Lighting a LED with ambient RF (was candle)

[Billyb97113]

To quote John Larkins:
"More practical would be to get usable LED lighting from ambient RF"

I tried several different types of diodes and Ferrite rods and coils and
could never get a measurable voltage across a .01 uf cap. I used a 10X scope
probe as the only load. I would be happy if the LED occasionally blinked
using any of the inductive circuits found on the internet.

Any ideas?? -bill

[daku...@gmail.com]

Hello,
There are apparently a few working RF energy harvesters,
one by a Joe <Something> who in fact has a patent on it.
This basically uses a voltage doubler, and if I remember
correctly a 30 feet antenna. Another design was discussed
on this newsgroup, and this person was having some
grounding issues. Be warned though that the output is
very low, in the milliWatts.

[upsid...@downunder.com]

Did you try to activate your cell phone close to your system ?

The capture area of a dipole is proportional to the square of the
wavelength. Thus, much more power is available at lower frequencies
with the same field strength (V/m).

Repeat your measurement close to a medium wave (AM) broadcast station
using a tall vertical antenna. Since your antenna most likely will be
much shorter than 1/4 wavelength, it will be very low impedance and
highly capacitively reactive, thus a series inductance is needed, in
order to get any power delivered to the load.

A LED at 2 V and 1 mA would be a 2 kohm load, while an electrically
short antenna would have a few ohms or even milliohms radiation
resistance, thus some impedance step up is needed.

A full size dipole would have 50-75 ohm impedance and might drive
nicely a back to back LEDs at 20-30 mA. At 100 MHz, this would require
about 2 V/m field strength, thus the system must be quite close to the
transmitter.

Any systems using ferrite rods on the MW band would be practically
useless, since the typical ferrite rod antenna gain is -30 to -50 dB
below the dipole.

[miso]

I vaguely remember something about fluorescent tubes glowing near TV
transmitters. Maybe Sutro tower.

[Jan Panteltje]

On a sunny day (Tue, 02 Oct 2012 02:29:39 -0700) it happened miso
<mi...@sushi.com> wrote in <k4ec7i$haj$1...@speranza.aioe.org>:

>
>I vaguely remember something about fluorescent tubes glowing near TV
>transmitters. Maybe Sutro tower.

Fluorescent tubes also glow when stuck in the ground under a HV power line.
I used to walk around with a glowing neon next ot my 7 MHz xmtr antenna.

[Michael A. Terrell]

miso wrote:
>
> I vaguely remember something about fluorescent tubes glowing near TV
> transmitters. Maybe Sutro tower.

I saw that at the VOA station at their 'Bethany' facility in Mason
Ohio. The original W.W. II Crosley transmitters were the worst
offenders, with large glass doors in front of the output tubes. Of
course, having up to 1 MW of HF RF in a half acre made it simple to
light the lamp. They pulled a 4' tube out of a locker to show off the
effect to a bunch of teenagers who were there on a tour from our
highschool's ham radio club.

[Bruce Varley]

"Billyb97113" <billy...@yahoo.com> wrote in message
news:k4dohb$at9$1...@dont-email.me...

People apparently don't build crystal sets any more, what a pity. 10s of Km
from a MW broadcast transmitter, 10-20 metres long wire antenna into an LC
circuit provides enough voltage to push through a low foward voltage diode,
and generate a low level but audible signal in hi-Z earphones. That provides
a rough indication for what can be achieved. Accessible energy density is
proportional to distance squared, unless you're very close, and the receive
antenna efficiency is likely to be a dominant factor.

[John Larkin]

On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
<billy...@yahoo.com> wrote:

I have no idea how much power can be extracted from a ferrite rod
antenna. Googling produces vague results.

A "longwire" antenna would probably pick up more power. Any reasonable
antenna will be a fraction of a wavelength for AM, so it looks sort of
like a capacitor driven by a current source. That could drive a
resonant circuit, of which the LED capacitance itself is part of the
LC capacitance. That wouldn't need a separate rectifier diode, which
would just waste power. That could be Spiced easily. Antenna, L,
variable C, LED. One of those trimpot-looking ceramic variable
capacitors would work. It's an impedance matching problem.

When I was a kid, I had a crystal radio with proper high-impedance
headphones. With roughly a 40 foot longwire antenna on the roof, it
was *loud*, maybe milliwatts of audio power. A good LED is visible in
room light at a microwatt.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

[John Fields]

On Tue, 02 Oct 2012 09:12:59 -0700, John Larkin
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
><billy...@yahoo.com> wrote:
>
>>To quote John Larkins:
>>"More practical would be to get usable LED lighting from ambient RF"
>>
>>I tried several different types of diodes and Ferrite rods and coils and
>>could never get a measurable voltage across a .01 uf cap. I used a 10X scope
>>probe as the only load. I would be happy if the LED occasionally blinked
>>using any of the inductive circuits found on the internet.
>>
>>Any ideas?? -bill
>>
>
>I have no idea how much power can be extracted from a ferrite rod
>antenna.

---
Why, then, is that admission of ignorance germane?
---

>Googling produces vague results.

---
GIGO?
---

>A "longwire" antenna would probably pick up more power.

---
Since you've previously stated that you don't know how much power can
be extracted from a ferrite rod antenna, and you're now stating that
a "longwire" antenna would probably pick up more power, it strikes me
that what you're dealing in is conjecture.
---

>Any reasonable
>antenna will be a fraction of a wavelength for AM, so it looks sort of
>like a capacitor driven by a current source. That could drive a
>resonant circuit, of which the LED capacitance itself is part of the
>LC capacitance.

---
That would be self-defeating since, if the intent is to harvest power
from ambient RF, the band captured by resonance would exclude the
other sources rejected by the Q of the tank.
---

>That wouldn't need a separate rectifier diode, which
>would just waste power. That could be Spiced easily. Antenna, L,
>variable C, LED. One of those trimpot-looking ceramic variable
>capacitors would work. It's an impedance matching problem.

---
Unless you're prepared to post the "easily SPICED solution" to which
you allude, your commentary is just lip service.
---

>When I was a kid, I had a crystal radio with proper high-impedance
>headphones. With roughly a 40 foot longwire antenna on the roof, it
>was *loud*, maybe milliwatts of audio power. A good LED is visible in
>room light at a microwatt.

---
Apples and oranges.

--
JF

[Jim Thompson]

On Sun, 07 Oct 2012 17:10:59 -0500, John Fields
<jfi...@austininstruments.com> wrote:

>On Tue, 02 Oct 2012 09:12:59 -0700, John Larkin
><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>
>>On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
>><billy...@yahoo.com> wrote:
>>
>>>To quote John Larkins:
>>>"More practical would be to get usable LED lighting from ambient RF"
>>>
>>>I tried several different types of diodes and Ferrite rods and coils and
>>>could never get a measurable voltage across a .01 uf cap. I used a 10X scope
>>>probe as the only load. I would be happy if the LED occasionally blinked
>>>using any of the inductive circuits found on the internet.
>>>
>>>Any ideas?? -bill
>>>
>>
>>I have no idea how much power can be extracted from a ferrite rod
>>antenna.
>
>---
>Why, then, is that admission of ignorance germane?
>---
>

[snip]

Larkin took another crap, and his hat covered his eyes ?:-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

[John Larkin]

On Sun, 07 Oct 2012 17:10:59 -0500, John Fields
<jfi...@austininstruments.com> wrote:

>On Tue, 02 Oct 2012 09:12:59 -0700, John Larkin
><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>
>>On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
>><billy...@yahoo.com> wrote:
>>
>>>To quote John Larkins:
>>>"More practical would be to get usable LED lighting from ambient RF"
>>>
>>>I tried several different types of diodes and Ferrite rods and coils and
>>>could never get a measurable voltage across a .01 uf cap. I used a 10X scope
>>>probe as the only load. I would be happy if the LED occasionally blinked
>>>using any of the inductive circuits found on the internet.
>>>
>>>Any ideas?? -bill
>>>
>>
>>I have no idea how much power can be extracted from a ferrite rod
>>antenna.
>
>---
>Why, then, is that admission of ignorance germane?
>---
>
>>Googling produces vague results.
>
>---
>GIGO?

If you are so good at web searching, tell us how much power a ferrite
rod can extract from AM stations.

>---
>
>>A "longwire" antenna would probably pick up more power.
>
>---
>Since you've previously stated that you don't know how much power can
>be extracted from a ferrite rod antenna, and you're now stating that
>a "longwire" antenna would probably pick up more power, it strikes me
>that what you're dealing in is conjecture.

Or maybe "engineering." You wouldn't understand.

>---
>
>>Any reasonable
>>antenna will be a fraction of a wavelength for AM, so it looks sort of
>>like a capacitor driven by a current source. That could drive a
>>resonant circuit, of which the LED capacitance itself is part of the
>>LC capacitance.
>
>---
>That would be self-defeating since, if the intent is to harvest power
>from ambient RF, the band captured by resonance would exclude the
>other sources rejected by the Q of the tank.

It's not self-defeating if it lights the LED. I think I mentioned
broadband power harvesting in a recent post.

You don't know or care about this, you just want to whine.

>---
>
>>That wouldn't need a separate rectifier diode, which
>>would just waste power. That could be Spiced easily. Antenna, L,
>>variable C, LED. One of those trimpot-looking ceramic variable
>>capacitors would work. It's an impedance matching problem.
>
>---
>Unless you're prepared to post the "easily SPICED solution" to which
>you allude, your commentary is just lip service.

Are you so helpless that you can't connect four parts in Spice?

