What Is Your Internal Resistance?

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  • Sunking
    Solar Fanatic
    • Feb 2010
    • 23301

    #16
    Originally posted by FunGas
    A 12.85V, 9.6V, 2.75A
    B 13.0V, 12.70V, 3.25A
    C 13.07V, 12.38V, 3.25A
    D 13.09V, 12.34V, 3.25A
    E 13.02V, 12.33V, 3.25A
    F 13.07V, 12.23V, 3.20A

    So, the approximate internal resistances are:-

    A 9.35Ω
    B 9.09Ω
    C 9.26Ω
    D 9.29Ω
    E 9.23Ω
    F 9.25Ω
    Might want to recheck you math. If any of the batteries really had 9 or more Ohm;s would be boat anchors and completely useless. 3 amps x 9 ohms = 27 volts (not possible on a 12 volt battery). A dead bolted fault would only draw 1.3 amps @ 0 volts. For example battery A [13.0 - 12.7] / 3.25 amps = .0923 Ohms. you only missed it by a factor 100.

    The good news is the batteries look A-OK for their size.

    EDIT NOTE:

    What this tells you is the max load the batteries can realistically handle in a working application. You want to limit voltage sag to no more than 2% of open circuit voltage. 2% of 12 volts = .24 volts. That tells you the max load is .24 volts / .09 Ohms = 2.67 amps. Or put another way 12 vols x 2.6 amps = 31 watts. The other thing it tells you is when your batteries have seen their better days and need replaced when at full charge you see the Ri has increased to 25%.
    MSEE, PE

    Comment

    • FunGas
      Member
      • Aug 2012
      • 99

      #17
      Originally posted by Sunking
      You want to limit voltage sag to no more than 2% of open circuit voltage.
      Really? that little? I'll let it charge again for a few days and see how much the stack Voltage sags when I turn on the load, but I think it's more like 0.7V

      Tested them again just now after sitting on the bench resting for a bit and the internal resistances are quite different:-

      A = 0.0904Ω
      B = 0.0923Ω
      C = 0.0931Ω
      D = 0.0925Ω
      E = 0.0917Ω
      F = 0.0925Ω

      Funny how C has the greatest value and is the newest battery...
      Dem

      Comment

      • Sunking
        Solar Fanatic
        • Feb 2010
        • 23301

        #18
        Originally posted by FunGas
        Really? that little? I'll let it charge again for a few days and see how much the stack Voltage sags when I turn on the load, but I think it's more like 0.7V

        Tested them again just now after sitting on the bench resting for a bit and the internal resistances are quite different:-

        A = 0.0904Ω
        B = 0.0923Ω
        C = 0.0931Ω
        D = 0.0925Ω
        E = 0.0917Ω
        F = 0.0925Ω

        Funny how C has the greatest value and is the newest battery...
        Well my calculator says they are the same as last time. That is why I said check your math. Looks like you had forgotten to move the decimal point first time around. For the AH rating of the battery looks good to me.

        Reason the sag is important is the voltage by the time it reaches the utilization device. Take for example a Inverter. Most are set to trip off line at 10.5 volts. You can take one of your batteries fully charged up and full rated capacity and connect it to say a 250 watt inverter. Connect a device that pulls 250 watts, and instantly the inverter will trip off line from under voltage despite the fact the batteries are fully charged up. Not only do you have to account for the voltage sag on th ebatteries, but you also have to account for voltage loss on the wiring between the battery and inverter. They all add up real quick. That is why you limit discharge to roughly C/8 to keep the sag to 2% or less.
        MSEE, PE

        Comment

        • FunGas
          Member
          • Aug 2012
          • 99

          #19
          Originally posted by Sunking
          the utilization device
          It's just a 12V Peltier that takes any real power, plus a few other low current items (timer, temperature controlled relay) - anyway, my end-of-day Voltage is now a respectable 13.2V
          The run time was decreased to 4 hours for the "winter". We still get crazy sun here but the temperature goes down (better for the solar panel?)

