# Best Batteries/Cells for the Buck



## danmansuttmeier (Mar 7, 2011)

Hey guys so as far as drag racing goes I am wondering what is the best battery/cell for the dollar. 

Lead is powerful but very heavy. 

Calib power seems a little expensive.

I found a guy who sells HiPower Lithium cells for cheap.

But the best it seems for racing are the headway cells. They can handle a very high c ratings.

Looking at a 40ah cell from my lithium guy, (specs: http://electriccarinternational.com/media/HiPower/HP-PW-40AH.pdf)

It seems for a 15sec drag race this cell puts out 200A at 5C. I would need 90 cells in series to get my 288V for my warp11hv. They would weigh 315lbs. The cells would cost $4500.

Looking at 8ah headways, it says they can handle a 20C discharge for 3 minutes. (http://currentevtech.com/Lithium-Batteries/Headway/Headway-38120P-8ah-cell-LiFePO4-p41.html)

That's 160A discharge. Now lets say I put 3 cells in parallel sets. Then 90 sets in series. So I would have my 288V and 480A discharge at 20C. They would weigh 179lbs. These cells would cost $4860.

So are headways the best drag racing battery/cell for the buck? They are cheaper and lighter.


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## Bowser330 (Jun 15, 2008)

The brick style cells (hipower, calib, thundersky, GBS) will never be the best for drag racing, those cells are for range & cruising, with occasional short bursts of power....

You need to specify what your max amperage targets are....do you want to be able to push out 1000A from the batteries? 

Well then with those hipower 40AH cells, if you want to stay at 5C 200A, then you will need 5 in parallel and 90 in series, so a total of 450cells * 3.3lbs = 1,485lbs. @ 50$/cell = 50$*450 = $22,500, 57.6kwh (@300wh/mile range = 192 miles)

headway power cells (p) 8ah, 20C is safe so that's 160A each, you will need 6 cells in parallel and 90 cells in series again, so thats 540cells * 300g = 356lbs., 18$*540 cells = $9,720, 13.8kwh (@300wh/mile range = 46 miles)




danmansuttmeier said:


> Hey guys so as far as drag racing goes I am wondering what is the best battery/cell for the dollar.
> 
> Lead is powerful but very heavy.
> 
> ...


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## danmansuttmeier (Mar 7, 2011)

well 1000amps would be nice. thanks for the range calculations. Seems like headways are the best option for what I need.


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## drgrieve (Apr 14, 2011)

It's not as simple as that.

It's all about voltage sag, but there are no voltage sag specifications available.

To make it worse the cell voltage sag is only part of the equation, what you should be truly interested is the pack voltage sag.

At 1000 amps depending on battery and battery pack construction techniques you could lose from 10% to over 50% of your power. 

Which would make your drags times very poor.

So to compensate you need to have a controller and a battery pack that can provide enough power after voltage sag to produce your 288 volts @ 1000 amps. 

Otherwise when your only getting 200 volts @ 1000 amps and your top speed is limited you'll be a sad puppy.

By the way I'm very interested myself in knowing the voltage sag from cell packs. To help make the same decision as you are (not for drags but street performance). But sadly I've collected very little information so far.

http://www.diyelectriccar.com/forums/showthread.php/cell-comparisons-please-contribute-62326.html


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## danmansuttmeier (Mar 7, 2011)

drgrieve said:


> It's not as simple as that.
> 
> It's all about voltage sag, but there are no voltage sag specifications available.
> 
> ...


it says on the headways and hipower operating voltage is 2.5-3.65. So i assumed 2.5 is as low as it would go with sag.


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## dougingraham (Jul 26, 2011)

If it is drag racing you are interested in then the pouch cells are probably king. The two I know of that have been used are:

Kokam SLPB98188216 which are 30Ah cells with dimensions of 180 x 9.8 x 218 and mass 840 grams. These can do 20C continuous and 30C peak. That is 600A and 900A peak. I believe these are the cells use by White Zombie.

Haiyin P68100120F-50C which are 6AH cells with dimensions of 100 x 6.8 x 20 mm and weigh 170gms. These can do 50C continuous and 67C peak. That is 300A and 402A peak. I believe these are the cells used by Warp Factor II.

Being a little conservative if I wanted a 2000A pack at 340 volts I would parallel 4 of the Kokams to make a 120AH wide pack and then series those to get 340 volts. That would be 80 of those 120AH bricks in series. This would be 320 cells total at a mass of 269kg (591 lbs) that would give an honest 2000A with fairly minimal sag. This would be a 35.5kwh pack. For the Haiyin it would need 7 cells in parallel giving a 42AH brick and the same 80 bricks in series giving a total of 560 cells. The mass of 560 cells in this case is 95kg (209 lbs) which is a lot less. This is a 12.4kwh pack

There is also an A123 systems 20AH cell that could be promising.

