4

I want to calculate the lifetime of a battery operated device. Without self discharge of the battery, it is easy to calculate the lifetime, it's just the battery capacity \$Q\$ (given in mAh) divided by the average current \$I\$ (given in mA).

In my case I have for example 300 mAh and average current of 3 uA. I get a lifetime of 100,000 h or 11.41 years. The self discharge of the battery is given as % per year, e.g. 5% per year. The battery capacity is then:

$$Q = Q_0 * D^T$$

Where:
\$Q_0\$ = capacity at the beginning
\$D\$ = the discharge rate (in my example 0.95)
\$T\$ = the time expressed in years

The lifetime in hours is:

$$t = \dfrac{Q}{I}$$

I replace \$Q\$ by the above equation:

$$t = \dfrac{Q_0 * D^{\genfrac (){1pt}{1}{t}{8760}}}{I}$$

My first question: Is my equation for \$t\$ OK or is there a failure?

My second question: If the equation is OK, how to solve it to get \$t\$?

I asked that question on MathOverflow as well.

Community
  • 1
Rick
  • 41
  • 1

2 Answers2

5

In practice reality steps in to make this much simpler and much harder than it may seem. Battery capacity varies quite with current and environmental conditions. For example, at 25 degrees Celsius a 2500 mAh low self disharge NimH cell may have about 2500 mAh capacity at the 10 hour rate (250 mA). It may have 2000 mAh capacity (or less) at 2.5 A. It may have a 3000 mAh capacity at 10 uA or even at 100 uA. Operate the battery at 250 mA and 0 degrees Celsius and capacity may drop to around 1500 mAh or even less.

SO unless you know the conditions accurately AND have a much better than average battery data sheet, it is usually good enough to calculate an approximate effective current for the self discharge losses and add that to the load current. If you know the effective self discharge current directly (as you appear to) the so much the better.

eg 2500 mAh battery. 5 year shelf life. Say 140 uA load current mean. Assume self discharge that effects shelf life is linear. Self discharge current = mAh / current So SDC = 2500/(5 years x 8765 hours) ~~= 2,500,000 uA hours / (4382 hours) ~= 60 uA.

Now add load current. Effective current = 60 uA leakage + 140 uA load = 200 uA

Run time = Battery capacity / effective current = 2500 mAh / 200 uA = 12500 hours or ~= 1.5 years. Actual is liable to be longer as battery capacity will be longer at very low currents.

eg

Russell McMahon
  • 150,303
  • 18
  • 213
  • 391
4

The self-discharge rate is given as a fraction of battery capacity, not a fraction of its current charge.

So, if the capacity is 300 mAh and you lose 5%, or 15 mAh per year, the self-discharge current is on the order of 1.7 µA.

Dave Tweed
  • 172,781
  • 17
  • 234
  • 402