How do I build an inductive charger? Answered
Sorry for making the initial question so generic. The basic concept I'm addressing is inductive charging but in perhaps an interesting fashion. I'm working on a school project and my idea is to build an inductive charger for my cell phone (and other devices) that is powered by my car's 12V accessory outlet. I know that I need to have a time-varying signal (AC) to create a time-varying magnetic field in order to transfer energy between the two inductive coils. I'll lay out what I've considered / calculated so far and any help would be greatly appreciated.
I am going to use a cheap DC-to-AC inverter to go from 12VDC to 120VAC, single phase, modulated sine wave (per the inverter's specs). For the inductive coils, I'm using magnet wire from Radioshack and Fry's. The primary coil has 40 turns with a radius of 3.45 cm. It is wound in a solenoidal shape with a length of 2.5 cm. The wire gauge is 24 which has an approximate current limit rating of 0.577 A. The secondary coil has 30 turns with a radius of 2.9 cm. It is also wound in a solenoidal shape with a lenth of 2.2 cm. The wire gauge is 22 which has an approximate current limit rating of 0.92 A. Both coils are wound around PVC pipe couplings as they were the only round, cheap forms that I could find to wind my coils on. I have been unable to find any data online regarding permeability of PVC or how it would / will impact magnetic field induction. Any information on this matter would be greatly appreciated.
Continuing on to the secondary coil side, if I understand it correctly, due to the turn ratio of 40 over 30, the output voltage would be approximately 90 VAC. Is this correct? For the output current, I've found that it is dependent upon # of turns, distance from primary coil, and loop radius. I know that orientation of the secondary coil relative to the primary coil also greatly influences this but I intend to place the secondary inside of the primary.
I used the simplified magnetic field equation of B = µ I / 2 R to calculate the approximate field strength at the center of the secondary coil. Based on this calculation, if my input current is 500 mA I get the following: (12.57e-7)(40*0.5)/(2*0.0345) = 3.643e-4 T --> 3.643e-4 = (12.57e-7)(30 * Io )/(2*0.029) --> Io = 560 mA. Do these calculations look on par with what I should be getting? So my Vi = 120 VAC Ii = 500 mA and Vo = 90 VAC and Io = 560 mA. From the secondary coil I plan to use a bridge rectifier to convert to DC and then drop down to 5VDC. Ultimately I would like to get 5VDC @ 500+ mA on the output to meet USB standards. How does this plan appear? Are there things that I am missing? Are there things that I don't need? Again, any and all help is welcomed and appreciated. Thanks.