1. Hi all - is it possible to induce a charge of 5 Volts and very low amps (milliamps) in an electrical device which at maximum distance will be 15 meters away using EM Waves such as Radiowaves?

My presumption is that it is possible but I do not know how much power it will require at the transmission side and if the emitted waves will be safe for human contact.

Will this require much battery power? Anyone know of any links to some of the Maths for this?

Thanks! Much obliged!

2.

4. Originally Posted by MacGyver1968
I plan to power a small SMD piezo speaker via EM Waves only. No power source at receiver as it will be powered by the sender which will be a hand held device like something you could fit on a key chain - at least that is the plan...

5. It might be hard to make the transmitter small enough to fit on a key chain. What is the purpose of the speaker, are you trying to scare trick or treaters? How loud does it have to be?

6. Oh...one other thing...what's the budget for this project?

7. Originally Posted by MacGyver1968
Oh...one other thing...what's the budget for this project?
Budget = low. Tricks do come to mind Its also a learning thing, trying to see what it can do too...Doesn't have to be too loud which is why I was going for around 5V @ milliamps - slight sound coming from a corner of the room which quickly turns off when they look at it (controlled by me) sort of thing maybe :P

8. I was looking at Piezo buzzers...they ran on 5v at 10mA...which doesn't sound like a lot, but trying to transmit that much power through the air might be difficult. I haven't done the math, but it might require 100's of watts of transmitting power with fairly large directional anteana.

9. Originally Posted by MacGyver1968
I was looking at Piezo buzzers...they ran on 5v at 10mA...which doesn't sound like a lot, but trying to transmit that much power through the air might be difficult. I haven't done the math, but it might require 100's of watts of transmitting power with fairly large directional anteana.
thats why I thought it might be able to work - know of any equations or links to them? The distance will only be 15 meters at max...!

10. I'd have to look it up...but considering your talking about millions of times the power that normally is received on an antenna...this circuit will not be small or cheap. It be much easier to use a 9volt battery to power the receiving circuit.

11. Originally Posted by MacGyver1968
I'd have to look it up...but considering your talking about millions of times the power that normally is received on an antenna...this circuit will not be small or cheap. It be much easier to use a 9volt battery to power the receiving circuit.
So what Voltage & Amps would you usually find in a TV Antenna from a TV signal?

12. Somewhere around 10-100mV with microamps of current. "Millions of times" may be a bit of an exaggeration...but it's a fuckload (technical measurement) more.

13. Why don't you just use a wireless transmitter to turn power on and off, and just use something like a battery to power the speaker?

14. Originally Posted by TheObserver
Why don't you just use a wireless transmitter to turn power on and off, and just use something like a battery to power the speaker?
That would be the easiest way - but I am also trying to learn a thing or two in the process...

If it is 10-100mV - taking the middle value of 50 if I want to induce a voltage of 5V then I need to increase the transmission power by a factor of 100 (50 x 100 = 5000), no?

Next question then - what is the normal transmission power?

15. One other thing to consider is crystal radios. They do something similar using a long antenna and a crystal - that with a little earphone and you have an AM radio receiver.
No enough power for a speaker. But you could probably attach a small amp with a small solar panel. That would be a fun project.

16.

17. thats why I thought it might be able to work - know of any equations or links to them? The distance will only be 15 meters at max...!
Let us start with an optical analogy. Suppose you have a perfect 100 W bulb radiating light in every direction, without any loss. What part on this power will you be able to collect at a distance of 10m with a receptor (a photo-electric captor for instance) which dimensions are 10cm x 10 cm ?
At a distance of 10m, the light will be uniformly distributed on a sphere of 10m of radius.
The surface of this sphere is S = 4 Pi RČ = 1256 m2
The surface of the captor is s = 0.1 x 0.1 = 0.01 m2
Thus, the part of the overall power, 100 W, that you will collect is 100 s / S = 0.008 W
If you convert this power in electricity, and supposing that the captor is able to convert the light without any loss, with a voltage of 5V, the current will be some 1.6 mA.
So, the part of the power of the source that you are able to collect is less than 0.01% of the power of the transmitter and it decreases with the square of the distance.

Now, what happens if we use radio waves instead of light ? Light and radio waves are electromagnetic waves and the same reasoning applies. Antennas with a perfect spherical radiation are called isotropic, They don't exist in reality, but we will suppose we have one, for simplification.
The physical surface of the reception antenna is not really meaningful, but a similar notion was introduced by analogy, it is the “surface of capture” of an antenna.
For instance, I was told that the surface of capture of a parabolic antenna is something like the surface of the reflector.
More generally, the surface of capture is equal to G LČ / 4 Pi, where G is the gain of the antenna, and LČ the square of the wavelength.
Let us assume that
• the reception antenna is also isotropic because we want this antenna to be omnidirectional. Then G = 1.
• the wavelength is 0.1 m , in order to have dimensions acceptable. It makes a frequency around 1500Mhz.
Then the Sc will be 0.01 / (4 x 3.14) = 8 x 10-4 m2
It is 150 times smaller than the surface we computed in the optical example. So the power we will receive will be 150 times smaller, too, some 0.05 mW.
If we consider that a transmitter of 100W in the 1500Mhz is not exactly a cheap toy and that it may cook your brain, I am sorry, Fatman47, but it doesn't sound to me as a very good project.

