# Battery Power Calculation



## Jan (Oct 5, 2009)

ikarus said:


> Hi all,
> 
> I am new about EV. I want to use LiFePO4 battery. But I am confused about battery power calculation. For example it is nominal voltage between 3.0V-3.3V and it is charged votage 3.6V and it is 50Ah (suppositional value). Which one of them is true formula below?
> 
> ...


None of them. The charge voltage is only important for charging. And a cell has only one number for Nominal voltage. It’s not a range. The nominal voltage is a lot like the average voltage. And there is only one average voltage. And the Ah rating is not enough to know the nominal and peak power a cell can produce.

This is how it works:

For power you need to know the continuous and peak C rate also. For LifePO the continuous C rate is usually 3C and peak is somewhere, depending on the exact chemistry, between 5C and 20C.

That C value means that a 50Ah cell can produce (3x50=) 150A continuously and between (5x50=) 250A and (20x50=) 1000A peak for a few seconds. 

Power is U x I or more common: V x A. The voltage of a full cell is much higher than a nearly empty cell, so you can calculate the power for every SOC (State of charge). But usually the nominal voltage is taken, because that’s the voltage you have most of the time. By far. So, the continuous nominal power is (150*3.2=) 480W and peak somewhere between (250*3.2=) 800W and (1000*3.2=) 3.2kW.

Now you know what nominal and peak power your cell can produce. Multiply that by the amount of cells, and you know what continuous and peak power your pack can produce. 

The continuous power is for your max cruising speed. This determines (minus all the losses) what your maximum continuous speed will be.

The peak power is for accelerating. This determines what your maximum acceleration will be.

That is, if your battery pack is the weakest link in the drive chain. Your controller and/or motor could be much more limiting.

And for energy (kWh) you simply multiply Ah*V = 50*3.2= 160Wh. That's not much. But you have to multiply that with the amount of cells in your pack. 

Millage is usually calculated as Wh/mile. And depends heavily on your kind of vehicle: Its mass, the efficiency of the transmission and all other parts involved, and most important at high speed, the Cw value for wind resistance.


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## major (Apr 4, 2008)

ikarus said:


> Hi all,
> 
> I am new about EV. I want to use LiFePO4 battery. But I am confused about battery power calculation. For example it is nominal voltage between 3.0V-3.3V and it is charged votage 3.6V and it is 50Ah (suppositional value). Which one of them is true formula below?
> 
> ...


Hi ik,

Welcome to the forum. You're busted by the math police  3.6 * 50 = 180, not 18. 

Next: Use the units. P = V * I or Watts = Volts * Amperes, which is voltage times current, not voltage times charge, which is what you did, apparently, but I can't tell because you didn't use the units in your equations.

Read up some on power and energy here http://www.diyelectriccar.com/forums/showthread.php?t=11708 

The 50 Ah figure represents the battery charge. To figure the power you need to know the current along with the voltage. The current depends on the load at the instant of concern. From the nature of your inquiry, it seems you wish to know the "rated" power or power capability of the battery. And that can be quite different that the actual power. Often such figures are stated or calculated as "nominal" using nominal figures for voltage and current.

So for LiFePO4 cells, most times 3.2 Volts is used as the nominal voltage. Power then would be nominal voltage times nominal current. But what is nominal current? Tough question. Battery industry came up with a scheme to confuse you called C. C is charge. Charge is Ampere hours or Ah. It is current times time. It actually amounts to the number of electrons you put into the battery. 

Current is the rate at which you use those electrons, or rate of charge, or C-rate. So you divide the charge by time to get current. Ah / h = A. (Ampere hours) / hours = Amperes.

I told you it was confusing. So if you used all the charge in your battery in 2 hours, and it was a 50 Ah battery, it would have a current flowing for those 2 hours of 25 Amperes. 50 Ah / 2 h = 25 A. Now use that current to figure the power. P = V * I. P = 3.2 V * 25 A = 80 W. So your battery can deliver a nominal power of 80 Watts for 2 hours. That would be a C-rate of ½. Or C/2.

The power of the battery is relative to this term "C-rate". A typical 50 Ah LiFePO4 cell might have a continuous C rating of 2 and short duration (burst) rating of 6C. So its continuous current rating would be 2 * 50A = 100 A. The continuous nominal power is 100 A * 3.2 V = 320 W. The short duration rated current is 6 * 50 A = 300 A, so burst rated nominal power = 300 A * 3.2 V = 960 W.

A different type of 50 Ah LiFePO4 cell may have a burst rating of 20C. Then the 20C current is 1000 A and the nominal burst rated power is 3.2 kW. Notice that I use the terms nominal and rated. This is because these calculations often produce values not seen in real life. The 20C calculation is an example. It is almost certain that at 1000 A the voltage from a 50 Ah cell will be below that 3.2 Volt nominal figure used for that calculation. So one must take care when playing the rating game using specifications and nominal numbers.

The power you get from the battery in real life is always the actual voltage times the actual current at the instant in time you are concerned with. That actual battery voltage is dependent on the internal resistance of the battery, called R_int or Ri for short. Actual voltage = open circuit voltage - (current * internal resistance) or Vb = Voc - I * Ri. 

Using the actual calculations, one can deduce a formula showing the actual peak power from a battery and that is P = V²/4R or the open circuit voltage squared divided by 4 times the internal resistance. For that 50 Ah cell, that may reduce to P = 3.55V² / (4 * 0.002Ω) = 1575 Watts. Such a peak power number is real but rarely used due to the rapid heating and stress on the cell.

So, battery power, probably more than you wanted to know 

major


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## Jan (Oct 5, 2009)

You scared him off.


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## ikarus (Mar 19, 2013)

@Jan, @major thanks for your explanations.

@major



> So, battery power, probably more than you wanted to know


You are right. Your detailed explanations more than I wanted. Thanks again.




> You scared him off.


I am here to face the facts 

Regards.


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