THIS INSTRUCTABLE IS NOT YET COMPLETE. I have been trying to build a wireless charger but have come across a problem. I have documented what I have done and learnt so far. Hopefully from this you can see my errors and help me out. Please read the last page to see what I think the problem is and suggest any ways to get around this. THANKS

The Project
I am trying to build a circuit that will allow gadgets that are usually charged by USB to be charged wirelessly. As an example I am reverse engineering an A4tech battery-less mouse. However it is too great a challenge for me and I am seeking help from you. I thought it would be better for me to turn this into a group effort than to ditch the project. I will give a detailed description of what I have built and learnt and hopefully you can tell me where I went wrong.

Step 1: Background Information

Transferring power wirelessly is reasonably simple. If you think about it, all transformers are wireless. But we want something that's truly wireless. Like the Splashpad (see picture). It is pretty much a transformer with an air core.
The mouse I was talking about (see other picture) in the introduction is exactly the same as the Splashpad it uses induction to transfer power across an air gap. This is the same technology as RFID; in fact it uses this to communicate with the pad. To make our own wireless device we need to know more about induction.

<p>send me wirless power transfer project circuit diagram input ac 230v 50 HZ and secondary coil side 12v 40 khz or other valid circuit daigaram on this mail haylaz7@hotmail.com or look www.brautfrisurens.com project category...</p>
Please send me wirless power transfer project circuit diagram input ac 230v 50 HZ and secondary coil side 12v 40 khz or other valid circuit daigaram on this mail <br>nirajmane9@gmail.com
<p>Hi , i want to build wireless transfer coil antenna to get 12 v and 5 v , how do i calculate the number of winding of the coil and the diameter and thickness of the coil , whats the right formula to be calculate to get the right coil please.thanks</p>
<p>I've the same doubt.. </p>
<p>So do I :p</p>
<p>Provided you have matching tuned LC circuits, you need to excite the primary coil with an alternated supply. If the magnetic field don't alternate you can't pick up the power on the secondary coil. Only some weak spikes caused by the induction feedback... From what I saw, I think that your problem is on the excitation circuit. A simple 555 can't do it. Also, the problem is not about using sine or square waves... Just alternate it from negative to positive and you will have a good surprise :-)</p><p>Try using a small micro controller (e.g. arduino is easy) and an h bridge to drive a coil with a square alternated signal. You can do this with a 555 and some gates and the fets... But will require more know how.</p><p>Have fun :-)</p>
<p>Regarding your question about voltage across the primary, I assume you mean a.c. voltage. The drop across the primary is proportional to the impedance of the primary at the signal frequency. This will depend partly on the reflected impedance from the secondary. And, I couldn't find the refrence to your operating frequency, but check the specs on your meter to see if it measures at that frequency.<br><br>One solution If your portable device can be kept in one position during power transfer and it does not need to be movable during transfer like the mouse, then you might want to consider taking a ferrite core and saw it in half (probably need a carbide blade for this). On one half wind your primary, on the other half wind your secodary. Mount the halves of this transformer in their respective devices so that you can re-contact the two halves by placing one device on top of the other. <br><br>This way, using a ferrite core, you would have better coupling and you could operate at a lower frequency than with an air core transformer.</p>
interesting I dont know how I could be of help private message me if you have any thoughts on this
I think you would be better off using a Crystal Oscillator instead of a 555 based timer. For one, you'll be generating a much much higher frequency than a 555 could. They're limited to about 500Khz depending on manufacturer. Check out http://www.vk2zay.net/article/262 <br>Alan Yate's site has a very nice outline for this entire project. Highly recommend checking his stuff out. About higher frequency, the higher your frequency the less inductance you'll need.
I'm just new to the electronics world so if anyone could answer me. I have a 1k and 10 ohms resistor, and I need to have a default of 1 minute on 555 timer, did I have the right value? I would appreciate any reply. thank you
