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Old Aug 07, 2012, 04:19 AM
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Multimeter & IR

Hi,

My apologies if this has been tackled before, but I searched and searched and could not find a simple, straightforward answer to this question. So here goes.

My equipment is barebones (250mAh packs, dedicated balance charger of the simplest kind) and I would rather not buy any new equipment for the time being.

My questions:
- Is there a safe way to measure/calculate the internal resistance of a Li-Po battery pack by using a digital multimeter?
- If there is, what is the method to follow (please understand that my background is in mechanics, and that was over 40 years ago, so make it simple and step-by-step, please...)

Thanks
G.
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Old Aug 07, 2012, 04:41 AM
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Lipo IR cannot be measured with a multimeter

Cheers
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Old Aug 07, 2012, 05:08 AM
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Thanks

G.
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Old Aug 07, 2012, 07:34 AM
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IR can be easily calculated with a multimeter.

You need only a low ohm resistor of known value.

The IR you will measure may or may not be the same as from another IR meter. They seem to not agree with each other more than agree with each other from what I've read. But the important thing to the user is that they will be consistant if the same meter is used at the same temperature.

This is the way I would do it.
#1 - Pick a resistor that will draw about a 5-10C current draw.
#2 - Measure the cell voltage without a load on it.
#3 - Connect the resistor load and measure the voltage after 3 seconds.
#4 - Calculate the IR

IR = (Vu - Vl) X (R/Vl)
Divide this number by the number of cells in the pack to get IR/cell.

Vu = unloaded voltage
Vl = loaded voltage
R = load resistance

This will give you an IR that can be used to grade and monitor your packs health. The value may be off a little or a lot from the "real" IR but it will provide a usable vlaue.

See my set up.

Resistors are avialable from ebay. http://www.ebay.com/sch/i.html?_trks...istor&_sacat=0
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Old Aug 07, 2012, 08:17 AM
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Thank you very much.

So, it would be ok to use a resistor in the range 3 - 5 Ohm?

G.
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Old Aug 07, 2012, 08:25 AM
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Sorry for double post, I tried to edit my previous one, it seems without success.

Why a preload? Is it necessary?

Incidentally - I just want thgis to have a way of following the behaviour of my batteries, not for accurate figures.

G.
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Old Aug 07, 2012, 12:03 PM
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Quote:
Originally Posted by GLFaria View Post
Thank you very much.

So, it would be ok to use a resistor in the range 3 - 5 Ohm?

G.
That would depend on the voltage of your pack.

R = V/I

So for a 240mah pack you want current to be 1.4 amps (.240 * 5)

Then the resistance for a 3 cell pack would be:
R = 12.6 Volts divided by 1.4 amps, or 9 ohms.

The power of the resistor would be
P = I * E or 12.6 * 1.4 or 17.64 watts.

So for a 240 mah 3S battery you would need a 9 ohm 20 watt resistor.

Someone please check my math. I haven't had to do this in a while.

Glen
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Old Aug 07, 2012, 12:03 PM
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Looks good Glen.

Preloading didn't make any difference in my tests. I have deleted that step from my measurements. You just don't want to test a pack right off of the charger. Let it settle down for several hours or do the measurement on semi discharged packs.

How many cell lipos are you testing? Like Glen said, the R will depend on the V.
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Old Aug 07, 2012, 02:37 PM
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The resistor you use is quite important. It must be a very robust one (have a very high wattage rating) or your measurements will vary all over the place as the resistor can change value quite rapidly and drastically if not of the proper construction or if it is to small a wattage rating and has a low mass. The temperature coefficient of the resistor must be very small. You may get more accurate results if you use two resistors and check out the change of IR as you vary the current from around 1c and 10c for the two resistors.
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Old Aug 07, 2012, 05:35 PM
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It's a two cell pack

G.
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Old Aug 07, 2012, 06:15 PM
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Ok, let's assume a pack voltage of 8V from a 250mah pack.
A 5C draw would be 1.25A.
R= E/I
R=8/1.25 = 6.4 ohm resistor.

