Electric bike: diagnosis of a lithium-ion battery 24V

[Christophe]

OK so much the better !

CPGE I went there, so I had to learn that ... a distant day! But we are not all lucky to be able to stay at school like you Remundo

Out of curiosity, I measured the Ri of the 1 element on 3 couples of U / I value and I get:

1.42 Ohm
1.42 Ohm
1.63 Ohm

Looks like it does ... a lot!
It's serious ?

[Remundo]

yes, it is very high for Lithium.

Are you sure of your measurements?

[Christophe]

Ben unfortunately yes, that's why I made several measures ... but, small detail: they are U / I couples in charge, not in discharge, it's good too?

In discharge I have to find a charge to 4V ...

[Remundo]

without schema, not easy to answer you,

let's say that the charging voltage is Ucharge = E + R xi

where E is the electromotive force of the battery (voltage measured at no load)

thus R = (Ucharge -E) / i

maybe you have a sign error on E?

You should also know that the batteries are "living beings": there are internal reorganizations (migration of chemical species) which vary the no-load voltage E just after a charge or a discharge, in particular if they were strong. .

[Christophe]

No need for a diagram: a power supply 1 Lithium element ...

And I applied this formula: https://www.econologie.com/forums/post208024.html#208024

With your formula I get 1.3 Ohm ... (almost constant)

In any case it's good too, I'll compare with a healthy element ...

[Remundo]

Ah OK, you avoid measuring E by doing 2 measurements under 2 different powers.

You are lucky because the formula remains valid in charge AND in discharge

The idea is:
in discharge: V1 = E - R i1 and V2 = E - Ri2
in charge: V1 = E + R i1 and V2 = E + Ri2

In 2 cases, by subtracting and then extracting R, we have R = | V1 - V2 | / | I1 - I2 |

Formulas, one must always know where they come from to know if one can or not apply them.

Otherwise, your battery has suffered a lot, the internal resistances of Lithium are rather 0.1 Ohm, or less

[Christophe]

There was no reason that it does not apply according to the curve

By cons I do not explain the 0.1 difference ... surely tolerance on the measures? (E not measured with same voltmeter as U and I measured them on the power supply)

But I will still do the same measures in discharge because there I was above 4.1 V (not long but above anyway) ... which increases can be R ??

Otherwise, your battery has suffered a lot, the internal resistances of Lithium are rather 0.1 Ohm, or less

Yet she did not do much ...

So for the rest, possible solutions:
a) change the element (s) lithium HS
b) hybridize the battery
c) 2

[Remundo]

There was no reason that it does not apply according to the curve

By cons I do not explain the 0.1 difference ... surely tolerance on the measures? (E not measured with same voltmeter as U and I measured them on the power supply)

Your formula is a report of 2 differences.

If you have 2 or 3% error on each measurement (typical with a good multimeter) of I or U, in difference, you cumulate about 5%

then, the ratio of the 2 differences doubles the relative uncertainty again.

So you arrive at a value of R +/- 10%

Since R is in the 1 Ohm, 0.1 Ohm remains in the measurement error, there is nothing alarming.

[Christophe]

Not alarming and yes it must come from this ... I like to understand ...

[Christophe]

Article to read on the reliability of automotive lithium batteries: https://www.econologie.com/forums/post208169.html#208169 (by a researcher at the CNRS)