Charging a capacitor to 6kV

EMF is EMF, and a capacitor will charge to ANY peak voltage if it is applied to the cap correctly and the cap can handle the pressure. It is all a matter of time and resistance to puncture (punch through/blow through).

HV caps that do not have their leads shorted can even grab ions from the air, and eventually "charge up".

Yes it is possible to use an audio transformer as a flyback. On this page

Perhaps you could use something like that as a 1st stage and use it to pulse a flash tube trigger transformer. You might even be able to use a 4 lead trigger transformer instead of the phone transformer.

Mike

"In view of such harmony in the cosmos which I, with my limited human mind, am able to recognize, there are yet people who say there is no God. But what really makes me angry is that they quote me for the support of such views." Albert Einstein (theoretical Physicist)

Just for the sake of insufferable pedantry, a "spark" and an "arc" are qualitatively different things. A spark is an incandescent flying chunk of some material, like the iron that's struck off the flint, or grinder sparks. The only ionization involved would be incidental as the hot spark contacts air molecules.

An arc, on the other hand, involves current flow through ionized air, which is an entirely different animal.

Now, one disclaimer here - a arc can make sparks, by blasting blobs of incandescent metal away as the contacts blow up. :-)

I saw a thing on TeeVee once where this guy claimed to make "ball lightning". All it was was a mongo bank of mongo batteries, and a big heavy steel set of contacts, which when they hit, arced, and the arc was so energetic it blew little balls of incandescent molten iron out, and they kind of skittered around on the floor, like a water drop in a hot skillet.

Cheers! Rich

have:

cycle you could store about 0.1 millijoules in the

Capacitors in series might work to achieve a net high voltage rating, BUT... ...the voltage will divide according to the capacitance; the larger the capacitance, the lower the voltage - and that is starting from all being at zero volts and assumes sufficent time for full charging (of all).

Be careful. In rectifier circuits the voltage also divides up according to individual leakage and that can be grossly different. The imbalance can reach the point where one cap takes the lion's share and then ...

very

or

=A0I

ect a

t 2kV

to put

ck

n each cycle you could store about 0.1 millijoules in the

ond

the

ure

some

job

ime

d text -

it occurs to me that the power supplies in electrostatic air filters might do it. those little desktop items now being sold as oreck, for instance; i've had about 5 of them over the decades as they've been manufactured by various companies, and the fan motors always die in a year or two but the power supply lives on, so maybe you can find one such on freecycle or craigslist or something. i'd give mine to you but i finally threw them away when i moved.

I guess I should ask this before I go much further. How dangerous is this? I intentionally choose a small size of capacitor for safety (actually, I'm just using the capacitance present in the Villard cascade). I am correct in saying I don't have to worry too much about shocks because the size of the capacitor is so small?

ANY TIME you're dealing with more than about 40 volts, there is a shock hazard. You don't "have to" "worry" about it - just pay attention to what you're doing, and when probing put your other hand in your pocket. Take off ALL jewelry. and ground the little finger of your probe hand. (that way, if you do get zapped, the current stays in your hand, rather than going through your vital organs.)

One of their favorite things in tech school was to say, "If you get up against a 12V filament supply, get a tickle and flinch, and you snatch your hand back into the 3KV, 100 mA supply, which one killed you?"

Cheers! Rich

Figure out how much energy there is for a discharge. The amount is equal to the amount of energy stored in a given value cap at that voltage plus the amount of stored energy delivered to that cap by the charging pump.

The best way to limit what is available on an output lead as a jolt delivered as a shock is to put a current limiting resistor in series with the output led. The voltage drop is small, but the current limit insures that any downstream discharge is limited to a specific amount.

E=0.5*C*V^2 = 1E-09 * (6E3)^2 = 36mJ (that's pretty small - like the same as a static shock surely?).

Surely the resistor have to be quite large to have an effect on the current delivered? Skin is between 5MOhms and 2kOhms, so I'd guess the resistor would have be to 100kOhms to make a difference?