# How do I build an inductive charger?

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.

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.

active| newest | oldestEssentially, what you're building is a DC-to-DC converter, but one where the transformer is split into two separate coils, and which is not bothering with trying to boost the voltage. 12VDC into oscillator (through power amplifier if necessary), driving first coil with 12V at sufficient current. Second coil picks up field of first coil, feeds rectifier/regulator circuit to bring it back down to 5V. Adjust number of turns in each coil so output of second coil is at a signal level appropriate to power your regulator . You'll want to get at least 2.5W of power out of the secondary, since your goal is 5V at 0.5A; that means you need to pump that much wattage through the primary (plus a bit for inefficiencies).

This is assuming the two coils can be placed adjacent to each other. If you're trying to do this over a greater distance, you start needing to get into resonant coupling to avoid excessive losses; the tricks for doing that efficiently were discovered relatively recently.

See "related" at right for a few inductive/wireless chargers. I haven't examined them in detail so I have no opinion about any one of them in particular, but "if it happens, it must be possible".

Note that if you just want to make it work, you could determine a reasonable ratio of turns for the two coils experimentally rather than calculating it. But for school they're going to want you to explain the decision, so I'd suggest calculating, building, adjusting, then explaining why experimental result doesn't match calculated result (if you can give a good analysis of why theory and practice don't produce identical results, that will probably earn you bonus points).

And at this point I'm going to drop out -- it's your project, after all. Have fun.

https://sites.google.com/site/ddmcintosh2projects/inductive-charger

Just change the turns ratio between coils and

get 90 VAC output from a 12 volt Sq wave oscillator.

Oh and you don't need sine wave, the Sq wave will do, because those

sharp square corners get strongly rounded in the pick-up coil inductance.

Resulting in a sort-of sine wave.

( Its actually difficult to transfer Sq to Sq wave )

You actually need 90VAC ??

Most designers would rectify the secondary put a regulator on it .

Then run their electronics from the DC.

( A Sq wave does bot require as much "C" filtering as a sine wave does ! )

A

I design and build for a living.

The starting point for a project is to look and see what's been done before Re-inventing the wheel is a waste.

Then go and make it better.

Here is a start point look at Simple DC to AC Inverter,

about 7th down.

Replace the transformer with your coils rectify the AC and regulate to 5V.

A

unless you want a more sinusoidal output voltage.

In the simple DC to AC inverter the magnet iron is replaced with a spacing of air.

14 Gauge is way more heave then you need. 22 Gauge is a good wire size.

Not knowing your DC oscillator circuit,

I would be trying a primary coil 20 to 100 turns on a 3" diameter and 3 times as many turns on the secondary output coil on a 2.5" or 3.5" diameter.

All coils create a flux centered as a toroid around them. A solenoid is too narrow for air energy coupling.

Adding a magnetically permeable material to a coil concentrates the flux.

but It is not useful for air transfer.

A

I need a way to charge a small 12 volt battery without any connections. The battery is in a sealed plastic compartment and I also need a switch that is magnet controlled that will be incased (water proof).

www.expocell.com

The battery is a P21/13 12v

http://www.expocell.com/hLeadAcidBattery/LeadAcidBattery/P21213.aspx

Are you interested in doing a job for a fee?

Kennedale, Texas

Reno, NV

A

Eric

www.permitprint.com

please send me an email,,

You are not using bare wire are you ??

Use "magnet wire" = copper wire with an enameled insulation [ see pic 1]

Scratch the insulation wire end with a knife to connect to it.

You can get at magnet wire at Radio Shack.

Wind 3" round coils like [ pic 2 ]

No ! you are not going for resonance in this project.

You can test your coil by putting an incandescent light bulb in series

with your coil ( Best Fuse Ever ) and a 60 Hz low voltage to start.

On the other coil should measure some voltage ( proof of concept ).

A