Newbie in need of technical brains .. and don`t laugh .

Discussion in 'General Electronics Chat' started by HaventgotaClue, Aug 8, 2012.

  1. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    Ok Guys
    Have read a few threads and seems you guys are the whizz kids and makes me realise i know so little ..

    Now this will probably be way to simple for you guys but for me its doing my head in and looking for a simple easy fix.

    I have been playing around with wind generation and have the DC motor which pushes out in from DC 25v to 100v.in strong winds , which is good , but now all i want to do is harness this power into 12v battery so i need to regulate it at say 14v to charge the batteries , and obviously it must NOT feed back to the generator or it will spin the thing . am i correct ??

    Now have searched and this is where it got technical , is there not a simple diode and a resistor or something that can just go on the power lead ?

    Simple cheap solution would be appreciated with what to buy in to do it etc, as over here where i am in Spain they are crap at electronic supplies so would need to buy them mail order ..

    cheers in advance
    Pete
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,439
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    You do not need to bring the voltage down to 14V.
    The battery will take as much current as it needs to charge itself.

    What you really need is a battery management system.

    Firstly, you need a current regulator so that you do not exceed the charge current.

    Then when the battery is partially charged you switch to a constant voltage charge.

    Once the battery is fully charge, you can lower the voltage and switch to a maintenance (trickle)charge.

    A diode will prevent the battery from feeding back into the charger.
     
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  3. PackratKing

    Well-Known Member

    Jul 13, 2008
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    Welcom to AAC.....
    I'm in the same relative boat you are, and will be watching this closely.
    I have a prime motor that will serve for a generator, a blower motor out of a fancy PWM DC Gas furnace, which is a rectified 3-phase winding, tho' have yet to put the wind to it to determine whether it will put out what another brief test indicated.
    I put a 2- inch pulley on the shaft, and ran that against a wire wheel just to spin it, and it put up 175 volts.............I just don't know yet how much current I can parlay that into.........
    If it will put up 120 volts at 10 amps, I will be happier than a hippie in a pot farm :D
     
  4. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    So i can let the battery whack 50 volts in and it will be ok ? :eek: wow now that suprises me.

    Remember i want the cheapest option not the all singing auto control thingy magigies, management sytem sounds expensive ? is there not a simple thing i can just put in the line ?

    is there a part number or diode number or something so i can get the right diode ?
     
  5. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
    17
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    thanks for the welcome. yes i have just started playing around with testing grabbing power from the wind, i started as i had an old car fan motor and was messing around spinning it and got power from it , so built the frame and wind blades and got it to spin and got hooked .. all i would be pleased with is having the outside lit up FREE all night etc , then take it from there , now i am at the stage where i got hold of a good DC 120v motor that seems to whack the power out at lowish revs as getting the speed from the wind is a big problem .. but i think i have that sussed and now i am at the how to control what does come out, chances are i will get 5 volts knowing my luck but !! If a massive wind gets up it could hit the 75 volts .. and we do get some strong wind , hence why i got into it.. BUT !! as you can tell i am not techy minded on rectifiers regulators batteries etc , hence why i joined..
     
  6. Austin Clark

    Member

    Dec 28, 2011
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    I'm not sure if I understand your reasoning here. The battery will only take as much current as it needs? I can only see that being true (to an extent) if the input voltage was around the full-charge voltage of the battery. If you tried to charge the battery with a higher voltage you'll get a voltage drop, and so long as the current isn't enough to burn the battery, it's ok, but once the battery fully charges... it won't stop, not without regulation.

    Can you explain your statement?
     
  7. Externet

    AAC Fanatic!

    Nov 29, 2005
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    Bienvenido, paisano.
    Better do it right. A DC to DC regulated converter capable of the maximum current the generator can put out, should be used.
    Whatever voltage the wind generates, 15 to 100 V, will apply say 13.5V for the battery to take the amperes it needs depending on its charge state.
    Look for such converter modules at Digikey.com or Mouser.com; there is hundreds to choose, and ask them to send air mail to your location. They will.
     
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  8. MrChips

    Moderator

    Oct 2, 2009
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    I explained this in the rest of the post.
    The main concern here is we don't want to fry the battery.
    Low wattage solar chargers do not use regulators.
    The problem here is that a wind turbine produces a lot more wattage. Hence there is a danger of pumping too much current into the battery. What you need is a current regulator.
    One way is to put a whopping big (high wattage resistor) in series (with the diode) to limit the current.

    For part numbers, we would need a lot more info.
    What is the battery model number?
    What is the capacity?
    What is the recommended charging sequence?