>---
>
>>When I was a kid, I had a crystal radio with proper high-impedance
>>headphones. With roughly a 40 foot longwire antenna on the roof, it
>>was *loud*, maybe milliwatts of audio power. A good LED is visible in
>>room light at a microwatt.
>
>---
>Apples and oranges.

No. Milliwatts and microwatts are the same dimensional units.

[John Larkin]

On Sun, 07 Oct 2012 15:28:35 -0700, Jim Thompson
<To-Email-Use-Th...@On-My-Web-Site.com> wrote:

>On Sun, 07 Oct 2012 17:10:59 -0500, John Fields
><jfi...@austininstruments.com> wrote:
>
>>On Tue, 02 Oct 2012 09:12:59 -0700, John Larkin
>><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>>On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
>>><billy...@yahoo.com> wrote:
>>>
>>>>To quote John Larkins:
>>>>"More practical would be to get usable LED lighting from ambient RF"
>>>>
>>>>I tried several different types of diodes and Ferrite rods and coils and
>>>>could never get a measurable voltage across a .01 uf cap. I used a 10X scope
>>>>probe as the only load. I would be happy if the LED occasionally blinked
>>>>using any of the inductive circuits found on the internet.
>>>>
>>>>Any ideas?? -bill
>>>>
>>>
>>>I have no idea how much power can be extracted from a ferrite rod
>>>antenna.
>>
>>---
>>Why, then, is that admission of ignorance germane?
>>---
>>
>[snip]
>
>Larkin took another crap, and his hat covered his eyes ?:-)
>
> ...Jim Thompson

You keep repeating the same childish attempts at insults, and you
can't post circuits that work. Sounds like senility to me.

I don't suppose you have anything intelligent to say about lighting
LEDs from ambient RF. JF, also getting senile, doesn't either.

[whit3rd]

On Monday, October 1, 2012 8:54:20 PM UTC-7, Billyb97113 wrote:
> To quote John Larkins: "More practical would be to get usable LED lighting from ambient RF" I tried several different types of diodes and Ferrite rods and coils and could never get a measurable voltage across a .01 uf cap.

The key here, is that RF thermal noise isn't harvest-able (unless
your receiver is cryogenic). You need to identify a frequency at which
the RF is strong (depends on your area, of course) and then build the
equivalent of a crystal radio - a tuned antenna with a rectifier that
isn't too lossy. It should be possible to use a voltage-multiplier
and neon lamp in most urban environments, if you tune to a local AM
station. The neon lamp will blink after enough charge gets deposited.

If the neon lamp lights when you're holding it in your fingers, there's
a real transmitter nearby. If an incandescent lights when you put
a rabbit-ear TV antenna onto it, the transmitter is perhaps
right there in the room... I've seen that happen.

[Jim Thompson]

On Sun, 07 Oct 2012 16:42:36 -0700, John Larkin

You have no idea of "effective area" ??:-)

>
>I don't suppose you have anything intelligent to say about lighting
>LEDs from ambient RF. JF, also getting senile, doesn't either.

Senility is due to shit tightening your hat band so no blood reaches
your brain... has nothing to do with age... I can cite all kinds of
octogenarians with far more smarts than Larkin.

[John Larkin]

On Sun, 07 Oct 2012 16:55:51 -0700, Jim Thompson

More word salad.

[John Larkin]

On Sun, 7 Oct 2012 16:46:23 -0700 (PDT), whit3rd <whi...@gmail.com>
wrote:

I had in mind an AM station and a longwire antenna. Just hang the LED
across an LC tank, and let it rectify the RF. LED capacitances are
often in the 10 pF range. A good crystal set is LOUD! in a city
environment.

[Jim Thompson]

On Sun, 07 Oct 2012 17:21:20 -0700, John Larkin

When it's your ignorance, it's "word salad" ?:-) Bwahahahahaha! You
just keep getting deeper and deeper into it. Keep trying >:-)

[upsid...@downunder.com]

On Sun, 07 Oct 2012 17:21:20 -0700, John Larkin

The effective area of a half wave dipole is about 0.12 square
wavelengths, thus at 1 MHz with 300 m wavelength, the capture area is
about 10000 m² or 1 ha.

The typical ferrite rod antenna has a gain of -30 to -60 dB below the
full size dipole, thus the effective area is somewhere between 10 m²
and 1 dm². Multiply this by the local power density [W/m²] and you get
a ballpark figure of the available power.

[John Fields]

On Sun, 07 Oct 2012 16:40:02 -0700, John Larkin

Red herring.

Your alleging my proficiency in searching the web has nothing to do
with your lackluster results.

---

>>>A "longwire" antenna would probably pick up more power.
>>
>>---
>>Since you've previously stated that you don't know how much power can
>>be extracted from a ferrite rod antenna, and you're now stating that
>>a "longwire" antenna would probably pick up more power, it strikes me
>>that what you're dealing in is conjecture.
>
>Or maybe "engineering." You wouldn't understand.

---

Ad hominem.

---

>>>Any reasonable
>>>antenna will be a fraction of a wavelength for AM, so it looks sort of
>>>like a capacitor driven by a current source. That could drive a
>>>resonant circuit, of which the LED capacitance itself is part of the
>>>LC capacitance.
>>
>>---
>>That would be self-defeating since, if the intent is to harvest power
>>from ambient RF, the band captured by resonance would exclude the
>>other sources rejected by the Q of the tank.
>
>It's not self-defeating if it lights the LED.

---
So say you, but so far we've seen nothing to support your claim but
vagaries.
---

>I think I mentioned
>broadband power harvesting in a recent post.

---
And that has what to do with this one???
---

>You don't know or care about this, you just want to whine.

---

Ad hominem.

---


>>>That wouldn't need a separate rectifier diode, which
>>>would just waste power. That could be Spiced easily. Antenna, L,
>>>variable C, LED. One of those trimpot-looking ceramic variable
>>>capacitors would work. It's an impedance matching problem.
>>
>>---
>>Unless you're prepared to post the "easily SPICED solution" to which
>>you allude, your commentary is just lip service.
>
>Are you so helpless that you can't connect four parts in Spice?

---

You just can't stop the ad hominem attacks, can you?

The point is - since you're the one who alluded to the "solution" -
the responsibility for fleshing it out is yours alone.

---

>>>When I was a kid, I had a crystal radio with proper high-impedance
>>>headphones. With roughly a 40 foot longwire antenna on the roof, it
>>>was *loud*, maybe milliwatts of audio power. A good LED is visible in
>>>room light at a microwatt.
>>
>>---
>>Apples and oranges.
>
>No. Milliwatts and microwatts are the same dimensional units.

---
Indeed, so that couldn't possibly have been what I meant, which was
that the juxtaposition of headphones and LEDs is like that of apples
and oranges.

--
JF

[John Fields]

On Sun, 07 Oct 2012 16:42:36 -0700, John Larkin

---
Have you already forgotten that you replied to posts of mine on the
subject?

--
JF

[Martin Brown]

On 08/10/2012 01:24, John Larkin wrote:
> On Sun, 7 Oct 2012 16:46:23 -0700 (PDT), whit3rd <whi...@gmail.com>
> wrote:
>
>> On Monday, October 1, 2012 8:54:20 PM UTC-7, Billyb97113 wrote:
>>> To quote John Larkins: "More practical would be to get usable LED lighting from ambient RF" I tried several different types of diodes and Ferrite rods and coils and could never get a measurable voltage across a .01 uf cap.
>>
>> The key here, is that RF thermal noise isn't harvest-able (unless
>> your receiver is cryogenic). You need to identify a frequency at which
>> the RF is strong (depends on your area, of course) and then build the
>> equivalent of a crystal radio - a tuned antenna with a rectifier that
>> isn't too lossy. It should be possible to use a voltage-multiplier
>> and neon lamp in most urban environments, if you tune to a local AM
>> station. The neon lamp will blink after enough charge gets deposited.

I don't see how you can make this work short of driving a micro Telsa
transformer and prayer. There just isn't enough total power except very
close to the transmitter or voltage available even in the tuned circuit
tank and as soon as you try to draw any current the Q falls rapidly.

You have to get to about 80v for a neon lamp to break down and glow. By
comparison getting 2-3v at 0.1uA or less for a white LED to just start
glowing should be a lot easier but still a bit of a challenge.

>> If the neon lamp lights when you're holding it in your fingers, there's
>> a real transmitter nearby. If an incandescent lights when you put
>> a rabbit-ear TV antenna onto it, the transmitter is perhaps
>> right there in the room... I've seen that happen.
>
> I had in mind an AM station and a longwire antenna. Just hang the LED
> across an LC tank, and let it rectify the RF. LED capacitances are
> often in the 10 pF range. A good crystal set is LOUD! in a city
> environment.

A high long wire antenna isolated from ground will quite often have
enough DC potential at very low current to light a neon screwdriver just
from the atmospheric potential gradient. Uselessly low current except
during thunderstorms when the LED would be at risk of damage.

My TV aerial is high enough up that on most winter days it will light a
neon screwdriver at ground level! A book value atmospheric gradient of
100V/m is typical but some days it is much more and at high enough
current to notice. If you try to measure it with an ordinary DVM rather
than an ultrahigh impedance electrometer you tend to get much less.

I would expect it to be able to provide enough current to light one of
the more sensitive LEDs provided it was protected from the overvoltage.
Never tried it though.

You can drive electrostatic toys from the potential difference across
two decent sized plates at different heights in the same way as with a
Zamboni pile like the perpetual ringing clock in Oxford.

http://en.wikipedia.org/wiki/Oxford_Electric_Bell

A simple variant also serves as a thunderstorm alarm.

