at 15:51
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    \$\begingroup\$ I once built an "anti intrusion system" on the same principle by fitting a hidden momentary LT switch to the top of my bedroom door and wiring the HT to the door handle. This was before the family went on a 2 week holiday. On my return, I had completely forgotten I had set this trap ... and went straight into my room !!!! \$\endgroup\$ – MikeJ-UK Jun 13 '13 at 16:25

protected by clabacchio Jun 14 '13 at 7:24

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