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Author Topic: Very Easy question about batteries?  (Read 5454 times)

Offline Magnethos

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Very Easy question about batteries?
« on: November 24, 2008, 05:23:27 PM »
Think that I have 1 battery (3 Volts @ 1mAmp), of course, think now that I have 10 batteries like the first one:
10 (3Volts @ 1 mAmp).... In parallel connection = 3 Volts @ 10 mAmp.

I know, I can get more volts or more amps... but If I get more volts I will get Less amps.
My question is... the total output is 3 Volts @ 10 mAmp... Can I get more Amps if I get less volts? 1Volt @ 30mAmp? How?

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Very Easy question about batteries?
« on: November 24, 2008, 05:23:27 PM »

Offline mscoffman

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Re: Very Easy question about batteries?
« Reply #1 on: November 24, 2008, 06:08:32 PM »

@Magnethos

The voltage on a single cell of a battery is set by the chemical reaction that runs
the battery and you generally cannot easily change that voltage. So simply
by stacking batteries - you are always going to have to have a multiple of
the voltage of a single cell. Useful chemical electro-negativites are in the range
of .75volts to 3volts on a single cell.

This though can be easily handled by what is called a dc-to-dc converter circuit. The circuit
produces AC by switching and the uses a transformer get to the desired voltage.  Rectify,
filter, and regulate and viola you have your desired output voltage. A converter can step up,
step down or simply isolate grounds. Correct designs make high efficiencies possible.

:S:MSCoffman

Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #2 on: November 24, 2008, 06:18:55 PM »
So... It's possible if I use a DC-to-DC converter unit... I won't use chemical batteries. I'm going to use an antenna receptor... in other words... energy from thin air. I use an antenna and a circuit.

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Re: Very Easy question about batteries?
« Reply #2 on: November 24, 2008, 06:18:55 PM »
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Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #3 on: November 24, 2008, 06:49:11 PM »
And now think... if I have 1300 Volts @ 20 mAH... I can get 12 Volts @ 2160 mAH?
(1300/12 = 108..... 20 * 108 = 2160)

I think I would need to build my own DC-DC converter, Right? But... this could be possible using my DC-to-DC?

Offline pese

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Re: Very Easy question about batteries?
« Reply #4 on: November 24, 2008, 06:54:16 PM »
Think that I have 1 battery (3 Volts @ 1mAmp), of course, think now that I have 10 batteries like the first one:
10 (3Volts @ 1 mAmp).... In parallel connection = 3 Volts @ 10 mAmp.

I know, I can get more volts or more amps... but If I get more volts I will get Less amps.
My question is... the total output is 3 Volts @ 10 mAmp... Can I get more Amps if I get less volts? 1Volt @ 30mAmp? How?
do you mean
3V 1mAh  that say the Battery can be unload (dissipate 3volt for duration of 1 hour
or 0,1mA for 10 hours  -  2ma for an hal hour and so so on
-

IF you use 10 batteries in parallel so you have the 10 times Power
(because 10 batteries.

This is: 3volt with 10mAh- ("h" for one hour)


Of you place the 10 batteries in SERIE so you
have 30volt with over-all 1mAh


If you calculate
1 Battery  3V time 1mA (h)  so is the dissipation
is:         3mW (h)

10 batteries

 3V 10mA   = 30mW
30V  1mA   =30mW      Both is 10 time more
                     power that with one single battery

I hope this was your question
GP

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Re: Very Easy question about batteries?
« Reply #4 on: November 24, 2008, 06:54:16 PM »
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Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #5 on: November 24, 2008, 07:03:42 PM »
@pese
Thanks for the explanation but that is not my question.

I'm going to explain it better... you know we can get energy from thin air using some capacitors, germanium diodes, etc...
Each "cell" draws 3 Volts @ 0,2 mAH (200 microAmps). So, If I build 100 cells the TOTAL output = 100 (3 Volt @ 0.2 mAH) = 6 Volts @ 0,6 mAH.

I want to increase the current, so I need to reduce voltage and increase amperes.
I have: 6 Volts @ 0,6 mAH
I want: 1 Volt @ 3,6 mAH

This is the explanation. Increase current, reduce voltage. Of course, the watt will be the same in both cases. The Volts and mAH are random values. I think we can get 13 Volts @ 0,3 mAH from each cell.

Can I increase the current? Using DC-DC converter is the only solution?

Offline Koen1

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Re: Very Easy question about batteries?
« Reply #6 on: November 24, 2008, 07:36:36 PM »
Yeah, I think you'll need to use some form of transformer.

Whether you use batteries or some other current source,
if you have a fixed input voltage and amperage and you wish
to decrease the voltage while increasing the amperage,
the easiest method that springs to mind is transforming.
Unfortunately transformers only work with changing
current, so that's AC or pulsed DC, simply inputting DC
current directly into a trafo will give you nothing.
Transformers are quite simple in this respect, for example
a trafo primary coils with 20 windings and an input current
pulse at 6V at some given amperage, will induce a current
pulse in a secondary coil with 10 windings that will be at
6 x (10/20) = 6 x 0,5 = 3Volt, at respectively increased
amperage. But you probably knew that. ;)
And of course you'll need to either use a circuit that will
turn your input DC current into DC pulses for use in your
transformer, or you'll need to use an inverter type circuit
to produce AC for use in the trafo. And you'll need the proper
circuitry behind the secondary to turn the output pulses or
AC back into DC current.

