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22.2V li-ion battery level indicator?

Discussion in 'Power Electronics' started by homemadefrog, Mar 23, 2012.

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  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If you're talking about the hand-drawn circuits, all the resistors are shown and their values shown.

    Some values may need to be tweaked to get an appropriate LED brightness (the 1k resistors are safe, but could be reduced significantly in value.
     
  2. homemadefrog

    homemadefrog

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    Sep 15, 2011
    Hi Steve. You mention it here

    "If you have a 3 terminal dual LED (common anode would be best) you can build this so the colour slowly changes from green, to amber, to red as the battery voltage falls by placing the anode to the +ve rail, and the cathodes to the ends of the 1k resistors.

    Because of the 10k base resistor, one LED will always stay very slightly on though."


    Is this incorrect? It seemed odd to me that the cathodes would go to the ends of the 1k resistors since the resistors come from the + lead but then again it worked for lower voltages.

    Thanks!
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, this is what I meant.

    [​IMG]

    I don't believe it will result in either LED coming to grief
     

    Attached Files:

  4. homemadefrog

    homemadefrog

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    Sep 15, 2011
    AWESOME! Yes of course that makes much more sense, annnnd it WORKS!

    I am so thankful for your help man.

    I just have one last question before I go and solder this thing back together.

    Do you have any idea how much this little circuit might draw? Is there a simple way of calculating it? Or should I hook it up for an hour and test the difference in my battery pack.

    The only reason I ask is because efficiency would be cool and I've seen some circuits out there that use chips instead of transistors. But if the difference in efficiency is negligible then I of course appreciate the simplicity of your design far over one with chips.

    Thanks again Steve.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    With an x Volt power supply, the most current it could draw would be 2x mA, and almost certainly less (since that requires both LEDs to be on and have no voltage drop across them or the transistors (and that isn't going to happen) 1x is more likely.

    note that the value depends on the 1k resistors.

    Calculate it as (2 * Vx) / Rled -- if you use volts and ohms, you get amps.

    Since the purpose of the circuit is to only light a single LED the "2 *" is overestimating the current. Ignoring the other voltage drops is not a real issue since the input voltage is fairlt high and these are relatively small (say a combined value of 2 to 4 volts).
     
  6. homemadefrog

    homemadefrog

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    Sep 15, 2011

    Awesome, thanks! apx. 22mA is nothing.

    So, I'm assuming decreasing the 1k resistors would slightly raise the circuit's consumption. I just checked out this neat led calculator page and it seems 1k is just fine. I'm doing my best to wrap my head around the math you are using. What does Rled stand for? It seems like the led is the only thing drawing current, meaning my assumption that the circuit itself would draw current (via resistors/transistors connecting +&-) is wrong. I'm afraid my ignorance is starting to show again, no big deal. I will solder this baby together!

    Thanks again!
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Rled is the resistor in series with the LED (1k in this case)

    You may find that the 10k resistor going to the base of the second transistor can be usefully increased in value. With a common anode LED it may cause one LED (the one on the left) to be slightly illuminated at all times. You may be able to increase it to between 50k and 100k which will minimise this.

    I wouldn't go any lower than 1k for the LED resistors if it's being operated at 22.5V.
     
  8. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    The circuit given to you will probably do the job but a comparator would have been the best approach. More complex but not very though.
     
  9. homemadefrog

    homemadefrog

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    Sep 15, 2011
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