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FrozenGate by Avery

Lithium Iron Phosphate- The Safer Alternative

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Jun 12, 2010
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So, many of you remember me from the Lithium Ion explosion thread. Well, I'm back with more battery news. And surprisingly, this time it's good news too.

Many of you here would be familiar with your lithium-ion cells, mainly in the form of 18650s. These blighters run at 3.7V, powering your FlexDrives and MicroBoosts day in, day out. But every now and again, they go bang. And when they do go bang, they usually end up taking quite a fair bit of equipment and surroundings out with them, and sometimes even causing injury.

Now, what if I told you there were lithium cells that couldn't explode, even if abused? Or lithium cells that have a 25% charge time for a given capacity than a lithium-ion cell? Or lithium cells that are immune to thermal runaway, even when shorted? Or that all this is done by an unprotected cell?

You'd probably think I was lying, or I had too much to smoke again. Either that, or I'm being paid by some cheap company to advertise their goods. But I'm not. Behold, the lithium iron phosphate LiIon cell.

Exactly the same form factor as standard 18650s, they can also be found in other common form factors such as CR123A and 10440/14500.

And at first sight, you wouldn't be able to tell it apart from a standard Li-Ion 18650.

sku_5105_4.jpg


The difference between standard Li-Ion cells and LiFePO4 cells is the composition of the cathode. While normal lithium ion cells use a lithium cobalt oxide as a cathode, lithium iron phosphate cells use lithium iron phosphate as a cathode, hence their name.

It's change in cathode that is responsible for the thermal and chemical stability that renders the cell virtually inert, even when abused and misused.

The reason behind the explosions commonly seen in Li-Ion cells is caused by the release of oxygen from the decomposition of lithium cobalt oxide. This outgassing can occur under high temperatures, overcharging, or reverse polarity charging. Eventually, an explosion (resulting from an overpressure condition) and fire will result.

Lithium cobalt oxide breaks down at temperatures above 200C. Although it seems high, in a thermal runaway condition, it is possible that this will go unnoticed until it is too late. Also, the decompositon is an exothermic reaction (giving out heat), which will eventually make the condition worse. Lithium iron phosphate, however, requires temperatures exceeding 800C to break down. Even when decomposition does occur, lithium iron phosphate isn't affected as badly as lithium cobalt oxide is, and the fire (if any) will be significantly eaiser to contain. The reason for this? The iron and phosphorus atoms hold onto the oxygen atoms much tighter than the cobalt can, hence reducing reactivity.

Another major advantage is shelf life- lithium iron phosphate has a significantly longer shelf life than Li-ion cells, nor are they affected by storage on a full charge. The power density difference between lithium iron phosphate and lithium ion is quickly offset after a year or two, through usage or storage.

Unfortunately, these benefits don't come free. With Lithium iron phosphate cells, you're trading off two things- power density, and voltage.

You'd have noticed that on the cell, it's marked at 3.2V. It's not a typo, LiFePO4 cells only run at 3V nominal voltage. They are charged at 3.6V, and 2.8V is considered empty. Although not all portables will be able to take a sudden drop in voltage, anything running from a FlexDrive or MicroBoost will be unaffected. This means the cell is capable of delivering less power than standard Li-Ion cells.

The second disadvantage is the low energy density. You'd have noticed that most LiFePO4 cells have a little more than half the capacity of an equivalent-sized Li-ion cell. This translates into shorter runtimes between charges, and less current deliverable by a cell. At this stage, lithium-ion is still the superior option for builds running >1W diodes.

There's the tradeoff- power density versus safety and fast charge times. It's up to you- if you want power, or peace of mind.
 
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I buyed some from DX (the blue ones) sadly they have a lower capacity (1350mAh) and voltage (2.8 cutoff, 3.2 common and 3.6 charged). The good side are they discharge a lot more than common cells (mine "chinese cells" discharge ~8 amps.. :p
good for lighting up stuff with a short-circuit haha :p
j/k..
 
I suppose the lower power density itself is also a safety factor. Considering both lower voltage and capacity, there is only about half as much energy in the same volume compared to lion cells.

The energy density of these cells is comparable to that of NiMH batteries. Those are also pretty safe - i've never seen them actually explode. The only risk with NiMH cells is short circuit, which makes them heat up and potentially melting anything plastic that contains them... but they will not usually run hot enough to start a fire.

Considering all that, these cells are interesting, but imho not superior to 3 NiMH in series.
 
they charge faster.. VERY FASTER as I read on other sites (some have seen to full charge in 15 minutes)
 
for all these reasons that's why they're a favorite battery in the rc-scene.
 
15 minute charging is pretty impressive - thats something you will not be able to do with NiMH's unless you actively cool them.

