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is battery drain linear with resect to Mah?

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  • #61
    I may try that but they are hard to find w/o paying too much or "long lead time" or must buy 1000 etc etc. No hurry, and not even sure i will need to now. Kinda sucks having moved fairly rural for the first time in my life. Used to live in L.A. till now and i could often find stuff like this local for pennies.


    • #62
      I took three different used 9V batteries and connected then to a 9V power supply.

      #1 Drew negligible current. This was a very tired battery.
      #2 Drew 10mA, steady
      #3 Started at 70mA, fell to 20mA after 30mins then steady

      So you can see that the drain on the feeder battery is significant. Thus, in general, the only safe and reliable method is not to directly parallel
      You missed the essential test: you did not put them in parallel.

      And which one do you call a "feeder battery"?

      Just curious.
      Juan Manuel Fahey


      • #63
        I agree with JMF that the chemical potential difference between the battery cell elements is a constant.
        And I agree with nickb that the simple model of a constant chemical EMF in series with a resistance is not sufficient.
        A logical extension of the model would be an additional parallel resistance that takes care of leakage (self-discharge).
        This would mean a built-in voltage divider which could explain the dropping no-load voltage at the battery terminals with use.

        I assume that the series resistance increases while the leakage resistance decreases with use.

        But most probably this model is still too simple.
        Last edited by Helmholtz; 11-21-2021, 08:18 PM.
        - Own Opinions Only -


        • #64
          Originally posted by J M Fahey View Post
          You missed the essential test: you did not put them in parallel.

          And which one do you call a "feeder battery"?

          Just curious.
          I take your point.

          This experiment was to see what would happen to the current in the battery that was being 'charged' and in particular would it rapidly go to a small number. Using a battery instead of a power supply would simply discharge the stronger battery until no current flows which exactly the reason not to put them in parallel. The "feeder" battery in this case is the power supply i.e the one that is least discharged. Using a real battery for the source would answer the question as to how much charge is lost for just one set of batteries so not much use in general. The experiment does prove that significant energy can be lost to the discharged battery and that is what matters.

          I didn't have a stack of identical batteries to try (or the inclination) but it would be interesting if you did enough tests. Now, if you have a bunch of batteries and time you wish to sacrifice....
          Experience is something you get, just after you really needed it.