--
Regards,
Martin Brown

[k...@att.bizzzzzzzzzzzz]

On Mon, 08 Oct 2012 09:06:13 +0100, Martin Brown
<|||newspam|||@nezumi.demon.co.uk> wrote:

>On 08/10/2012 01:24, John Larkin wrote:
>> On Sun, 7 Oct 2012 16:46:23 -0700 (PDT), whit3rd <whi...@gmail.com>
>> wrote:
>>
>>> On Monday, October 1, 2012 8:54:20 PM UTC-7, Billyb97113 wrote:
>>>> To quote John Larkins: "More practical would be to get usable LED lighting from ambient RF" I tried several different types of diodes and Ferrite rods and coils and could never get a measurable voltage across a .01 uf cap.
>>>
>>> The key here, is that RF thermal noise isn't harvest-able (unless
>>> your receiver is cryogenic). You need to identify a frequency at which
>>> the RF is strong (depends on your area, of course) and then build the
>>> equivalent of a crystal radio - a tuned antenna with a rectifier that
>>> isn't too lossy. It should be possible to use a voltage-multiplier
>>> and neon lamp in most urban environments, if you tune to a local AM
>>> station. The neon lamp will blink after enough charge gets deposited.
>
>I don't see how you can make this work short of driving a micro Telsa
>transformer and prayer. There just isn't enough total power except very
>close to the transmitter or voltage available even in the tuned circuit
>tank and as soon as you try to draw any current the Q falls rapidly.
>
>You have to get to about 80v for a neon lamp to break down and glow. By
>comparison getting 2-3v at 0.1uA or less for a white LED to just start
>glowing should be a lot easier but still a bit of a challenge.

The issue is power. ...and impedance matching.

>>> If the neon lamp lights when you're holding it in your fingers, there's
>>> a real transmitter nearby. If an incandescent lights when you put
>>> a rabbit-ear TV antenna onto it, the transmitter is perhaps
>>> right there in the room... I've seen that happen.
>>
>> I had in mind an AM station and a longwire antenna. Just hang the LED
>> across an LC tank, and let it rectify the RF. LED capacitances are
>> often in the 10 pF range. A good crystal set is LOUD! in a city
>> environment.
>
>A high long wire antenna isolated from ground will quite often have
>enough DC potential at very low current to light a neon screwdriver just
>from the atmospheric potential gradient. Uselessly low current except
>during thunderstorms when the LED would be at risk of damage.

An LED will also leak off charge without lighting (visibly).

>My TV aerial is high enough up that on most winter days it will light a
>neon screwdriver at ground level! A book value atmospheric gradient of
>100V/m is typical but some days it is much more and at high enough
>current to notice. If you try to measure it with an ordinary DVM rather
>than an ultrahigh impedance electrometer you tend to get much less.

Yet above you express a problem getting enough voltage to light a neon?

>I would expect it to be able to provide enough current to light one of
>the more sensitive LEDs provided it was protected from the overvoltage.
>Never tried it though.

Power. Impedance matching.

>You can drive electrostatic toys from the potential difference across
>two decent sized plates at different heights in the same way as with a
>Zamboni pile like the perpetual ringing clock in Oxford.
>
>http://en.wikipedia.org/wiki/Oxford_Electric_Bell
>
>A simple variant also serves as a thunderstorm alarm.

An AM receiver works better. A weather radio, even better.

[John Larkin]

On Mon, 08 Oct 2012 02:50:01 -0500, John Fields

If a crystal set can pump milliwatts of electrical power into a
headphone, something very similar should be able to light up an LED.
No mixed fruits are involved.

[John Larkin]

On Mon, 08 Oct 2012 09:06:13 +0100, Martin Brown
<|||newspam|||@nezumi.demon.co.uk> wrote:

I wasn't thinking electrostatics (hard to get DC from) but picking up
energy from an AM station.

[lang...@fonz.dk]

On 8 Okt., 16:15, John Larkin

<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
> On Mon, 08 Oct 2012 09:06:13 +0100, Martin Brown
>
>
>
>
>
>
>
>
>
> <|||newspam...@nezumi.demon.co.uk> wrote:
> >On 08/10/2012 01:24, John Larkin wrote:

> >> On Sun, 7 Oct 2012 16:46:23 -0700 (PDT), whit3rd <whit...@gmail.com>

apparently you can get ics for it, http://www.powercastco.com/PDF/powercast-overview.pdf
though power levels and distances a quite far from you usual AM
station

-Lasse

[John Fields]

On Mon, 08 Oct 2012 07:12:06 -0700, John Larkin
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:


>If a crystal set can pump milliwatts of electrical power into a
>headphone, something very similar should be able to light up an LED.
>No mixed fruits are involved.

---
So build one.

--
JF

[John Larkin]

I might some day. I already have a lithium-battery night light that I
built, glowing on the bookshelf near my bed. I figure it will last
20-30 years, always on. It has an interesting hands-free variable
brightness feature, where I can find it by its glow, but crank it up
as needed for finding my way around, after an earthquake or whatever.

Regular LED flashlights should do that, namely glow a little all the
time, so you can find them in the dark.

Given the energy density of lithium batteries, "energy harvesting"
rarely makes sense.

[amdx]

On 10/2/2012 9:28 AM, Bruce Varley wrote:
>
> "Billyb97113" <billy...@yahoo.com> wrote in message
> news:k4dohb$at9$1...@dont-email.me...
>> To quote John Larkins:
>> "More practical would be to get usable LED lighting from ambient RF"
>>
>> I tried several different types of diodes and Ferrite rods and coils and
>> could never get a measurable voltage across a .01 uf cap. I used a 10X
>> scope probe as the only load. I would be happy if the LED occasionally
>> blinked using any of the inductive circuits found on the internet.
>>
>> Any ideas?? -bill


> People apparently don't build crystal sets any more,

That is not correct!
Here's a very active group.

http://theradioboard.com/rb/index.php


And here are some well researched papers, by Ben Tongue of
Blonder-Tongue fame.

http://www.bentongue.com/xtalset/xtalset.html

Here's Dave's site with about 80 crystal sets you can build.

http://makearadio.com/crystal/index.php


Mikek

[amdx]

On 10/8/2012 10:21 AM, John Larkin wrote:
> On Mon, 08 Oct 2012 10:11:36 -0500, John Fields
> <jfi...@austininstruments.com> wrote:
>
>> On Mon, 08 Oct 2012 07:12:06 -0700, John Larkin
>> <jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>
>>> If a crystal set can pump milliwatts of electrical power into a
>>> headphone, something very similar should be able to light up an LED.
>>> No mixed fruits are involved.
>>
>> ---
>> So build one.
>
> I might some day. I already have a lithium-battery night light that I
> built, glowing on the bookshelf near my bed. I figure it will last
> 20-30 years, always on. It has an interesting hands-free variable
> brightness feature, where I can find it by its glow, but crank it up
> as needed for finding my way around, after an earthquake or whatever.
>
> Regular LED flashlights should do that, namely glow a little all the
> time, so you can find them in the dark.
>
> Given the energy density of lithium batteries, "energy harvesting"
> rarely makes sense.
>
>

The first LED flashlight I ever bought was like that. It had a faint
glow so you could find it in the dark.
It came in handy, I bought it at the Orlando Hamfest, on the way home
that night I had a flat tire, my kids had been playing with the light
and didn't know where it was. I turned off the lights in the van and
found it glowing under the rear seat.
The flashlight has slightly larger than the 9 volt battery that
powered it. It had a pretty developed surface mount circuit that
developed 3 different intensities for the LED.
Mikek

[Jeff Liebermann]

On Mon, 08 Oct 2012 07:12:06 -0700, John Larkin
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>If a crystal set can pump milliwatts of electrical power into a
>headphone, something very similar should be able to light up an LED.
>No mixed fruits are involved.

The 30 to 100pF zero bias capacitance of an LED is going to look like
a short circuit compared to the less than 1pF capacitance of a decent
detector diode. However, if the capacitance is known, it can be
resonated with a series inductor to light up the LED:
<http://www.edn.com/design/led/4368392/An-LED-s-intrinsic-capacitance-works-in-a-650-mV-LRC-circuit>

Here's an RF energy scavenger experiment that successfully lit up an
LED (Fig 15) using a voltage multiplier:
<http://www.hindawi.com/journals/apec/2010/591640/>
However, he cheated and used a nearby transmitter to power the device.
Still, the use of a voltage multiplier might be useful.

I don't think a crystal radio does not deliver "milliwatts" to the
headphones. See:
<http://www.crystal-radio.eu/enluidsprekertest.htm>
which offers a table of minimal earphone power levels commonly used on
crystal sets. Operating levels will be somewhat higher.
Headphone Sennheiser Model HD433 9.6 pW
Headphone Sennheiser Model HD330 0.78 pW
2x 2000 Ohm headphone Telefunken EH333 0.022 pW
2x 2000 Ohm headphone Omega 0.033 pW
Crystal earplug "Taiwan" 0.13 pW
Driver unit Adastra Model: 952-207 0.0078 pW
Looks more like fractions of a pico watt, which is what I would expect
from the signal levels found on a crude antenna.

Basic crystal radio calcs:
<http://www.crystal-radio.eu/engev.htm>

Here's the consensus on the original question:
"Crystal radio to power an LED?"
<http://answers.yahoo.com/question/index?qid=20071201183427AAwom9C>
Accumulating the scavenged energy and using it to build an LED flasher
might be a good way to make it work.


