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Using a 12v battey — Parallax Forums

Using a 12v battey

mosquito56mosquito56 Posts: 387
edited 2008-01-25 16:37 in General Discussion
·I have a very expensive 12v lion battery used in r/c electric motors. I want to use this on my board which is setup for 9v.
·1. Can I connect it straight to the 5v chip?
·2. The prop runs on 3.3 v meaning the voltage is stepped down by a factor of 4. If the prop draws 100ma does that mean that 300ma is being converted to heat by the voltage regs?
·· ·I am hoping to use this in an automobile and don't want to heat up the curcuit dropping the voltages.
Thanx

BTW: I was an electronic tech. I know how to READ a schmatic but forget DRAWING one. LOL

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······· "What do you mean, it doesn't have any tubes?"

······· "No such thing as a dumb question" unless it's on the internet

Technologically challenged individual, Please have pity.

Comments

  • David H.David H. Posts: 78
    edited 2008-01-23 17:21
    mosquito56,
    From what I understand, you have to have a 5V regulated for the BasicStamp. If the prop runs on 3.3V, I'm assuming you need a regulator that will drop the voltage to 3.3V, that will handle the 12V input. You will probably need a heatsink on the regulator to disipate that heat from dropping the voltage. Without a regulator, you will probably fry the chip or a BasicStamp.
    I'm sure others that know more perticulars will answer also.
    Good luck

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    David


    There are 10 types of people in this world,...
    Those that understand binary numbers, and those that don't!!!
  • FranklinFranklin Posts: 4,747
    edited 2008-01-23 22:49
    Take a look at switching regulators. They are more efficient than linear devices.

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    - Stephen
  • mosquito56mosquito56 Posts: 387
    edited 2008-01-23 23:05
    Thanx for the idea. I was asking a very simple "I thought" question regarding power loss in regulators. Since most normal electronics runs at 5 volts I need to step the car battery down to 5v.

    Will I always get warm regulators?

    I will check on the swithcing regulators. Thanx

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    ······· "What do you mean, it doesn't have any tubes?"

    ······· "No such thing as a dumb question" unless it's on the internet

    Technologically challenged individual, Please have pity.
  • DiablodeMorteDiablodeMorte Posts: 238
    edited 2008-01-24 02:15
    There's a very nice power regulator out there, i saw it on hack-a-day, let me go check:
    Ok, it was a switching regulator: http://www.dimensionengineering.com/DE-SW050.htm
  • DgswanerDgswaner Posts: 795
    edited 2008-01-24 15:08
    Yes you will always have Heat generated on a linear regulator, based on how much higher the input voltage is compared to the regulated output voltage.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    "A complex design is the sign of an inferior designer." - Jamie Hyneman, Myth Buster

    DGSwaner
  • phil kennyphil kenny Posts: 233
    edited 2008-01-24 15:51
    Mosquito56 said...

    2. The prop runs on 3.3 v meaning the voltage is stepped down by a factor of 4.

    From a purely mathematical perspective, your statement is technically true.

    But to determine the power loss, you just have to take the difference
    between the regulator input voltage and the load voltage. In this case you
    have 12v - 3.3 v or 8.7 volts dropped across the regulator. If you are
    using a linear regulator and the load current is 100 ma, the power
    lost as heat in the regulator is 0.1 * 8.7 = 0.87 watts or 870 mw.
    Mosquito56 said...

    If the prop draws 100ma does that mean that 300ma is being converted to heat by the voltage regs?

    No.

    Post Edited (phil kenny) : 1/24/2008 4:18:03 PM GMT
  • mosquito56mosquito56 Posts: 387
    edited 2008-01-24 18:57
    Considering the average soldering iron is 15W 870mw is a chuck of power.· I lose 8w of power for every watt I use in the curcuit.

    I hope the switching regulator with a large cap might cut the heat. Power efficiency isn't a problem, just the heat and vibration.

    Finally a couple guys who actually answered the question. Thanx

    Have a very linear mind, it's hard to think about·Question B when Question A is unanswered.

    Question B: Can I hook the 12v battery to a normal 9v system. The system has a 5v regulator and a 3.3 behind that. Would the datasheet on the regulator answer this question?


    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    ······· "What do you mean, it doesn't have any tubes?"

    ······· "No such thing as a dumb question" unless it's on the internet

    Technologically challenged individual, Please have pity.

    Post Edited (mosquito56) : 1/24/2008 7:03:06 PM GMT
  • Mike GreenMike Green Posts: 23,101
    edited 2008-01-24 19:11
    Most regulators can handle significantly more than 12V as long as the heat produced by the voltage drop is adequately dissipated. Some regulator circuits use an electrolytic capacitor on the input side and that has to have a voltage rating high enough to cover the input voltage.

    If you're using a Parallax Protoboard or Demo Board, there's not adequate heatsinking for voltage inputs greater than 9V near the maximum current load of the regulator. If you've built your own board, you can include adequate heatsinking.
  • mosquito56mosquito56 Posts: 387
    edited 2008-01-25 16:37
    Thanx, I am using the breadboard right now. I was thinking of finding a switching reg, put it on a seperate board with switch and curcuit breaker and then to the input of the education board. Haven't had time to do any of this, working on another project right now so I just use a recharable 9v battery.
    The curcuit breaker was recommended because of the old tendency for lions to catch fire. I don't think this is true anymore but better safe than sorry.

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    ······· "What do you mean, it doesn't have any tubes?"

    ······· "No such thing as a dumb question" unless it's on the internet

    Technologically challenged individual, Please have pity.
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