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Charger: How would you make a Mobile device charger using charging mechanism of a shake flash light? pls help!!!!!!!? Answered

mobile charger using faradays law of induction?


The flashlight is almost all you need. Think of the light as the device you'll charge, the load on the circuit while discharging, and your there. Just test your output, voltage and current, and match to your phones specs.

''Just test your output, voltage and current, and match to your phones specs.''

how exactly will i match it to my phones specs?

Look on your phones ac wall charger. It will have dc voltage and current specs.

Sorry maybe I wasn't clear
I meant what will my circuit look like to get USB power (5v)
I'm really new to all this and don't know how to make this
Sorry if i have been a hassle

No problem.

The flashlight works by charging either a battery or capacitor using the current induced coil by shaking the magnet through the coil. Since the inductor is generating AC, as you can see from the .gif, there is an AC to DC bridge rectifier between the inductor and the battery/capacitor. Then the battery/cap is connected, through a switch, to the light bulb, possibly with circuitry to limit current and/or voltage. You will replace the light bulb with a power plug that matches your phone. Attached below is a circuit block diagram. The switch is shown in the position used to charge the storage battery. Throwing it to the other position allows the battery to discharge through the USB plug into your phone.

Before breaking the flashlight, charge it up, turn it on, and use a multimeter to measure the voltage across and current through the light. It might be good to measure these values while shaking as well as there may be a spike. The battery/cap should smooth this but better safe than sorry. Once you build a circuit, test it rigorously before hooking it to your phone and risk damaging it.

Inductive Charger.jpg

thanks alot man
but one more thing
should i have a 5v cap, keeping in mind the voltage will drop, or should i use a, lets say 10v cap and then a 5v regulator (so that my votage never gets too low to charge my phone)

I think you should consider using a rechargeable battery not a cap. Using caps is complicated and they don't just work like a battery. Here's some quick quick details. A cap is measured not in voltage but in capacity, the unit is farads. A farad is a certain number of electrons, a coulomb (6.25e18) to be exact, at one volt. So a 1 farad cap holds 1 coulomb of electrons at 1 volt. This means a 1 farad cap holds 1 amp-second at 1 Volt. in comparison a standard AA holds about 2.8 amp-hours at 1.5 Volts. This means you need a room full of caps to have the same capacity as a AA battery. The only practicle way to store that much on smaller caps is at a high voltage, dangerously high.

Instead of a cap you need to get a rechargeable 9V and add a DC to DC conversion after the switch, before the USB plug.

last question
Dont batteries need a highish voltage to charge?
because the charging mechanism only gives off a low voltage ( I think around 0.5v-1v)
will that be enough to charge any type of battery?
thats why i was going to use a capacitor, basically to raise the votage high enough.

Thnaks for all the help and sorry for being such a bother.

You can change the voltage from the inductor by changing the coil design. Optimal charging voltage is slightly above it's rated discharge voltage, say 1.4 V for a 1.2 V AA, and a current about 10% * 1 hour of the the capacity, say 270 mA for a 2700 mAh battery. You can get away with lower currents but they will charge slower.

Using caps and diodes to build a voltage multiplier is possible but takes a lot of space and is better suited for much higher voltages I believe. Also since your AC current and voltage would vary wildly while you were shaking it that would probably cause problems.
I'm treading near the edge of my solid knowledge base with some of this capacitor theory so take it with a grain of salt.

look into low voltage DC-DC step-up convertors. They are usually small chips that convert low voltage DC to higher voltage DC at the cost of current.

Thanks for every thing
very helpfull feedback
& sorry for being such a noob ;)

Well, you know the basic principle, what do YOU think you need ? A coil of wire, a magnet, a tube ?