Samsung ICR18650-22F Cell Specifications

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Warning: The information in this thread was obtained from various sources on the Internet, including any datasheets linked below, and is provided for reference only. It is not guaranteed to be accurate. To prevent fire or personal injury, never charge or discharge a cell before verifying the information yourself using the original specifications sheet provided by the manufacturer.

Brand:Samsung
Model:ICR18650-22F
Capacity:2200mAh Rated
Voltage:3.60V Nominal
Charging:4.20V Maximum
1100mA Standard
2200mA Maximum
Discharging:2.75V Cutoff
440mA Standard
4400mA Maximum
Description:Green Cell Wrapper
White Insulator Ring
18650 Form Factor


Data References:
http://www.tme.eu/de/Document/71b610093f14a53a8417944008628183/ICR18650-22F.pdf

Pictures:

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Dear Friend
I removed my Toshiba laptop battery for replace due expiring of battery. But the matter is above battery difficult to findin the market now.
Pls let me know any compatible battery type by Panasonic and LG for Samsung ICR18650 22f.[diy]undefined[/diy]
 
These cells are commonly found in Aldi Infinty 20 V Drill Batteries - 5S2P
I have recovered 70 from new batteries and Capacity/Resistance tested them 3 times with the Lii500.
Average Capacity tested from4.1 volts - 3 volts - 4.1 volts x 3= 2144 Mah
Average Internal Resistance =54 mOhms
Average final voltage = 4.12 V

These are going into a4S16P battery for 5050 LED strings and security camera.

Hope this helps someone.
cheers
emmsee :D
 
standard discharge 2200mA, not "440mA Standard"
 
These cells are commonly found in Aldi Infinty 20 V Drill Batteries - 5S2P
I have recovered 70 from new batteries and Capacity/Resistance tested them 3 times with the Lii500.
Average Capacity tested from4.1 volts - 3 volts - 4.1 volts x 3= 2144 Mah
Average Internal Resistance =54 mOhms
Average final voltage = 4.12 V

These are going into a4S16P battery for 5050 LED strings and security camera.

Hope this helps someone.
cheers
emmsee :D
Hi I'm just starting out gathering batteries to make a power wall or something along those lines, can you mix different types of 18650 cells??
 

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Hi I'm just starting out gathering batteries to make a power wall or something along those lines, can you mix different types of 18650 cells??
Yes you can mix different types and for low-quality packs its perfectly fine. However its not ideal and you'll encounter various issues.

Not every 18650 is the same, they have various discharge curves and just matching mAh will result in a badly matched pack. You'd want to match the mWh of each cell in your used voltage range, but such testers are expensive, so most go for mAh matched instead. But that will loose you roughly 5% useable capacity.

Here you can compare two cells with each other: https://lygte-info.dk/review/batteries2012/Common18650comparator.php

You must also keep the weakest link of the chain in mind. If you build a pack with 1x 500mA discharge and 9x 1000mA discharge cells in parallel, your maximum current should be limited to the weakest cell * the amount of cells in parallel.

Also, dont forget to use a BMS with atleast a weak Balancer, or else you'll risk ruining your batteries by overdischarging them or possibly starting a fire by overcharging them.
 
Yes you can mix different types and for low-quality packs its perfectly fine. However its not ideal and you'll encounter various issues.

Not every 18650 is the same, they have various discharge curves and just matching mAh will result in a badly matched pack. You'd want to match the mWh of each cell in your used voltage range, but such testers are expensive, so most go for mAh matched instead. But that will loose you roughly 5% useable capacity.

Here you can compare two cells with each other: https://lygte-info.dk/review/batteries2012/Common18650comparator.php

You must also keep the weakest link of the chain in mind. If you build a pack with 1x 500mA discharge and 9x 1000mA discharge cells in parallel, your maximum current should be limited to the weakest cell * the amount of cells in parallel.

Also, dont forget to use a BMS with atleast a weak Balancer, or else you'll risk ruining your batteries by overdischarging them or possibly starting a fire by overcharging them.
Thanks for info, I've got an opus Bt-c3100 charger/tester still finding my way in how to use it.
I have been charging all the batteries first and leaving them for a week to see which ones loose the most voltage, maybe not very scientific, but thought it was the best approach to start with..
 

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You must also keep the weakest link of the chain in mind. If you build a pack with 1x 500mA discharge and 9x 1000mA discharge cells in parallel, your maximum current should be limited to the weakest cell * the amount of cells in parallel.
This is not true
Are you confusing parallel with serial?
When connected in series, the capacity of the battery will be equal to the capacity of the weakest cell.
But with parallel capacitance and currents simply add up.
 
