Wireless Power Project need Help (Its not same old thing)

  • Thread starter harshnisar
  • Start date
So, i have decided to make this wireless power transmission project and it is not through mutual induction but microwave transmission.

I know , Microwave transmission ain't feasible and stuff and it has been discussed and all here, but i just want to make a small prototype , like a meter or even less.

I want to charge my cell phone using that..
Charger Specs
800mA and 5-7 V.

Now, i don't know alot about communication but i plan to learn through this project.
According to my basic understanding,
I plan to get the required wave using a Function generator of CRO.
Transmit it using an Antennae, recieve it... Convert to DC and then connect to the battery.

How do i select the wave frequency i want?
How do i get the preferred voltage ?

I can calculate the watts i require for the charging, but how do I use the information for selecting a wave. Please help.

Thanks in advance.
If you are thinking of "Tesla" or recent wireless power concepts, you need to research "Near Field propagation" because that's what all these use. Near Field is different from regular radio propagation also known as Far Field propagation.

Generally Near Field only occurs close to the antenna (relative to wavelength). Hence the name.

"So, i have decided to make this wireless power transmission project and it is not through mutual induction but microwave transmission."

Note, this discussion assumes far-zone.
You might start your design with the Friis transmission formula for received power:

P[itex]_{rec}[/itex] = [itex]\frac{P_{t}G_{t}G_{r}\lambda^{2}}{(4\pi r)^{2}}[/itex]

"Microwave transmission ain't feasible"

Not so. Your smart phone is one of many examples of devices that transmit power in the microwave region.

"How do i select the wave frequency i want?"

There are a lot of details to consider here. First, you need to know how much power you will be transmitting. This is important since your higher power signal might jam any receivers operating at nearby frequencies, including your smart phone. This could get you in serious legal trouble as the RF spectrum is already packed and the FCC maintains strict regulations concerning in-band and out-of-band interferers.

Next, you have to consider your antenna design. If you're going to use a dipole, you will most likely size it to be slightly less than [itex]\frac{\lambda}{2}[/itex] in length. So, your choice of frequency will have to be such that a half wave dipole is of reasonable size.

"How do i get the preferred voltage ?"

Once you have settled on an antenna design, you can determine the open circuit voltage at the terminals of your receiving antenna. In order to do this, you must be able to calculate three quantities:
1.) the electric field, [itex]\vec{E}[/itex][itex]^{inc}[/itex], that is incident on your receiving antenna. This is the far-zone electric field radiated by your transmitting antenna.
2.) the electric field, [itex]\vec{E}[/itex][itex]_{a}[/itex] of your receiving antenna. This is the far-zone field your receiver would radiate if it were a transmitter instead.
3.) the vector effective length (aka vector effective height), [itex]\vec{l}[/itex][itex]_{e}[/itex]. This is quantity is derived from [itex]\vec{E}[/itex][itex]_{a}[/itex].

Then, the open circuit voltage V[itex]_{OC}[/itex] = [itex]\vec{E}[/itex][itex]^{inc}[/itex] * [itex]\vec{l}[/itex][itex]_{e}[/itex] (note that this is a dot product).

jsgruszynski suggested considering the near field. There are important reasons for this. In general, the total [itex]\vec{E}[/itex] and [itex]\vec{H}[/itex] fields have terms that are proportional [itex]\frac{1}{r}[/itex], [itex]\frac{1}{r^{2}}[/itex] and [itex]\frac{1}{r^{3}}[/itex]. These higher power of r terms represent energy that is stored within the fields near to the radiating element. To see this for yourself consider an infinitesimal electric dipole (l << [itex]\lambda[/itex]) with constant current directed along [itex]\hat{z}[/itex]. You can use the vector potential [itex]\vec{A}[/itex]([itex]\vec{r}[/itex]) to solve for [itex]\vec{E}[/itex] and [itex]\vec{H}[/itex] using:

[itex]\vec{A}(\vec{r}) = \frac{\mu}{4\pi}\int^{l/2}_{-l/2}\vec{J}_{l}(\vec{r}')\frac{e^{-jkr}}{r}dz'[/itex]

[itex]\vec{E}[/itex] = -j[itex]\omega\vec{A} - j\frac{1}{\omega\mu\epsilon}\nabla(\nabla \cdot\vec{A})[/itex]


[itex]\vec{H} = \frac{1}{\mu}\nabla\times\vec{A}[/itex]

If you've done the math correctly, you will see that the [itex]\theta[/itex] component of [itex]\vec{E}[/itex] will have all three of the aforementioned [itex]\frac{1}{r^{n}}[/itex] terms.

Once you have the fields, you can consider the Poynting vector:

[itex]\vec{S} = \frac{1}{2}\vec{E}\times\vec{H}^{*}[/itex]

You will notice that S will not be entirely real nor will it be entirely directed along [itex]\vec{r}[/itex].

The power [itex]P_{t}[/itex] from the Friis formula is given as

[itex]P_{t} = \int\int_{S}\Re(\vec{S} )\cdot\hat{n}dA[/itex]

As you can see, only the real part of the radial component of the power density, S is radiated.

If you want to learn more, I suggest having a look at "Antenna Theory" by Balanis.


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