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Help in stepping down voltage from a battery

Discussion in 'Power Electronics' started by Mindcrime13, Jan 6, 2010.

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  1. Mindcrime13

    Mindcrime13

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    Jan 6, 2010
    Hello and happy new year!

    Im working on an airsoft replica and the issue im having is that the battery i have is way to powerfull but is the only battery that fits my replica propperly. The battery is lipo 11.1 volts, 1600mah, 20C i need to take the voltage down to at least 8 volts, i know that there is a formula to find what resistor i can use to step down the voltage but i really dont know, so any input , help is appreciated

    thanks!
     
  2. Resqueline

    Resqueline

    2,848
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    Jul 31, 2009
    Need to know: What's it driving? What's the current drain?
     
  3. Mindcrime13

    Mindcrime13

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    Jan 6, 2010
    Thanks for the help!

    Im not sure if this is what you are asking but this is some info i found related to the issue:

    The battery's voltage determines how much power the motor will have. Stock AEGs take an 8.4 volt battery. Upgraded AEGs sometimes need higher voltage battery packs, such as 9.6, 10.8, and the insane 12 volt pack. The higher voltage will make the motor turn faster (higher ROF) and/or harder (to drive those bigger springs). Please also note that a higher voltage battery will put much more stress on your motor and the mechbox internals. You can easily tell what voltage a battery has by counting the number of cells; each NiCd or NiMH cell is nominally 1.2 volts. A 10 cell pack has 12 volts. A 7 cell pack is a standard 8.4 volt pack. The power of the motor (measured in either watts(W) or Horsepower(HP), 746 W = 1 HP) is a function of the voltage it is being fed and the Current it is drawing. Watts equals Current (in Amps, 1000 Milliamps = 1 Amp) times voltage. Please note that the motor will draw more amps running at a higher voltage since it acts somewhat (not quite, but good enough for a simple comparison) like a resistor. The motor follows Ohm's law: Current(in amps) = Voltage divided by Resistance (in ohms). The resistance of the motor stays somewhat constant. So, the amount of current the motor draws will increase linearly with the voltage. Since the power of the motor is volts multiplied by current (which also increases with the voltage), the motor's power increases with the square of the voltage, while the current increases linearly with the voltage increase.

    Let's assume that we have a motor, running at 8.4 volts, and drawing 10 amps (a guestimate...nice round numbers). This motor puts out 84 watts of power, and has a resistance of about 0.84 Ohms. Now, if we were to increase the voltage of the battery pack to 9.6 volts, the motor will draw about 11.43 Amps, and will put out about 110 watts of power. A 10.8 volt battery pack will cause the motor to draw 12.9 amps and produce 138.86 watts of power. The somewhat insane 12 volt battery will cause the motor to draw 14.29 amps, and produce 171.43 watts of power (that's about twice the power! but we are only using 43% more voltage. Remember, double the voltage, quadruple the power, increase the voltage by the square root of 2, and you double the power). Note from this that the motor will drain the battery faster if a higher voltage battery pack is used, since it will draw more amps. For this reason, an 8.4 volt, 1500 Mah battery will last much longer than a 12 volt, 1500 mAh battery pack.
     
  4. Mindcrime13

    Mindcrime13

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    Jan 6, 2010
    Aeg is automatic electric gun,
     
  5. Mindcrime13

    Mindcrime13

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    Jan 6, 2010
    I also read that the ideal amperage should be about 30 amps
     
  6. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Oh-oh, that's gonna take more than just a resistor..
    You'll want a PWM based motor speed controller with a number of mosfet transistors in parallell.
    Those are among other things found in radio control gear for electric cars & planes.
    There are also motor controller kits & stuff to be had. Some ready-made and some self-build.
     
  7. Mindcrime13

    Mindcrime13

    8
    0
    Jan 6, 2010
  8. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    The challenge is the amperages mentioned. A 3V drop times 30A means 90W is going to be dissipated in a resistor or a zener..
    The item you found there is only capable of 1A so it can't be used no matter what.
    You need something like this: http://www.dimensionengineering.com/Sabertooth2X10.htm
     
  9. Mindcrime13

    Mindcrime13

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    0
    Jan 6, 2010
    So bassicly i need to disipate 90watts?
     
  10. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    No, that's the beauty of a PWM controller, it does the job with practically no dissipation at all.
    It simply turns the full battery voltage on & off very quickly. If it conducts 72% of the time then the motor will behave as if it is receiving 8V.
     
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