          Thanks for fixing my batteries!
          Dem

          Comment

          • yewsuck
            Junior Member
            • Nov 2014
            • 9

            #20
            Thanks for the Lifepo4 knowledge.

            -YS

            Comment

            • Jemplayer
              Member
              • Jun 2014
              • 77

              #21
              Originally posted by Sunking
              OK you might be saying to yourself "huh"? But it is a number you can put to great use, especially as a barometer to tell you are on borrowed time and won't be long till you are replacing batteries. But this thread is all about me.

              Basically a call to post your batteries internal resistance for me to crunch data. It will not take long measure it, nor is it difficult requiring a lot of test equipment, just a couple of simple steps. What you get out of it is priceless.

              So here is how you do this.

              1. First you want the batteries to be at or near 100% charged up, anything above 90% is good and rested a bit if possible.

              2. Turn off all loads so there is ZERO current flowing and panels are off like just after sunset. Once you are sure the system is dead, measure and record the battery voltage right on the battery positive and negative post. Monitor for a minute or so until it stabilizes. Label this voltage as V1

              3. Now leaving your meter connected to the battery turn on all the load devices you can without over loading anything.

              4. This maybe the show stopper for most of you, but note and record the current with everything turned on.

              6. Simultaneously while you record current, note loaded voltage on battery and record that voltage. Label current as Il, and voltage as V2

              7. Test concluded, put everything back to normal.

              OK to find your battery or batteries internal resistance we use the data you collected. Ri = [V1-V2] / Il

              So let's run through an example. 12 volt system loaded with 40 amps. Open circuit battery voltage or V1 = 12.5 volts, Loaded voltage or V2 = 11.86 volts. So Ri = [12.5 - 11.86] / 40 = .016 Ohms.

              How do you use that number? It works best on brand new batteries to establish a Base Line, but still useful on older batteries. When you start noting the Ri is growing means you have lost capacity and the batteries are on there way out.

              I have a battery tester that loads the battery with 10 amps. It uses a analogue meter which is not easy to read exact voltages, but could use it in conjunction with a digital multi meter, this would do the same job as described above, correct? On a 6 v 232 ah batt.

              Comment

              • Sunking
                Solar Fanatic
                • Feb 2010
                • 23301

                #22
                Originally posted by Jemplayer
                I have a battery tester that loads the battery with 10 amps. It uses a analogue meter which is not easy to read exact voltages, but could use it in conjunction with a digital multi meter, this would do the same job as described above, correct? On a 6 v 232 ah batt.
                Sort of but I bet that testes is made for a 12 volt battery, and on a 6 volt battery is only going to draw 5 amps, and 5 amp sis not really enough to give you an accurate measurement. The more current you can test with, the more accurate the result. On a 230 AH battery one would like to put at least a 100 amp or more load on. One way to do that is with an inverter with something like a toaster or blow dryer as a load. A 1000 to 1200 watt load should draw around 100 amps on a 12 volt battery. Actual current is not important, what is important is you know what it actually is. Example if you were drawing say 111 amps and your battery voltage drops from 12.6 down to 10.5 volts then you know the Ri = [12.6 - 10.5] / 111 = .0189 Ohm's. If were exactly 100 amps then the voltage would only drop down to 10.71 volts and you get the exact same result.

                Ideally in a lab you load the battery down until the voltage is half, or 6.3 volts on a 12.6 volt battery. But there is no safe way for DIY to do that.
                MSEE, PE

                Comment

                • Jemplayer
                  Member
                  • Jun 2014
                  • 77

                  #23
                  Originally posted by Sunking
                  Sort of but I bet that testes is made for a 12 volt battery, and on a 6 volt battery is only going to draw 5 amps, and 5 amp sis not really enough to give you an accurate measurement. The more current you can test with, the more accurate the result. On a 230 AH battery one would like to put at least a 100 amp or more load on. One way to do that is with an inverter with something like a toaster or blow dryer as a load. A 1000 to 1200 watt load should draw around 100 amps on a 12 volt battery. Actual current is not important, what is important is you know what it actually is. Example if you were drawing say 111 amps and your battery voltage drops from 12.6 down to 10.5 volts then you know the Ri = [12.6 - 10.5] / 111 = .0189 Ohm's. If were exactly 100 amps then the voltage would only drop down to 10.71 volts and you get the exact same result.

