Is it possible to charge a battery from a distance of ~10 feet without contacting the battery?
How much power would be needed to transmit, receive, and rectify enough VA to charge a 6 volt battery at about 40 mA?
Do you know how to do it?
Is it possible to charge a battery from a distance of ~10 feet without contacting the battery?
How much power would be needed to transmit, receive, and rectify enough VA to charge a 6 volt battery at about 40 mA?
Do you know how to do it?
Read a biography of Tesla.
** Easy - done using solar cells and sunlight all the time. ** If there is no sunlight, a spotlight will do the trick.
Forget the RF idea.
....... Phil
Phil,
Sorry, I should have said indoors.
Next problem: It must be 'invisible". I want to charge a battery that is within a plastic container. At first, I considered an inductive charger as in a toothbrush, etc.
Unfortunately the battery is on the other side of the moat.
** Moat ????????
Time to pull up the drawbridge - the barbarians have arrived .....
BTW:
The solar cells can be positioned anywhere an the spot light can be filtered to give only infrared.
........ Phil
Depends on your definition of "invisible". Perhaps a 250 Kw transmitter operating at 2 MHz? You'll need a full size radio tower (maybe 600 feet tall?) on your side of the moat.
I will charge the batteries, as well as everything around it. Such as you, your wife and kids.
Add a light to your helmet for that cool look!
Scuff your feet on the carpet and point at the battery.
6V*40mA = 240mW. Allow a bit more for inefficiencies in the rectifier and charging circuit. Perhaps 400mW is enough.
Pr = Pt * Gt * Gr * [lambda/(4*pi*r)]^2
where Pr = received power, Pt = transmitted power, Gr and Gt are transmit and receive antenna gains, lambda is the wavelength, r is the distance in the same units as lambda, and the path is assumed to be non-absorptive line-of-sight. This also assumes "far field" where r>>lambda and r>>antenna size. Assuming you can use directional antennas on both ends, you can play around with the requirements to get what suits your need. Say, for example, Gt=Gr=10dB = 10, lambda = 1/10 foot, r = 10 feet. Then you'd only need about 6.3 kW of transmitted power to get the desired 400mW at output of the receive antenna.
Does that help?
Cheers, Tom
Yeah. With LOTS of transmitted power, or with VERY directional antennas (difficult to keep aimed properly).
Like I was gonna say, Yeah, it's doable, but if you walk in front of the transmitting antenna, your blood will boil. =:-O
Cheers! Rich
On a sunny day (Fri, 24 Nov 2006 23:18:02 GMT) it happened Rich Grise wrote in :
At even higher frequencies you could consider using a 1kW lightbulb and a solarpanel.
Makes you see better too :-)
"Jan Panteltje"
** Wow - who would have ever thunk of that ???..... Phil ..... totally amazed ....
Thanks guys,
Good answers, I will take your advice.
Plan B: I could possibly reach the unit with a long pole to attach an inductive charger.
AFAIK All I would need is an air coupled transformer, a diode, and a timer.
(It could get more complicated with regulation and limiting, but as a minimum-ist approach it would work.)
Correct?
If both ends are resonantly coupled the efficiency goes way up.
-- Dirk http://www.onetribe.me.uk - The UK\'s only occult talk show Presented by Dirk Bruere and Marc Power on ResonanceFM 104.4 http://www.resonancefm.com
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