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Battery Voltage getting dropped during discharging

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I'm having a 4.2V 2A Li-ion battery.

Just to understand how the battery behaves when discharging, I connected the 4.2V Li-ion battery across this DC Electronic load and set the current limit to 500mA on the electronic load.

When I turned on the DC Load, I saw the voltage on the Electronic load to be around 2.4V while discharging and the current consumed by the load is 500mA.

I was confused. This should not happen, right?

The battery voltage should be constant right since the current limit is less than 2A (the battery capacity)? Maybe, I would have accepted the 2.4V if the current was set more than 2A. But it is not the case.

Please let me know whether this behavior is correct or if there's something wrong with my understanding.
 

Hi,

I guess 2A is wrong at all. It should be 2Ah. It's not capacity, it's not current, it's charge.

500mA is the current.
There are batteries with 2Ah, which can easily supply 500mA, and there are 2Ah batteries where you are not allowed to draw 500mA.
As so often: the datasheet should tell.

Connecting a load to a battery will always drop the voltage. In short time (seconds) according Ohm's law (battery internal resistance and wiring resistance) ... long time (over hours) according the charge state.

All this is well known and you can find a lot of information in the internet. High quality information is provided by the battery manufacturers.
(They know best how their batteries work)

Klaus
 

I agree, the question is almost useless without a battery specification. If it's an industry standard cell, I will allow 1C, more likely 2C discharge current with moderate voltage drop. You'll refer to the datasheet in case of doubt. Means your cell is either discharged, defective or non-standard.
 

Why do you wrongly think that the discharging battery voltage will stay constant?

You have ruined the battery cell to allow its voltage to drop too low.
Below 3.2V its capacity is reduced permanently.
Below 3.0V metallic lithium might form inside and cause it to catch on fire and/or explode if it is fast charged.
 

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