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Simple Conversion Circuit

Discussion in 'General Electronics Discussion' started by nesba263, Sep 20, 2010.

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  1. nesba263

    nesba263

    5
    0
    Sep 20, 2010
    Hello all-

    Thank you in advance for any help here....

    I was hoping someone with some electronics knowledge could help me out with a basic circuit diagram to replace a battery setup I have in a small gadget I have. The device uses (2) LR41 batteries in series, to power a small motor. They wearout all the time extremely fast and its annoying as all hell. I was hoping some minor soldering to the terminals to an external battery source could be a longer lasting replacement.

    Current Circuit Diagram

    [​IMG]

    Requested Updated Circuit Diagram. (Need help filling in the ? areas)

    [​IMG]

    What electronic parts, resistors, etc, do I need to put in the circuit that I can buy at Radio Shack or order online that will provide the correct current and voltage, without melting the motor? I put blank spots A and B in the new circuit diagram to make explanation easier. Help is greatly appreciated. This is something I would get a ton of use out of if I could get it working.

    A configuration that would allow for 2 D-Cell Alkaline in series (or) 2 D-Cell series in parallel (4 batteries total) would be amazing if I could reduce the mAh to a correct level with correct voltage. I know the basics, so other recommendations are appreciated. I am aware that I will have the make the power supply external and solder to the LR41 terminals a small wire to feed the circuit.

    Thanks in advance to anyone who responds. I'm sure this is an easy one for the experts out there.

    To save time in reference, the specs for a LR41 cell are as follows:
    Alkaline, 1.5v, 25-45 mAh, depending on the quality and material (silver oxide or alkaline) of the cell.
    Diameter x Height: 7.9 × 3.6mm

    Tom Ward
     
  2. davenn

    davenn Moderator

    13,877
    1,964
    Sep 5, 2009
    hey tom

    if you want to run the motor of external batteries ... then why not just forget about the internal ones ?? keeping the internal batteries and parallelling external ones across them isnt really going to do much to significantly increase the life of the small internal ones.

    if you can run it on external batteries, that implies that compact portability isnt really an issue ?
    what is the motor being used for ?,
    for what periods of time does it have to be running for ?

    if you are going to use external and internal batteries together then neither of your diagram A or B locations is correct as that is series and will double the voltage to your motor
    The external batteries need to be parallel across the internal ones

    Dave N
     
  3. Leighcusack

    Leighcusack

    15
    0
    Sep 9, 2010
    Have to agree.
    Rip out the button batteries, then install two AAA,AA,C,D batteries in series.
    The button batteries and any of the ones mentioned above are 1.5 volt so no other componates are needed.
    You can get AA batteries these days with 2000mAh capacity. Google eneloop. They are made by sanyo.

    Install the new batteries as per your original circuit diagram.

    Hopefully you get 20 times more batterey life the the originals.

    Cheers
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    Be careful that the motor current isn't limited by the impedance of the power source.

    If the motor goes a lot faster with bigger batteries, you ma have another problem to solve.
     
  5. nesba263

    nesba263

    5
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    Sep 20, 2010
    Thanks for the responses

    My thoughts are exactly what 'Leighcusack' responded and I guess there was some confusion in my explanation. My concern was that the 1.5v is correct, but that 2 D cell in series, in the place of the button LR41 batteries would push too much amperage.

    Davenn: Thank you for the comment, my intension was to remove the LR41's completely, soldering leads to an external power supply and portability is not a concern.

    (*steve*) : Yep, that is my concern... a motor spining itself apart or failing due to load or heat.

    So the problem lies with impedance it appears. Is there a way to control the impedance to mimic that of a pair of LR41's, using D/C/AA's etc? I know the voltage is correct, I just didnt want to fry the motor. I know that current relates to resistance and voltage via ohm's law.... 2 of those variables are fixed in the batteries I choose, correct?. It would seem logical to me to put an appropriately sized resistor or a variable resistor (rheostat) inline to keep the current down in the circuit. Is this correct or am I way off? Also, if the correct answer requires the resistance of the motor, I can measure it. (I have a digital multimeter......bah..... probably should have measured it now that I'm thinking about it, but I'm @ work for the next few hours. )

    I understand the basic physics here, actually embarassed to ask these questions. ( I have a B.S. minor in physics, but remember more about gluons and leptons than basic circuitry ) I build enterprise IT data networks now, and my physics classes have mostly faded into memory.

    Thank you very much to all who responded.
     
  6. NSklavos

    NSklavos

    8
    0
    Sep 5, 2010
    The motor will only draw as much current as it needs, whether you use 2 LR41's or 2 D cells. Adding a resistor will only slow down the motor. If you are concerned about excessive current draw due to motor failure, put a low value fuse in series with the external batteries.
     
  7. nesba263

    nesba263

    5
    0
    Sep 20, 2010
    Perfect, thanks for the response!
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Other than the fact that the response is wrong.

    The issue is (or may be -- you need to test) that the motor demands more current than the small batteries are able to supply, thus the voltage sags, and the current is limited to what the batteries can supply.

    It's not an ideal situation, but it's pretty much *exactly* how those little LED torches powered from tiny batteries work.

    Using larger batteries with the ability to supply more current and the motor will run faster, possibly too fast, and possibly drawing excessive current leading to overheating and possible burning out.
     
  9. Leighcusack

    Leighcusack

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    Sep 9, 2010
    *Steve* can't he measure the amps running through the circuit in the original design and compare the result to the AA batteries?

    Also, if the button cells were loaded that high wouldn't the voltage drop also?

    I don't mean to step on your toes so to speak, just interested in your response.
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Yes to both.
     
  11. nesba263

    nesba263

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    Sep 20, 2010
    final question

    So would putting a rheostat inline before the motor control the current, and allow me to prevent frying the motor?
    Its literally a simple, inexpensive DC motor. I can put a meter on it and take resistance or current measurements if required to get the correct result.

    Thanks a ton for all the participation. You guys rock.
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    No, get a multimeter and measure the open circuit voltage across a new set of the small batteries, then measure the voltage while the motor is running.

    Then measure the current drawn from the batteries when the motor is running.

    If the battery voltage doesn't fall more than (say) 10% you will have no problems.

    If it falls by about 40% or more, you would definitely need to do something.

    Between those figures, there is some grey area.
     
  13. nesba263

    nesba263

    5
    0
    Sep 20, 2010
    I'll try that, thank you.
     
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