This is not true
Are you confusing parallel with serial?
When connected in series, the capacity of the battery will be equal to the capacity of the weakest cell.
But with parallel capacitance and currents simply add up.
Current, amperage output, is dependent on the number of cells connected in parallel. This is what I think they are referring to. The weakest link when connecting multiple cells in parallel is really dependent on the cell(s) that have the lowest max current output capability. So the max current output should not really exceed the lowest capability cell(s) * the number in parallel.
For example, if the lowest is 0.5C of 2000mAh cells (which would be 1A max), and there's 10p and the others are rated at 1C of 2000mAh (which is 2A), then the max output amps for the whole pack should be no higher than 10 * 1A (the lowest), or 10A. Regardless if the other cells are capable of outputting at a higher ampacity.
This is because pulling too much will cause the lower rated cells to try to discharge harder than they are rated, or it could cause momentary imbalance in the pack. The lower rated cells would actually have a voltage drop compared to the other cells "during the load". Once the load is removed, then the other cells will flood into the weaker cell to make equilibrium again. So that weaker cell(s) will basically be having an inrush and outrush like waves into a dry dock when the doors are opened until balance is reached. This could lead to that lower rated cell(s) to come to end of life sooner and possibly make them to go short circuit becoming a heater.
 
Sorry, much-esteemed Korishan, You are not right
Current, amperage output, is dependent on the number of cells connected in parallel.
Yes, sure!
The output current in the example of scottmx would be 9.5 amps. A weak cell will not give more than its 0.5A without the voltage dropping across it, and it cannot drop because the connection is parallel.
This is easy to verify experimentally.

After removing the load, no equalizing currents also flow. Equalization currents flow, if there is a difference in the potentials of individual cells that are connected in parallel - only at the first moment - with a direct connection. By the way, the equalizing currents are not so large - within the limits acceptable for the majority cells - you can calculate and try experimentally (I specifically investigated this issue) -
no more than 5A in the worst case (3 and 4.2 volts for 18650). This is a lot, but not fatal.

ANY cells of the same technology (Li-Co, Li-Mn, Li-Ni-Co, Li-Co-Ni-Al) can be connected in parallel.
The capacitance and the maximum output current will be equal to the sum. Without any negative effects.
 
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A weak cell will not give more than its 0.5A without the voltage dropping across it, and it cannot drop because the connection is parallel.
This is easy to verify experimentally.
Then do the experiment where you are measuring the cell exactly at the cell, and also measuring the voltage where it connects to the connecting bar.
And also measure the voltage exactly at each cell. So you'll need a meter that can read at least 3-4 channels simultaneously.
Under a heavy load, the weak cell will drop by a few milliVolts compared to the others.

You can do the same experiment with several battery banks and applying an enormous load. Even if the batteries are connected in parallel, there "will" be a voltage difference during the heavy load. Even if it is only <100mV. It's still there. There is no perfect system. There are various differences in resistances that will affect the outputs.

you can calculate and try experimentally (I specifically investigated this issue)
Please provide documentation on these experiments and results. We'd all like to be educated and learn new information if provided.
 
Even if it is only <100mV. It's still there
And what?
This drop will be only on the connecting wires due to the flowing currents. It's obvious, I think.
How do these millivolts affect the output current of 0.5-1 Amperes?
So you'll need a meter that can read at least 3-4 channels simultaneously.
?
Two is enough, which I did, even though I knew in advance that I would receive.
Please provide documentation on these experiments and results
I have no. This is basic electrical engineering. This is done in a few minutes: two elements, ammeter and a thick wire.
 
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This drop will be only on the connecting wires due to the flowing currents. It's obvious, I think.
So you agree with my previous statement. There "is" a voltage drop compared to the other cells.

Two is enough
Not really, as this will only allow you to see 2 cells. Having more than channels would allow you to show not only two cells next to each other, but across the whole pack. Changing the connection points after a test could skew the results

I have no
Ok, that makes sense
two elements, two ammeters
Nope, using 2 different ammeters means there are a host of other variables to account for. Different resistances in the wires, probes, connections, solder points, etc, etc. By having multiple channels on the same device you drastically narrow the unknown variables.

Even 2 Flukes that have been calibrated could be off from each other by a few milliVolts. Moving a wire, or the unshielded ones could induce different values.