This 6 ohm 5% 100W resistor would work nicely. http://www.ebay.com/itm/100W-Power-5...item3378ab58e4
$5.42 shipped
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Old Aug 07, 2012, 06:19 PM
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Sorry again

Thank you, people, you have been most helpful and I am learning a lot.

It seems I got it right.
For a two cell pack, nominally (more on this below), and for C between 5 and 10, I calculate the resistor at between 6 (lower C) and 3 Ohm (higher C)

Now, I got my packs pre-charged, and then quite charged indeed - one of them reads 8.25 V, the other one 7.62 V. I waited about one hour for warming after taking them off the fridge (not the freezer!).
Although the manufacturer states "max. load not over 8.3 Volt", at two cell pack should (nominally at least) be at around 7.4V.

Is there any danger for the packs if I wait a couple of days until I have a reasonably weak wind so I can fly my plane (newbie, light plane, putting my faith in WindGuru...). After all, I don't have a clue to how long they have been on the shop's shelf...

Otherwise, I'll have to start collecting lamps - which around here, these days, are not easy to come by in the tungsten type because of environmental concerns.
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Old Aug 07, 2012, 06:40 PM
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OK, let's start at the beginning.

A lipo cell is 1/2 charged at 3.85V/cell and fully charged at 4.2V/cell.

Lipo packs when received should be right around 3.85V/cell. If they are much higher it is probably a used pack that has been charged. If it is much lower, it might be a bad cell.

A fully charged pack should read between 8.2 and 8.4V

Packs should be stored at 1/2 charge, 3.85V/cell.

Your pack at 8.25V is 100% charged. Your pack at 7.62V is a little less than 50% charged.

Good idea to wait for calm winds to fly. You might wish to discharge that 8.25V to less than 8V and stick it back in the fridge.

Questions?
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Old Aug 07, 2012, 06:54 PM
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Thank you, just one question if you don't mind:

"...pack at 7.62V is a little less than 50% charged."?

Isn't that 92%?

BTW, I bought the packs as being new, in sealed plastic bags.
Of course, if big banks make big scams, then maybe not so big manufacturers or shops feel free to make not so big scams...

G.
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Old Aug 07, 2012, 06:55 PM
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To add to what Hoppy said. Be sure to charge your batteries before use. The day of flight would be best. After flying charge/discharge the batteries to about 3.85 Volts per cell. This is referred to as Storage Charge.

Please read this:
https://sites.google.com/site/tjingu...arging-how-tos

Glen
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Old Aug 07, 2012, 07:14 PM
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Quoting myself:
"Isn't that 92%?"

Ah, sorry, we are talking Volts, not mAh

Now off to bed, it's past 1 AM here and I must be up at 7AM.

Thanks for all, you are a wonderful bunch

G.
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Old Aug 07, 2012, 07:15 PM
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Quote:
Originally Posted by GLFaria View Post
Thank you, just one question if you don't mind:
"...pack at 7.62V is a little less than 50% charged."?
Isn't that 92%?

BTW, I bought the packs as being new, in sealed plastic bags.
Of course, if big banks make big scams, then maybe not so big manufacturers or shops feel free to make not so big scams...

G.
Full charge is 4.2 v/cell
Full discharge is 3.0 v/cell
Pack at 7.62 = 3.81 v/cell = about 50%

I repeat...Please read this:
https://sites.google.com/site/tjingu...arging-how-tos

Glen
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Old Aug 07, 2012, 07:28 PM
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Quote:
Originally Posted by GLFaria View Post

BTW, I bought the packs as being new, in sealed plastic bags.
Of course, if big banks make big scams, then maybe not so big manufacturers or shops feel free to make not so big scams...

G.
Manufactures deliver packs at about 3.85v/cell. Never at 4V+. If that pack was at 8.25v, I'd take it back as it's been charged and leaving a pack fully charged damages it over time.

