The Peukert Effect: capacity and current drain

Calculation -|- Example -|- Implications -|- Refining -|

Many of these subjects are surrounded with urban folklore and myth, some of which have been around for so long it is the devil's own job to dispel them. (Smartgauge)

Concerning Capacity and Current Draw

The usable energy available from a battery depends very much on how fast it is taken from the battery. Slower energy draws usually mean more usable capacity from the battery. Energy draw is called power and is the product of current and voltage. Energy itself is a product of power and time. A battery's available energy capacity is how much power can be pulled from the battery for how long and in what way.

The capacity gained by changing energy draw or current drain can be determined by evaluating Peukert's Formula T = C / Iⁿ where T is how long you can drain current I from a battery that has a capacity C and an internal resistance characteristic n. The capacity constant C in this form is not a specification sheet constant but would need to be calculated (see below) from typical battery specification charts if looking for a solution at one point. For differences between two levels of power draw, it can be left as an arbitrary constant. Note that the battery voltage is also relatively fixed so current is directly proportional to power and that is why current is often used instead of power.

You can use pre-existing calculators (see Smartgauge Peukert Calculator or Uve's Battery Page) or you can roll your own. To gain a better feel for what is involved, you only need a handle on basic algebra. Here is an example for comparing capacity differences due to a change in current draw.

Calculating it out

For the case of a comparison when the current changes by a factor of x, the comparison would be expressed as the difference between the time the capacity C provides a reference current T_a = (C / iⁿ) and time at current xi such that T_b = (C / (xi)ⁿ) given the battery identified by the constants C (characteristic capacity) and n (Peukert coefficient). The problem here is to figure out the relative change in capacity due to a change in current. That is first approached by using Peukert's relationship to identify one of the capacities in terms of the other. Since the only variables in the relationship are time and current and we know the change in current (xi), that means solving for the time at one current based on the fact that the comparison value has x times that current.

Take the difference in times current is available T_a - T_b = (C / iⁿ) - (C / (xi)ⁿ)

factor the right hand side to C ((1 / iⁿ) - (1 / (xi)ⁿ) )

Determine that the lowest common denominator is (xⁿ iⁿ)

multiply the first term by xⁿ top and bottom to obtain a common denominator (xⁿ iⁿ)

and scale to C (( xⁿ / (xⁿ iⁿ) ) - (1 / (xⁿ iⁿ) ))

simplify to C ( xⁿ – 1) / (xⁿ iⁿ) or (C / iⁿ) (( xⁿ – 1) / (xⁿ ))

Substitute T_a = (C / iⁿ)

and we get the difference between the times for current draws differing by a factor of x as T_a ( (xⁿ - 1) / xⁿ )

So, when the time at increased current by factor of x, T_b= T_a - ( T_a ((xⁿ - 1) / xⁿ )) = T_a (1- ((xⁿ - 1) / xⁿ )),

Note that this is now simply a relationship between the time available, the proportional change in current, and the Peukert coefficient. The actual current or voltage or capacity of the battery are not at issue in this consideration as long as the comparison is for identical batteries of the same voltage, capacity, and Peukert coefficients.

Example

Without Peukert, twice the current should mean half the time or T_b= T_a / 2. What this calculation shows is that, instead of reducing the time by the same factor as the current is increased, for the case of twice current, the reducing factor for time is instead (2ⁿ - 1) / 2ⁿ . To see what this means, here is the number crunching for a 100 AH battery (like the typical 12v) and a 200 AH battery (like the typical 6v). The “n” column is a Peukert coefficient over a reasonable range for most wet cell lead acid batteries with the theoretical ideal 1.0 added to illustrate the non-Peukert case. The “2 exp” and “factor” columns are the evaluation of 2ⁿ and (2ⁿ - 1) / 2ⁿ ). The '20 hour' column uses a 20 hour, such as the common specification chart time as a reference and shows the time you'd be able to draw twice the current. The '10A capacity' shows how that extra draw reduces the battery capacity from a reference of 100 AH. The '%diff' is how much the capacity was reduced by percentage for a particular Peukert Coefficient when doubling the current. Available capacity and %diff are also shown for a reference battery with twice as much capacity. The 'R' column is the value discussed below.