I'm not sure if this is an important factor when lasers are concerned though. We don't need excessive charge or discharge rates generally. You might be looking at 2, maybe 3 amps or so when powering the 445's off a single cell, and that's about it.
 
yeah: absolutely right Benm. Maybe 8 amps for a LM317 would be too much.. :p:p:p

btw: the DX ones are rated 30 minutes for full charge, but obviously we can push these things :p

Maybe charge them with a full metal charger (to handle the heat and not melt like plastic) with 5 amps would solve the problem.. :)
 
I have noticed with a fully charged 18650 @4.2V a flexdrive running 1.5A will be pullig 2A+ from the battery. When the voltage on the batteries drop the current draw goes up. I will run a test tonight with my bench supply @3.6V-2.8V with a flex driving a 445 @1.5A but I would assume it will be close to 3.5-5A as the battery gets to the 2.8V cut off. And that would give you a very short run times. Using with a linear driver would probably be more feasible.
 
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My LM317 test driver got thermal shutdown using these batteries (a pair for a red :p) in 20 seconds
Assuming 9v battery barely warms the LM317, is the current that heats the linear drivers, so, these would be the worst idea to use on a linear drived laser
I'm no professional, just my 0.02 cents again.
 
That's one of the major disadvangates of LiFePO4 cells- lower forward voltage.

But like I said, if you're going for >1W, it's better to stick to Li-Ion.

LiFePO4 is for reliability and safety, not for max power.

Also, I'm in the process of re-writing the OP, to include the stuff about charge times, now I've found a citation for it.
 
No, it's better for max power as well. They can evidently supply 30A which is a power of 90W of power compared to a LiIon's 4A or so which is a power of 14W. LiIon may be more unstable, but it's capacity is far higher.

When the voltage on the batteries drop the current draw goes up exponentially.

It should be linear since the flexdrive draws a constant power to give constant current. If voltage goes down linearly, current should go up linearly otherwise you have a net increase in power. If it's drawing 4W at 3.6V, the current will be 1.1A. If it's drawing 4W at 3V, the current should be 1.3A.
 
No, it's better for max power as well. They can evidently supply 30A which is a power of 90W of power compared to a LiIon's 4A or so which is a power of 14W. LiIon may be more unstable, but it's capacity is far higher.



It should be linear since the flexdrive draws a constant power to give constant current. If voltage goes down linearly, current should go up linearly otherwise you have a net increase in power. If it's drawing 4W at 3.6V, the current will be 1.1A. If it's drawing 4W at 3V, the current should be 1.3A.

Definitely better for higher current demand but sacrifice capacity. It is always a give and take.:beer:

Exponential may not be the right word. But it goes up as the voltage goes down. If it went up Exponential it would probably be drawing Amps in the double digits.:eg: So when you say linear should it draw 1A more for every 1V less or how would that be figured?
 
goninanbl00d;

The best companies in Lithium Ion battery manufacturing, have also been making their cells safer:

Panasonic/Sanyo successfully achieved safety and high capacity by using its unique Heat Resistance Layer (HRL) technology that forms an insulating metal oxide layer between the positive and negative electrodes . . . . . . . . . . . . The layer prevents the battery from overheating even if a short circuit occurs.

I have used several hundred Lithium-Ion 18650 batteries for the last five years.

I have never even caused one cell to vent, even using them in some RC trucks my son and I raced.

I do use ONLY "brand-name" cells.

LarryDFW
 
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It should be linear since the flexdrive draws a constant power to give constant current. If voltage goes down linearly, current should go up linearly otherwise you have a net increase in power. If it's drawing 4W at 3.6V, the current will be 1.1A. If it's drawing 4W at 3V, the current should be 1.3A.

Thats the way it works indeed, although losses in switching elements are relatively higher as the input voltage drops, so you could see a slight increase in total input power with dropping voltage.

At some point the driver will simply stop working for lack of input voltage, but that point could be far below a safe discharge point for the battery you are using.

The best way to deal with this would be modifying the switch mode converter to have an adjustable cutoff voltage. Depeding on the cell chemistry you are using you could adjust that to, for example, 2.8 or 3.3 volts.
 
At some point the driver will simply stop working for lack of input voltage.

Any idea what point that is for a flexdrive?

So when you say linear should it draw 1A more for every 1V less or how would that be figured?

I should have been more clear. It is close to linear (but not technically) until you get to really low voltages.

Power = Current × Voltage

^In most cases. You know two of the values, so you can solve for the third. Current as a function of voltage would be:

Current = Power ÷ Voltage

If we hold power to 4, we can plot:

attachment.php


With volts on the X-axis and amps on the Y-axis. Now I know what you're thinking. "That doesn't look linear at all!" and that's true. If you had a 0.1V battery and wanted to get 4W out of it, you'd need to draw 40 amps! But if we zoom in to actual values you would find in a lithium battery, we get something that looks a lot more like a smooth line:

attachment.php
 
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The graphs don't show here, but the math sounds good.

I'm not sure what the cutoff point is for a flexdrive... does anyone have a proper schematic for it?
 





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