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How can i make a transofer/whatever to step up the voltage of a 9 volt battery. This will also make the current go down won't it.?.

2007-03-16 06:37:58 · 7 answers · asked by Sean G 3 in Science & Mathematics Engineering

7 answers

Depending on the application, the industry uses one of the following three methods, listed below in the order of increasing power level:

1) A charge pump: lets the potential energy built up in a row of parallel capacitors then switches to connect the capacitors in series to add up the voltage.

2) A boost converter: lets the kinetic energy built up in an inductor then switches to discharge that energy to a capacitor to pump up the capacitor voltage.

3) A transformer: oscillates the DC to an AC, runs the AC through a step up transformer, then rectifies the secondary AC back to a DC.

2007-03-16 08:03:29 · answer #1 · answered by sciquest 4 · 0 0

It would have to be converted to ac (oscillator), put into a transformer then converted back to dc (rectifier). Usually this is done at high frequencies as it saves on the transformer.

These things are commercially available and would not be cost effective to build.

Check out rswww.com for dc to dc converters.

I don't know what you mean about the current going down. The current drawn is dependent on the voltage applied and resistance connected (ohms law). 9v PP3 batteries don't have much power anyway. Why not use a 6v lantern battery if you are going to increase the voltage anyway?

2007-03-16 07:29:39 · answer #2 · answered by Poor one 6 · 0 0

You can put batteries in series to raise the voltage, If you have 2-9volt batteries, put them in series and you have an 18 volt. Source, I think you will have to use similiar batteries. If you want less voltage, you can put a zener diode on the circuit and this will drop the voltage. So take a look at your application and what voltage do you need? Radio Shack has a lot of things that you can use to play around with.

2007-03-16 06:49:08 · answer #3 · answered by Anonymous · 0 0

Transformers will not work for DC, so you will have to build an oscillator and feed that into the low voltage coil of the transformer. You can easily build an oscillator with a relay, but it will be noisy and the relay will probably wear out eventually. Obviously, you could build a solid-state oscillator, but it is more difficult...

Yes, current will go down.

2007-03-16 06:42:29 · answer #4 · answered by computerguy103 6 · 0 0

What you need is a boost converter. These use a MOSFET switched at 100kHz or greater to energize a coil. The output is filtered by a capacitor. You don't need a transformer, just a single coil.

Yes, the output current will be lower. Power out is slightly less than power in due to inefficiencies.

Search for DC-DC boost converter for more info.

2007-03-16 07:26:26 · answer #5 · answered by vrrJT3 6 · 0 0

you would need some sort of circuit interrupter in the circuit going to the primary side of the transformer in order to induce current in the secondary coil. Since a 9v battery is d.c. it would not oscillate like a.c would with out one. And yes the current would go down.

2007-03-16 06:44:08 · answer #6 · answered by Anonymous · 0 0

Look into DC-DC converters. Yes, you will give up current to increase voltage.

2007-03-17 13:33:28 · answer #7 · answered by joshnya68 4 · 0 0

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