Neither A123 Systems or Dow/Kokam really want to sell you these batteries. I haven't tried to buy the Haiyin cells but I have seen them priced somewhere at $29.99 each.

You parallel the cells first and then combine those into series strings. This way a single weak cell only drops the voltage a little bit. If you series the cells and then parallel just at the ends a single weak cell will affect that whole string. If you have a huge bucket of money to spend you might consider load testing a lot of cells and picking the best out. You are never going to run the cells down in a 10 second race to the most important factor will be minimizing the voltage sag at load. Slightly elevating the temperature of the cells will probably improve performance as the chemical reactions work better and average voltage goes up.

My big problem with the pouch cells is the effort that goes into connecting them into a usable battery. There are so many interconnects and the currents are so high that a mistake can ruin your day pretty quickly. And all that complexity does add mass and mass is the enemy of acceleration.

Good luck!


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## drgrieve (Apr 14, 2011)

danmansuttmeier said:


> it says on the headways and hipower operating voltage is 2.5-3.65. So i assumed 2.5 is as low as it would go with sag.


Sounds like you need to do a lot of reading. I'e found EVTV videos to be quite informative in learning about lithium iron chemistry - especially lifep04.

Voltage sag is unlimited (though I'm not sure if it can go negative??). In fact maximum power from the cells is when voltage sag is half your starting voltage (3.3) which is 1.65 volts. That sort of discharge will impact your cycle life, but I guess for the ultimate drag time that is secondary.

This is why C rates are meaningless. You need to know for each C how much sag does it incur. And as mentioned above the warmer the cells the lower the sag (until you hit the failure point of course). This is also variable according to the cell makeup and is also unpublished information.

It's the same for the motors. There is little published information on how much you can push into them. How much voltage can you go over, how much amps, how long can they take the abuse, what modifications were done to achieve these amazing numbers - I've heard of a warp 9 taking 1800 amps during a drag.


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## danmansuttmeier (Mar 7, 2011)

Interesting that there isnt more info. Or the companies don't want it shared.


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## blackslax (Oct 11, 2011)

I am in the same predicament. We are doing an E-Race Bike. The problem is obviously not the amps, its the volts. I am not a battery scientist so I an speaking in general terms to illustrate the point (so please let's not have every slide rule guy out there nit pick this apart. I am merely making an example) With a volt drop considered, we need +-150 cells to get 360v well 20ah cells x 150cells x 20C is 60,000 amps for 3 minutes theoretically on hand. I think we could launch the space shuttle with that power.

i had a good chat with a bicycle guy yesterday and he suggested smaller ah batteries and I am now investigating it. He said it will be more work to assemble, but it will get costs down a lot.


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## major (Apr 4, 2008)

blackslax said:


> ....We are doing an E-Race Bike. The problem is obviously not the amps, its the volts. I am not a battery scientist so I an speaking in general terms to illustrate the point (so please let's not have every slide rule guy out there nit pick this apart. I am merely making an example) With a volt drop considered, we need +-150 cells to get 360v well 20ah cells x 150cells x 20C is 60,000 amps for 3 minutes theoretically on hand.


Hi slax,

Call it a nitpick if you like and I don't even need a slide rule to tell you don't have a clue as to series parallel cell connection. 360volts and 60,000 amps. You think maybe you have something wrong with that calculation 

150 cells in parallel might get you 60,000A, but at single cell voltage minus sag, so at just a couple of volts. Hook those same cells in series for your 360V and at 20C you could see 400A. It is an either or situation. Voltage adds OR current adds, not both.

Regards,

major


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## frodus (Apr 12, 2008)

150 cells in series at 20Ah is 3kwh, thats it.

if the cells were 20C rated, that's 400A at 360V (with voltage drop you said), which is 144,000W peak.

Not sure where 60,000Amps came from.


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## blackslax (Oct 11, 2011)

meant to quote see below


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## blackslax (Oct 11, 2011)

major said:


> Hi slax,
> 
> Call it a nitpick if you like and I don't even need a slide rule to tell you don't have a clue as to series parallel cell connection. 360volts and 60,000 amps. You think maybe you have something wrong with that calculation
> 
> ...


OK so arithmatic beats the slide rule. thank you. I have had time to start this project since having 4 discs fused last week. So needless to so I probably have more chemicals on board than Charlie Sheen right now. thanks for your kindly mannered response. Back to the drawing board.


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## major (Apr 4, 2008)

blackslax said:


> OK so arithmatic beats the slide rule. thank you. I have had time to start this project since having 4 discs fused last week. So needless to so I probably have more chemicals on board than Charlie Sheen right now. thanks for your kindly mannered response. Back to the drawing board.


