## Monday, June 28, 2010

### In batteries, 2+2=1. Actually more like 1/2. Well... maybe a bit less.

This is a blog post I've wanted to write for a decade. The reason I haven't (other than the obvious problem that a decade ago, I did not know what a blog was!), is because its a tough post to write. But, folks tell me that I have a gift for explaining things (I use the world "folks" is a generic sense to indicate a number greater than 0), so I shall try.

Everyone wants to make a better battery. What they mean when they say "better" is a battery that has more energy. This is what many (not all) battery researchers are trying to do, and this is what every user wants. If you read my post titled "A Moore's law for batteries? Maybe not", you will know that the game is to find new materials that make up the anode and cathode of a battery.

The idea here is to find a new material that has more capacity than the existing material and/or find one that operates at a higher voltage. Capacity is a measure of the amount of charge (electrons) you can get per gram. More is obviously better. Capacity times the voltage is energy; for real-world applications what matters is the energy. Typical numbers for capacity for lithium-ion batteries would be 140 mAh/g for the cathode and 330 mAh/g for the anode. The typical voltage of a lithium-ion battery is 3.7 V.

A lot of research in lithium-ion batteries is focussed on increasing the capacity. There is also an active area of interest in increasing the voltage to above 3.7 V. Increasing the voltage is going to be hard (very hard), so increasing the capacity appears to be the way batteries will improve, atleast in the short-term.

Simple enough.

Those of you who are paying attention have probably noticed that for every gram of material, you only have ~1/2 the capacity in the cathode compared to the anode. If you want to make a battery with a capacity of, say 330 mAh, then you have to take 1 gram of the anode, but you need 2.35 g of the cathode (330/140). What this means is that you have a total weight of 3.35 g to get a capacity of 330 mAh. So the capacity of your battery is actually 98 mAh/g (330/3.35). So you started with a anode at 330 mAh/g, a cathode at 140 mAh/g and you get 98 mAh/g for the battery. A 2 mAh/g cathode with a 2 mAh/g anode give you a 1 mAh/g battery. 2+2 is actually only 1. Certainly not 4. Not even 2! Welcome to batteries.

If you have a new anode with say 10 times the capacity (so 3300 mAh/g) you can do the same math and you will get a cell capacity of 134 mAh/g (for a battery of capacity 3300 mAh the weight is 24.5 g). You go to all this effort to make something 10 times better and you get to use your iPhone for an extra 30% talk time. A bit disappointing! On the other hand, if you had an cathode that was, say, twice as good, at 280 mAh/g (with an anode at 330 mAh/g), then your cell capacity goes up to 151 mAh/g. Much better. 50% better. This is why most researchers want to find a better cathode. Its more bang for the buck.

All this is pretty simple. All battery folks know this. 2+2=1. End of story.

Or is it? There is another small factor that even battery researchers sometimes miss. This factor is the dead weight in a battery.

If you really want to use the anode and cathode, you need some extra real estate. Things like separators to keep the electrodes apart, current collectors to collect the current, and packaging to make sure you contain it in a neat little package. All these add weight and volume. It doesn't matter if you have 10 times the capacity in a new anode, you still have to carry this dead weight.

This is a lot like a gasoline-powered car. Only the gasoline has any useful energy in the car. But to use the gasoline, you need a tank, an engine, the wheels, the drivetrain.... you get the point.

Obviously, if you can make the weight of the rest of car as light as you can (no seats?), it helps you get more from your tank of gas. Similarly, if you can minimize the amount of unwanted weight, it helps a lot in the battery. What this means is that you try to increase the ratio of the active materials (the anode and cathode) to that of the inactive material (the separators, current collectors etc).

But there is a catch. Turns out that you can't increase the amount of the active material willy-nilly. It has to do with losses in a battery. If you put extra active material in, you have to add a bit of the inactive with it. And increasing the amount of active materials involves making the anode and cathode thicker and there is a limit to how thick these can be made before losses become prohibitive. One needs to account for these factors.

*Geek meter on*

Remember the example above where we calculated 98 mAh/g using a battery of capacity 330 mAh with a weight of 3.35 g? If you do the math on the extra weight for the inactive material, you have to add an extra ~3.35 g. You can do the math to convince yourself of this number or you can trust me. I would suggest doing the latter. So you actually only get a capacity of 49 mAh/g (1/2 of 98)! 2+2=1/2!