--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

[Jeff Liebermann]

<jjla...@highNOTlandTHIStechnologyPART.com> wrote:

>I wasn't thinking electrostatics (hard to get DC from) but picking up
>energy from an AM station.

Well, let's do the math.

An LED will barely light at 1.4VDC and 2ma.
1.4V * 0.002A = 2.8mw
That's how much power needs to be produced by this contrivance.

For the transmitter, I'll use KSCO, which is conveniently nearby.
Daytime power is 10,000 watts.
<http://www.radio-locator.com/cgi-bin/pat?call=KSCO&service=AM&status=L&hours=D>
The three contours are 2.5, 0.5, and 0.15 mV/m field strength.

At the 2.5mV/m contour a fair size dipole antenna will pickup about
100uV into 75 ohms.
<http://www.nd2x.net/calculators/FS.html>
Notes:
1. Plenty of typo errors on this page, but the numbers seem correct.
2. A simple dipole still has 0dBd gain down to about 1/10th
wavelength. The impedance becomes small, but the gain remains at
about 0dBd.

From the above calculator:
Receive-Power = E^2/R = (100*10^-6)^2 / 75 = 133*10^-12 watts
= 133 picowatts

That's not anywhere near enough power needed by the LED (2.8
milliwatts). However, it's more than enough for the <1 picowatt
required to drive a crystal radio earphone.
<http://www.crystal-radio.eu/enluidsprekertest.htm>

The impedance of the LED is 1.4V / 0.002A = 700 ohms. 30pF of diode
capacitance at 1MHz is about 5K ohms which isn't going to have a big
impact on the 700 ohms. I'll call it 750 ohms because it makes the
numbers come out neatly. A tapped coil resonant at about 1MHz should
suffice for matching the 75 ohm antenna to the 750 ohm LED. Turns
ratio is:
sqrt(750/75) = 3.2
With 100uV at the antenna, the coil will deliver an inadequate 320uV
to the LED. The LED needs 1.4V, not 320uV.

[John Larkin]

On Mon, 08 Oct 2012 12:39:52 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:

>On Mon, 08 Oct 2012 07:12:06 -0700, John Larkin
><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>>If a crystal set can pump milliwatts of electrical power into a
>>headphone, something very similar should be able to light up an LED.
>>No mixed fruits are involved.
>
>The 30 to 100pF zero bias capacitance of an LED is going to look like
>a short circuit compared to the less than 1pF capacitance of a decent
>detector diode. However, if the capacitance is known, it can be
>resonated with a series inductor to light up the LED:
><http://www.edn.com/design/led/4368392/An-LED-s-intrinsic-capacitance-works-in-a-650-mV-LRC-circuit>

Why not put the LED right across the crystal set LC tank? The tank
needs capacitance anyhow. We have some high-efficiency LEDs in the low
10s of pF. The classic crystal set cap was a 365 pF variable.

>
>Here's an RF energy scavenger experiment that successfully lit up an
>LED (Fig 15) using a voltage multiplier:
><http://www.hindawi.com/journals/apec/2010/591640/>
>However, he cheated and used a nearby transmitter to power the device.
>Still, the use of a voltage multiplier might be useful.
>
>I don't think a crystal radio does not deliver "milliwatts" to the
>headphones. See:
><http://www.crystal-radio.eu/enluidsprekertest.htm>
>which offers a table of minimal earphone power levels commonly used on
>crystal sets. Operating levels will be somewhat higher.
> Headphone Sennheiser Model HD433 9.6 pW
> Headphone Sennheiser Model HD330 0.78 pW
> 2x 2000 Ohm headphone Telefunken EH333 0.022 pW
> 2x 2000 Ohm headphone Omega 0.033 pW
> Crystal earplug "Taiwan" 0.13 pW
> Driver unit Adastra Model: 952-207 0.0078 pW
>Looks more like fractions of a pico watt, which is what I would expect
>from the signal levels found on a crude antenna.

Those are audibility threshold powers. They say nothing about how much
power a crystal set can deliver.

>
>Basic crystal radio calcs:
><http://www.crystal-radio.eu/engev.htm>
>
>Here's the consensus on the original question:
>"Crystal radio to power an LED?"
><http://answers.yahoo.com/question/index?qid=20071201183427AAwom9C>
>Accumulating the scavenged energy and using it to build an LED flasher
>might be a good way to make it work.

Seems silly. A good 50 KW AM station might provide 100 uw/sq meter
roughly five miles away. Collect some of that with a longwire antenna.
1 uw is enough to light a good led visibly in room light.

[John Larkin]

On Mon, 08 Oct 2012 18:46:25 -0700, Jeff Liebermann <je...@cruzio.com>
wrote:

><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>
>>I wasn't thinking electrostatics (hard to get DC from) but picking up
>>energy from an AM station.
>
>Well, let's do the math.
>
>An LED will barely light at 1.4VDC and 2ma.
> 1.4V * 0.002A = 2.8mw
>That's how much power needs to be produced by this contrivance.

The Osram LEDs that we use are often too bright at 2 mA. They are
clearly on in normal room lighting at 1 uA. Dark adapted, under
optimal conditions, I could just barely make out light at about 800
picoamps.

My night light is an Avago green LED, a Tadiran lithium cell, and a 1
meg resistor. It should last 20 or 30 years.

According to my good'ol red Radiotron book, one might expect roughly
100 uW/sq meter a few miles from a 50KW AM station. You'd only need a
couple of uW to light an LED. So it's just an impedance matching
problem. It looks doable.

[John Larkin]

On Tue, 02 Oct 2012 02:29:39 -0700, miso <mi...@sushi.com> wrote:

>
>I vaguely remember something about fluorescent tubes glowing near TV
>transmitters. Maybe Sutro tower.

I used to drive past Sutro Tower every day in my ratty old Ford
Fiesta. The speakers would make awful sounds, even with the radio off.
I guess the output transistors were rectifying the RF picked up by the
speaker wires.

[gregz]

Jan Panteltje <pNaonSt...@yahoo.com> wrote:
> On a sunny day (Tue, 02 Oct 2012 02:29:39 -0700) it happened miso
> <mi...@sushi.com> wrote in <k4ec7i$haj$1...@speranza.aioe.org>:

>
>>
>> I vaguely remember something about fluorescent tubes glowing near TV
>> transmitters. Maybe Sutro tower.
>

> Fluorescent tubes also glow when stuck in the ground under a HV power line.
> I used to walk around with a glowing neon next ot my 7 MHz xmtr antenna.

I need to try something like that up the road, where there are no houses,
but feed from power station goes through. Three phase might screw it up.

Greg

[John Larkin]

So at 100 uw/sq meter (a few miles from a 50 KW AM station) that
computes to 1 watt! Of course, that's a pretty big antenna.

>
>The typical ferrite rod antenna has a gain of -30 to -60 dB below the
>full size dipole, thus the effective area is somewhere between 10 m²
>and 1 dm². Multiply this by the local power density [W/m²] and you get
>a ballpark figure of the available power.
>

Even the low end 10 m^2 times 100 uw/m^2 is a milliwatt! A good LED is
visible at a couple of microwatts, so even a ferrite rod antenna
should visibly light up a good LED a few miles from an AM station.

[gregz]

John Fields <jfi...@austininstruments.com> wrote:
> On Tue, 02 Oct 2012 09:12:59 -0700, John Larkin
> <jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>
>> On Mon, 1 Oct 2012 20:54:20 -0700, "Billyb97113"
>> <billy...@yahoo.com> wrote:
>>
>>> To quote John Larkins:
>>> "More practical would be to get usable LED lighting from ambient RF"
>>>
>>> I tried several different types of diodes and Ferrite rods and coils and
>>> could never get a measurable voltage across a .01 uf cap. I used a 10X scope
>>> probe as the only load. I would be happy if the LED occasionally blinked
>>> using any of the inductive circuits found on the internet.
>>>
>>> Any ideas?? -bill
>>>
>>
>> I have no idea how much power can be extracted from a ferrite rod
>> antenna.
>
> ---
> Why, then, is that admission of ignorance germane?
> ---
>

>> Googling produces vague results.
>
> ---
> GIGO?

> ---


I was just comparing results googling vs blekko. Google just returns me to
here.
I'm just trying to learn blekko. No ehow wiki on blekko.


Greg

[gregz]

http://blekko.com/

Greg

[Jeff Liebermann]

On Mon, 08 Oct 2012 19:24:00 -0700, John Larkin
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 08 Oct 2012 18:46:25 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>>
>>>I wasn't thinking electrostatics (hard to get DC from) but picking up
>>>energy from an AM station.
>>
>>Well, let's do the math.
>>
>>An LED will barely light at 1.4VDC and 2ma.
>> 1.4V * 0.002A = 2.8mw
>>That's how much power needs to be produced by this contrivance.
>
>The Osram LEDs that we use are often too bright at 2 mA. They are
>clearly on in normal room lighting at 1 uA. Dark adapted, under
>optimal conditions, I could just barely make out light at about 800
>picoamps.

Ok, let's use your numbers and see what happens.

I didn't know LED's would work at 1uA. Do you have the Osram part
number handy? I found some "low current" LED's on the Osram web pile,
but the spec sheets were for 2ma.

For the LED:
1.4VDC * 1uA = 1.4 microwatts
1.4VDC / 1uA = 1.4M load impedance

>According to my good'ol red Radiotron book, one might expect roughly
>100 uW/sq meter a few miles from a 50KW AM station. You'd only need a
>couple of uW to light an LED. So it's just an impedance matching
>problem. It looks doable.

The ambient level in US metro areas is about 50 uWatts/sq-meter from a
study that I can't seem to find.

