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How healthy are your batteries? (How to measure internal resistance)

Hemlock_Mike

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Perhaps this is a time to introduce the unit Mho. ie: 1/R -----
7 to 13 Mho , the cell is excellant and 2.9 to 4 Mho, it's getting bad.

Using this measure gives a high numerical rating for good cells vs low numbers for degraded cells..

HMike

I found a 3 Ohm 5 Watt ceramic resistor. This will give me a ~1.3 Amp load. This is short of the 2+ Amp loads often required in 445's and big greens but suitable for this cell testing.
 
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Hemlock_Mike

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A follow-up on the above question------------ Using a 5 Watt 3.1 Ohm WW resistor----

I tested two older unprotected cells and read : 200 and 180 mOhms. Not bad

Two newer protected cells read : 270 and 220 mOhms.

Does the protection circuit affect the readings? Can they affect the reading by 50 mOhms?

HMike

In retrospect, 50 mOhms is about right for good MosFets. While observed, this is not likely to be a deciding factor on the cell quality.
 
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LarryDFW

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Mike;

Protection circuits can easily add 50 mOhms to 80 mOhms.

They will cause more voltage drop on high current lasers.

LarryDFW
 
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HIMNL9

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I have few of these protection circuits (from DX), and all of them uses a power mosfet as switch, connected in serie to the negative pole of the battery ..... 50 milliohm is a common value, for a mosfet RdsON value .....
 

Hemlock_Mike

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DrLava --
Your equation bothered me for over a day. It didn't seem to fit Ohm's law.
Today I realized --- multiplication by the inverse of I (* R / v2)..

Tricky of you!!!

HMike
 

Cyparagon

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Wow, I've got quite a variety. Tested with 2.7Ω 10W power resistor:

Powerplus: 160mΩ
ultrafire: 120mΩ
2xTenergy (protected): 300mΩ and 370mΩ
26 Cells pulled from old laptop batteries range from 370mΩ to 780mΩ averaging maybe 500mΩ
A123 LiFePO: as low as 65mΩ
 

drlava

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DrLava --
Your equation bothered me for over a day. It didn't seem to fit Ohm's law.
Today I realized --- multiplication by the inverse of I (* R / v2)..
You're right; it's a simplification to make the equation easier to write. I (battery) = V2/R
The equation could also be:
Ri = (V1/V2 - 1)*R


Awesome dude, +1 for you! :D
thank you!

Wow, I've got quite a variety. Tested with 2.7Ω 10W power resistor:

Powerplus: 160mΩ
ultrafire: 120mΩ
2xTenergy (protected): 300mΩ and 370mΩ
26 Cells pulled from old laptop batteries range from 370mΩ to 780mΩ averaging maybe 500mΩ
A123 LiFePO: as low as 65mΩ
Interesting about the LiFePo! suggests that a 1W+ 445 could run from a single A123 sized cell host!
 

Toke

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The LiFePo4 18650 is supposed to handle up to 30A in peaks, and their use is RC gear with discharge times in minutes support that.

It is quite likely that if a 18650 can deliver up to 100W, a A123 can easily handle the 6-10W of a 445nm. The only downside is that the LiFePo4 have a lower (½) capacity than the regular lithium batteries.
 

HIMNL9

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.....
Ri = (V1-V2)*R/V2
Then, it can be derived also for the difference of the voltages, divided for the current under load, right ?

I mean, as (V1-V2) / (V2/R)

Was wondering to build an instrument, just for the curiosity, but the double cycle reading present some problems, for analog instruments (analog sample and hold is a PITA :p) ..... maybe is much more easy with a microcontroller ..... must be enough one with a pair of A/D input and an output ..... and a high current relais, maybe a mercury-wetted reed relais, or a power hexfet, taking in consideration also its RdsOn (also if, for some models, it can be ignored, being less than 10 milliohms ..... i have still some IRF3205 with 8 milliohms of RdsON) ..... For precision maniacs must be took in consideration also the temperature of the battery, i know (Ri increase when temp decrease), but at our level, i suppose it can be ignored .....

2 seconds reading cycles must be enough for left the load curve stabilize, starting with a 5 / 10 mA or similar load for "stabilize" the battery ..... let me see how can be an automatic measure cycle for the MCU ..... detect the connection of the battery (input voltage rise over 0,5V , as example) -> wait 1 second at low current load for stabilize -> take the v1 measure and store it -> close the high current load on the battery -> wait 1 second -> take the measure of V2 and store it - take the measure of the current (on the high current load) and store it -> disconnect the load -> calculate and show on the display ..... must not be so difficult, for the microcontrollers-maniacs .....

(just another of my crazy ideas, don't care about it :p :D)
 

drlava

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Yep, it could be done with a micro and display. The common 10-bit A/D's built in would give about a 5mOhm resolution with a simple setup and <1mOhm with some scaling.
 

LtKernelPanic

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If you aren't measuring an 18650, choose a load resistance that will load the cell but not overload it. Capacity/3 should be ok.
I hate to ask such a noob question but what do you mean by that? Does that tell me how many ohms of resistance that I should use? What would I use for 350mA 10440s?

I understood everything else you mentioned but that part threw me for a loop.
 

Hemlock_Mike

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I've logged back in for a moment.
My 10440 cells are rated at 500 mAHr. C/3 means just that -- 500/3. Use Ohm's law to determine the resistor to provide 1/3 rd the current capacity of your cells at full charge voltage.
It's that simple.

HMike
 

drlava

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Mike is correct, to make it real simple, assuming you are using a fully charged '3.6V' battery, to target C/3 use a resistor around:

R = 12.6/C

Where C is the battery capacity in amp-hours (in his example, C would be 0.5 and R would be about 25 Ohms)
 

Cyparagon

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Interesting about the LiFePo! suggests that a 1W+ 445 could run from a single A123 sized cell host!
I think you're confusing CR123A (the size) with A123 (the company).

But yes, assuming 500mAh with a discharge of 30C, a CR123A LiFePO battery would provide up to 15A (continuous 50 watts or so?) But it would last less than 2 minutes under that load. Powering a 445 at 1A, it would last about 20min if my calculations are correct.
 

Toke

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I think you're confusing CR123A (the size) with A123 (the company).

But yes, assuming 500mAh with a discharge of 30C, a CR123A LiFePO battery would provide up to 15A (continuous 50 watts or so?) But it would last less than 2 minutes under that load. Powering a 445 at 1A, it would last about 20min if my calculations are correct.
Pretty close, you are just overlooking the voltage requirement of 5-6V.
The CR123 would have to provide app. 2A at 3.3V to a boost driver to get 1A at 5-6V.

So maybe 10min.
 




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