You might try to use a form of voltage divider circuit to
lower the voltage, in a sort of reversed voltage multiplier
setup, but to be honest I cannot give you any good
advice on that. I can tell you about a "Marx generator"
voltage multiplier which is sort of a solid state
version of a step-up transformer without the electromagnetic
part and it doesn't need pulsing. But how to do that trick
in reverse is not something I can easily explain to you.
Sorry.
Perhaps Mscoffman can explain that one? :)

Regards,
Koen

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Re: Very Easy question about batteries?
« Reply #6 on: November 24, 2008, 07:36:36 PM »
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Offline spinner

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Re: Very Easy question about batteries?
« Reply #7 on: November 24, 2008, 07:43:36 PM »
...
I'm going to explain it better... you know we can get energy from thin air using some capacitors, germanium diodes, etc...
Each "cell" draws 3 Volts @ 0,2 mAH (200 microAmps). So, If I build 100 cells the TOTAL output = 100 (3 Volt @ 0.2 mAH) = 6 Volts @ 0,6 mAH.

I want to increase the current, so I need to reduce voltage and increase amperes.
I have: 6 Volts @ 0,6 mAH
I want: 1 Volt @ 3,6 mAH

This is the explanation. Increase current, reduce voltage. Of course, the watt will be the same in both cases. The Volts and mAH are random values. I think we can get 13 Volts @ 0,3 mAH from each cell.

Can I increase the current? Using DC-DC converter is the only solution?

If you have a source, capable of providing "only" 0,6 mA (without H...) at 6V (high impedance current source) , then you can directly connect your 3,6 mA load - the Voltage will automatically drop to 1V.... Of course, depends on what kind of source and load you really have..

Electricity from a thin air, eh?  I remember having a HV mica type capacitor, which got charged "all by itself, from a thin air" overnight....
A few Joules for free, and even more when the weather was changing..

My best FE "thin air type" device was a tuned LC detector with a large frame coil/antenna - I got some real Watts, for free.

Of course, having a big power AM transmitter in the neighbourhood was very helpfull...
Cheers!

Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #8 on: November 24, 2008, 07:59:42 PM »
@spinner

Very interesting you device... Any plans or explanation?
I found the schematic of the 'Electricity from air' circuit in youtube.

Some people can get good amount of volts... and I read this: (the first picture)

So, I need more amperes... the idea would be use a wide rod and 'connect' all the circuits to the rod to get enought energy. If I connect 1000 I could get 260 Watt (in theory).

What do you think?

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Re: Very Easy question about batteries?
« Reply #8 on: November 24, 2008, 07:59:42 PM »
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Offline Anothertruthfinder

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Re: Very Easy question about batteries?
« Reply #9 on: November 24, 2008, 10:23:09 PM »
Hi folks ;D thought id throw a little spanner in the works - slight tangent from original question/subject but linked nonetheless,
 
 i was messing around with circuit 'electricity from thin air circuit' its basically the tate ambient power module schematic if im correct anyway i put my walkie talkie near literally the circuit with no antenna, i even got rid of the ceramic caps in the circuit and on transmission (440mhz i believe) it would climb to above 20 volts @ 50ma - approx one watt? slight fluctuance cause i was holding it and wobbling, the walkie talkie in transmit drew 4.2 volts @ 300ma - 1.260 watts approx and it sometimes went down in power to transmit the more the power module drew from the walkie talkie or just stayed exactly the same power consumption near the module or not. a mini tesla localised transmitter thingy? lit 48 blue leds! anyway my batteries were going flat so eventually i couldnt carry on or could i? i built another module and managed to loop one of modules back in to the battery compartment and it would then run for longer each time on flat batteries - initially youd turn it on away from modules and start transmitting and it would go switch itself off in two seconds each time, near the modules i was getting finger ache holding down transmit! well over a minute and a half each time again (some better than others cause of the crudity of the exp.)  and still lighting the leds - not quite as bright though. at some point i believe if i add enough of these modules in the right array/conditions etc. i will gain overunity at some point and power a load and so on. a simple experiment anyone can do and in principle quite simple and easy to try and develop - maybe shades of a tpu principle but is nearly made for us already!
i had problem increasing current which is why also this particular thread is educational for me too and thanks for all the info folks, i hope i havent bored you too much lol!

eel  ;D

Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #10 on: November 24, 2008, 10:34:53 PM »
@Anothertruth

Your info is very interesting. The problem is the current you get from the device, in the videos I see 200 to 300 microAmps (0,3 miliamps!  :( ) But we can get a good amount of volts. So, if we could get 1300 Volts from 100 or more circuits, then we could drop the voltage and increase the current,

1300 Volts from 100 circuits/ 12 = 108
if 1 circuit = 0.3 mAmps, then 100 circuits = 30 mAmps * 108 = 3240 mAmps and 12 VDC... around 26 Watt!! not bad at all!!


