How long will the battery last? Estimating it properly

Capacity over average current is the easy part. Here is what the cell's rated drain, its internal resistance and the cutoff voltage do to that answer.

The estimate everyone starts with is right as far as it goes:

$t = \frac{Q_{usable}}{\bar{I}}, \qquad \bar{I} = I_{active}D + I_{sleep}(1-D)$

A sensor that sleeps at 2 µA and wakes for 10 ms every 10 seconds at 20 mA averages 22 µA, and on a 235 mAh coin cell that is 10 683 hours — 1.22 years. The battery runtime calculator does this, including self-discharge as a parallel drain. The Li-ion charge calculator covers the other direction: how long a charge takes, and what the charger dissipates.

The reason field trials come back shorter is not that the arithmetic is wrong. It is that the capacity in the numerator was measured under conditions your circuit does not meet, and that a cell can stop working while it still holds charge. Both are quantifiable, and both are in the manufacturers’ own documents.

A current waveform that sits at a few microamps and rises to twenty milliamps for a short burst every ten seconds, with the time-weighted average drawn across it as a horizontal line barely above the sleep level. — Fig 1 — The model every runtime estimate starts from. The average is the time-weighted mean of the two states. At this 0.1 % duty cycle the burst still supplies 20 µA of the 22, so the burst is what to attack first — but that balance flips as the wake period lengthens, which is the next figure.

Which of the two currents to work on

In the example above the burst contributes 20 µA and the sleep contributes 2 µA, which is close enough to make either look worth optimising. That balance shifts fast with the wake period, and the shape of the trade decides where effort goes.

Wake less often and the burst term falls in proportion — until it drops below the sleep current, at which point the curve flattens onto a ceiling that no amount of firmware work can lift. The knee is where the two terms are equal, at a wake period of $I_{active}\,t_{active} / I_{sleep}$ : for a 2 µA sleep that is 100 s, and for 10 µA it is 20 s. Past it, the next order of magnitude of runtime is a hardware problem: a lower-leakage regulator, a part with a genuinely low-power sleep mode, or one fewer always-on peripheral.

There is a second ceiling behind that one. Energizer’s lithium coin handbook gives the shelf life of a lithium coin cell as 10 years at room temperature, with about 1 % of capacity lost per year “due to ingress and egress of vapors through the seal”. A design that computes to fifteen years is a design that will be replaced at ten.

Runtime plotted against wake period on log axes, for several sleep currents. At short periods the curves converge because the active burst dominates; at long periods each flattens onto a ceiling set by its own sleep current. — Fig 2 — Where the effort belongs. Waking less often helps only until the sleep current takes over, and after that no amount of duty-cycle work moves the answer. The knee — where the burst and the sleep contribute equally — is at I_active·t_active / I_sleep: 100 s for a 2 µA sleep, 20 s for 10 µA.

The capacity was measured at a current you will never use

This is the largest single gap between the datasheet and the application, and it is stated plainly rather than hidden. The Energizer CR2032 datasheet gives its 235 mAh “to 2.0 volts (Rated at 15K ohms at 21°C)” — that is 0.19 mA, continuous, at room temperature. The handbook’s table gives the same for the whole range:

CR1025    30 mAh    rated at 68 kΩ   (~43 µA)
CR1220    40 mAh    rated at 45 kΩ   (~64 µA)
CR2016    90 mAh    rated at 30 kΩ   (~97 µA)
CR2032   240 mAh    rated at 15 kΩ   (~190 µA)
CR2430   290 mAh    rated at 10 kΩ   (~290 µA)
CR2450   620 mAh    rated at 7.5 kΩ  (~390 µA)

Every one of those tests runs down in three to ten weeks — 24 days for the CR1616, 66 for the CR2450. The rating drain was chosen to make the measurement finish, not to resemble a product. The handbook says as much:

The capacity of a battery in an application will depend on the device drain rate and the cutoff voltage.

Two directions of error follow. A design averaging 22 µA draws less than the rating current, and lithium coin cells are generally a little more efficient at lower drains — so the capacity is, if anything, slightly better than the nameplate. A design averaging 2 mA draws ten times more than the rating current, and gets substantially less. Neither case is described by the number on the front page.