For the antenna:
100 uWatts/sq-meter is the energy density. To convert into detected
energy, the effective aperture of the receive antenna will be needed:
<http://vk1od.net/antenna/concepts/Ae.htm>
<http://vk1od.net/software/fsc/index.htm>
For a 1Mhz dipole (143 meters long), that would be about 7000
sq-meters effective aperture.
100 uWatts/sq-meter * 7000 sq-meters = 0.7 watts

That will work, but who is going to install a 143 meter long half wave
dipole just to light up an LED? Using a more reasonable 0.05
wavelength dipole, detected voltage will be 1/10th of the dipole,
resulting in 1/100 the detected power as 0.007 watts = 7 milliwatts.
That's considerably more than the 1.4 microwatts needed for the Osram
LED, so it's quite possible that it will work.

Assuming I add loading coils to the 1/10th wavelength antenna to bring
the dipole back up to 75 ohms, the input tank will need a turns ratio
of:
sqrt(1.4*10^6 ohms / 75 ohms) = 136:1
which is buildable but will probably need a big air core coil.

If an outdoor dipole antenna is too much, an indoor loop might work:
<http://www.instructables.com/id/Medium-Wave-AM-broadcast-band-resonant-loop-antenn/>
<http://www.angelfire.com/mb/amandx/loop.html>
<http://www.bobsamerica.com/2-foot-loop.html>

There are some additional losses, which haven't been considered. Balun
loss at the antenna, half wave rectification only recovers half the
power, decrease in antenna gain due to proximity to the ground, and
resistive (Q) losses in the tank circuit.

>My night light is an Avago green LED, a Tadiran lithium cell, and a 1
>meg resistor. It should last 20 or 30 years.

The Tadiran batteries used in some SmartMeter applications have lasted
25 years. I'm fairly sure I won't live long enough to see such a
battery die.
<http://www.tadiranbat.com/pdf.php?id=waterworld-article>

Incidentally, for finding things in the dark, I use phosphorescent
paper and a UV LED flashlight. I paste various shapes cut from the
paper to my calculator, cell phone, TV remote, door knob, light
switches, etc. They're not self lighting, but when hit with a UV
flashlight, they're VERY bright and easy to find.

[upsid...@downunder.com]

On Mon, 08 Oct 2012 19:38:47 -0700, John Larkin
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 08 Oct 2012 10:24:49 +0300, upsid...@downunder.com wrote:

>>The effective area of a half wave dipole is about 0.12 square
>>wavelengths, thus at 1 MHz with 300 m wavelength, the capture area is
>>about 10000 m² or 1 ha.
>
>So at 100 uw/sq meter (a few miles from a 50 KW AM station) that
>computes to 1 watt! Of course, that's a pretty big antenna.

Sounds about right. That 100 µW/m² is in good agreement with old CCIR
(now ITU-R) field strength diagrams above average soil for a few
kilometers at 1 MHz.

With an outdoor antenna, just tune out the capacitively reactance with
some loading coil and use some 1:100 step up transformer and you might
get some usable LED currents.

I remember seeing some articles (long before the Internet) about a
transistorized tunable receiver powered by rectifying the signal from
a strong local broadcast station :-)

>>The typical ferrite rod antenna has a gain of -30 to -60 dB below the
>>full size dipole, thus the effective area is somewhere between 10 m²
>>and 1 dm². Multiply this by the local power density [W/m²] and you get
>>a ballpark figure of the available power.
>>
>
>Even the low end 10 m^2 times 100 uw/m^2 is a milliwatt! A good LED is
>visible at a couple of microwatts, so even a ferrite rod antenna
>should visibly light up a good LED a few miles from an AM station.

For those not so familiar with the metric system 1 m² = 100 dm².

At the low end, the available power is only 1 µW.

Just googled around and found a measurements of a small (5 cm)
loopstick with -80 dBi gain, thus, the available power would be 100 nW
or 50 nA LED current. Perhaps an eye, well adapted to darkness for
half an hour, might be able to see something :-).

[Martin Brown]

On 09/10/2012 02:46, Jeff Liebermann wrote:
> <jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>
>> I wasn't thinking electrostatics (hard to get DC from) but picking up
>> energy from an AM station.
>
> Well, let's do the math.
>
> An LED will barely light at 1.4VDC and 2ma.
> 1.4V * 0.002A = 2.8mw
> That's how much power needs to be produced by this contrivance.

Your figures are over 3 decades out of date. The best modern white and
some green LEDs are just about visibly lit on the die in normal room
lighting at 1uA. When dark adapted you can drop that by a factor 100 or
even more. ISTR the voltage drop is nearer 2v though and only green or
white ones are worth trying since you need peak scotopic sensitivity.

Actual power requirement is about 2V * 0.01uA = 20nW in total darkness.

It is getting the 2v potential difference that is hard.

Selecting the brightest diode from a batch would be worthwhile...

>
> For the transmitter, I'll use KSCO, which is conveniently nearby.
> Daytime power is 10,000 watts.
> <http://www.radio-locator.com/cgi-bin/pat?call=KSCO&service=AM&status=L&hours=D>
> The three contours are 2.5, 0.5, and 0.15 mV/m field strength.
>
> At the 2.5mV/m contour a fair size dipole antenna will pickup about
> 100uV into 75 ohms.
> <http://www.nd2x.net/calculators/FS.html>
> Notes:
> 1. Plenty of typo errors on this page, but the numbers seem correct.
> 2. A simple dipole still has 0dBd gain down to about 1/10th
> wavelength. The impedance becomes small, but the gain remains at
> about 0dBd.
>
> From the above calculator:
> Receive-Power = E^2/R = (100*10^-6)^2 / 75 = 133*10^-12 watts
> = 133 picowatts
>
> That's not anywhere near enough power needed by the LED (2.8
> milliwatts). However, it's more than enough for the <1 picowatt
> required to drive a crystal radio earphone.
> <http://www.crystal-radio.eu/enluidsprekertest.htm>
>
> The impedance of the LED is 1.4V / 0.002A = 700 ohms. 30pF of diode

Actually for a sensibly chosen modern high intensity LED suited to the
task it is more like an impedance match to 2V/0.01uA = 200M.

> capacitance at 1MHz is about 5K ohms which isn't going to have a big
> impact on the 700 ohms. I'll call it 750 ohms because it makes the
> numbers come out neatly. A tapped coil resonant at about 1MHz should
> suffice for matching the 75 ohm antenna to the 750 ohm LED. Turns
> ratio is:
> sqrt(750/75) = 3.2

Except it that should be sqrt(2x10^8/75) = 1700 turns

> With 100uV at the antenna, the coil will deliver an inadequate 320uV
> to the LED. The LED needs 1.4V, not 320uV.

With 100uV on the antenna this gives 0.17V still not enough on its own,
but a clever boost converter might be able to store enough energy on a
low leakage capacitor for the occasional flash.

If you take the average current drawn down to below 1nA then it looks to
me like you would be in the right ballpark for ordinary transistors.
The 5000t resonant coil will be hard to make though.

--
Regards,
Martin Brown

[upsid...@downunder.com]

On Mon, 08 Oct 2012 19:15:06 -0700, John Larkin

<jjla...@highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 08 Oct 2012 12:39:52 -0700, Jeff Liebermann <je...@cruzio.com>
>wrote:
>
>>On Mon, 08 Oct 2012 07:12:06 -0700, John Larkin
>><jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>>>If a crystal set can pump milliwatts of electrical power into a
>>>headphone, something very similar should be able to light up an LED.
>>>No mixed fruits are involved.
>>
>>The 30 to 100pF zero bias capacitance of an LED is going to look like
>>a short circuit compared to the less than 1pF capacitance of a decent
>>detector diode. However, if the capacitance is known, it can be
>>resonated with a series inductor to light up the LED:
>><http://www.edn.com/design/led/4368392/An-LED-s-intrinsic-capacitance-works-in-a-650-mV-LRC-circuit>
>
>Why not put the LED right across the crystal set LC tank? The tank
>needs capacitance anyhow. We have some high-efficiency LEDs in the low
>10s of pF. The classic crystal set cap was a 365 pF variable.

The resonant circuit impedance in a typical MW receiver is about 100
kOhms, while a LED circuit impedance is in the order of 100-1000 ohms.

[MrTallyman]

AM radios receive femtowatts.

Sheesh!

[Jan Panteltje]

On a sunny day (Tue, 9 Oct 2012 02:33:00 +0000 (UTC)) it happened gregz
<ze...@comcast.net> wrote in
<112229436371442754.34...@news.eternal-september.org>:

http://www.doobybrain.com/2008/02/03/electromagnetic-fields-cause-fluorescent-bulbs-to-glow/

Have fun!

[John Devereux]

John Larkin <jjla...@highNOTlandTHIStechnologyPART.com> writes:

> On Tue, 02 Oct 2012 02:29:39 -0700, miso <mi...@sushi.com> wrote:
>
>>
>>I vaguely remember something about fluorescent tubes glowing near TV
>>transmitters. Maybe Sutro tower.
>
> I used to drive past Sutro Tower every day in my ratty old Ford
> Fiesta. The speakers would make awful sounds, even with the radio off.
> I guess the output transistors were rectifying the RF picked up by the
> speaker wires.

We had that with a hydraulic servovalve!

"this is the BBC world service...."

--

John Devereux

[upsid...@downunder.com]

On Mon, 08 Oct 2012 21:35:05 -0700, Jeff Liebermann <je...@cruzio.com>

To be precise 70 m tall (vertical polarization).