It may also have many uses under it's belt.
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Old Aug 07, 2012, 08:02 PM
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Another way to measure it, if you have a reasonable charger, is:

a) Charge at 1A until about mid-charge (3.8Vish. IR changes with charge state, so try to be consistant)
b) Note the voltage. Increase the charge rate to 2A, and note the voltage
c) Drop the current back to 1A, and again note the voltage.

Using V=IR, dV=dIR too, so:
From step a, V=3.80, I=1A.
b, V=3.91, I=2A
c, V=3.82, I=1A

The reason we take two changes is because the battery is charging the whole time, so this will somewhat cancel that out.

From a to b, dV=0.11V, dI=1A. From b to c, dV=-0.11, dI=-1.0A. a-b gives an IR of 0.11oms, b-c gives 0.11ohms. Take an average, and away you go.

Again, this may or may not line up with the IR other methods give. I only use my charger's inbuilt IR measurement now, but it's only useful to compare to other PL8 user's results...
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Old Aug 07, 2012, 09:37 PM
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Neat, I'll give that a shot and compare to the method I've been using.
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Old Aug 08, 2012, 06:00 AM
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Alas, I do not have a reasonable charger... Maybe later.

I guess for now the only way I have of (very roughly) giving my batts a given charge will be by charting V against time as they are being charged, starting from a given level. Rough, but it may just work.

Glen,
"I repeat...Please read this:
https://sites.google.com/site/tjingu...arging-how-tos"

thanks for the advice and the tip, I started reading it yesterday, but it was too late to read it all, and besides my personal available memory is no longer what it was some years ago. Not that I am so old, it must be the mileage
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Old Aug 08, 2012, 01:39 PM
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One more comment GLF,
The 100% charge = 4.15-4.2V/cell is with cell/pack off the charger for awhile.

The voltage of the pack under charge will reach 8.4V for a 2s well before the pack is fully charged. These lipo chargers operate under the constant current/constant voltage principle.
They will charge at the full current setting until the voltage reaches 4.2V/cell at which time the current will start to drop but the voltage will stay at 4.2v/cell until the current drops to a low level.
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Old Aug 08, 2012, 02:37 PM
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Quote:
Originally Posted by Odysis View Post
Another way to measure it, if you have a reasonable charger, is:

a) Charge at 1A until about mid-charge (3.8Vish. IR changes with charge state, so try to be consistant)
b) Note the voltage. Increase the charge rate to 2A, and note the voltage
c) Drop the current back to 1A, and again note the voltage.

Using V=IR, dV=dIR too, so:
From step a, V=3.80, I=1A.
b, V=3.91, I=2A
c, V=3.82, I=1A

The reason we take two changes is because the battery is charging the whole time, so this will somewhat cancel that out.

From a to b, dV=0.11V, dI=1A. From b to c, dV=-0.11, dI=-1.0A. a-b gives an IR of 0.11oms, b-c gives 0.11ohms. Take an average, and away you go.

Again, this may or may not line up with the IR other methods give. I only use my charger's inbuilt IR measurement now, but it's only useful to compare to other PL8 user's results...
My comparison test -
Resistor - Charger Charger/5
12______67________13.4
14______70________14
6_______45________9
46_____103________21
40_____110________22

Like the resistor test, it differentiates between the good and bad packs. Like Odysis said, just use the numbers compared to each other rather than compared to other methods.

Thanks for the idea.
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Old Aug 09, 2012, 06:01 AM
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Originally Posted by hoppy View Post
One more comment GLF,
The 100% charge = 4.15-4.2V/cell is with cell/pack off the charger for awhile.

The voltage of the pack under charge will reach 8.4V for a 2s well before the pack is fully charged. These lipo chargers operate under the constant current/constant voltage principle.
They will charge at the full current setting until the voltage reaches 4.2V/cell at which time the current will start to drop but the voltage will stay at 4.2v/cell until the current drops to a low level.
Thanks. I waited for two or three minutes after taking the pack off the charger before taking that mesurement, just to be sure.

G.
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