Connect with us

Squeeze that battery

Discussion in 'Power Electronics' started by StaticCharge, Dec 20, 2012.

Scroll to continue with content
  1. StaticCharge

    StaticCharge

    8
    0
    Dec 20, 2012
    Hello,I am new here.This site is awesome! I came here to learn about electronics,and I already have a question! I was wondering...uhh I don't know how exactly to put this...
    Ok what I want to know is there a way to make one battery give 2x more energy,but of course last half of it's normal life.I need more power,and less weight.So battery would give 100% more energy,but drain 100% faster.Is this possible? Do voltage multipliers do this?
     
  2. donkey

    donkey

    1,293
    56
    Feb 26, 2011
    ok... define energy. are you talking voltage,current or watts?
    have you ever googled joule thief?
     
  3. StaticCharge

    StaticCharge

    8
    0
    Dec 20, 2012
    I am talking about watts.I want to "drain" more watts.Of course,battery would discharge faster.I saw joule thief but it's just a voltage booster.
     
  4. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    If its voltage you want search for DC-DC converters or boost converters, Some sites like sparkfun sell tiny ones that fit inside a battery holder.

    If its more current youre after you need some juicy capacitors, but output cannot be constant since it takes more time for the battery do charge them than it takes the load to discharge them.
     
  5. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Let me see if I understand what you are asking. You have a battery driving some load. The battery lasts for some number of hours. But what you want is to use the same battery to drive more power into the load (brigher light, faster running motor, louder sound, etc), and have the battery last 1/2 as many hours.

    Basically the way to cause the same load to draw more power is to up the voltage supply to it. But is the device you are driving capable of handling more voltage? If not you will likely destroy it or limit its life.

    Are you tied to this particular kind of battery? Typically, if you want move voltage, the best way to do it is use a higher voltage battery. Or you can use a boost DC-DC converter, but this is not 100% efficient.

    If you tell us more specifcally what you are trying to do, we could be of more help.

    Bob
     
  6. donkey

    donkey

    1,293
    56
    Feb 26, 2011
    ok let me putit very simply
    watts =amps *volts.
    if you increse the amps you half the volts.if you increase thevoltsyou half the amps. the thing youCANNOT do isincrease the watts without increasing either the volts (withoutdecreasing the amps) or vice versa
     
  7. StaticCharge

    StaticCharge

    8
    0
    Dec 20, 2012
    Yep that's what I want,but I wanted output to be constant.I see now that it's not possible.Thank you everyone for your replies! I am not an expert electrician so expect some more questions from me xD
     
    Last edited by a moderator: Dec 21, 2012
  8. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    It is possible. For constant current, you need to combine both approaches, with a few constraints:

    Use a blocking oscillator or something similar to charge a capacitor (or a bank of them) to a relatively high voltage, then regulate the output to the voltage required by the load.

    Depending on the efficiency of the circuit, it should run and dump energy into the capacitors until the battery's potential is near zero (as opposed to say a 9V battery being considered dead at 7V).

    The main constraint is that the load cannot continuously draw more current than can be drawn from the battery. Eventually the voltage across the capacitors will drop below the regulator's threshhold and they must be allowed to recharge before the next use. The time required to recharge will increase as the battery is depleted (just like a camera flash which runs on exactly the same principle - only difference being that the flash uses all its stored energy at once).

    For longer running times, use bigger (and / or more) capacitors. Also note that energy will be wasted in both the booster and the regulator circuits.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,838
    Jan 21, 2010
    Another issue is that due to the internal resistance of the power source (the battery), drawing power at twice the rate will result in slightly LESS than half the lifetime.
     
  10. StaticCharge

    StaticCharge

    8
    0
    Dec 20, 2012
    Yeah...a schematic for this would be awesome! You guys must know where to get one,and there must be a name for such a circuit rght? :D
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,838
    Jan 21, 2010
    The simple answer is to halve the resistance of the load.

    But what is the load?
     
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    652
    May 8, 2012
    This should handle any battery you can fit between the ram and the anvil.

    Yes, we need a Devil smilie. :p

    Chris
     

    Attached Files:

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads
There are no similar threads yet.
Loading...
Electronics Point Logo
Continue to site
Quote of the day

-