You're welcome, slax. You could get both your 360V and 60,000A by using 22,500 cells in a 150P150S series parallel combination.  

Get well soon,

major


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## blackslax (Oct 11, 2011)

major said:


> You're welcome, slax. You could get both your 360V and 60,000A by using 22,500 cells in a 150P150S series parallel combination.
> 
> Get well soon,
> 
> major


Ithink you mis understood, I wanted 60,000 amps and 4rpm on the motor


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## coulombKid (Jan 10, 2009)

major said:


> Hi slax,
> 
> Call it a nitpick if you like and I don't even need a slide rule to tell you don't have a clue as to series parallel cell connection. 360volts and 60,000 amps. You think maybe you have something wrong with that calculation
> 
> ...


The maximum power transfer theorem indicates that the maximum power is transmitted when the internal resistance of the source (batteries) equals the resistance of the load (DC motor). I doubt any of the packs have an internal resistance as low as the DC motor, especially at low RPM. To simulate all this in MATLAB you would need back-emf data for your motor over speed and load as well as the I-V behavior of a single cell. I've never seen comprehensive I-V data released for any of the cells that are allowed to be sold at present. In simplest terms the internal resistance is the negative reciprocal of the slope of the load line on the I-V graph. We like linear relationships but none of them will be.


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## blackslax (Oct 11, 2011)

coulombKid said:


> The maximum power transfer theorem indicates that the maximum power is transmitted when the internal resistance of the source (batteries) equals the resistance of the load (DC motor). I doubt any of the packs have an internal resistance as low as the DC motor, especially at low RPM. To simulate all this in MATLAB you would need back-emf data for your motor over speed and load as well as the I-V behavior of a single cell. I've never seen comprehensive I-V data released for any of the cells that are allowed to be sold at present. In simplest terms the internal resistance is the negative reciprocal of the slope of the load line on the I-V graph. We like linear relationships but none of them will be.


I was working an the maximum smart ass theorem Wrong again


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## major (Apr 4, 2008)

coulombKid said:


> The maximum power transfer theorem indicates that the maximum power is transmitted when the internal resistance of the source (batteries) equals the resistance of the load (DC motor). I doubt any of the packs have an internal resistance as low as the DC motor, especially at low RPM. To simulate all this in MATLAB you would need back-emf data for your motor over speed and load as well as the I-V behavior of a single cell. I've never seen comprehensive I-V data released for any of the cells that are allowed to be sold at present. In simplest terms the internal resistance is the negative reciprocal of the slope of the load line on the I-V graph. We like linear relationships but none of them will be.


Hi coul,

Actually the relationship of battery (or cell) voltage and current is linear. V = Voc - Ri * I. Where V = actual battery voltage at the terminals, Voc = open circuit voltage, Ri = battery internal resistance and I = battery current. The plot of this function (V vs I) is a straight line starting at Voc, 0 A and descending to 0 V at Isc. Isc = short circuit current for the battery and by definition occurs at V = 0. So Isc = Voc / Ri.

The power delivered measured at the battery terminals is P = V * I. This P function can be calculated and drawn as a curve as P vs I. At I = 0 (open circuit), P = 0. At Isc, P = 0 because V = 0. If we substitute (Voc - Ri * I) for V in the power equation, we get P = Voc * I - Ri * I². This is the classic quadratic equation, a parabola. 

The power function has a singular maximum. You can draw it and plainly see it or you can use math. The maximum point on the curve occurs when the slope = 0, which is when the derivative = 0. dP/dI = Voc - 2 * Ri * I. Set this = 0 and solve for I to get I = Voc / (2 * Ri) for the current at which maximum power occurs. Notice that this is equal to ½Isc. 

We can now substitute the formula for current at maximum power into the basic power equation. This results in: Pmax = Voc² / (4 * Ri). So if you know the battery internal resistance, it is pretty easy to calculate the maximum power. And it is clear that the maximum power will occur at ½Voc and ½Isc. A double check would be to use Pmax = ½Voc * ½Isc. We know Isc = Voc / Ri, so Pmax = ½Voc * (½ * Voc / Ri) = Voc²/4Ri. 

When one looks at the power function plotted against current for the battery (the parabola), it is obvious that: 1) You want to be left of the peak for efficient power delivery from the battery. 2) The rate of power increase with respect to current increase diminishes rapidly as you approach peak power. So typically it is not wise to load the battery to more than about 40% of the short circuit current value.

Regards,

major


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## MN Driver (Sep 29, 2009)

Good info, I'm going to have to track down that video where those guys at the battery were short circuiting their cells and had an ammeter on them. I though it was somewhere around 3000 amps or so but that is actually my WAG but I thought short circuit current (also note huge wires were used) would be much higher for a cell we see many people ripping 1000 amps out of.