For you battery geeks, you can verify these numbers by calculating the theoretical energy of the battery using the 98 mAh/g and multiplying by 3.7 V to get 360 Wh/kg (the theoretical capacity of a graphite/LiCoO2 cell). You can calculate the practical capacity by multiple 49 mAh/g by 3.7 V to get ~180 Wh/kg (A typical value for a 18650 cell using cobalt oxide). Well well well... the math works, does it not?

Here is the rub. Remember the example where we had an anode that has 10x the capacity. We had a cell capacity of 134 mAh/g. If you do the calculation for the extra weight and recalculate the capacity you get only 55 mAh/g.

You have to think about this a little bit, but it turns out that if you have more capacity you will need less of the anode, so now the inactive weight becomes a larger fraction of the total weight of the battery. You could have compensated for this by taking the same weight of the anode and just talking a lot more cathode, but like I was saying, this is impossible because it increases the losses in the battery to a point where it would be useless.

So we have an anode with 10x the capacity and we gain 12% in cell capacity (and don't get me started on the voltage! That is for another post).

If you don't believe me, do the math. If you don't know how to do the math; I guess you have to believe me! It would be embarrassing if someone spots an error; but then again I'm assuming that no one has actually made it this far.

If you do the same math on the battery where instead of the anode being better, the cathode is twice the capacity, where we calculated a cell capacity of 151 mAh/g without the inactive weight you will calculate a cell capacity of 68 mAh/g with the inactive material. So you made a cathode of twice the capacity and your cell capacity actually went up by 38% (remember we calculated this to be 50% better before).

Turns out that even if you make a battery with 3300 mAh/g for the anode (in a sense, this is close to the best Li-ion anode we know of) and a cathode of 280 mAh/g (the best Li-ion cathode we know of) we get a cell capacity of 110 mAh/g. 2.2x the present-day battery. But this assumes the voltage of the two are the same. In reality the materials that have this capacity have a lower voltage, which means that the energy is not really that high. Turns out that this best base scenario battery is better by maybe a factor of 1.8 to 1.9. Meaning, this battery will approach 340 Wh/kg.

*Geek meter off*

As a matter of fact, everything else being equal (i.e., amount of inactive material), the best Li-ion battery we can dream of making in the future, based on what we know as of late June 2010, will have a energy density of ~340 Wh/kg. If you want something better, you pretty much have to work on the inactive material. All these are for cell-level numbers. If you go to a battery pack, things get even worse, but that is for another post. If someone tells you that they can make a battery where the energy is greater than this, you better dig.

Most people, including battery researchers, don't think about this extra weight. Its actually a very important factor in a battery. There is such a great focus on new materials that folks forget that reality may be as good as your simple math leads you to believe. In addition to new materials, we have to think about ways to decrease the inactive materials in a battery. There is far too little research on this and a lot to be gained from doing something about it.

So next time you hear about a new material with more capacity, ask not how much more theoretical capacity you can get, instead ask how much practical energy you actually get. Remember that you can't just divide theory by 2 to get practical; it could be less (a lot less). And don't forget the voltage. Its also critical.

To help you, here is a link to a excel spreadsheet that has a battery simulator specifically for a lithium-ion battery. I hope the mac version of excel is compatible with a PC. The sheet that has the calculations is protected. Contact me to unprotect. Let me know if you catch errors. The simulator makes a LOT of assumptions. If you want them all relaxed, contact me.

Venkat

## Sunday, June 20, 2010

### You say potato, I say...battery?

Saw something interesting in the news on a potato battery. Its a Zn-Cu battery with the two rods inserted into a potato. Why is this new? The authors say that they boiled the potato and were able to get less resistance and more power! I swear I'm not making this up.
Check out http://jrse.aip.org/jrsebh/v2/i3/p033103_s1 for the abstract. They say its cheaper than a AA battery and can be used to power a low-power LED light. I have not dug into the numbers to see if the cost claims are right. It would seem that with the low power that one gets from each cell you will need a lot of potatoes; the authors say they need 5.
My dad is visiting us from India and is sitting across from me. He tells me that 5 potatoes in India costs Rupees 15 (~30 cents). He tells me that a AA battery is either a bit cheaper or comparable! And I still need to buy some zinc and copper (and a pot to boil the potatoes, and spend some money heating it)
I get the impression that this is sort of a cute press release of something we already know and I wonder if it solves anything. I have to boil the potato and so need energy; I gain a means of making electricity, but I lose food to do that. And we really need someone from countries in the developing world to tell us if the cost numbers are what my dad tells me they are in India.
Anyway... could not resist.

Venkat