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# What is the most efficient and portable way to convert 3 volts (Battery power) to around 400 Volts?

I need a portable and efficient way to convert battery power (3 volts) to around 400 volts.

I will be using alkaline batteries...

This will be used to charge a high voltage capacitor bank.

## Discussions

It's like Ork said, this is essentially what you get in a camera flash charger. Such a circuit is typically powered by 1 or 2 AA batteries, and charges a capacitor to a voltage of around 300 volts or so. There are a number Instructables which make use of camera flash.

https://www.instructables.com/tag/?sort=none&limit%3Atype%3Aid=on&q=camera+capacitor

I also found this next link, which looks at camera flash charger circuits in detail, including circuit diagrams and explanations of how these circuits function.

http://www.talkingelectronics.com/projects/XenonFlasher/XenonFlasher.html

Red mentined voltage multipliers. The Wiki article on voltage multipliers is pretty good, if you think a multiplier stage might be needed.

http://en.wikipedia.org/wiki/Voltage_multiplier

Many of the other answerers have tried to say something about the meaning of energy and power (as the product of voltage and current) and maybe about the impossibility of building something

truly powerful, that runs on just two AA batteries. I agree with everything they have said, but instead of just repeating what they've said, I offer you the following koan:Understand this koan and you will be a step closer to enlightenment.

8 AA batteries cannot jump start a car because there is not enough current and there is ESR.

Very good grasshopper. Eight AA batteries wired in a series cannot start a car because this arrangement cannot supply enough current. That's what the textbooks say. Moreover this ESR, this series resistance, is part of the model used to explain what a battery is like, electrically. A battery

is likean ideal voltage source in series with a resistor.A more philosophical way of saying the same thing, is that real devices have limits. There is only so much current a AA battery can supply. There is only so much power (the product of voltage and current) it can supply in any particular instant. Moreover there is only so much total energy in the battery.

At its most simple level the

start-your-car-with-AA-batterieskoan is intended to illustrate the truth that voltage is not equal to power.There is a misconception many people have regarding electricity. They see the measurement of voltage as sort of

the unitfor measuring electricity, and for comparing one electricity source with another. This misconception leads people to believe the voltage of a electrical power source is an indicator of how much work it can do, how much power it can deliver, or even how dangerous it is.For example a 1.5 volt alkaline battery is thought to less powerful than a 12 volt car battery, and a 12 volt car battery is thought to be less powerful than a 110 volt wall outlet. It's sort of like there's a rule that says, "The electricity source with the greater rated voltage is the more powerful one." However this rule is only approximately true, and then only some of the time. So really it's not a very good rule.

Of course the part of the story that is missing is current, measured in amperes. A specification for voltage and a specification for current, will maybe tell you something about power, if you multiply those two numbers together. Maybe.

Anyway, I think what is missing from your story, from your question: numbers which speak of power.

How much energy will your capacitor bank hold?

(Recall the energy in a charged capacitor is U = (1/2)*C*V

^{2})How quickly do you want to charge it?

If you want to charge it in time T, then the average power being delivered to your capacitor bank while charging it is: Pavg = U/T, where U was the answer to the previous question.

I think this is a meaningful way to think about your capacitor bank charger, and a good way to get started. Think about it in terms of energy and time.

Hmmm.

My capacitor bank is kind of strange though.

The diodes are there because the capacitors must discarge at different times.

The "ideal" diode would have no voltage drop across it, but i'm wondering if they limit current too much. In that case i geuss i will just use them in parralel.

Each capacitor is 1500uF with 20% tolerance, which gives us a total of 3000uF(Or 3 mF).

I know this is a lot, that is 3.00*10 (exponent)-3.

And for how fast, I need to think about that for a second.

(calculation, the exponents are the numbers in the brackets)

U=((0.5)*3.00*10(-3))*400(2)), so U is 0.24 apparently (I think i did something wrong).

Oh yes, also, I will be using C or D type batteries arranged in the picture.

(Sorry about the pic quality; it was done using microsoft paint).

It worries me that someone who so far seems to be very inexperienced is going to try to charge up potentially lethal levels of electricity!!! I am not a believer in the Nanny state but I thing you need to be more careful.

At the very least, someone interested in this should read Instructables such as "Work Safely With 1High Voltage", and get supervision from someone who knows what they're doing.

Done.

I've given up worrying about them.

What you need is a voltage multiplier. Google "Voltage Multiplier" and read about them. You'll need some way to create or simulate ac voltage but you can do that with a 555 timer. Them more you step it up the more time it will take to charge the cap. In otherwords, the more voltage the less current you have.

That works, and is essentially how electronic flash units charge up. But remember that what you gain in voltage you lose in amperage. Batteries can only supply so many watt-hours before they go dead.... and standard commercial alkies really don't contain all that much power.

At least ist better than the poster who expects to use a battery to charge a Leyden jar and extract big sparks from it.