Maker Pro
Maker Pro

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.
 
B

Bob Eld

Jan 1, 1970
0
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.

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.
 
C

Charles

Jan 1, 1970
0
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.

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

Bob Eld

Jan 1, 1970
0
`
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".

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.
 
J

Jasen Betts

Jan 1, 1970
0
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?

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

yeah that's it exactly.
 
Top