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Ive built a simple circuit on some strip board, just a motor and a micro controller and LED and powered by 6 AAA batteries. The motor takes its power straight from the battery and the micro and LED go through a 5V LDO. I also have a slide switch to turn off the power.

The problem is when the power is turned on the voltage I measure across the batteries drop at a steady rate, about 0.01V a second, if I turn the switch off the voltage jumps up and also starts to climb back up to the real voltage across of the batteries. If I wait long enough the voltage is back to the real voltage of the batteries. If I switch it on the process starts again.

Will there is enough voltage for the LDO the circuit works perfectly so im sure there isnt a short, and from what I can see there isnt a short anywhere. Nothing is getting hot at all either. Im not quite sure what to look for or what it could be..

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  • \$\begingroup\$ Have you got a meter? Can you measure the current? \$\endgroup\$
    – Andy aka
    Mar 19 '14 at 21:44
  • \$\begingroup\$ Batteries do have a practical current limit. Any higher and the internal reaction can't keep up. \$\endgroup\$ Mar 19 '14 at 21:51
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That's how batteries work. As you draw current from them, you deplete the chemical reactants near the plates, and it takes time for new reactants to take their place. This creates a gradient in the reactant concentration within the volume of the battery.

The terminal voltage is a function of both the reactant concentration at the plates and the voltage drop across the internal resistance of the battery (which rises a bit with reactant depletion). If you let it run long enough, the terminal voltage should stabilize at some lower value, although the overall depletion of the battery will eventually become a factor.

When you switch off, the voltage drop across the internal resistance disappears immediately, but it takes some time for the reactant concentration gradient to disappear, which is why you see the sudden jump up followed by the slow recovery to the original terminal voltage.

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  • \$\begingroup\$ Thanks Dave for that!. So does that means my circuit is just drawing to much current for these batteries? The 6 batteries should be around 9V but they are at 7.5V now as they have been used a bit. Would not being at full capacity have an effect in the regeneration, or do I simple need to a new power source? What is this effect called if there is a name and what quantifies it if I was to look in a datasheet. \$\endgroup\$ Mar 19 '14 at 22:08
  • \$\begingroup\$ This effect occurs in all batteries, but you can reduce its effects by choosing batteries that have the capacity to supply your average current consumption for at least 10 to 20 hours. And yes, it sounds like your current set of batteries is nearly depleted, which means that both the overall reactant concentration is low and the internal resistance is high with respect to a fresh set. \$\endgroup\$
    – Dave Tweed
    Mar 19 '14 at 22:13
  • \$\begingroup\$ What I mean is would there be something on the data sheets to show this. Such as maximum currentbdraw per voltage graph. As if your saying when the batteries are at full capacity I wouldn't have this effect possibly. How would I check if the battery is capable \$\endgroup\$ Mar 20 '14 at 9:27
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    \$\begingroup\$ No, this information isn't generally published, because these characteristics are not tightly controlled in the manufacturing process, and the manufacturer does not want to have to guarantee performance in this area. \$\endgroup\$
    – Dave Tweed
    Mar 20 '14 at 12:04
  • \$\begingroup\$ Ah interesting. As everything is soldered to strip board I tried to measure the current by taking a battery out of the holder so there's 5 batteries and then use my meter in place of the 6th battery. I can't remember exactly as I did this yesterday but it sure never went over 100mA, I'm sure it was always under 10mA...just trying to remember the setting I had it on....As I said as well 6 AAA 1.5v and the pack was at 7.5v \$\endgroup\$ Mar 20 '14 at 13:35

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