                  Ideally in a lab you load the battery down until the voltage is half, or 6.3 volts on a 12.6 volt battery. But there is no safe way for DIY to do that.
                  Yes I see. What I can tell you is I used this load tester on a band new and an old batt and got the exact same reading, so that's a good sign. I think it is 10amp @6 volt, but don't have it in front of me, will check. It seems to use an element of sorts to load the batt.

                  Just to confirm, I do this test as you describe using a microwave etc, on the entire string correct? Micro pulls 1750 watts. Can use toaster and micro to load 2.5 kW odd can then add a vacuum clear another 1300 watts and really get them down. Or do I need apply this large of a load to one batt at a time.

                  The load tester mention before was done on individual batteries..

                  Must say Sunking, you are a great resource on this forum , and I thank you for all the info shared to us nubs..

                  Comment

                  • Sunking
                    Solar Fanatic
                    • Feb 2010
                    • 23301

                    #24
                    Originally posted by Jemplayer
                    Yes I see. What I can tell you is I used this load tester on a band new and an old batt and got the exact same reading, so that's a good sign. I think it is 10amp @6 volt, but don't have it in front of me, will check. It seems to use an element of sorts to load the batt.

                    Just to confirm, I do this test as you describe using a microwave etc, on the entire string correct? Micro pulls 1750 watts. Can use toaster and micro to load 2.5 kW odd can then add a vacuum clear another 1300 watts and really get them down. Or do I need apply this large of a load to one batt at a time.

                    The load tester mention before was done on individual batteries..

                    Must say Sunking, you are a great resource on this forum , and I thank you for all the info shared to us nubs..
                    Thanks

                    No you can do it all connected together as that is the idea. Just a note of caution so no one gets the wrong idea, DO NOT OVER LOAD your system and wiring as that is a great plan for a fire. My recommendation and what I do is have a Current Shunt or Hall Effect Current Transducer permanently installed between the battery and load device like an inverter if the inverter or load device does not come with it built in. Just make sure you are working with fully charged up batteries because as the battery discharges, resistance goes up. Also note temperature also significantly effects Ri. As temp goes up, resistance goes down. So try to do this when the batteries are at normal operating temps which should be around room temp. This is why even with a new battery in your car the engine cranks slowly when it gets really cold.

                    The paradox is cold batteries last longer, stores less energy, and cannot deliver high currents. Warm batteries can store more energy, deliver higher current, but loose cycle life. We call it Dancing on the Head of a Needle. Batteries are spec and rated @ 77 degrees F or 25 C.
                    MSEE, PE

                    Comment

                    • inetdog
                      Super Moderator
                      • May 2012
                      • 9909

                      #25
                      Originally posted by Sunking
                      Sort of but I bet that testes is made for a 12 volt battery,
                      Is that perhaps a Freudian slip?

                      "Freudian Slip: When you say one thing but mean your mother."
                      SunnyBoy 3000 US, 18 BP Solar 175B panels.

                      Comment

                      • Sunking
                        Solar Fanatic
                        • Feb 2010
                        • 23301

                        #26
                        Originally posted by inetdog
                        Is that perhaps a Freudian slip?

                        "Freudian Slip: When you say one thing but mean your mother."
                        Either that or a typo. My keyboard is acting crazy lately. I digress.
                        MSEE, PE

                        Comment

                        • russ
                          Solar Fanatic
                          • Jul 2009
                          • 10360

                          #27
                          Originally posted by Sunking
                          Either that or a typo. My keyboard is acting crazy lately. I digress.
                          Maybe you were thinking about the pickle slicer?
                          [SIGPIC][/SIGPIC]

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