I would like to light up a path in the woods with hundreds of LEDs using RF. The idea is that I would carry a RF transmitter powerful enough to light LEDs. As I approach the LEDs would get brighter, showing the path, and fade out as I pass out of range. Similar to the effect of carrying a lantern. I would like the LEDs to become noticeable, though dim, at 15 to 30 feet.

Is this possible?

I have never attempted anything like this before and I would really appreciate any tips or help with this project.

Update* Parameters:

Transmitter weight as heavy as 20 lbs

Forward voltage up to 3v

No solar pannels

No long wires linking LEDs

Passive LEDs (no battery)

Parabolic dish if necessary

Also, is safety an issue when holding a transmitter so close?

  • 2
    \$\begingroup\$ get framiliar with Free-space path loss. you can't escape physics. then work backwards from your load to your source \$\endgroup\$ – hassan789 Jun 11 '15 at 4:02
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    \$\begingroup\$ Do the LEDs have a battery and solar re-charge during the day? If not then it's probably best you forget this idea. \$\endgroup\$ – Andy aka Jun 11 '15 at 7:34
  • \$\begingroup\$ @hassan789 Get familiar wit inductive power transfer and what has been achieved.Note that "Near field" RF like systems have some advantages. \$\endgroup\$ – Russell McMahon Jun 11 '15 at 7:57
  • \$\begingroup\$ @Andyaka I'm avoiding batteries and solar panels for this project. The challenge is to use only RF to power them. \$\endgroup\$ – Ryan Jun 11 '15 at 18:54
  • \$\begingroup\$ Setting a challenge that can be calculated out to be not rewarding in terms of power that can be gotten at 30 feet seems to me strange. Maybe you have a trick up your sleeve? \$\endgroup\$ – Andy aka Jun 11 '15 at 22:43

It feels intuitively like this might be possible. If you carried a 10 watt omnidirectional light bulb, in a dark place, it would certainly light up a white object at 30 ft. So why not at RF?

RF spreads out the same as visible light. The capture area of a dipole is about 0.1*wavelength^2.

So I calculate that at 145 MHz, a 5 W transmitter will deliver 5 mW to a dipole antenna 10 m away. The trick would be rectifying the RF without wasting too much of it. If you manage 50% efficiency that will produce 1 mA in the LED, which should be visible.

5 W is easily available from a handheld radio, but you'd need a decent antenna, not the rubber duck it ships with. VHF (136-174 MHz) is a reasonable choice of frequency, (you'd need a license to transmit). SMD LEDs are bright enough at 1 mA. So the hard parts are the rectifier, and getting the voltage high enough to drive the LED. A full-wave dipole has a high impedance, which might generate enough voltage (5 mW into 1000 Ohms is just over 2 V).

The finished path-light would be a string of wire perhaps 1m long, with a epoxy coated circuit in the middle. LED, diode and matching circuit. You could hang them from the trees.

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    \$\begingroup\$ No, transmitting 5 or 10 watts for a non-communication purpose is not legal under an amateur radio license. \$\endgroup\$ – Chris Stratton Dec 14 '20 at 10:02
  • \$\begingroup\$ True, strictly it's between amateurs. No one way transmissions. You certainly couldn't use this to power your Christmas lights all December. But there's a long history of experimental use of radio by amateurs, so as long as your overs are short, you stay away from busy frequencies and you identify, you could quite reasonably do on-air tests. It's a technical question and technical answer, saying it wouldn't work because it's illegal isn't really helping :) The stated purpose is beyond an experiment of course. \$\endgroup\$ – tomnexus Dec 14 '20 at 19:11
  • \$\begingroup\$ No, that's not even legal as a test. \$\endgroup\$ – Chris Stratton Dec 14 '20 at 19:14

It's doable, but there is a realness / hardness tradeoff which makes other methods more attractive.

  • Powering LEDS with RF at 30 feet is difficult and inefficient.

    Cheating so that LEDs 30 feet away are power by RF is doable.

