Instructables
Picture of Disposable Camera Nixie Tube Driver
Before I get too far in this instructable, I would like to say that this was not my original idea. You can see two implementations of this idea already on Flickr. The links are:
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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Step 1: Gather Materials and Tools

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For this instructable, you will need:

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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ccarpio26 months ago

The capacitor just blown up in my face :S

felixphew8 months ago

What would be the easiest way to step this (mine outputs ~260V) down to 200V? I am looking to drive 3 IN-1 (ИН-1) tubes.

the_don125 (author)  felixphew6 months ago

Easiest: increase the series resistance. I don't know the properties of your tubes in particular, but if you model them as a 200V ideal voltage sink, then use Ohm's law to find the resistance needed to get your tubes rated current at 60VDC.

Is it elegant? No. Not at all. Will it work, sure!

Thanks, that was helpful, but it turns out the power supply is so piss-weak it steps down to below 200V as soon as you hook anything up to it. No adjustment required!

P.S. You could also just use a trimpot as a quick-and-dirty voltage divider, if needed.

dumle293 years ago
ive accedentialy shorted one of those caps with my finger.

i guess as long as its on the same hand, theres no worry (not across the heart) but it HURTS LIKE HELL
I was burned by a transformer from a scanner fluorescent lightbulb. Luckily it was just across my finger. Hooked up to a computer power supply, ouch... I had charred spots on my finger.
BrunoG dumle293 years ago
I only got shocked by a small transformer, nothing big, it was the transformer that comes with the "Radioshack Learning Lab" kit.
Hey! I have that kit. It's pretty useful. I'm still learning from it. I also made a shocking circuit from it and you could adjust how much the shock hurt. I don't think that kit is still being sold, is it?
the_don125 (author)  dumle293 years ago
Its true, if you only work with one hand, and zap yourself only through that one hand, you are less likely to stop your heart via electrocution.

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 got a shock of a car battery once, despite my beliefs that there isn't enough potential difference to across our body
the_don125 (author)  eXtremeSomething3 years ago
All depends on the person, some can feel shocks from sources as low as 10V, some don't notice till much higher.
I got shocked by an EL wire inverter and it wasnt pretty. It's output was 200 volts ac but it only ran off 2 AA cells. So, i guess it is a matter of current, not volts. My slayer exciter outputs 4000 volts and it burns my finger!
Current and voltage play parts in how dangerous it is, but current is the main factor.
Yes. If you don't have enough voltage, you cant pull the current needed, however high voltage doesn't meen that it will pull a large current. the power supply high not provide enough of a puch.

U=R*I

so if R is high and U is low, I needs to be low or the math wouldn't add up
tsagert dumle293 years ago
Pretty much any wall wart can kill you, hence why GFIs trip at about 4 milliamps (but wont work unless current isnt returning, like using two hands to stick two bare wires into the hot & neutral of a GFI, it wont ever trip if Iin=Iout).

Higher voltage makes the current able to travel along or penetrate your skin.
While that's true voltage is not what kills you. Tazers operate at 1mill+ volts! It's the amperage that is dangerous.
A few people have been killed in my old line of work, 48v DC systems in telecom offices.

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.
agm882 years ago
you could DIE
forte19942 years ago
Um... its not the voltage that kills its the amps... and realistikly think about it. Can u touch both ends of the 1.5V battery with out dying? V=ir therefore if you increase the voltage u decrease the current thus making it not deadly. HOWEVER THIS DOES NOT APPLY IF U HAVE A PACEMAKER!! That zap can stop it... basicly: Yes use caution when working with electricity No u wont die if you touch the cap. It will hurt a LOT tho. BTW a car battery can kill even if it only has 12V because it can out put MASSIVE amout of current!
the_don125 (author)  forte19942 years ago
What you said does not make sense. "V=ir therefore if you increase the voltage u decrease the current thus making it not deadly." Basic algebra states that increasing V while holding R constant will result in an increasing I.

You are correct, that it is current that kills, not voltage. However, the human body has a skin resistance of approximately 47kOhms, when dry, so you need a reasonably high voltage to generate a dangerous amount of current across the chest. Basic V=IR shows you only need about 1V of potential difference to generate the generally regarded to be "lethal" current of 10uA across the chest, but don't forget the cross-section of the chest is rather large in comparison to the heart, so not all of the current flows through the heart.

Once you exceed about 36V though, that cross section difference is not always enough to prevent a lethal current from flowing through the heart. Some people are more sensitive to it than others (some people can feel a jolt with as little as 10V), sure, but you won't see me grabbing two terminals at a 36V potential difference with opposite arms. Because these capacitors can let out a jolt at upwards of 300V, you see how they can be dangerous; only one pulse of "enough" current is needed to do very weird things to the heart.

This is why when working with high voltage, you want to use only one hand, and pocket the other. This way, current is not inclined to flow through the chest; it will stay in the arm, maybe cause external and/or internal burns, but that's a far better fate than stopping the heart.