>Using a more reasonable 0.05
>wavelength dipole, detected voltage will be 1/10th of the dipole,

at what impedance level ?

>resulting in 1/100 the detected power as 0.007 watts = 7 milliwatts.
>That's considerably more than the 1.4 microwatts needed for the Osram
>LED, so it's quite possible that it will work.
>
>Assuming I add loading coils to the 1/10th wavelength antenna to bring
>the dipole back up to 75 ohms, the input tank will need a turns ratio
>of:
> sqrt(1.4*10^6 ohms / 75 ohms) = 136:1
>which is buildable but will probably need a big air core coil.

The radiation resistance drops inversely proportionally to the square
of wavelength below 1/4 wavelengths, thus the matching network not
only needs to tune out the antenna capacitively reactance, but also
transform the very low (a few ohms or less) to the standard 50/75 ohm
impedance levels.

[upsid...@downunder.com]

Definitively _NOT_

While the provincial US organization "IHF" tried to introduce the dBf
(femptowatt decibels above 1 W) in order to make some sense into
advertisement, those 10 dBf figures apply _only_ to receivers in the
100 MHz band with +/- 75 kHz FM deviation.

Due to the band noise around 1 MHz, those dBf figures are useless.

[WoolyBully]

There is no "p", idiot.

[John Larkin]

Oh, yeah. We engineers use powers of 3, mostly. Deci and hecta and all
those are less common here.

A longwire, or a big loop, sounds more promising.

>
>At the low end, the available power is only 1 µW.
>
>Just googled around and found a measurements of a small (5 cm)
>loopstick with -80 dBi gain, thus, the available power would be 100 nW
>or 50 nA LED current. Perhaps an eye, well adapted to darkness for
>half an hour, might be able to see something :-).

1 nw is about the threshold with a good green LED.

[John Larkin]

Hey, we're engineers. We know how to match impedances.

[Michael A. Terrell]

John Larkin wrote:
>
> Why not put the LED right across the crystal set LC tank? The tank
> needs capacitance anyhow. We have some high-efficiency LEDs in the low
> 10s of pF. The classic crystal set cap was a 365 pF variable.

That would reduce the 'Q' to an unusable level since a parallel L/C
circuit is high impedance.

[John Larkin]

The Q would be reduced in proportion to how much power is extracted by
the LED. Delivering power into the LED is the whole point, so
naturally that will reduce Q.

Of course, the impedance match should be optimized for maximum power
transfer. Crystal sets often used tapped inductors for that reason, so
thet might be reasonable here.

[Michael A. Terrell]

John Larkin wrote:
>
> On Tue, 09 Oct 2012 10:38:17 -0400, "Michael A. Terrell"
> <mike.t...@earthlink.net> wrote:
>
> >
> >John Larkin wrote:
> >>
> >> Why not put the LED right across the crystal set LC tank? The tank
> >> needs capacitance anyhow. We have some high-efficiency LEDs in the low
> >> 10s of pF. The classic crystal set cap was a 365 pF variable.
> >
> >
> > That would reduce the 'Q' to an unusable level since a parallel L/C
> >circuit is high impedance.
>
> The Q would be reduced in proportion to how much power is extracted by
> the LED. Delivering power into the LED is the whole point, so
> naturally that will reduce Q.
>
> Of course, the impedance match should be optimized for maximum power
> transfer. Crystal sets often used tapped inductors for that reason, so
> thet might be reasonable here.

And the voltage would be lower at that tap.

[gregz]

Jan Panteltje <pNaonSt...@yahoo.com> wrote:
> On a sunny day (Tue, 9 Oct 2012 02:33:00 +0000 (UTC)) it happened gregz
> <ze...@comcast.net> wrote in
> <112229436371442754.34...@news.eternal-september.org>:
>
>> Jan Panteltje <pNaonSt...@yahoo.com> wrote:
>>> On a sunny day (Tue, 02 Oct 2012 02:29:39 -0700) it happened miso
>>> <mi...@sushi.com> wrote in <k4ec7i$haj$1...@speranza.aioe.org>:
>>>
>>>>
>>>> I vaguely remember something about fluorescent tubes glowing near TV
>>>> transmitters. Maybe Sutro tower.
>>>
>>> Fluorescent tubes also glow when stuck in the ground under a HV power line.
>>> I used to walk around with a glowing neon next ot my 7 MHz xmtr antenna.
>>
>> I need to try something like that up the road, where there are no houses,
>> but feed from power station goes through. Three phase might screw it up.
>>
>> Greg
>
> http://www.doobybrain.com/2008/02/03/electromagnetic-fields-cause-fluorescent-bulbs-to-glow/
>
> Have fun!

That's how I pictured it !!

Greg

[John Larkin]

On Tue, 09 Oct 2012 12:35:49 -0400, "Michael A. Terrell"

Depends on what you connect to which tap.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com

http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators

Custom laser drivers and controllers

Photonics and fiberoptic TTL data links

VME thermocouple, LVDT, synchro acquisition and simulation

[John Larkin]

On Tue, 9 Oct 2012 02:42:35 +0000 (UTC), gregz <ze...@comcast.net>
wrote:

Yup, we are famous, often just minutes after we post.

[mike]

I've heard stories (urban legends?) about people who were prosecuted for
running a wire down
their fence and stealing power from the power company.
If the fence is there, how does hooking a light bulb across it cost the
power company additional money? Aren't you just diverting the losses
through a light bulb as they make their way into the ground?

[Michael A. Terrell]

John Larkin wrote:
>
> On Tue, 09 Oct 2012 12:35:49 -0400, "Michael A. Terrell"
> <mike.t...@earthlink.net> wrote:
>
> >
> >John Larkin wrote:
> >>
> >> On Tue, 09 Oct 2012 10:38:17 -0400, "Michael A. Terrell"
> >> <mike.t...@earthlink.net> wrote:
> >>
> >> >
> >> >John Larkin wrote:
> >> >>
> >> >> Why not put the LED right across the crystal set LC tank? The tank
> >> >> needs capacitance anyhow. We have some high-efficiency LEDs in the low
> >> >> 10s of pF. The classic crystal set cap was a 365 pF variable.
> >> >
> >> >
> >> > That would reduce the 'Q' to an unusable level since a parallel L/C
> >> >circuit is high impedance.
> >>
> >> The Q would be reduced in proportion to how much power is extracted by
> >> the LED. Delivering power into the LED is the whole point, so
> >> naturally that will reduce Q.
> >>
> >> Of course, the impedance match should be optimized for maximum power
> >> transfer. Crystal sets often used tapped inductors for that reason, so
> >> thet might be reasonable here.
> >
> >
> > And the voltage would be lower at that tap.
>
> Depends on what you connect to which tap.

Have you ever seen a _working_ crystal radio with the capacitor
connected to the tap?

[Michael A. Terrell]

mike wrote:
>
> I've heard stories (urban legends?) about people who were prosecuted for
> running a wire down
> their fence and stealing power from the power company.
> If the fence is there, how does hooking a light bulb across it cost the
> power company additional money? Aren't you just diverting the losses
> through a light bulb as they make their way into the ground?

How many wire fences are grounded? At one time, they were used for
telephone service in rural areas. The other side of the phone was
grounded.

[rickman]

On 10/9/2012 9:48 AM, WoolyBully wrote:
> On Tue, 09 Oct 2012 12:14:58 +0300, upsid...@downunder.com wrote:
>
>> Definitively _NOT_
>>
>> While the provincial US organization "IHF" tried to introduce the dBf
>> (femptowatt decibels above 1 W)
>
> There is no "p", idiot.

In this whole conversation the only thing you chose to comment on is a
typo and you have to call the guy an idiot at that...

What??!!!

Rick

[John Larkin]

On Tue, 09 Oct 2012 18:01:06 -0400, "Michael A. Terrell"

No, but we're making a night light, not a crystal radio. And I didn't
say that the cap had to be connected to the tap, but it might make
sense.

[Tim Williams]

"mike" <spa...@gmail.com> wrote in message
news:k5248n$lgr$1...@dont-email.me...

> I've heard stories (urban legends?) about people who were prosecuted for
> running a wire down
> their fence and stealing power from the power company.
> If the fence is there, how does hooking a light bulb across it cost the
> power company additional money? Aren't you just diverting the losses
> through a light bulb as they make their way into the ground?

Normally, the fence wires are either open circuit (e.g., barbed wire on
wood posts) or short circuit (e.g., chain link on steel posts). Either
way, the "radiated" power is reflected, not absorbed. A small phase shift
occurs, but that's all. A resistive load, however, really does drain
power from the lines.

In principle, every radio station can know exactly how many radio
receivers are tuned to their station, though getting quite that much
accuracy (nW out of kW, while modulating with program information) is
impractical.

It's slightly easier for a power company to notice watts, or hundreds of
watts, out of gigawatts, at a surprisingly constant, unmodulated
frequency.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

[rickman]

Really?

How do they know if I am listening vs. having a receiver turned off vs.
a tree absorbing the signal?

All this time I thought the signal was radiated away from the antenna
and if no one received it the signal just kept going...

If a transmission is sent and no one is there to receive it, was it ever
really sent at all?

What!!??

Rick

[kengi...@interlux.com]

On Tue, 09 Oct 2012 10:55:52 +0300, upsid...@downunder.com wrote:

Why not just use a long vertical antenna and tap into the ES field of
the Earth?