Basically, you haven't disproved my previous statement. Under heavy loads there can be a noticeable different in voltage between the weaker cell(s) than that of the other cells.
And this...
A weak cell will not give more than its 0.5A without the voltage dropping across it,
I can't agree with. Even a cell is rated at a particular current discharge, that doesn't mean that it won't try its best to keep up with the other cells. If there's energy stored, it will do what it can to give what it has until the voltage drops too low to push the power out effectively.
 
So you agree with my previous statement
Certainly! There is always some voltage drop - due to the difference in the length of the connecting wires
Having more than channels would allow you to show not only two cells next to each other, but across the whole pack
I don't see a difference when considering this question
using 2 different ammeters
Sorry, I described myself there and got better, but you already answered.
One ammeter is enough, of course - we measure the equalizing current - between two differently charged elements :)
Under heavy loads there can be a noticeable different in voltage between the weaker cell(s) than that of the other cells
It cannot under any load - well, a PARALLEL connection - which automatically implies the equality of voltages on the elements. Both in theory and in practice.
The difference can be at the battery terminals if they are connected by tires of different lengths - due to the voltage drop on the tires. And this is a negligible amount.
Even a cell is rated at a particular current discharge, that doesn't mean that it won't try its best to keep up with the other cells.
That mean

We have two cells with an open circuit voltage of 4.00 volts. One is capable of delivering a current of 1 Ampere, the other - 0.5. If the elements are of the same type (of the same technology), this clearly means that the second one has more internal resistance (approximately 2 times). Do you agree with this?
Under a load of 1 A, the voltage on the first one drops to 3.5 volts, and on the second - to 3,
0 (approximately - what is important here is much more; at a current of 0.5 A, the voltage on the second will drop to 3.5 V, as on the first).
We connect in parallel. Loading.
The first gives out everything it can - 1 A. The second - 0.5 A. The voltage on both drops to 3.5 V.
What is apart, what is together - both are comfortable.
WHY would the second try to lower it even more?!
 
A cell that is rated for 0.5A is rated as such as being its "safe" and "nominal" current rate. This does not mean that it is it's "max" rating. And just like an internal combustion engine is designed to run around 2000-3000 rpms, doesn't mean that it is not capable of running at 7000rpms.
And running it at that high discharge/rpm rating can cause damage to them.

So, if you put a load of 2A on those 2 cells, one has a C rating of 1, the other 0.5 (with 1000mAh assumed capacity), then they *both* will try to do 1A. It's actually more probable that the 1C rated cell will do closer to 1.2A, and the 0.5C rated cell 0.8A. But it won't just "stop" at 0.5A just because it's "rated" at that.
 
Engine yes, battery no.
It's actually more probable that the 1C rated cell will do closer to 1.2A, and the 0.5C rated cell 0.8A.
Not probable.
Above it was about the maximum current (as I understand it). But it is not significantly. The specified ratio will be fulfilled at ANY currents: if the load requests 300 mA, the first will give 200, and the second 100. The same is true in the short circuit mode (well, just try for one second - nothing will be done with the battery in this mode).
Just because that the internal resistance is the basic characteristic of the battery and under equal conditions (and they are equal by definition), if it changes, then synchronously, while maintaining the ratio.
 
Please provide documentation on these experiments and results
Let we have two cells with voltage U1 and U2 and internal resistance R1 and R2. Then, with a parallel connection, an equalizing current I = (U1 + U2) / (R1 + R2) will flow between them.
Typical internal resistance of a new cell is 50 mOm.
I = (4.2-3.0) / (0.05 + 0.05) = 12 Amperes. This, of course, is a lot.
But here the resistance of the wires and the shunt (if the ammeter is connected in series) is not taken into account, which is almost always much greater than the internal resistance (by several times) - about 0.5 Om. Therefore, in reality, I have never managed to fix a current of more than 5 amperes. And this is the ultimate difference. In reality, it is on a couple of cells, one of which is fully charged, and the other WAS discharged not only just now, but some time ago, much less. Usually, when discharged, the voltage quickly rises to 3.4-3.5 volts due to depolarization. Well, no one specifically connects cells with such a big difference.
Here, with the simultaneous connection of, say, 7 fully charged with one fully discharged, it will be very bad for it - yes! :).
So this is an almost incredible combination :)
 
Salvaged 6 of those from a 15 years old HP laptop battery. 2 of them had 2.2v, 2 of them had 2.1-2.15v, the remaining 2 had exactly 2v's on them. Lii500 couldn't recognize the batteries while Dragon VP4 Plus did it right away. Charging them right now. Will make an update with the results.
 

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