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I am experimenting with large currents from ultra capacitor discharge.

For example with 500 A (at 2.8 V) you get a very impressive demonstration of the magnetic field of a straight conductor using compass needles or iron chips (compare: http://iopscience.iop.org/article/10.1088/0143-0807/31/1/L03/pdf).

Another example is the Thomson ring experiment http://www.rose-hulman.edu/~moloney/Ph425/0143-0807_33_6_1625JumpingRing.pdf where you may get up to 9000 A for a very short time.

Suppose all voltages used are below 60 V. What do you need to consider about safety in this case?

Here is what I think:

  • Since the voltage is too low, there should be no danger from current through the human body.
  • There might be a danger from sparks and lightening when if there are contact problems.
    • This may be dangerous because of UV light
    • and because of sparks hitting directly the eye
  • Also the there might be heat problems which makes thinks to vaporize which you may inhale
  • A capacitor discharge generates an EMP which may affect pacemakers for example

I am not sure if I mentioned all possible dangers concerning this. My question is:

  • Under which conditions (minimal current, discharge time...) which danger will become relevant
  • What to do to make it safe
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    \$\begingroup\$ 60V is high enough to be a potential hazard. Even car batteries, at no more than 12V, are to be treated with care, because they can output a lot of energy very quickly into whatever they're connected to. This is the sort of thing that, if you have to ask, you really shouldn't be doing it. \$\endgroup\$ – Hearth Apr 26 '17 at 12:37
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    \$\begingroup\$ High current can generate very high voltage and substantial energy with just a bit of inductance when switched so you can't ignore that. Burn hazards and possible blindness from molten metal and eye or cancer-causing UV skin damage (you can get a bad 'sunburn' from a low voltage arc). Even a car battery can cause loss of a finger if you accidentally weld it across a conductive ring. Try to find University health and safety guidelines or corporate ones. Some of us have written them. \$\endgroup\$ – Spehro Pefhany Apr 26 '17 at 12:48
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    \$\begingroup\$ Don't forget to add "have fire extinguishers at hand" and "buy some insurance" to your safety checklist. \$\endgroup\$ – Enric Blanco Apr 26 '17 at 14:12
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    \$\begingroup\$ I think Spehro's point is V=L*di/dt. If you have some L and switch the current very rapidly (high di/dt) you can get very large V. So the basic inductance formula quantifies the point nicely. And all cables and wires have some L. \$\endgroup\$ – John D Apr 26 '17 at 14:47
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    \$\begingroup\$ Things I can think of that aren't in your list, mostly collected from YouTube videos of risk analysis by people who like to blow things up: Accessible fire extinguisher; fire resistant carpeted flooring if you're working with things that may spray molten metal (sounds counter-intuitive but carpet prevents molten bits from bouncing into unnoticed/inaccessible places); never work alone always have a buddy; beware of things that create x-rays; have a remote location where you can cut power from if you need to escape; clear the room of tripping hazards; keep the area clean and tidy. \$\endgroup\$ – Jason C Apr 27 '17 at 4:37
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If we consider this model of a high energy switching circuit we can simulate the induced voltage on a nearby conductor. This provides a simple simulation of the type of electromagnetic interference that would be coupled into nearby conductors or electronic devices.

schematic

simulate this circuit – Schematic created using CircuitLab

As any switch opens, or you simply tap wires together to briefly conduct that 1,000 amps, when you have 1u (1 micron, 10,000 Angstroms, or 1/25th of a mil) separation of the wires, the 3 volt potential causes an arc.

The 100pF across 1micron separation (3mm by 4mm ---- heavy wire --- contact) will resonant with the 1uH (~~ 1 meter) wire in your high-current path. Fring will be 15MHz. What is the dI/dT of 1,000 amps ringing at 15Mhz?

100,000 MegaAmps/second.

Place a wire 4" from the high current, that wire formed into 4" by 4" loop; expect 2,000 volts across the ends of that 4" loop.

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  • \$\begingroup\$ Nice, but I guess that the 200 V will be there only a very short amount of time (how much?). Then the deposited energy will be the relevant parameter. How much will the energy be if I touch the contacts of your 200 V wire? \$\endgroup\$ – Julia Apr 26 '17 at 16:10
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    \$\begingroup\$ I had a typo; the dI/dT is 100,000 amps/uSec or 100Billion (10^+11) amps/second; thus Vloop is 2,000 volts; note the Vinduce formula --- if totally accurate --- needs some integration and/or natural_logs. However, for ratios as I used, with Distance and Loop edges approx. the same, there is little error. So expect 2,000 volts. \$\endgroup\$ – analogsystemsrf Apr 26 '17 at 16:19
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    \$\begingroup\$ @JRE: As I said, it depends on the time scale and then on the energy. I often touched a demonstration parallel plate capacitor (low energy!) or a whimshurst machine (30 kV), nothing happened (however I calculated the energy content before using it). \$\endgroup\$ – Julia Apr 27 '17 at 5:25
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    \$\begingroup\$ The OP wanted to know how to be safe. This answer illustrates the risk. \$\endgroup\$ – analogsystemsrf Apr 27 '17 at 14:08
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    \$\begingroup\$ If the ringing frequency is 15MHz, the skin depth (where 63% attenuation occurs) is 17 microns. But at 2,000 volts versus 20 milliVolts (neural potentials), you need 100,000:1 attenuation; at 8.6 dB per skindepth (neper) you need 100dB (100,00:1) / 8.6 dB or 12 skin depths or 200microns. Seems like the scalp tissue will keep the energy out of the neurons. But your mileage may vary on these numbers. \$\endgroup\$ – analogsystemsrf May 3 '17 at 3:51

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