http://www.flickr.com/photos/mdweezer/322631504/in/set-72157594420700670/
http://www.flickr.com/photos/samwibatt/1610784412/
On with the instructable! This documents how to turn a regular disposable camera into a high voltage power supply capable of driving 2 or 3 medium-sized nixie tubes, for roughly $8.
***Disclaimers***
This instructable works with voltages in excess of 250V. This is more than enough to give you a potentially fatal electric shock if handled incorrectly. If you are unfamiliar with how to work with high voltage, please refrain from performing this instructable. Exercise caution throughout the following steps to avoid electrical dangers. If you choose to undertake this instructable, you do so at your own risk.
This instructable involves soldering. A soldering iron becomes very hot during its use, to the point where it can cause instant second-degree burns. Exercise caution throughout the following steps to avoid burns. If you choose to undertake this instructable, you do so at your own risk.
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Signing UpStep 1Gather Materials and Tools
A disposable camera
A potentiometer of 100Kohms or higher
A resistor of 50Kohms or higher
Wire cutters
A small screwdriver (may not be needed, depends on your camera)
A multimeter
A soldering iron
Solder
Red wire
Black wire
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(When I was younger, I was messing with one of these boards. I knew better than to touch the large menacing cap, what I didn't think about was avoiding the solder points it of which it was attached to other parts of the board. Lesson learned.)
i guess as long as its on the same hand, theres no worry (not across the heart) but it HURTS LIKE HELL
That being said, enough of a zap can create internal burning, putting the victim into shock, so there is always the chance of a fatal shock when working with voltages above 36V (some sources say as low as 12V).
Always remember, electricity is our friend, and just like any friend, you better treat it with respect, or else you will get burned.
I only got ring voltages on my forearm when reaching in to hook up wirewrap connections on the AT&T switch.
You dont always know that cap energy wont go in your finger, up a nerve of vessel and navigate its way to a vital organ and not right out again.
U=R*I
so if R is high and U is low, I needs to be low or the math wouldn't add up
Higher voltage makes the current able to travel along or penetrate your skin.
I presume you shorted your multimeter with connecting it directly to the output? I didn't know these can provide currents over 200mA (these usually have fuses rated at 200mA).
Other than that, it's a nice instructable you've got there!
If we assume that the circuit will output 300V open-circuit, and by putting a 50k ohm load across the voltage source, the output then sags to 230V, you should be able to use ohm's law to figure out the internal resistance. (230V / 50k = current; 70V / current = internal resistance). Obviously this is flawed because we don't know if the open-circuit voltage is 300V, but its the best I've got for now.
As for my meter, I did connect it across the output with a fully-charged capacitor in place. While the circuit itself can probably only supply 10mA or so short-circuit, the capacitor can dump a fair amount of current, which it did.
I did exactly the same thing a couple years ago. Fuse in the meter blew so hard that I could see a huge spark flash through the nontransparent meter casing! But fuses are quite cheap, so thats not a problem.
Also, do you think a 1/4 watt resistor can do? Im not sure, coz, lets say the current is 10mA, voltage drop on a resistor 120V, so that's 1.2W.
Or am I wrong here?
Why?
Well, the diode would prevent current from flowing as the circuit powers up. However, nixie tubes like to be driven around 180V, more than that and you might put too much power through them, shortening their lifespan. The diode will start conducting backward when the circuit is at 200V, ensuring the circuit stays at or below 200V. This does improve safety somewhat as well.
Now, with this circuit's approximate internal resistance of 15k, that means the internal resistance must dissipate 100V (the diode drops the other 200), for about 6mA of current. This results in the diode dissipating 200V * 6mA ~ 1.3 W of power, so get yourself a diode rated for 2 or more Watts of power dissipation, and you'll be set!
Alright, i'll try it sometime.