    Activating LEDS with RF from 30 feet is a doddle.

    Activating LEDS some other way may be even easier.

RF or Inductive power transfer over 30 feet requires high Q resonant circuits, generally needs larger than smaller antennas and is liable to be hard and expensive to do well at a size and cost that is acceptable.

Inductive power transfer only transfers power to a matching load when the load is "in range" and well designed circuitry draws far less power when not loaded. RF transmitters "throw the power out into the aether" [tm] regardless of whether a load is present or not so must work at the maximum power level ever needed for the application. (You could arrange an interactive system that determines proximty and raises power level, butr complexity rises rapidly).

If you MUST transfer power then having the receiver coils close to your path will reduce the transmit power and/or coil sizes. You can then run wires to LEDs or other loads as desired.

Having the LEDs powered by eg batteries and detecting the proximity of a small portable transmitter is far easier than either of the above two choices.

Use of proximity detectors along the path is also easy. These could be ultrasonic (modules with 5 to 10 metre range cost under $10). Or light-beam-breaking or capacitive proximity or ...

As a guide to what can be achieved with range search for what MIT did sa few years ago.

Korea Advanced Institute of Science and Technology 2014 - 5 metres (KAIST](http://www.sciencedaily.com/releases/2014/04/140417124509.htm)

Wikipedia - Inductive Charging

Wikipedia - Wireless Power )

  • 3
    \$\begingroup\$ This might work (12 magnetrons mounted on a horn antenna). Doubles as a death ray. Since it draws 15kW (9.6kW of RF at 2.45GHz) it needs a serious power supply. The LEDs would need a microwave rectenna. \$\endgroup\$ – Spehro Pefhany Jun 11 '15 at 4:53

A bit late answer.., but some years ago I was able to light up a led at 1400 meters from an fm broadcaster. Power was 20 kw erp. The clue is that you have to transformer the energy to a very high impedance. Antenna was a 1/2 dipole.

  • \$\begingroup\$ So at 10m you would only need 1W erp? Can you show us a drawing and/or a photo of the device? \$\endgroup\$ – Bruce Abbott Jan 4 '20 at 5:17
  • \$\begingroup\$ Yes that works, but on UHF and higher you need more power. \$\endgroup\$ – Bert jan Jan 4 '20 at 18:22
  • \$\begingroup\$ Here are some links. I picked up from a transmitter on 103.5 Mhz at Ugchelen, NL. With a signal generator at home I designed a voltage step up circuit. A voltage of 30 mV at 75 ohms appeared to be the limit. I searched for a sensitive led. The one I used already lights up with a 1 megaohm resistir on 12 volt DC. \$\endgroup\$ – Bert jan Jan 4 '20 at 18:26
  • \$\begingroup\$ Here are some links youtube.com/watch?v=ojt9MW0_yAc pa1bj.blogspot.com/2014/01/led-ontvanger.html later I used a 1/1 wave dipole. This one has already a high impedance on itself. Sorry it's in Dutch language \$\endgroup\$ – Bert jan Jan 4 '20 at 18:31
  • \$\begingroup\$ OP didn't specify a frequency or antenna size, so 100MHz is fine. Can you include schematic and/or construction details for the step up circuit in your answer? \$\endgroup\$ – Bruce Abbott Jan 4 '20 at 18:32

Yest it is possible.

Assuming your transmitter is not directional you might be able to get 1% efficiency at 15 feet, so if you have a few 1 watt LEDs you will want a 100 watt transmitter. So a 5 pound battery might give you 1 hours use. That might be off by an order of magnitude so someone should do the math, but the point is it would be much more efficient to just hack in a control unit to some off the shelf solar garden lights and use a low power transmitter (maybe a mobile phone Wi-Fi Direct ping would be strong enough).

  • 1
    \$\begingroup\$ I calculate that the RF power received at 15 feet from a 1 watt transmitter using 433MHz is about 0.15 mW. That is about 1.5% efficient. \$\endgroup\$ – Andy aka Jun 11 '15 at 8:10

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