I'm just new to the electronics world so if anyone could answer me. I have a 1k and 10 ohms resistor, and I need to have a default of 1 minute on 555 timer, did I have the right value? I would appreciate any reply. thank you
Great post, thanks for all the accompanying info. Is it possible to dim the secondary by way of reducing the voltage through the primary or does that mess up the tuning of teh coils?
i need to charge my mobile by wireless system using wifi radiations &amp; signals
very well...
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how many microfarads is 181K ? <br>
180 pF<br>or<br>.00018 uF<br>here's a handy link:<br>http://www.electronics2000.co.uk/calc/capacitor-code-calculator.php
So I tried using your circuits in a circuit simulator. I couldn't get them to work properly.
i need to connect one of the circuits to a transformer without a centre tap how can i do it
&nbsp;I think your problem is you're having 2 different resonating frequencies. the primary may be at one, but the secondary is at another, thus a low, almost 0 voltage. the easiest thing would be to play with inductor lengths. or figure out which capacitors on the mouse, with the inductor, create the resonator, figure out the capacitor value and calculate the needed inductance for your secondary.
From your video, the circuit seems to be working greatly. But less power are transferred. What voltage did you get at your secondary coil?.
I dont think my meter can measure such low voltages at such high frequencies. I got a reading of 0.015mV while the LED was glowing.
Can you tell me which transistors you used.... its no. and what is your wire gauge for primary secondary coil.<br />
Yep...i think you're right. Maybe you should use an oscilloscope to measure the exact voltage value..
Your problem is simple: you have an impedance mismatch. This means that your primary coil is not well matche to the signal source, and to make it work you will need to match the signal source impedance to the coil impedance. Also you can consider the coil and the parallel cap as an LC resnance circuit acting as once circuit. You need a transistor driver and voltage multiplier, do not worry too much about the current now but please lift up the voltage as of now so you can get more transfer of power. Also remeber that the USB is limited in power and the manufacturer have made accurate calculations on how much power to transfer vs matching the impedance. What I would do is the following: The antenna coil must be driven by higher voltage higher current signal source first, since the USB cant provide higher current lets focus on higher voltage. Now charge up&nbsp;this high voltage in a cap and drive it to the coil via a MOSFET transistor with low ON rDS resistance and proper gate charge. I bet once you&nbsp;do that&nbsp;you should be&nbsp;able to then measure much higher voltage than what you reported. The secret to solving your problem is impedance matching... Please let&nbsp;me know how it goes and call me if you have a problem. Good luck... By the way I worked with high voltage security circuits and TV and power transfer circuits and I can help you&nbsp;as my time allows me.&nbsp;
I don't know if you still check this instructable, or if you're even an active member of the community, but I think I may have found the problem.<br /> <br /> You say throughout the instructable that you want a high frequency. For an inductive coupling application, however, high frequency is not a good thing. To fix this you could try adding a capacitor in series with the coupled signals, as sort of a filter. You could also try lowering the frequency.<br />
&nbsp;Can anybody explain the part of the schematic he added in step 10 I can't for the life of me understand it!! <br /> PLEASE!
Check out this person's instructable!&nbsp;&nbsp;&nbsp;&nbsp; http://www.instructables.com/id/Wireless-Power-Transmission-Over-Short-Distances-U/ &nbsp;&nbsp;&nbsp; It's pretty cool!<br />
It is possible your circuit is using more milliamps than your charging circuit is delivering...&nbsp; Ohms law works only when your source of AMPERAGE is UNLIMITED...&nbsp; In the real world, you are limited by what your SOURCE can deliver.&nbsp; For example, if you place two tiny 9volt batteries in series, you will have 18 volts... connect this to your automobile in place of your car battery.&nbsp; Ohms law says it should work... but you know there are not enough AMPERES&nbsp;AVAILABLE to do that job.&nbsp; The same is true on little circuits that only draw milliamperes.&nbsp; If the current source is too low, then amperes will not be delivered and THEN you see the voltage sag down low as result.
