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I have a battery that states a 2.75v battery discharge cut-off voltage. Is that voltage measured at the poles of the battery while it is connected to the load or they are talking about the electromotive voltage?

For example. A Sony battery I have, when I measure it with a multi-meter it displays 3.61v EMV but when connected on the load it only gives 1.81V at the poles of it. Which voltage I should consider to stop discharging it?

Why such a large difference in voltage?

EDIT: Since you asked, the battery is a Sony US18650GR - O7. It is connected on a custom made circuit with 2 parallel 4.7ohm (5 watt) resistors (2.35 in total) plus some mOhms (maybe 0.5-1 Ohm) due to cables, shunt resistor (current sensing) and connectors resistance. At start it draws something like 1.2 Amps and when the voltage dips at something like 1.81v it draws 600-700 mA. No cut-off (over-discharge) circuit.

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  • \$\begingroup\$ Different cell chemistries have differing charge and discharge requirements. Look up the specific recommendations for that battery. \$\endgroup\$ – DrMoishe Pippik Aug 28 at 21:28
  • \$\begingroup\$ Yes, but they don't state if that voltage is the electromotive force or the voltage when connected to a load. \$\endgroup\$ – ekalyvio Aug 28 at 21:43
  • \$\begingroup\$ @ekalyvio Are you referring to the unloaded (open-circuit voltage) vs the loaded voltage (voltage while supplying current?) If so, it still varies from battery to battery. If it is under load you just have to know what the "dead" voltage is for a given current draw. \$\endgroup\$ – DKNguyen Aug 28 at 22:10
  • \$\begingroup\$ Your battery sounds somewhat "troubled". See test results here for a typical Sony 18650 cell - NOT your one. You c an see the sort of voltage drop you'd expect at various loads - yours is dropping to what sounds like a dangerously low voltage under reasonably modest load. | Measurement conditions are specified by manufacturer (or should be) but "under specified load" is the usual expectation. secondlifestorage.com/image/lqdvib. \$\endgroup\$ – Russell McMahon Aug 29 at 12:13
  • \$\begingroup\$ I am not saying the my cell is in good condition. It was removed from an old laptop battery that I replaced the cells. I also tried an other 'good' semi-charged Samsung ICR18650-26H cell and this again drops the voltage to 3.0-3.1v under load. I guess that it is because the circuit I am using. But... my question is not about the battery but on if the voltage cut-off is measured with the battery under load or it is the electromotive force of it. \$\endgroup\$ – ekalyvio Aug 29 at 12:47
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The assumption is that you are measuring the battery voltage while it is connected to the load. It isn't practical to expect someone to periodically remove the battery from the load to measure its voltage.

As for your Sony battery, you haven't really told us what kind of battery it is or how much current the load is drawing from it. We can't give a good answer without that information. You may be overloading the battery...trying to draw too much current from it. Or, the battery may be faulty.

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    \$\begingroup\$ I wouldn't go so far as to say that's the general assumption. I would generally assume the dead voltage is an open-circuit voltage if it isn't specified one way or the other since the loaded dead voltage requires specifying a current to be of any use. If this is a lithium type battery then 2.75V is far too low to be an open-circuit voltage so I would assume it's a loaded voltage just based off of that. \$\endgroup\$ – DKNguyen Aug 28 at 22:15
  • \$\begingroup\$ A voltage sag of almost 50% under load sounds pretty terrible though. \$\endgroup\$ – Dampmaskin Aug 28 at 22:17
  • \$\begingroup\$ @DKNguyen I would argue that the under-load voltage is the only voltage that really matters, and is the voltage that the battery manufacturer needs to specify. If I'm using a battery I care about its voltage while in use, not after removal from the load. And in my experience, battery manufacturers will specify a load current along with the specification of capacity and terminal voltage. \$\endgroup\$ – Elliot Alderson Aug 28 at 22:21
  • \$\begingroup\$ @ElliotAlderson I agree. But on a lot of the batteries I have nothing is listed at all. \$\endgroup\$ – DKNguyen Aug 28 at 22:34

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