Ken Gillmore

[John Larkin]

On Tue, 9 Oct 2012 18:56:17 -0500, "Tim Williams"
<tmor...@charter.net> wrote:

>"mike" <spa...@gmail.com> wrote in message
>news:k5248n$lgr$1...@dont-email.me...
>> I've heard stories (urban legends?) about people who were prosecuted for
>> running a wire down
>> their fence and stealing power from the power company.
>> If the fence is there, how does hooking a light bulb across it cost the
>> power company additional money? Aren't you just diverting the losses
>> through a light bulb as they make their way into the ground?
>
>Normally, the fence wires are either open circuit (e.g., barbed wire on
>wood posts) or short circuit (e.g., chain link on steel posts). Either
>way, the "radiated" power is reflected, not absorbed. A small phase shift
>occurs, but that's all. A resistive load, however, really does drain
>power from the lines.
>
>In principle, every radio station can know exactly how many radio
>receivers are tuned to their station, though getting quite that much
>accuracy (nW out of kW, while modulating with program information) is
>impractical.

How does the transmitter know that a receiver is there, as opposed to
the signal passing on into space?

>
>It's slightly easier for a power company to notice watts, or hundreds of
>watts, out of gigawatts, at a surprisingly constant, unmodulated
>frequency.

Not really. A mile of wet mud under the line would extract a lot of
power, and not be accountable. Line losses will vary with temperature,
and they can't know the temperature of every point on the transmission
line. And they can only meter to a reasonable fraction of a per cent
accuracy, to detect line losses. A kilowatt out of a gigawatt would
never be detected... 1 PPM.

Hey, this is cool...

http://www05.abb.com/global/scot/scot245.nsf/veritydisplay/fa4150f852382867c12577f8004c0d5d/$file/br_hv-tg(en)c_2gja708402-1012.pdf

There are huge potential transformers, too.

[Tim Williams]

"rickman" <gnu...@gmail.com> wrote in message
news:k52euc$kg8$1...@dont-email.me...

>> In principle, every radio station can know exactly how many radio
>> receivers are tuned to their station
>

> Really?

Yes!

In principle.

You might've shot off a reply before noticing the "impractical" part
though.

> How do they know if I am listening vs. having a receiver turned off vs.
> a tree absorbing the signal?

They don't. I said "tuned", implying the antenna is a net absorber, not
just a phase-shifting reflector (i.e., reactive, but unmatched loading).

> All this time I thought the signal was radiated away from the antenna
> and if no one received it the signal just kept going...

It does, but the antenna either absorbs or re-radiates it. Actually, if
the antenna is resonant, it does that anyway regardless -- this has
applications in passive RADAR, of course.

This is a trivial consequence of the boundary conditions imposed by a hunk
of metal. Fields don't just pass willy-nilly, they interact in an
analytical and in-principle-detectable manner. Whether you can detect ppm
or ppt in practice is for the engineers to figure out.

> If a transmission is sent and no one is there to receive it, was it ever
> really sent at all?

Absolutely!

If a heat source at 500K is radiating into a room at 500K, is it really
radiating?

Physics tells us, yes, and the room is likewise radiating. The net
balance is zero, so no heat transfer occurs (they are in equilibrium).
But that doesn't mean there's nothing there.

What really amazes me is, John doesn't know this.

[mike]

On 10/9/2012 9:41 PM, Tim Williams wrote:
> "rickman"<gnu...@gmail.com> wrote in message
> news:k52euc$kg8$1...@dont-email.me...
>>> In principle, every radio station can know exactly how many radio
>>> receivers are tuned to their station
>>
>> Really?
>
> Yes!
>
> In principle.
>
> You might've shot off a reply before noticing the "impractical" part
> though.
>
>> How do they know if I am listening vs. having a receiver turned off vs.
>> a tree absorbing the signal?
>
> They don't. I said "tuned", implying the antenna is a net absorber, not
> just a phase-shifting reflector (i.e., reactive, but unmatched loading).
>
>> All this time I thought the signal was radiated away from the antenna
>> and if no one received it the signal just kept going...
>
> It does, but the antenna either absorbs or re-radiates it. Actually, if
> the antenna is resonant, it does that anyway regardless -- this has
> applications in passive RADAR, of course.

Ok, but unless that re-radiation is re-absorbed by the source, it can't tell
the difference. The coupling coefficient is very small.

>
> This is a trivial consequence of the boundary conditions imposed by a hunk
> of metal. Fields don't just pass willy-nilly, they interact in an
> analytical and in-principle-detectable manner. Whether you can detect ppm
> or ppt in practice is for the engineers to figure out.
>
>> If a transmission is sent and no one is there to receive it, was it ever
>> really sent at all?
>
> Absolutely!
>
> If a heat source at 500K is radiating into a room at 500K, is it really
> radiating?

No, it's not.
That's a closed system.
The only way for the room to be in equilibrium at 500K and the radiator
at 500K is if the
net power radiated is zero.

[Jan Panteltje]

On a sunny day (Tue, 09 Oct 2012 14:16:47 -0700) it happened mike
<spa...@gmail.com> wrote in <k5248n$lgr$1...@dont-email.me>:

If the fence is an open circuit not much power will be delivered.
If you take the ends and connect a light bulb maybe some power is delivered.
That *could* be the reasoning.

[John Devereux]

I often find my own posts when I search for stuff. Not that I even post
all that much. Must have very specialised interests... (or a specialised
way of putting things).

--

John Devereux

[Tim Williams]

"mike" <spa...@gmail.com> wrote in message

news:k535mq$q97$1...@dont-email.me...

> Ok, but unless that re-radiation is re-absorbed by the source, it can't
> tell
> the difference. The coupling coefficient is very small.

Exactly! I *did* use words like "in principle", "it's up to engineering",
and "10^-12" in there. :)

>> If a heat source at 500K is radiating into a room at 500K, is it really
>> radiating?
>
> No, it's not.
> That's a closed system.
> The only way for the room to be in equilibrium at 500K and the radiator
> at 500K is if the
> net power radiated is zero.

But I didn't say net power radiated. I said radiated, i.e. emitted alone.

All things radiate all the time!

[Michael A. Terrell]

Then build it, and show us.

[John Larkin]

On Wed, 10 Oct 2012 09:04:17 -0400, "Michael A. Terrell"

Not this month; I have a day job. The big hassle would be stringing up
the antenna.

There's little doubt, from the math, that it would work, at least
enough to get some visible light out.

--

John Larkin Highland Technology Inc

www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators

Custom timing and laser controllers

Photonics and fiberoptic TTL data links

[Michael A. Terrell]

Cat whiskers & Germanium diodes have a very low forward drop.

[John Larkin]

On Wed, 10 Oct 2012 12:18:03 -0400, "Michael A. Terrell"

Low barrier schottkies are good, too, and have lower series
resistance. But it might be better to let the LED rectify its own RF,
and have zero diode loss.

The best RF detectors are germanium back diodes, as far as I know the
only germanium devices fabbed using modern lithography. Except
photodides.

LED capacitances are all over the place, 10 pF to hundreds. A low c
part would be easier to use here.

[mike]

Ok, but where'd the differential power come from?
Does it increase the line loss?
Or does it merely reroute loss power through the light bulb
as it moves into the ground and radiate it as light?

[mike]

On 10/10/2012 5:55 AM, Tim Williams wrote:
> "mike"<spa...@gmail.com> wrote in message
> news:k535mq$q97$1...@dont-email.me...
>> Ok, but unless that re-radiation is re-absorbed by the source, it can't
>> tell
>> the difference. The coupling coefficient is very small.
>
> Exactly! I *did* use words like "in principle", "it's up to engineering",
> and "10^-12" in there. :)
>
>>> If a heat source at 500K is radiating into a room at 500K, is it really
>>> radiating?
>>
>> No, it's not.
>> That's a closed system.
>> The only way for the room to be in equilibrium at 500K and the radiator
>> at 500K is if the
>> net power radiated is zero.
>
> But I didn't say net power radiated. I said radiated, i.e. emitted alone.
>
> All things radiate all the time!
>
> Tim
>

Ok, but so what?
All things absorb all the time. In equilibrium, the absorption and
radiation
cancel.
If you have a covered bucket of water, you have evaporation and
condensation
happening in equilibrium. But there's no LOSS of water.

[lang...@fonz.dk]

On 10 Okt., 22:15, mike <spam...@gmail.com> wrote:
> On 10/10/2012 1:51 AM, Jan Panteltje wrote:
>
>
>
>
>
>
>
>
>
> > On a sunny day (Tue, 09 Oct 2012 14:16:47 -0700) it happened mike

> > <spam...@gmail.com>  wrote in<k5248n$lg...@dont-email.me>:

>
> >> On 10/9/2012 1:28 AM, Jan Panteltje wrote:
> >>> On a sunny day (Tue, 9 Oct 2012 02:33:00 +0000 (UTC)) it happened gregz
> >>> <ze...@comcast.net>   wrote in

> >>> <112229436371442754.348458zekor-comcast....@news.eternal-september.org>:

>
> >>>> Jan Panteltje<pNaonStpealm...@yahoo.com>   wrote:
> >>>>> On a sunny day (Tue, 02 Oct 2012 02:29:39 -0700) it happened miso

> >>>>> <m...@sushi.com>   wrote in<k4ec7i$ha...@speranza.aioe.org>:

>
> >>>>>> I vaguely remember something about fluorescent tubes glowing near TV
> >>>>>> transmitters. Maybe Sutro tower.
>
> >>>>> Fluorescent tubes also glow when stuck in the ground under a HV power line.
> >>>>> I used to walk around with a glowing neon next ot my 7 MHz xmtr antenna.
>
> >>>> I need to try something like that up the road, where there are no houses,
> >>>> but feed from power station goes through. Three phase might screw it up.
>
> >>>> Greg
>

> >>>    http://www.doobybrain.com/2008/02/03/electromagnetic-fields-cause-flu...