In the Real World, you're telling us what you observe on the outside of the battery.&nbsp; But inside the battery there is actually an _internal resistance_.&nbsp; If that were very, very low, then the 9V battery would deliver as much current as a car battery, but it's not low.&nbsp; So when you attempt to run your car with it, the voltage drops across the internal resistance, and all you get outside the battery is a very low voltage, what's left after the V drop across the internal resistance. <br /> BTW, if you try to run the car with the two 9V batteries, they will get hot because of the power dissipated in the internal resistance.<br /> <br /> There are these car starting packs that you can buy that are very small, and put out a very high current because their internal resistance is very low.&nbsp; Thus they are able to turn over the engine and start the automobile.<br /> <br />
Dont power it off the pc. Grab a more powerful adapter that plugs straight into the wall.
i agree with darstar u should just get a wall mount for it
Yeah I also agree. I might try to start my own project for my black berry. I will post my figures as soon as i start.<br />
HI, I am working on the same project.(for my final year). right now i am working on the primary coil and acc. to my calc. i need to transfer .24 henry though it. for that purpose i need a 32awg wire(copper). my question is do i have to use copper windg. or can i use aluminium wndg also??? i think puffin_juice is usin al wire in video. for rest of the circuit i am currently using function generator.... one more thing is it possible for you guys to post the full circuit diag.(or the one used in the video by puffin_juce) thanx
It doesn't make a difference, any conductor that is carrying current will produce a magnetic field.
the only difference with copper and aluminium wire will be resistance with heat i believe
not really when magnetizing&nbsp; any element and demagnetizing it again you get a energy loss called hysteresis loss. basically it takes energy to magnetize an element and energy again to demagnetize it again. copper has a very low hysteresis loss compared to other metals, in other words it doesn't stay strongly magnetized.<br />
The aluminium wire, (or the lower resistence wire) will draw a higher current, this will mean that a larger magnetic field is induced. if you cooled the primary circuit, you would notice an improvement (although alot of energy would be wasted on the refrigeration.
is it possible for u to send primary circuit diagram?? if yes pls send it to viking_rana@yahoo.com Thanx
A few observations. The smaller coil looks like it doesn't have enough turns. More turns should help. When I say more, I mean double or even more. The full wave bridge is a power waster. I would remove two of the diodes and double the number of turns in the smaller coil, turning it into a center tapped coil. This and the two diodes make a full wave rectifier. The diodes should be low loss, use 1N5817 schottky diodes for these.
Also, with what little I know of electronics...<br /> Would increasing the voltage and number of coils on the primary also provide a marked improvement?<br />
I'm currently working on a similar project.&nbsp; Have you looked into using a FET Driver instead of a 555 timer?&nbsp; This will lead you to much higher voltages and current, thus a stronger magnetic field.&nbsp; <br /> <br /> Also, how did you wind such nice coils?<br />
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&quot;A very well-trained cat&quot; -- Hahahahahah!&nbsp; Cats don't have masters, they have staff!!&nbsp; ;-)<br /> <br /> <br />
Hi<br /> <br /> Anyone has infomation on how to determine the mutual inductance or coupling factor? And also the equation to determine the inductance of the coils?<br /> <br /> hope if anyone can help me :(<br /> rgds<br /> alex<br />
Breadpig, this is simple an air-core transformer so M should follow as such:<br /> M=N1*N2*Phi=K * (L1*L2)^.5<br /> where N1 is the # of turns on the primary, N2 is the # of turns on the secondary, and Phi is the permeance of air.<br /> <br /> From this you can extract K, the coupling factor,<br /> K= (N1*N2*Phi) / (L1*L2)^.5<br /> <br /> As for determining the inductance of the coils, if you do not have a multimeter or some other device to measure this, you can sweep the frequency on a frequency generator looking at the voltage across a parallel cap and your coil.&nbsp; The peak voltage is then related to 2*pi*f = 1/(L*C)^.5<br /> <br />
&nbsp;Make sure you are perfectly aligned with the primary coil, at an angle of 90 degrees, you will get no power.<br /> <br /> also, where is the power regulation for the mouse?<br />
&nbsp;why are you using square wave?<br /> why not sine wave ?

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