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I am using AA LI cells for an embedded system project.This is a low power embedded project with various modules consuming current in pulses. Recently one of the prototype board stopped working. Although I measured the voltage across the AA measured 1.6V, it still does not work. How should I go debugging the problem and find the root cause. Can the AA cell be bad although it shows 1.6V?

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  • \$\begingroup\$ 1.6V is 1.6V. If your circuit is supposed to run on that, then the problem is elsewhere. As for how to debug, there is no way we can answer that. It depends on your circuit, of course, and it would be far to broad to talk about this in general. \$\endgroup\$ – Olin Lathrop Nov 4 '12 at 13:11
  • \$\begingroup\$ Thanks. Yes the circuit is supposed to work with the specified voltage. My question is ,in general,can a AA cell be bad although the measured voltage across it is perfect. \$\endgroup\$ – Elsa Adams Nov 4 '12 at 13:24
  • \$\begingroup\$ As I said, volts is volts. The only issue might be how much the volts goes down as load is applied. However, if the voltage is always 1.6 and your circuit is supposed to run on that, then the battery is not the issue. Make sure it is really 1.6V when the circuit is drawing its maximum current though. A bad battery could have good open circuit voltage but drop fast as load is applied. \$\endgroup\$ – Olin Lathrop Nov 4 '12 at 14:04
  • \$\begingroup\$ That is a very good point. Thanks for your suggestion. I will do and update you the result.Thanks again \$\endgroup\$ – Elsa Adams Nov 4 '12 at 14:07
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As a battery ages its internal resistance increases. Therefore, measuring the voltage on an 'idle' battery (i.e. 0 current) will not always yield the same as it does when current is actually drawn from the battery.

If you have 'modules consuming current in pulses' it may well be the case that, during those pulses, the battery's voltage will drop below its 'idle voltage' significantly.

Use an oscilloscope to measure the actual voltage during a pulse.

You may also want to put a capacitor in parallel to the battery. This decreases the overall impedance of your power source for short pulses significantly and will almost always yield a much longer usable time for any battery being used pulsed.

To get a rough (optimistic) estimation of a battery's response to a current pulse, one may measure the voltage during a somewhat 'prolonged' pulse of the same current. - Defining 'prolonged' here is up to the reader, but the point actually is: If a battery's voltage drops below a certain level under constant (current) load I, then it probably will do (even more) so for pulsed loads of the same current I.

If you know that your pulses are 1A max., just put a 1A constant current load (read: resistor) across the battery and take a voltage reading while the current flows. This will definitely reveal a dead battery better than its idle voltage.

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  • \$\begingroup\$ ok thanks.I will do it and give you updates. Is there any other tests that I can conduct to learn the effect on battery due to pulsed current consumptions. \$\endgroup\$ – Elsa Adams Nov 4 '12 at 14:00
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What makes you assume the problem is the battery in the first place?

From the datasheet of the L91, the nominal voltage is 1.5V (not sure where you get 1.7V) and the graphs show a maximum of 1.6V under a small load (50mA) which drops quickly to level out at around 1.5V. The cells are also capable of handling large current pulses (up to 5A)

Given the above it would appear the battery is likely to be okay, but as Hanno says it needs to be tested under load. If all looks well then it's a problem with your module. Also a simple way of making sure it's not the battery is to use a known good source e.g. a bench supply to power the suspect module.

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  • \$\begingroup\$ Sorry for the confusion. It is an Energizer L91 and it supposed to provide 1.7V \$\endgroup\$ – Elsa Adams Nov 4 '12 at 13:22
  • \$\begingroup\$ Okay, I updated the answer. \$\endgroup\$ – Oli Glaser Nov 5 '12 at 0:09

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