x=2				100 AH battery		200 AH battery
			20 hour	5A reference		10A reference
n	2 exp	factor	2I time	10A capacity	% diff	20A capacity	%diff	R
1.00	2.0	0.5	10	100	0	200	0	100
1.10	2.14	0.53	9.33	93.3	7.18	186.61	7.18	117
1.12	2.17	0.54	9.2	92.02	8.67	184.04	8.67	121
1.14	2.2	0.55	9.08	90.75	10.19	181.5	10.19	125
1.16	2.23	0.55	8.95	89.5	11.73	179.01	11.73	129
1.18	2.27	0.56	8.83	88.27	13.29	176.54	13.29	133
1.20	2.3	0.56	8.71	87.06	14.87	174.11	14.87	138
1.22	2.33	0.57	8.59	85.86	16.47	171.71	16.47	142
1.24	2.36	0.58	8.47	84.67	18.1	169.35	18.1	147
1.26	2.39	0.58	8.35	83.51	19.75	167.02	19.75	152
1.28	2.43	0.59	8.24	82.36	21.42	164.72	21.42	157

Compare the “factor” to 0.5 for the simplest case where n=1. As you can see, it rises from the non-Peukert 0.5 at the ideal to nearly 2/3 at coefficient values just a bit larger than you'd find in a typical battery. In other words, doubling current means you subtract from half to 2/3 of the time it is available depending upon the coefficient that describes the internal resistance of the battery.

The table also shows that this reduction in capacity is proportionally the same no matter the size or the voltage of the battery.

Implications for increasing the RV battery bank

If you turn this around to represent the reduced current draw from each battery in a bank you can find out what happens if you add batteries to your RV. For two batteries in parallel the current from each battery for your RV load is reduced by half so x= 0.5 and for three it would be x = 0.33. Substitute those values into the equation and, for a 1.22 typical coefficient, one additional battery would increase capacity in each battery by 14% and a third would get it to 22% - a fourth battery gets it to a quarter current with a 26% improvement over the single battery providing full current.

So the advantage of 2 batteries over 1 is 2.14 times the capacity and for 3 batteries it is 3.22 and for four it is 4.26 times the one battery capacity. When you add batteries to your RV battery bank, you will get more than a simple spec sheet might make you think by simply adding numbers, Note that this adding of batteries is for a 12v battery or sub bank as you have to keep the output of your bank at 12v for your RV.

Another factor to keep in mind is that if you substitute a pair of 6v in series to upgrade your one 12v battery different considerations apply. You will be replacing one battery with two and that will share the load and that will create reduced currents that increase available capacity above that you might expect from the specifications sheet. After that, there is what happens when you change loads in your RV. The current for that load is distributed to parallel parts of your bank but run through the entire string of serial parts. This implies that serial batteries will be less 'sensitive' to load changes as far as capacity is concerned. That could be a concern if you run larger loads for any amount of time, such as an inverter running an entertainment or computer system or a furnace on a cold night. One way to investigate this might be to look at the rules for adding resistance in parallel and series.

Refining the Equation

Wikipedia and Smartgauge describe how the Peukert relationship is defined and how it is adapted to conform to modern battery specifications. The relationship is defined for the 1 amp capacity rating C₁ , current I, time t, and constant k.

C₁ = Iⁿt or t = C₁/ Iⁿ

The adjustment to the usual amp hour rating uses the time of the rating H, and the current for that rating c in the following form.

t = H / (iH/c)ⁿor t = (H / (H/c)ⁿ) (1 / iⁿ) or t= (H/Hⁿ) (cⁿ) (1 / iⁿ)

That means the 1 amp hour rate C₁ is replaced by (H/Hⁿ) (cⁿ) to obtain a 'Peukerized' capacity at hour rating H. See the “R” column in the table above to see how this changes with a 20 hour rated 100 amp hour battery as the sample. It is essentially the same calculation as shown above scaling the current from 1 amp to the current needed to obtain the specified amp hour rate. You can see how it is used in the Smartgauge Peukert Calculator (an Excell Spreadsheet) . This spreadsheet also provides a graph of available amps versus discharge rate. You will need to unprotect the sheet to see the formulas but they are straightforward calculations of this adjustment and its consequences for particular hour values and its derivatives.

You can calculate a Peukert Coefficient from an AH specification and reserve minutes at Uve's Battery Page.