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## dougingraham (Jul 26, 2011)

MN Driver said:


> Good info, I'm going to have to track down that video where those guys at the battery were short circuiting their cells and had an ammeter on them. I though it was somewhere around 3000 amps or so but that is actually my WAG but I thought short circuit current (also note huge wires were used) would be much higher for a cell we see many people ripping 1000 amps out of.


It would be interesting to know what the battery terminal voltage was on that test since everything has resistance. You could minimize it to maybe a foot of wire but how would you attach it to the battery? I am guessing they were some distance away from the cell when they shorted it.


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## major (Apr 4, 2008)

MN Driver said:


> Good info, I'm going to have to track down that video where those guys at the battery were short circuiting their cells and had an ammeter on them. I though it was somewhere around 3000 amps or so but that is actually my WAG but I thought short circuit current (also note huge wires were used) would be much higher for a cell we see many people ripping 1000 amps out of.


It's not hard to do. Example: A cell with open circuit voltage = 3V. Internal resistance = 1 milliohm. Short circuit current = 3000A. With an external resistance of 2 milliohm, battery current = 1000A.

Yes, the Isc (short circuit current) is somewhat theoretical, but now-a-days, someone could actually put a superconductor across the battery terminals  I'm not sure why they would do that. Afterall, short circuit power is zero, so it is a condition of no work. Useless as far as I can tell. But I guess there is always the entertainment factor


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## coulombKid (Jan 10, 2009)

major said:


> Hi coul,
> 
> Actually the relationship of battery (or cell) voltage and current is linear. V = Voc - Ri * I. Where V = actual battery voltage at the terminals, Voc = open circuit voltage, Ri = battery internal resistance and I = battery current. The plot of this function (V vs I) is a straight line starting at Voc, 0 A and descending to 0 V at Isc. Isc = short circuit current for the battery and by definition occurs at V = 0. So Isc = Voc / Ri.
> 
> ...


 With your more in depth explanation we could use a power meter and a data logger to optimize the power out-put shift points on an automatic transmission for e-drag racing use. Make sure the ratio change for each shift just pushes you back to an equal power magnitude point on the left side of the curve. Same math for down-shift regen braking going from right to left instead (if you have enough traction).


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## blackslax (Oct 11, 2011)

major said:


> Hi coul,
> 
> Actually the relationship of battery (or cell) voltage and current is linear. V = Voc - Ri * I. Where V = actual battery voltage at the terminals, Voc = open circuit voltage, Ri = battery internal resistance and I = battery current. The plot of this function (V vs I) is a straight line starting at Voc, 0 A and descending to 0 V at Isc. Isc = short circuit current for the battery and by definition occurs at V = 0. So Isc = Voc / Ri.
> 
> ...


Not sure about the helmet, but major Tom definitely takes his protein pills and puts his thinking cap on. Great stuff. I think I learned a lot, but I don't have a wall big enought to write all of the equations on and bask in the glow. Thanks Major.


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## major (Apr 4, 2008)

major said:


> Hi coul,
> 
> Actually the relationship of battery (or cell) voltage and current is linear. V = Voc - Ri * I. Where V = actual battery voltage at the terminals, Voc = open circuit voltage, Ri = battery internal resistance and I = battery current. The plot of this function (V vs I) is a straight line starting at Voc, 0 A and descending to 0 V at Isc. Isc = short circuit current for the battery and by definition occurs at V = 0. So Isc = Voc / Ri.
> 
> ...


For some other guy I drew up a chart to help explain this so I thought I'd post it here.


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## MN Driver (Sep 29, 2009)

The maximum power point being at half of the nominal voltage would mean that you would need to raise the total voltage of the pack so that way your controller and motor has enough voltage available to not be in the back-EMF arena where power is limited. So basically if you have cells less capable of delivering the current, you need a higher voltage pack to make up for it. If this means there is extra capacity then the amp draw is lower. The bonus with smaller cells is the IR is lower so for the same capacity you'll get more power. I've been doing some comparing between the same capacities of 60Ah, 70Ah, and 100Ah LiFePO4 cells but in the end my capacity is high enough for power draw to not be a big issue when the weather isn't cold. At the maximum power point, the cells would heat themselves rapidly though and reduce or eliminate the need for heating. It doesn't work though when my job's parking is outdoors where we usually get a week under -10f and it's a half mile from the 20mph tighly wound onramp that winds out directly into an uphill with a very short lane to speed up in, eek! I don't have the data to be confident in not heating the pack. 20kwh pack, 20 to 55/60mph 1/4 mile of uphill in -10f weather with a 1900 pound(pre-conversion weight) car.


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