>
> >>> Have fun!
> >> I've heard stories (urban legends?) about people who were prosecuted for
> >> running a wire down
> >> their fence and stealing power from the power company.
> >> If the fence is there, how does hooking a light bulb across it cost the
> >> power company additional money?  Aren't you just diverting the losses
> >> through a light bulb as they make their way into the ground?
>
> > If the fence is an open circuit not much power will be delivered.
> > If you take the ends and connect a light bulb maybe some power is delivered.
> > That *could* be the reasoning.
>
> Ok, but where'd the differential power come from?
> Does it increase the line loss?
> Or does it merely reroute loss power through the light bulb
> as it moves into the ground and radiate it as light?

its basically an aircored transformer, a load on the secondary, your
wire and bulb, will load the primary, the line

-Lasse

[Tim Williams]

"mike" <spa...@gmail.com> wrote in message

news:k54l92$57b$1...@dont-email.me...

> Ok, but so what?
> All things absorb all the time. In equilibrium, the absorption and
> radiation
> cancel.
> If you have a covered bucket of water, you have evaporation and
> condensation
> happening in equilibrium. But there's no LOSS of water.

Exactly! And just as a sealed container, sitting in the refrigerator,
collects condensation on one side (which is actually due to net work, due
to a thermal gradient, but it needn't be much), so too, an antenna can
detect the nature of the EM waves around it, and anything affecting them
(the net power transfer, back and forth, likewise being potentially very
small).

[rickman]

On 10/10/2012 4:15 PM, mike wrote:
> On 10/10/2012 1:51 AM, Jan Panteltje wrote:
>> On a sunny day (Tue, 09 Oct 2012 14:16:47 -0700) it happened mike
>> <spa...@gmail.com> wrote in<k5248n$lgr$1...@dont-email.me>:
>>

>>> I've heard stories (urban legends?) about people who were prosecuted for
>>> running a wire down
>>> their fence and stealing power from the power company.
>>> If the fence is there, how does hooking a light bulb across it cost the
>>> power company additional money? Aren't you just diverting the losses
>>> through a light bulb as they make their way into the ground?
>>
>> If the fence is an open circuit not much power will be delivered.
>> If you take the ends and connect a light bulb maybe some power is
>> delivered.
>> That *could* be the reasoning.
>
> Ok, but where'd the differential power come from?
> Does it increase the line loss?
> Or does it merely reroute loss power through the light bulb
> as it moves into the ground and radiate it as light?

Yes, adding a load (the lightbulb) to the fence increases the power loss
from the transmission line. But if the fence had a resistor at the end,
such as a damp pole, it will do the same thing. I don't think anyone
can claim you were then stealing power. If the damp pole is replaced
with the lightbulb what is the difference? Intent, I suppose.

Rick

[rickman]

On 10/10/2012 10:21 AM, John Larkin wrote:
> On Wed, 10 Oct 2012 09:04:17 -0400, "Michael A. Terrell"
> <mike.t...@earthlink.net> wrote:
>
>>
>> John Larkin wrote:
>>>
>>> No, but we're making a night light, not a crystal radio. And I didn't
>>> say that the cap had to be connected to the tap, but it might make
>>> sense.
>>
>>
>> Then build it, and show us.
>
> Not this month; I have a day job. The big hassle would be stringing up
> the antenna.
>
> There's little doubt, from the math, that it would work, at least
> enough to get some visible light out.

I get tired of you guys arguing about your personal differences. I know
I'm not going to stop it, but maybe I can give it a little reality.

I want to learn about this and am willing to do the experimental stuff.
Give me the info on what circuit to use and how long a wire to string,
and I'll build it. I'll also need info on how close to the radio
station to be and what transmission level is needed.

Rick

[k...@att.bizzzzzzzzzzzz]

There is no damp pole. Your gain. Their loss. No compensation.

[Jan Panteltje]

On a sunny day (Wed, 10 Oct 2012 13:15:25 -0700) it happened mike
<spa...@gmail.com> wrote in <k54l1n$3ro$1...@dont-email.me>:

>> If the fence is an open circuit not much power will be delivered.
>> If you take the ends and connect a light bulb maybe some power is delivered.
>> That *could* be the reasoning.
>
>Ok, but where'd the differential power come from?
>Does it increase the line loss?
>Or does it merely reroute loss power through the light bulb
>as it moves into the ground and radiate it as light?

Depends on the fence.
I assume we are talking about inductive coupling.

=========================== HV lines

-------------------
| |
| gate
| |
--------------------

If the fence wires are on poles, and go round like this, replacing 'gate' with 'bulb' will perhaps produce light.
No blub no current no light.

If you close the 'gate' short, then current cause some dissipation in the fence wires,
how much 'power' is then consumed depends on the coupling and the wire resistance + inductance.
Inductance of a large loop of wire can be very big actually I tried that.

I would not expect a lot of power from this whole setup, but for sure a LED or small bulb would work.
If you use a fence with many wires and put the turns in series...

[mike]

On 10/11/2012 2:20 AM, Jan Panteltje wrote:
> On a sunny day (Wed, 10 Oct 2012 13:15:25 -0700) it happened mike
> <spa...@gmail.com> wrote in<k54l1n$3ro$1...@dont-email.me>:
>
>>> If the fence is an open circuit not much power will be delivered.
>>> If you take the ends and connect a light bulb maybe some power is delivered.
>>> That *could* be the reasoning.
>>
>> Ok, but where'd the differential power come from?
>> Does it increase the line loss?
>> Or does it merely reroute loss power through the light bulb
>> as it moves into the ground and radiate it as light?
>
> Depends on the fence.
> I assume we are talking about inductive coupling.
>
> =========================== HV lines
>































> -------------------
> | |
> | gate
> | |
> --------------------

_________________________________________________________________________Ground

>
> If the fence wires are on poles, and go round like this, replacing 'gate' with 'bulb' will perhaps produce light.
> No blub no current no light.
>
> If you close the 'gate' short, then current cause some dissipation in the fence wires,
> how much 'power' is then consumed depends on the coupling and the wire resistance + inductance.
> Inductance of a large loop of wire can be very big actually I tried that.
>
> I would not expect a lot of power from this whole setup, but for sure a LED or small bulb would work.
> If you use a fence with many wires and put the turns in series...

For me, the question is how much loss from the power line with the gate
open?
closed?
I edited the picture to show the ground that goes on forever, and
adjusted the scale...but I ran out of "returns" before I thought there
were enough.

A related question...
If the fence loop is lossless, how does that change the loss from the HV
line?
And how does that differ from the case with a lossy led added?

[Jasen Betts]

On 2012-10-11, mike <spa...@gmail.com> wrote:

> A related question...
> If the fence loop is lossless, how does that change the loss from the HV
> line?
> And how does that differ from the case with a lossy led added?

a lossess loop would only reflect power, this would decrease the
impedance of the power line nearby.


as someone else said, treat it as an air-core transformer

power line
--------wwwwwwwwwwwwwwwwwwwww-------------
--------wwwwwwwwwwwwwwwwwwwww-------------
--------wwwwwwwwwwwwwwwwwwwww-------------

___nnnn__
| |
| ___ |
`-[___]-' load, (resistor, open, or short).


--
⚂⚃ 100% natural

--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---

[mike]

On 10/12/2012 9:15 PM, Jasen Betts wrote:
> On 2012-10-11, mike<spa...@gmail.com> wrote:
>
>> A related question...
>> If the fence loop is lossless, how does that change the loss from the HV
>> line?
>> And how does that differ from the case with a lossy led added?
>
> a lossess loop would only reflect power, this would decrease the
> impedance of the power line nearby.
>
>
> as someone else said, treat it as an air-core transformer
>
> power line
> --------wwwwwwwwwwwwwwwwwwwww-------------
> --------wwwwwwwwwwwwwwwwwwwww-------------
> --------wwwwwwwwwwwwwwwwwwwww-------------
>
> ___nnnn__
> | |
> | ___ |
> `-[___]-' load, (resistor, open, or short).
>

The hardest part of a solution is defining the problem.
Your schematic leaves out the considerable loss to the ground underneath
the fence.

I never did get along with dot and cross products. I prefer a more
heuristic approach.

Given the frequency involved, a typical fence is WAY short of a wavelength.
A more practical model might be a capacitive divider.
The capacitance of the wire is tiny with respect to the total
capacitance between the power line and the ground.

Open, the wire does mostly nothing.
When loaded, the wire diverts some of the energy loss going into
ground through the load before it gets to the ground.
I claim that the net CHANGE in loss from the power line is miniscule...
maybe some fraction (ratio of distances?) of the loss to the ground
directly under the wire
and way less than the total ground loss.

One experiment might be to measure the current between one end of the
fence and ground. Open and short the other end. More when shorted >>
inductive coupling. More when open >> capacitive coupling.

I propose that the capacitance dominates. Someone with math skills
might have an opinion.

[josephkk]

The losses to ground (ground currents) aren't nearly as high as you think.
The wire resistance is usually _the_ dominant loss factor.

>
>One experiment might be to measure the current between one end of the
>fence and ground. Open and short the other end. More when shorted >>
>inductive coupling. More when open >> capacitive coupling.
>
>I propose that the capacitance dominates. Someone with math skills
>might have an opinion.

Well i have seen the solvers operate and watched how the equations build
up. I was at a high power seminar many years ago and magnetic induction
dominated in every line studied.

?-)