Why Li-po battery voltage drop

I am testing for a high discharge rate battery pack. The specification of the battery pack shows it is 3S (11.1V), 5000mAh with 35C max discharge rate. I tested it by fully charged it and then discharge it by an electronic load at 20A. I found that the voltage of the battery pack dropped about 1-1.5V when I discharge it at 20A. Then it return back to normal voltage when I stop discharging it.

I would like to ask is it normal? As the protection circuit will cut off the power when the battery dropped to about 9V, the battery power will be cut when the battery pack still has plenty of charge (the battery pack is in fact at voltage 10-10.5V). Can it be solved?

Thank you very much.

eepty said:

I found that the voltage of the battery pack dropped about 1-1.5V when I discharge it at 20A. Then it return back to normal voltage when I stop discharging it.

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It means the pack has internal resistance, a fraction of an ohm. Sending Amperes through a resistance causes a voltage to develop across the resistance. This voltage is subtracted from the voltage of the battery pack. You're losing 20 or 30W in the battery pack simply by drawing 20A. The wasted power generates heat, and heat is bad for batteries as a general rule.

BradtheRad said:

It means the pack has internal resistance, a fraction of an ohm. Sending Amperes through a resistance causes a voltage to develop across the resistance. This voltage is subtracted from the voltage of the battery pack. You're losing 20 or 30W in the battery pack simply by drawing 20A. The wasted power generates heat, and heat is bad for batteries as a general rule.

Click to expand...

Thank you. I have also tested the temperature of the battery pack when I discharge it at about 22A. The battery pack has a soft enclosure, I put the temperature sensors at 4 different positions. 3 of them on the soft enclosure, 1 of them on the protection board. I found the temperature at the soft enclosure rises about 8 degC, but the protection board rises about 50 degC.

Is there any technique to solve this problem? We cut the power when the battery voltage dropped below 9V to avoid damaging the battery. However it seems that the actual voltage of the battery is not really 9V, because when I stop discharging it, it goes back to 10.5V . I hope to use the power of the battery as much as possible so that I have a longer usage time.

Can I calculate the battery internal resistance by 10.5V - 9V / 22A = 0.681ohm. Then calculate the real battery voltage by Vbat = V(measured) + (discharge current x 0.681)?

Thank you very much.

Hi,

Your treating the battery may be risky. You risk explosion and fire.

It seems you need some basic knowledge, thus I recommend to read through some battery basics.
Every battery manufacturer should provide such informations.
There are websites with dedicated informations, like:

Also read every datasheet of every battery you (want to) use.
Every battery will have it´s own limits.

Klaus

The battery's voltage reading always drops when you attach a severe load.
We are not necessarily able to tell whether the drop is due to internal resistance, or whether the battery chemistry isn't robust enough to produce all the power you're asking.

The common solution is to try a battery pack with greater Amp-hour capacity.
It is feasible to add a second pack in parallel, however greater care is needed so that you avoid imbalance between the packs.

As it is your load appears to be 200 Watts which generates as much heat as two 100W light bulbs. No wonder your pcb goes up 50 deg C. It's important to keep all contacts clean, and to use wires that are thick enough gauge so they offer miniscule resistance to 22A.