18. thats why I thought it might be able to work - know of any equations or links to them? The distance will only be 15 meters at max...!
Let us start with an optical analogy. Suppose you have a perfect 100 W bulb radiating light in every direction, without any loss. What part on this power will you be able to collect at a distance of 10m with a receptor (a photo-electric captor for instance) which dimensions are 10cm x 10 cm ?
At a distance of 10m, the light will be uniformly distributed on a sphere of 10m of radius.
The surface of this sphere is S = 4 Pi RČ = 1256 m2
The surface of the captor is s = 0.1 x 0.1 = 0.01 m2
Thus, the part of the overall power, 100 W, that you will collect is 100 s / S = 0.008 W
If you convert this power in electricity, and supposing that the captor is able to convert the light without any loss, with a voltage of 5V, the current will be some 1.6 mA.
So, the part of the power of the source that you are able to collect is less than 0.01% of the power of the transmitter and it decreases with the square of the distance.

Now, what happens if we use radio waves instead of light ? Light and radio waves are electromagnetic waves and the same reasoning applies. Antennas with a perfect spherical radiation are called isotropic, They don't exist in reality, but we will suppose we have one, for simplification.
The physical surface of the reception antenna is not really meaningful, but a similar notion was introduced by analogy, it is the surface of capture of an antenna.
For instance, I was told that the surface of capture of a parabolic antenna is something like the surface of the reflector.
More generally, the surface of capture is equal to G LČ / 4 Pi, where G is the gain of the antenna, and LČ the square of the wavelength.
Let us assume that
• the reception antenna is also isotropic because we want this antenna to be omnidirectional. Then G = 1.
• the wavelength is 0.1 m , in order to have dimensions acceptable. It makes a frequency around 1500Mhz.
Then the Sc will be 0.01 / (4 x 3.14) = 8 x 10-4 m2
It is 150 times smaller than the surface we computed in the optical example. So the power we will receive will be 150 times smaller, too, some 0.05 mW.
If we consider that a transmitter of 100W in the 1500Mhz is not exactly a cheap toy and that it may cook your brain, I am sorry, Fatman47, but it doesn't sound to me as a very good project.

19. Have you checked out principles of 'crystal radios'? It's basically an AM wave to low V current amplifier.

20. Originally Posted by Write4U
Have you checked out principles of 'crystal radios'?
What possible difference would that make? caKus offered a detailed, quantitative analysis. Substitute your own numbers if you are analyzing a different scenario.

Crystal radios work exactly the same way as all other radios. There is no magic. You have an antenna whose job is to capture some fraction of the incident energy. The rest of the circuitry just converts the RF into audio, losing energy in the process, of course.

It's important to carry out calculations if you don't have a good feel for what the magnitudes are. A typical cellphone is able to function -- barely -- with tens of femtowatts intercepted by the antenna. That's femto.

A crystal radio, with a superb antenna, low-loss components, and world-record headphones, might be able to produce a barely audible signal with a nanowatt hitting the antenna. That's nano.

Yes, those powers can go up if you reduce the distance to the transmitter. But still, look at those numbers. You aren't going to power up anything substantial with those kinds of power levels. If you want to increase to even microwatt powers, you would need to increase transmitter powers by at least a factor of one thousand. So instead of a radio transmitter of tens of kilowatts, you'd need tens of megawatts or more (again, depending on distance).

This is the same math that doomed (and still doom) all Tesla-based fantasies of wireless power transmission on an industrial scale.

21. Originally Posted by tk421
Originally Posted by Write4U
Have you checked out principles of 'crystal radios'?
What possible difference would that make? caKus offered a detailed, quantitative analysis. Substitute your own numbers if you are analyzing a different scenario.

Crystal radios work exactly the same way as all other radios. There is no magic. You have an antenna whose job is to capture some fraction of the incident energy. The rest of the circuitry just converts the RF into audio, losing energy in the process, of course.

It's important to carry out calculations if you don't have a good feel for what the magnitudes are. A typical cellphone is able to function -- barely -- with tens of femtowatts intercepted by the antenna. That's femto.

A crystal radio, with a superb antenna, low-loss components, and world-record headphones, might be able to produce a barely audible signal with a nanowatt hitting the antenna. That's nano.

Yes, those powers can go up if you reduce the distance to the transmitter. But still, look at those numbers. You aren't going to power up anything substantial with those kinds of power levels. If you want to increase to even microwatt powers, you would need to increase transmitter powers by at least a factor of one thousand. So instead of a radio transmitter of tens of kilowatts, you'd need tens of megawatts or more (again, depending on distance).

This is the same math that doomed (and still doom) all Tesla-based fantasies of wireless power transmission on an industrial scale.
I haven't got the expertise to judge caKus detailed presentation, but it sounds fairly expensive to me.

The only difference I can see is that AM radiowaves are abundant at great power. If the goal is to generate a small voltage for a trigger mechanism at very low cost, this might be an option.

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