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Toy circuit safety question

My cousin's husband asked me about an electric shock circuit. He remembers at school some kid made joke suitcase, that if picked up zapped the recipient.

I know the circuit, and I strongly expect it to be the same one used in the practical joke pens and staplers. It just involves a DPDT relay, a 100uF capacitor, a small transformer and a 9v battery.

The basic idea is that the capacitor is charged up (to 9v) triggering the coil which then disconnects the battery and discharges the capacitor through a transformer which can step the voltage up. The transformer has primary/secondary impedances of 1.2kOhms / 3.2Ohms (leading to a step-up factor of about 20).

My question is just - how safe is this circuit? He wants to make it to demonstrate some electronics to his son (and I agree with him that there is a lot in it for a boy to learn).

I calculated the output energy of each pulse: E = 0.5 * C * V^2 = 4 mJ which is very low, so I told him it's fine if he's sensible, and keeps the output away from the face etc. I just want to check here that my logic is sound.

Thanks.

Reply to
fred.r.hatt
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It should be safe enough. The output voltage is very much higher than a 20:1 step up would indicate. It's probably several thousand volts and is due to the collapse of the magnetic field in the transformer when the switch (or relay) is opened. That is a different mechanism than simple transformer action.

You did not mention it but the circuit can be connected to continuously interrupt the current or buzz putting out a continuous train of high voltage pulses, fifty or so a second. Sounds cool and "shocking" but not dangerous. Bzzzzzt! Yikes. There's not enough energy in the system to do damage.

Reply to
Bob Eld

I made one like that in school. It worked fine, was fun and did not kill anyone.

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Reply to
RFI-EMI-GUY

Certain medical conditions can make this a dangerous road to travel.

Reply to
Charles

Well that is quite a vague statement. Obviously they won't be using it on people with pacemakers, or anybody they don't know well.

Reply to
erlend.davidson

I am interested in this. Both from the perspective of "how this thing works" and also because I am a physical scientist by training and by profession. Even if you know of a paper which explains this effect? I have a feeling it is to do with the inductance of the transformer, V=LdI/dt, and since with a square wave the dt is very small then the voltage across the inductor is very large?

Actually this is what the circuit does - when the capacitor has mostly discharged the relay switches off from the transformer to connect the capacitor to the battery again. I'm not entirely sure, but someone told me this is the "electric fence circuit".

Reply to
fred.r.hatt

As anal retentive as you are, its a wonder your head hasn't exploded.

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Reply to
Michael A. Terrell

20:1

to

(or

voltage

High voltage generated from a collapsing magnetic field is very common. That's the way induction coils, ignition coils, fly back transformers, and many other inductive circuits work. Look up induction coil, fly-back circuit, inductive kick and so on.

Yes V = L di/dt and V = Nd(phi)/dt. When current flows in an inductor it creates a magnetic field (phi). When the current is interrupted the field collapses trying to maintain the removed current flow. Since there is no load when a switch opens other than stray capacitance in the windings, the rate of current collapse, di/dt becomes very high. This creates the high voltage pulse, Ldi/dt.

The stored energy in the inductor is 1/2 LI^2. This energy comes out both in the high voltage pulse but also in the spark that occurs at the switch when it opens. A capacitor across the switch can recover some of this energy and by resonance re-apply it to the coil for a number of high frequency cycles enhancing the output. Automotive ignition works this way.

Reply to
Bob Eld

It's a "flyback converter" you can look that phrase up and get a better description than I can give.

yeah that's it exactly.

Reply to
Jasen Betts

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