How is this possible?

Discussion in 'General Electronics Chat' started by Rolland B. Heiss, Feb 6, 2015.

  1. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    As of this share the voltage is now 6.70 and rising with the voltmeter only connected to the battery and the battery connected to nothing else. 24 hours ago I connected this 'dead' battery to the circuit and the battery read 6.99 then. What is going on here?

     
  2. MrChips

    Moderator

    Oct 2, 2009
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    What you are observing is quite common.
    As you say the 9V battery is as dead as a door nail.
    When you draw current from it the voltage will drop over time. The main reason is that the internal resistance of the battery is increasing with time and usage.

    When you disconnect the battery and allow it to sit, the battery will recover slightly and hence the open circuit voltage will be higher that when loaded and will rise slowly.
     
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  3. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    Seems like it is recovering more than slightly MrChips. Right now it is at 6.71 and still rising and mind you, this is after powering the two LED's for over 24 hours.
     
  4. MrChips

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    Oct 2, 2009
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    As I said, you have to factor in the internal resistance of the battery.
    As soon as you reconnect the circuit, you will find the voltage has dropped.
     
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  5. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    Yeah, the voltage drops when connected but it rises back up once disconnected nearly to the point when it was connected in the first place. I can't figure that out and don't understand how it rises in voltage to such a point after being disconnected. Of course you know more than I do because I'm merely learning and experimenting so I don't mean to seem contentious. I just see what I see and what I am seeing amazes me.
     
  6. MrChips

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    Oct 2, 2009
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    The circuit is a load like a resistor.
    The battery has internal resistance.
    When you connect the circuit to the battery you instantly create a voltage divider.
    The voltage measured across the battery will immediately be less than the unloaded battery.

    Remove the load and you remove the voltage divider. The measured voltage of the unloaded battery will immediately read higher.
     
  7. #12

    Expert

    Nov 30, 2010
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    There is also the chemistry aspect. The chemicals in the battery have become depleted. It takes them more and more time to interact because the useful stuff is so diluted with the used up stuff. If you just lay it down and come back in a couple of days, the voltage will have risen as far as it's going to rise.
     
  8. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    I get that, but how does it keep increasing slowly but steadily nearly to the point or voltage the 'dead' battery held when I first connected it 24 hours ago? It's still rising on the meter and now reads 6.73. In the next minute or so it will be 6.74 and so on.
     
  9. #12

    Expert

    Nov 30, 2010
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    but how does it keep increasing slowly but steadily?

    The chemicals in the battery have become depleted. It takes them more and more time to interact because the useful stuff is so diluted with the used up stuff.
     
  10. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    If something is depleted then shouldn't it decrease as opposed to increasing? I'm now at 6.75 and increasing.
     
  11. #12

    Expert

    Nov 30, 2010
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    The facts are staring you in the face. The MOSTLY depleted chemicals continue having a little bit of activity until they are COMPLETELY depleted. It takes them more and more time to interact because the useful stuff is so diluted with the used up stuff.
     
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  12. #12

    Expert

    Nov 30, 2010
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    When I was 9 years old, I got a flashlight that had allegedly dead batteries. Each day after school I would turn the flashlight on and it would light up for about half a second. I came to the conclusion that the batteries must still have a little bit of good stuff left in them, but it is so little that it takes all day to build up enough energy to light the bulb for half a second.

    This conclusion was confirmed by really smart people that taught college level chemistry.
     
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  13. wayneh

    Expert

    Sep 9, 2010
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    You're seeing a diffusion-limited process. The chemicals are not fully depleted in the battery overall, but they are depleted near the electrodes. To get more reaction and power out of the battery, you have to wait for the reactants to find the electrodes. When current flows, a concentration gradient grows around the electrodes and the available power drops. Power returns as the gradient dissipates.
     
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  14. Rolland B. Heiss

    Thread Starter Member

    Feb 4, 2015
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    Thanks for helping me understand what I saw happening. At times I get overly excited because I'm merely learning from doing and observing as opposed to reading about something someone else already tried, whether success or failure. However, I still read all about what other people have tried as well! Y'all have much more knowledge than I in electronics and all of your input is helping me gain ground (no pun intended). At the moment I'm experimenting with two differing semiconductors in an attempt to create a solar cell but I'll leave that for another thread down the road. :)
     
  15. KLillie

    Member

    May 31, 2014
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    Measure the amperage. I'm sure you'll find IT doesn't keep increasing.
     
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  16. Lundwall_Paul

    Member

    Oct 18, 2011
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    Best answer yet!
     
  17. nsaspook

    AAC Fanatic!

    Aug 27, 2009
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    The charge recover factor in batteries can significantly change the total battery capacity in dynamic conditions. It's also something that overlooked in most capacity monitors. The code from my DIY monitor is from the two well bank diffusion Kinetic Battery model.

    http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CCcQFjAB&url=http://www.researchgate.net/profile/Marijn_Jongerden/publication/220386978_Which_battery_model_to_use/links/00b4952fb854957eeb000000.pdf&ei=yCrYVOyQGJSZoQTisIDwCw&usg=AFQjCNGbYMO9u0sUcKgSwFCM6sAGxJIsaA&bvm=bv.85464276,d.cGU&cad=rja

    Code (Text):
    1.  
    2. hist[z].rest_rate = 0; // stores energy in Ah than can be recovered by battery rest and other factors
    3. hist[z].rest_factor = RFACTOR; // Efficiency factor of battery recovery
    4.  
     
  18. WBahn

    Moderator

    Mar 31, 2012
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    The final unloaded voltage that the battery will recover to can be quite high. But as you drain the battery you will see a number of things: The voltage that the battery drops to for a given load is lower and it drops quicker, and also the voltage that it recovers once unloaded to will be lower and it will take longer.

    This is why determining the state of charge of a battery with an unloaded voltage measurement is almost meaningless for most battery technologies. You need to measure the voltage when under a suitable load.
     
  19. takao21203

    Distinguished Member

    Apr 28, 2012
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    Its only an intermediate stage, some time, the battery goes bad and reads 0.00

    It could leak out and create a shortening path, or it could dry out, or 2) after 1), or it could turn into kind of a Zamboni pile.
    The dry 9v kinds are easy to dismantle, you'll see theres not much reaction mass, most is carbon. It will diffuse out + dry out.

    Its actually true, the metal plates and the carbon as such could react for much longer.
     
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