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I'm an ee student and for the group project we have chosen to make a Wireless power transfer. Our goal is to charge a laptop. As for the result we have to demonstrate charging a laptop, so considering modern laptops I think we need at least 60-80 watts of power to be transferred to the receiving coil.

So for the first step we need to create the driver for the primary coil which will creates an alternating current using DC voltage, for the purpose we intend to use a H-bridge and a mosfet driver and an arduino. The purpose of using an arduino is to be able to vary the switching frequency. Also we intend to use resonant mode instead of inductive mode because the resonant mode can transfer energy more than inductive.

As our syllabus doesn't cover any of this wireless power transfer, we are gathering information and learning of it.

Is the use of h-bridge is suitable for this purpose? Input voltage of transferring is 12volts, on which factors we should choose the mosfets on the current rating.

We are also in a tight situation as we only have about 2 months to complete this.

Note:- the distance between two coils can be just few centimetres and the receiving coil won't be built to the laptop. For the demonstration purpose we will just use the secondary coil and laptop apart from each other.

Any help would be appreciated. Regards

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    \$\begingroup\$ 80 Watts over 30 cm ?! I think you should be more realistic. Consult with your professor before setting overly ambitious goals. \$\endgroup\$ – Maple Jan 14 at 11:32
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    \$\begingroup\$ What frequency you are going to use? Anyway, take in mind, that - without the directional antenna system - the eficciency of transfer on such distance will be probably few percents, so to transmit 60...80W you will need to have the transmitter output power more than 1kW! It can be dangerous for health, and can broke your local law about RF emission levels. \$\endgroup\$ – VillageTech Jan 14 at 11:36
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    \$\begingroup\$ At this power level, if the receive antenna is built into the laptop, the transmitter stands a pretty good chance of destroying the laptop. \$\endgroup\$ – Brian Drummond Jan 14 at 11:44
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    \$\begingroup\$ Exactly. If we expect that receiver coil will get 80W, the similar power will be received by all metal elements in laptop and in its environment, resulting in valuable heating of them - so there is a fire risk too... \$\endgroup\$ – VillageTech Jan 14 at 11:47
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    \$\begingroup\$ @Aimkiller well... repeat after me: "inverse cube distance" :) \$\endgroup\$ – Maple Jan 14 at 12:01
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Any help would be appreciated

The most significant recommendation I can give is to simulate the set up. This starts fairly easily with a transmit coil driven by a sinusoidal voltage source coupled to a receive coil that is resonant tuned. The coupling is dictated by the dimensions of the coils and this, in turn can be calculated by first understanding the top picture below (the transmitter): -

enter image description here

Previous answer where the diagram came from.

The receive coil can be assumed to be at point P and therefore it will collect a total flux of approximately B x Area and, knowing frequency of the field you can calculate induced voltage.

So develop some math that gives you induced voltage against the diameters of the coils. What you will always find is that bigger diameter coils makes the problem of distance easier.

Develop your coil size this way and then work out what the coupling factor is at the maximum distance you want.

Then go to a proper circuit simulator (I use micro-cap) and model the circuit using the coupling factor k. You will also need to add a bridge rectifier to the receiver along with smoothing capacitor and simulated load (representing the power consumed).

You can then see how much drive voltage is needed on your transmit coil to get the power output. You will also need to model coil losses (series resistance) for both transmit and receive ends. These will act against you.

At that point you will probably want to figure out how you will resonate your transmit coil and, if you are using a H bridge then series resonance is a likely candidate. Any driving circuit that produces "hard" voltages will need to be loaded by a series resonant circuit. A class A stage would be parallel resonant.

Practical advice: you might consider using ferrite plates behind the Rx coil to reduce the field travelling further and upsetting your laptop. And you will definitely not want your receive coil brought into close contact with your transmit coil because the total induced voltage might by hundreds of volts and blow up your charger circuits.

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  • \$\begingroup\$ This gave me a clearly idea of what we have to done. Appreciate your answer, thank you very much. \$\endgroup\$ – Aimkiller Jan 14 at 13:37
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    \$\begingroup\$ If you can get hold of litz wire for your coils you will help yourself. I've done a similar project for about 10 watts and got maybe 15 cm. Power in was about 25 watts with litz wire but about 35 watts for normal Cu. If you get to now your coupling efficiency you can add circuits that limit the power in to your transmit coil just in case things get too close. You will also find that the best coupling position is close to the Tx coil but any closer and the coupling reduces due to detuning effects. \$\endgroup\$ – Andy aka Jan 14 at 13:40
  • \$\begingroup\$ ah, I see. not sure about the litz wire will be available to buy in here thought -_- yeah, without any over-voltage protection device charger circuit will get fried. There are many things to learn as this project going on. Thanks for the help :) \$\endgroup\$ – Aimkiller Jan 14 at 16:31
  • \$\begingroup\$ "Any driving circuit that produces "hard" voltages will need to be loaded by a series resonant circuit. A class A stage would be parallel resonant." Please explain more about this \$\endgroup\$ – Aimkiller Jan 18 at 20:09
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    \$\begingroup\$ Series resonant produces a low impedance hence should not be overdriven too much by a sturdy voltage source. A class A stage basically needs DC bias currents and therefore only a parallel resonant circuit fits the bill. \$\endgroup\$ – Andy aka Jan 18 at 21:08

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