    What is the specs on the generator?
    What is the voltage and current output at various wind speeds?
    What is the typical and max windspeed in your area?

    AAC is probably not the best place to find information on wind turbines.
    There is much more available on the internet:

    http://www.omafra.gov.on.ca/english/engineer/facts/03-047.htm


    http://www.aeeolica.org/

    ¡buena suerte!
     
    Last edited: Aug 8, 2012
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  9. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    Loads of info on wind turbines , but hardly anything in simple terms for wiring to batteries etc, hence why i figured i will ask the experts..
    Thats what i was hoping . someone could tell me what to simply put in line, i am not intending running the house on the power just playing on lights at the moment .
     
  10. luvv

    Member

    May 26, 2011
    186
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    Think the reason you won't find a 1$ simple solution for your turbine is because they are a very variable power source.

    Like Mr chips said,there is a ton of things to consider,a turbine needs a lot of control over it output.

    Just last week I had the very same conversation w/ a green power tech about some of the systems involved w/ modern wind turbines.

    They involve all the basics like battery monitoring to more elaborate EM braking...
     
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  11. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    Actually, if you limit the current of a 50V DC source to the battery, there won't be 50V being applied to it. If something limits the current (a resistor for example) there will be a voltage drop over it, i.e. you will measure 50V before the current limiting device and less voltage after it.
     
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  12. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    Surely someone can give me a simple cheap way to do it ?

    what diode to stop the back feed from a 12v battery , a simple part number etc ?

    and a simple blocker to stop more than 14v hit the battery ?

    come on guys your the experts ??
     
  13. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    14V to 100V @ 0A = 0 Power; how about some currenf figures if you want part nos. For 3A 1N5402, 3A @ 200V,or, 1N1190A, 40A @600V- at least it is a number
     
    Last edited: Aug 9, 2012
  14. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    estimate around 6 amps tops..
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    On the current limiting side it might be possible to use a bank of cheap incandescent lamps in parallel as the series current limiting resistance. The high non linearity of the combined lamp resistance might just do the trick.

    Using the data for a 100W lamp from here ....

    http://donklipstein.com/incchart.html

    I obtain a lamp I[V] equation

    Power Fit: I=aV^b
    Coefficient Data:
    a = 7.10679521221E-002
    b = 5.13398238673E-001

    Using this in a simulation with a nine lamp in parallel bank model I obtain a charging current range from about 1A to 6.4A into a 12V battery + [0.7V Vforward] series diode with a 15V to 100V generator range.
     
    Last edited: Aug 9, 2012
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  16. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    well that sounds great but it just went straight over my head :confused:

    simple diode and a resistor is what i am looking for ? anyone ????
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    When a light bulbs filament is cool it is low resistance. When it is glowing white hot it is a much much greater resistance.

    What this means is if the battery is taking everything the power source has to offer the light bulb will be taking all the voltage, and glowing white hot, which will limit the current significantly. If the battery isn't taking a lot of current the bulb will be running much cooler, and not glowing, and its (light bulb) resistance will be much lower.

    You are wanting to oversimplify the problem, and it ain't simple.

    Me, I'm following the thread with interest. I might learn a couple of things here. Keep asking questions, but be willing to learn too. If you don't understand what someone said, focus on it until you wrap your head around it. Eventually you will arrive at a solution, and will understand the solution, which is the best of both worlds.
     
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  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    @ HaventgotaClue

    In principle - creating something like this as your "resistor"...

    http://www.instructables.com/id/Old-World-Light-Bulb-Load/

    If your wind generator can produce 100V at 6A [about 600 Watts] your 12V battery can only absorb 6A at 12V or about 72 Watts. That's about 528 Watts lost as heat. I guess you could string the lights in your tool shed as cheap source of lighting power whilst charging the battery. The stronger the breeze - the brighter the lights in the shed.

    Any alternative solution that doesn't waste all that unsaved energy as heat is going to be more complex and expensive.

    It's a bit like employing cheap labor. If you pay peanuts you get monkeys.
     
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  19. HaventgotaClue

    Thread Starter New Member

    Aug 8, 2012
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    Interesting seeings as i was looking to rig it up to run lights only for testing and take it from there .. so effectively the lights themselves will be doing a job of reducing the battery so long as i get the right diode ? back to the diode guys ?
     
  20. ramancini8

    Member

    Jul 18, 2012
    442
    118
    What is generating the power; I would guess it is a mechanically driven generator. If this is the case you might be able to regulate the generator voltage by regulating the generator field current. Check the generator paperwork to find out if its output is controllable this way.

    You can get away with a simple current source except that the current source power dissipation will be high at high generator voltages.
     
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