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Re: Very Easy question about batteries?
« Reply #10 on: November 24, 2008, 10:34:53 PM »
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Offline Anothertruthfinder

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Re: Very Easy question about batteries?
« Reply #11 on: November 24, 2008, 10:55:24 PM »
hi mag ;D sorry i think theres a little mix up - my fault, i should have said it like this i was getting 20+ volts @ 50 miliamps - the microscale cuts of on my meter hehe and the walkie drew 4.2 volts @ 300 miliamps, in further research i think i might need to insulate and or isolate/redirect  the modules as stacking them in parallel or series just seems to increase the voltage? so i think there's bleedover interferance or something occuring - ive taken one apart to cut power transmission costs in the sense of just disconnected the lcd backlight mic etc. i want to customize an insulated base which holds the antenna say horizontally and mount loads of isolated? modules along its path right in the circuit hotspots hehe there's stacks to be tested with this principle or idea different caps, diodes etc.
i was on this for a while and got steered by mr dbowlings replications hehe but this is an easy repeatable principle - and thats just one frequency cb ive tried! there must be a perfect reasonance for the circuits somewhere - and i havent even tried two or more different frequencies yet.
ill be on the case again very soon lol  ;D

p.s. please call me 'eel' folks its a bit shorter!  :D

Offline pese

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Re: Very Easy question about batteries?
« Reply #12 on: November 24, 2008, 11:02:49 PM »
So i understand more.
but uA micro /mycroAmp
ma milliAmps
to undertsand right.

so if you will have the half value of voltage , but doubled Amp ( even same mWatts.
AND the power comes from RF (radio frequencies)  (not from 60Hz powerlin "hum".
than you will use first ans small transformer  . over an ferritcore, and tap them in the mid for output.

Also Antenna ferrit core an coil from AM Radio is usefull.
so the diode 1N60  or other will become the half RF voltage.
with same power ( uWatts)
(sam system as an crystal-receiver) tuned (with vari-cap.
or untuned, to receive over-all frequencies , that ar an NOIS from the univers, including all radiostation that are in district (als on night time) long distance.

BUT :  !!
You willl have an problem!
nears 60 years ago (as child) i find out . IF i use more than one antenna or more then one receiver. The incoming power will divided  (over the receivers)!!
With ONE tuned Antenna on my crystal receiver , i cane tuned that another receiver(in near) will become less input signal.. So it ist shure that (also untuned). 100 receivers or antennas, that are bundeled together cant not receive 100 times more power.!!

My father was radio-repair-men and coudnot explain me, that what i have found out. I remember this in the last years and found also the answer.

It is sure , that you cant not use practical the suggested device.
GP

P.S.
I seen now your circuit  THIS is "nonsense".
It exist an lot of existing, practical to use receivers (RF/AC based) and also somes that brings static DC down).
aswell as tuned receivers that bring you some mycroWatts.
But in ANY WAY you can not dubbel or to 100 times forced this received power !
The IDEA is good- but not any value to look for the circumstance to try it

@pese
Thanks for the explanation but that is not my question.

I'm going to explain it better... you know we can get energy from thin air using some capacitors, germanium diodes, etc...
Each "cell" draws 3 Volts @ 0,2 mAH (200 microAmps). So, If I build 100 cells the TOTAL output = 100 (3 Volt @ 0.2 mAH) = 6 Volts @ 0,6 mAH.

I want to increase the current, so I need to reduce voltage and increase amperes.
I have: 6 Volts @ 0,6 mAH
I want: 1 Volt @ 3,6 mAH

This is the explanation. Increase current, reduce voltage. Of course, the watt will be the same in both cases. The Volts and mAH are random values. I think we can get 13 Volts @ 0,3 mAH from each cell.

Can I increase the current? Using DC-DC converter is the only solution?


Offline Magnethos

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Re: Very Easy question about batteries?
« Reply #13 on: November 24, 2008, 11:14:53 PM »
of course eel, :)

So, the ouput you obtained was 20 Volts 50miliAmps! Wow!

So, 100 cells.... 2000 Volts 5000mAmps!!! Man!

2000/12 = 166.7
5000 * 166.7 = 833500 mAmps  :o   and 12 VDC

This sounds very interesting to be true...


@pese
I think you're right... if we use more receivers, the amount of energy of each receiver will decrease proportionally. Because it's awesome that we could get 12VDC @ 8335 Amperes from a single radio or walkie talkie...
But we could obtain 260 Watt if we build 1000 receivers and the input is the earth energy field instead the walkie talkie.

Offline pese

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Re: Very Easy question about batteries?
« Reply #14 on: November 24, 2008, 11:22:15 PM »
Jes this was an good exampele to understand this.
Additional each antenna, tuned circuit and so on
that of an small amount OFF from the ambient  (RF)
waves-field. so  if that are divided to 100 receivers. the summary is not higher tan this from one singel receiver antenna circuit
Pese

 

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