It is actually known that a 9V battery can kill; if you manage to pierce the skin with both terminals in such a way that the current chooses to flow through your bloodstream into one arm, across the chest, and out the other, enough current is generated to stop the heart. This is because blood is rather low-resistance, so a high current is generated. Without your skin acting as a safety resistor, even surprisingly low voltages can be very dangerous.
the_don125 (author)  the_don1252 years ago
tl;dr
Current is the killer, but you need sufficient voltage to overcome resistance to cause harm.
Low voltage is dangerous in *some* cases.
High voltage is dangerous in *all* cases.
(The above is only true when working with DC.)
I see. Well i always thaught that because the power needed came from a 1.5V battery ment that no matter what it should not kill you but Its good to know now that I am wrong as im starting work with el wires. Wouldn't wana make this my last project! Thanks!
if used (in)correctly energy from a 1.5 volt battery may be able to do some harm, however if you remove the capacitor all you can get is a painful shock, the capacitor on the other hand could probably do quite alot of damage, i have been working on something a bit similar today.
the_don125 (author)  the_don1252 years ago
Also worth noting: the 47kOhms of dry skin resistance generally will completely overwhelm the internal resistance of a battery. That is why you can take (usually) 12V worth of AAA batteries, or a 12V car battery, and touch across the terminals with dry skin, and you won't know the difference; the current that will flow is almost identical, despite the car battery's significantly lower internal resistance.

Assuming a car battery to have an internal resistance of 0.2Ohms, and AAA batteries to have a series resistance of 100Ohms, we get:
12V / (47k + 0.2) = 255uA
12V / (47k + 100) = 254uA
Not much difference!

Of course, with wet skin (or in the case of pierced skin), this story changes dramatically. Let us assume we're talking pierced skin, so only the blood is providing resistance, and let us assume blood provides 10Ohms of resistance:
12V / (10 + 0.2) = 1.17A (yes, Amps...you'd be good and fried)
12V / (10 + 100) = 109mA (still very dangerous)

Our friend the capacitor, as discussed above, has a resistance similar to the car battery, extremely low, and therefore extremely dangerous when at high voltage.
GASSYPOOTS2 years ago
dude take the batterys out first >.< derp >.<
the_don125 (author)  GASSYPOOTS2 years ago
You can take the battery out (there is only one), but the capacitor will remain charged. Only taking the battery out and failing to discharge the capacitor is a great way to get yourself electrocuted.
You are exactly right! Except I was trying to repair a digital camera and one of the caps lit me up! I didn't know that cameras had components that required such high voltages and amps.

It is much worse than grabbing AC wires! Always discharge any cap before working on any circuit. I do now!
This just saved me A LOT of time! Thanks!

(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.)
FallenSub3 years ago
I always thought that these circuits provide around 300 volts. How did you get it to be around 180? Voltage divider? If so, can I use 1/4 watt resistors?

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!
the_don125 (author)  FallenSub3 years ago
Yes, I had a 200k ohm potentiometer in series with the nixie, so the nixie would drop 180V, and the pot would drop the rest. It seems these circuits have a fairly high internal resistance, so they cannot output very much current. In fact, if we take a look at step 4, we can somewhat figure out the internal resistance!

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.
Thanks for a fast response!
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?
the_don125 (author)  FallenSub3 years ago
Depends on what you're trying to power. If you're going to max out the supply, then in theory, yes, a 2W resistor is needed (though I feel like the camera might burn out first). If you're just trying to light a nixie or two, I've only needed to pass about 2mA, which is 230V (voltage source after adding safety resistors) - 180V (nixie tube voltage) = 50V * 2mA = .1W, so well within the abilities of a 1/4W resistor.
Okay, thanks for your time!
Does the charge button need to be shorted?
the_don125 (author)  Halt! I am Reptar3 years ago
This camera actually had an on-off switch for the flash, rather than the more common "button" design. I didn't have to short anything, but rather used the on-off switch as a power switch for this power supply. I liked having that feature rather than having an "always on" supply that would've resulted from shorting the charge button.
I had to solder the charge button closed to get my Nixie to light up, i dont think you mentioned that. Maybe yours doesnt use a charge button? I got some modern Kodak ones from Walgreens' recycle box.
sci4me5 years ago
So I can just discharge the cap and then cut it off as long as the wires are connected to the circuit?
the_don125 (author)  sci4me5 years ago
Sure, that is probably a much safer way to use this circuit.
Wouldn't holding the shutter button down for a few seconds be enough to discharge the cap so that it would be safer to open?
the_don125 (author)  grstearns3 years ago
Not always. The gas discharge tubes that make up a camera flash work on the same principle as the nixies in that they will operate down to a certain voltage, then stop conducting. The point where they stop conducting will often leave a fairly high voltage still in the capacitor.
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