Power reduction - help

Discussion in 'General Electronics Chat' started by dragonfly_sg, May 1, 2012.

  1. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    I have very limited knowledge of electronics but I am trying to build a battery power USB port for charging. I have a general idea and some schematics skim off this site but I have some questions that I need help on.

    I am thinking of hooking on a 9V 2000mAh rechargeable battery to a 7805 to power a female USB port.

    1) if I do the above I know the output of the 7805 is 1A and therefore I am assuming that the USB is also going to provide 1A when it's hookup. Am I correct to assume this?

    2) I assume if I am charging something that draws 1A then the battery will last about 2 hours. But if I am charging something that need 300mA does the battery last about 2 hours or 6 hours (ball park)?

    3) Is there anyway I can restrict the USB to output a max on only 500mA.? What do I need to reduce the power, should it be done before the 7805? Does reducing the power extend the hours of drain I can get from the battery since my equipment needs a max of 500mA.?

    Thanks and any advise will be greatly appreciated.
     
  2. toninph

    New Member

    Dec 10, 2010
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    Maybe this is helpful: http://pinouts.ru/Slots/USB_pinout.shtml. Be careful not to blow up your USB device.
    Current draw is determined by the receiving end so a current limit is normally not required. An 7805 will deliver up to its limit. Above that it will shut down.
    If you still want a current limiter, consider an LM317.
     
  3. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    Thanks, from the link it seems the standard draw is 500mA. That why I was thinking of how to do that.

    Took a look at the LM317 but it seems rather complicated for me :D. Correct me if I am wrong the model LM317M is the one I need?
     
  4. toninph

    New Member

    Dec 10, 2010
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    LM317M is fine.
     
  5. MrChips

    Moderator

    Oct 2, 2009
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    Using a 7805 5v voltage regulator is fine. Your load will take as much current as it needs.
    I cannot imagine that there is any USB device that has been designed to take more than 500mA.

    Imagine if I designed, built and tried to sell a computer that ran on 150VAC only. No one would buy it because all the outlets are 120VAC (in the USA).
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,237
    1) Yes

    2) Because of it's internal resistance a 2000mAH will not deliver a full 2 hours with a 1A draw. The AH rating is typically for a 10 hour load e.g. 200mA for a 2000mAH battery.

    3) There's no need to restrict the current to 500mA and it won't help the battery life if you add a current limit. The equipment only draws what it requires.
     
  7. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    Thanks guys, I really appreciate the information you've provided. Time to take out the bread board :)
     
  8. BillB3857

    Senior Member

    Feb 28, 2009
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    You need to be aware that some units that charge via a USB connection require a data handshake from the port to verify the proper charger is in use. If they don't get a proper response, the charge cycle will not start. A lot of cell phones are like that.
     
  9. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    Roger and acknowledge. Will hook it up with registers for the 2nd and 3rd pin for that, thanks for the reminder.
     
  10. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    Got another question. If I use a series of AA to make a 9.6V battery pack and I want to hook it up with 2 small motor 3V each and a LED white. I intend to hook them up in series so they are makes up a total of about 9+ V, do I need a add a resistor or anything to this circuit? I don't know the draw of the motors but the LED is standard 20mA.
     
  11. toninph

    New Member

    Dec 10, 2010
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    Can you draw a schematic of that? Is helpful to understand what you want to do.
     
  12. MrChips

    Moderator

    Oct 2, 2009
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    If the motors take more than the maximum allowed current of the LED you will end up blowing the LED.
     
  13. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    MrChips, you're are right.

    Change of design. Toninph attached is my circuit diagram.

    Now its just 9V 2000mAh battery, a 7806 to bring it down to 6V and powering 2 units of V3 motors in series (I don't know how many amps they draw. Small motor that drives toycars.) I have the caps recommended in the Datasheet for the 7806. Do I need anything else or is this thing fine, don't want to burn out the motors.

    [​IMG]
     
  14. BillB3857

    Senior Member

    Feb 28, 2009
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    Unless you can insure that both motors are seeing the same load, the voltage on the one with the least load will climb. If one motor is loaded to the stall point (stops turning) then the other will see full voltage. You may be better off putting the motors in parallel with the proper voltage applied.
     
  15. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    Putting it in parallel means I need to lower the voltage to 3V right.?

    Hmmm..... mentally it seems very hard to drop to 3 voltage. Can I use a LM317? I don't know how low I can go with a LM317, the datasheet state the drop range ie. Vi - Vo (3V to 40V) but said nothing about the what the min Vo it can support.

    Or I can check that the motors are of the same make and model, then I can keep the 6V.

    I'll chew on that for awhile.
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    It doesn't matter whether you make them the same make and model. If they are in series, then the load seen by one motor will affect the operation of the other motor. I can pretty much guarantee that you won't like the result.

    How are you going to be using these motors? Just putting power to them and then they will turn at whatever speed they can? Or are you going to need some kind of control so that you can adjust the speed?
     
  17. dragonfly_sg

    Thread Starter New Member

    May 1, 2012
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    These motors are actually motor for two aerators I am going to use to keep baits alive on fishing trips. Their performance individually doesn't really matter, but I can't afford to have both fail on me at the same time.

    So I guess I have answered my own question, stepping down to 3V is my only option it seems. Question is, can the LM317 do that?

    Do remember that I am an electronic dummy as I state in my first post in this thread, so go slow on me :D
     
  18. paulktreg

    Distinguished Member

    Jun 2, 2008
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    I presume weight isn't a problem? Why not just run the motors from two D Cells? Get rechargeables and carry a few spares?
     
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