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.

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orksecurity6 years ago
If you're talking about going up to 120VAC, and then back down to 5V, you are definitely doing this the hardest way possible.

Essentially, 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".
Also, websearching for "DC to DC converter" will find many descriptions of that circuit. Usually it's used to trade off volts against amps by boosting voltage, but there's no reason you can't run it the other way.

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.
dmcintosh34 years ago
I have been working on a DIY inductive charger (from scratch). You can reference it for low-level details:

Also a block diagram.
iceng6 years ago
You don't need to go up to 120 VAC.
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 ! )


AMNunnally (author)  iceng6 years ago
Thank you for your comments. Actually I don't need 120 VAC or even 90 VAC on the output. The only reason I'm using 120 in and 90 out is because I'm using a readily available dc inverter to get an AC signal. In doing this, I don't need to build one. I guess a good question at this point would be, is it easier to design and build a sq wave oscillator and then rectify the signal on the output side or do what I'm doing and use a pre-built inverter and then rectify on the output? I'm aiming for the simplest form with this project so your suggestions are greatly appreciated and taken into consideration. The only for sure limits I have for this project are an input voltage of 12 VDC, input current of 10 A max, output voltage of 5 VDC and output current of 500 mA min. How I get there isn't terribly important as long as it is by induction across a small gap. The primary and secondary coils can't be wrapped on the same form or share the same core (other than air). Thanks again for your help.
Liking your reasoning and you write clear.
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.

AMNunnally (author)  iceng5 years ago
Sorry for the delayed response. Thank you again for your advice. I was reviewing the schematics that you linked me to. Do I need to add a capacitor across the primary coil for this set up? I am assuming that I don't since the schematic doesn't show one. Going back to the 120 VAC inverter idea, if I were to use this as simply a conceptual step (i.e. proof of voltage/current being induced on the secondary coil) what would I need between the inverter and the coil? Since I already have this portion of the project done, I wanted to at least show a proof of concept to my professor before switching to the other circuit you recommeded. I upgraded my coils to 14 Ga. magnet wire as the current rating for the 22 Ga. didn't seem adequate. I'm also trying two different coil geometries. One is using solenoidal coils. The second is using a spiral wound coil where the coil is one wire wrapping thick but expands outward. If I understand the coil geometries correctly, the solenoidal coil will create a field that is more "directional" than the spiral coil right? Last question/though I have is what benefit would there be if I added a small ferrite core inside one or both of the coils? Since the cores wouldn't be connected, would there still be a benefit? Thanks again for your advice.
No you don't need a capacitor across the secondary output coil,
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.

towshark iceng5 years ago
Dear iceng,
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).
The battery is a P21/13 12v
Are you interested in doing a job for a fee?

Kennedale, Texas
iceng towshark5 years ago
Sure always interested with a longtime member. How do you envision this;
  • Trickle type unregulated charge ?
  • Power transfer across an air-gap with a regulator on the battery side ?
  • No problem on the an in-cased magnet controlled water proof switch.
We can PM details.
Reno, NV

towshark iceng5 years ago

please send me an email,,
AMNunnally (author)  iceng5 years ago
I was wondering if you could clarify what you said about the solenoid being too narrow for air energy coupling. All of the coils that I have tried so far have been solenoidal and didn't work well. I've been learning a lot more about this as I've been reading various things and am now going to try spiral coils. If I created a coil with a diameter of 3" and it had 50 turns it would ultimately be 1.27 inches wide if wound to be be only one wire's thickness high. In the end, the coil would have a 3" diameter hole in the middle and would measure approx. 5.5" in diameter across the entire coil. Does this sound correct? Also, while I understand that I don't need a capacitor across the output of the circuit you recommended, if I were to be simply connecting this coil to an AC source, in that instance should I add a capacitor? If so, having an inductor and a capacitor together would create a resonant frequency for the "circuit", how would this resonant frequency be affected, if at all, by the 60 Hz frequency of the AC source? Again, thank you for all of your help and advice. I have learned a lot thus far and hope to learn more. I haven't yet been successful in what I've tried so far and I'm sure that what you've helped to explain will get me farther than I've been able to do myself.
You said it yourself "didn't work well" it's a fact I know, you got to accept.
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 ).

drmike9585 years ago
I have a related question. Since HP dumped 100's of thousands of Touchpads into the world and they use an inductive charger called the touchstone that I assume will no longer be made along with the tablets, how can we make our own? I'm thinking of a stand similar to the Touchstone that the Touchpad can sit on either in landscape or portrait mode. It would probably be made out of wood with the induction coil behind. I know that the touchstone requires 5 Volts and 2 amps to charge. What would be the most efficient way to build the circuitry? Thanks!