12v to 9.6v

Discussion in 'The Projects Forum' started by seesoe, Jan 8, 2009.

  1. seesoe

    Thread Starter Active Member

    Dec 7, 2008
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    hello,

    i have a paintball gun that has a 9.6v 170 mah battery that im changing out with a 12v source feed.
    battery - http://www.pbreview.com/products/reviews/3108/

    how can i build something nice and simple that will regulate 12v to 9.6v?
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Depends upon how much current it draws, but four 1N400x diodes in series will have about a 2.4v drop across them.
     
  3. seesoe

    Thread Starter Active Member

    Dec 7, 2008
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    not sure.. i guess not much, its a simple board that chargers a cap and has dip switch bus to change the rate of fire and when to fire the cap that triggers a solenoid.
     
  4. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
    5
    Hey Why Dont you try a LM317 regulator circuit?? Requires only abt 4 components.. Give it a google..
     
  5. seesoe

    Thread Starter Active Member

    Dec 7, 2008
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    i like:), thanks for the idea, but what about resistor value? lowest watt possible?
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    The diodes will probably be more efficient for this application, but either will work. Be prepared for the LM317 to get hot.
     
  7. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
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    I know Diodes are good but a very exact voltage is hard to obtain, and you need higher current diodes, not 1N400x.. It will be good initially but voltage will not be stable as it heats up... LM317 can stand 1500mA if properly used.. Try this one and calculate the exact voltage using the formula.. you will get the resistor.. Verify it with a multimeter.. And be sure to mount a heatsink..
    http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    Depends on how you define stable, I doubt this circuit will pull much, not with batteries rated at 170mah. At an amp their lifespan is around 10 minutes. This is a 9V equivalent, which isn't a very high current source (100ma would be pretty high), so those diodes will be accurate to within .1V over the range most likely.

    LM317 is a good part with good regulation, but it is major overkill for this application. The other issue is those diodes will be very wirelike, which means they will fit in a very small space.

    Truth to tell though, I don't really care either way. I have used both with good results.
     
  9. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
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    Agreed Eitherway...
     
  10. seesoe

    Thread Starter Active Member

    Dec 7, 2008
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    thank you jj for all the suggestions, im the type of guy that likes to make electronics applications complicated, and i really want to use the LM317 but the application is a very simple little project and the diodes will work just right i think. bill is correct it will be wire like and i can just heat shrink it, but the lm317 will need heatsink and stuff im pretty lazy haha.

    anyways thanks all for the help ill report back with the results:)
     
  11. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
    5
    Thanks for that seesoe, I came to know only recently that the LM317 doesn't need a heatsink if the load is below 500mA. Thats why I suggested it. We all have situations requiring a supply voltage and resistors don't seem to work well with load so i found LM317 as a risk free solution thought a bit costly. I haven't had tried diodes yet thats why I didn't recc that.. ANyway thanks everyone for the info and I'll try diodes next time.. Thanks Bill for the suggession.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    This is not correct!
    The important part is how much power dissipation is occuring in the regulator itself. Ambient temperature is important too...

    Let's do an example. Let's say you made a 10v supply using an LM317 that itself was getting power from a 12v supply. You have a 20 Ohm load on the output (total) for a 500mA output current.
    So, you have a 2v drop across the regulator itself, and 500mA going through it.
    P=EI, or Power(Watts) = Voltage * Current
    P = 2V * 500mA = 2*.5 = 1 Watt of power to dissipate as heat.

    Now let's say we're going to use that same regulator circuit, and load but power it from a 30v supply instead of a 12v supply. Now we're dropping 20V across the regulator!
    P = 20V * 500mA = 20*.5 = 10 Watts of power to dissipate as heat, 10 times as much as before!
     
  13. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
    5
    Thats correct.. It Depends on Load And Drop volatge. But I have found that it rarely heats up when I drive 4 leds on a supply of 12v regulated to 2.8v. Here his application drops 2.4v only and requires some more than a 100mA... But I agree to what you pointed.
     
    Last edited: Jan 9, 2009
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