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

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by Christophe » 12/07/11, 20:24

Remundo wrote:the risk is that the internal resistances of the batteries are low.

Eg 0.1 V difference for 0,01 Ohm will generate 10A

If you have 1V, it will be 100 A, it can grill one of the batteries, and at least melt the connection wires.

We can overcome this problem by placing a resistance of 10 Ohm, the batteries will balance gently (0.1 A if 1V difference), this resistance can be shunted thereafter as long as the batteries remain connected in parallel.


Oh yes it's a big risk! Can you briefer us a little bit on this?

Am a ball in this area, how is it measured?

I tested directly to the ohmmeter and I have this:

On the lithium I have on the general output between the 0 and 24V:
Infinite when the battery is on
0 by reversing the polarity

10.46 kOhm when off (????)

On 4 lead-acid batteries, I have infinity or zero depending on the polarity of the measurement.

Infinite when the polarity is respected (red on the +)

It's good I can wire? : Cheesy:
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by Macro » 13/07/11, 08:21

Christophe wrote:
Oh yes it's a big risk! Can you briefer us a little bit on this?

Am a ball in this area, how is it measured?

I tested directly to the ohmmeter and I have this:

On the lithium I have on the general output between the 0 and 24V:
Infinite when the battery is on
0 by reversing the polarity

10.46 kOhm when off (????)

On 4 lead-acid batteries, I have infinity or zero depending on the polarity of the measurement.

Infinite when the polarity is respected (red on the +)

It's good I can wire? : Cheesy:


You measured the internal resistance of batteries to the ohmetre ??? And he did not slam ???
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by Christophe » 13/07/11, 10:13

Uh thin ... I doubted it was not so easy ... : Mrgreen: : Mrgreen:

No it has not slammed, it's a semi pro: ITC-777 auto mode range, so there may be specific protections on this model (for neuneus like me :D ) ??? No idea...

So my values ​​are good or not?
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by bamboo » 13/07/11, 10:17

Christophe wrote:
Remundo wrote:the risk is that the internal resistances of the batteries are low.

Eg 0.1 V difference for 0,01 Ohm will generate 10A

If you have 1V, it will be 100 A, it can grill one of the batteries, and at least melt the connection wires.

We can overcome this problem by placing a resistance of 10 Ohm, the batteries will balance gently (0.1 A if 1V difference), this resistance can be shunted thereafter as long as the batteries remain connected in parallel.


Oh yes it's a big risk! Can you briefer us a little bit on this?

Am a ball in this area, how is it measured?

I tested directly to the ohmmeter and I have this:

On the lithium I have on the general output between the 0 and 24V:
Infinite when the battery is on
0 by reversing the polarity

10.46 kOhm when off (????)

On 4 lead-acid batteries, I have infinity or zero depending on the polarity of the measurement.

Infinite when the polarity is respected (red on the +)

It's good I can wire? : Cheesy:


Christophe, Macro is right, internal resistance is not measured like that and you're lucky that you multi-meter is still this world ...

To measure the internal resistance, you have to charge the battery and measure voltage and intensity. Then measure the no-load voltage.

No time to make a diagram, but try to represent the internal resistance and the external resistor in series on the battery.

Take the following conventions:
Ui= internal voltage (not measurable directly)
Ri= Internal resistance (not measurable)
Ue= voltage on an external load Re
Ie = load intensity

The no-load voltage gives the internal voltage of the battery (Ui) since the internal resistance (Ri) does not take up voltage (because I = 0)

Then measure Ue and Ie
Ui = Ri * Ie + Ue
So Ri = (Ui-Ue) / Ie
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by Christophe » 13/07/11, 10:36

Ok thank you Indy, I understood the method except how to have Ui? You say that Ui is not measurable but that it must be measured by the internal tension? Hu?

Is there a special value and indicated for Re (in relation to the capacity of the battery?) Or can we take what we want?
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by bamboo » 13/07/11, 10:55

Christophe wrote:Ok thank you Indy, I understood the method except how to have Ui? You say that Ui is not measurable but that it must be measured by the internal tension? Hu?

Is there a special value and indicated for Re (in relation to the capacity of the battery?) Or can we take what we want?

Ui is, beforehand not measurable because it's internal, but in fact, when you put your multimeter, you have Re = infinite and Ie = 0
=> suddenly, Ui = Ue empty

Clear, not clear?

The load resistance Re depends mostly on what you have in stock :D
The ideal is that it is relatively small (close to Ri), but you will have too many watts to clear and so you will burn it ...
example: 24 ohms is already huge compared to Ri and yet it will release you 24W, but it would surprise me that you have a resistance able to clear it without blackening ... : Cheesy:

EDIT : if you take the elements 1 by 1 you will have a better indication. First of all you will have more choices to take Re and moreover it will tell you if an element has a stronger internal resistance.
What will be a good indication of the disruptive element
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by I Citro » 13/07/11, 11:51

indy49 wrote:The load resistance Re depends mostly on what you have in stock :D
The ideal is that it is relatively small (close to Ri), but you will have too many watts to clear and so you will burn it ...
example: 24 ohms is already huge compared to Ri and yet it will release you 24W, but it would surprise me that you have a resistance able to clear it without blackening ... : Cheesy:
I use to unload my batteries (balancing elements) an assembly based on 12V iodine ampoules. Yes, it gives off heat ... and it lights up.
: Lol:
Dirk Pitt uses, for its part, threaded rods 1 meter connected in series, as a resistance. : Idea:
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by dirk pitt » 13/07/11, 12:04

small picture of my battery bank SAFT 6V
it's about 50A 300W
it is 4 M4 threaded rods in series hanging from the ceiling of my garage.
Image

to be more precise on the internal resistance, I advise to take between two values ​​of different discharge current rather than the vacuum voltage as one of the two points because it is not linear.

Rinterne = (U1-U2) / (I1-I2)

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by Christophe » 13/07/11, 12:26

dirk pitt wrote:to be more precise on the internal resistance, I advise to take between two values ​​of different discharge current rather than the vacuum voltage as one of the two points because it is not linear.

Rinterne = (U1-U2) / (I1-I2)

Image


Ah, that's much clearer! Thanks Pitt!

Will do the manipulation in the afternoon.
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by dirk pitt » 13/07/11, 12:29

take the opportunity to measure the individual tensions of the elements under discharge, you may be surprised ....
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