The published capacity of each coin cell in a range plotted against the drain current its rating was measured at, showing that every rating uses a current of tens to hundreds of microamps and that the resulting test lasts between about three and ten weeks. — Fig 3 — Every coin cell capacity in Energizer's handbook, with the drain the number was measured at. They are all in the tens-to-hundreds of microamps, and every test runs down in 24 to 66 days. That is a convenient test duration, not a description of your circuit.

The cell can fail while it still has charge

Here is the mechanism that ends most coin-cell products. The handbook is direct about the scale of it:

In general, the IR of lithium coin cells is significantly higher than what is found in other common battery chemistry systems. For example, the starting IR of a 2032 battery is near 10 ohms, the starting IR of an E92 AAA alkaline battery is near 0.3 ohms.

Thirty times the internal resistance of an alkaline AAA, when new. And it does not stay there:

The battery IR will typically increase during discharge due to the impact of the reaction by-products on the battery chemistry.

The consequence is arithmetic. During a burst the terminal voltage is $V_{OC} - I R_{int}$ , and the cell is considered flat when that falls below the cutoff its capacity was rated to — 2.0 V. So the largest burst the cell can supply is

$I_{max} = \frac{V_{OC} - V_{cutoff}}{R_{int}}$

fresh, 3.0 V open circuit, 10 Ω     100 mA
                            30 Ω      33 mA
                            50 Ω      20 mA
later, 2.7 V open circuit,  50 Ω      14 mA

A 20 mA radio burst that worked on a fresh cell on the bench fails on a half-used cell in a cold enclosure, with most of the capacity still in it. Nothing about the remaining charge changed — only the cell’s ability to deliver it quickly.

Terminal voltage during a current pulse plotted against pulse current, for several values of internal resistance. With a fresh cell the voltage stays above the cutoff to a hundred milliamps; with an aged cell it crosses the cutoff below twenty. — Fig 4 — Why a coin cell dies with capacity left in it. The handbook gives the starting internal resistance of a 2032 as "near 10 ohms", and states that it rises through discharge. The terminal voltage during a pulse is simply OCV − I·R, and the cutoff is crossed at a current that falls as the cell ages.

The largest pulse current a coin cell can supply while staying above the cutoff, plotted against its internal resistance, for two open-circuit voltages. The ceiling falls hyperbolically as resistance rises. — Fig 5 — The same relation solved the other way: the biggest burst the cell can supply without falling through the cutoff. A fresh cell at 3.0 V handles 100 mA; the same cell later in life, at 2.7 V open-circuit and 50 Ω, handles 14 mA. A radio burst that worked on the bench fails in the field for this reason and no other.

What the handbook’s own worked example shows

The handbook computes an internal resistance from a pulse measurement, and the numbers are worth reading carefully:

The IR calculation would be (3.279V - 2.429V) ÷ (.097A - .000003A) = 9Ω.

That is a fresh cell — and 97 mA has already dragged it from 3.279 V down to 2.429 V, less than half a volt above the cutoff. A pulse that size is right at the edge on a new cell. It is also why the handbook’s pulse-effects section notes that a high-pulse discharge curve “would meet a 2 volt cutoff much sooner than the average drain CCV due to the voltage drop during the pulse”.

The fix is a capacitor, and it is bigger than you expect

Put a capacitor across the cell and it supplies the burst while the cell refills it slowly in between. The cell then sees the average current, which is what the runtime equation assumed all along. The size follows from the charge the burst removes:

$C \ge \frac{I\,t}{\Delta V}$

For the 20 mA, 10 ms burst above and a 0.2 V droop allowance, that is 1000 µF. Not a decoupling capacitor — a tantalum, a supercapacitor, or one of the hybrid-layer capacitors sold specifically for this. Shorten the burst to 1 ms and it becomes 100 µF, which is the single most useful thing to know here: burst duration and capacitor size trade one for one, so a radio that transmits faster is cheaper than one that stores more.

Two practical constraints come with it. The capacitor’s ESR is in series with the cell’s tens of ohms, so it has to be small by comparison or it simply moves the problem — and the same DC bias and temperature loss that class 2 ceramics suffer applies here, where a nominal 100 µF X5R at 3 V may be well under half that. The other is charge time: the capacitor has to refill through the cell’s own resistance before the next burst, and $5RC$ with 50 Ω and 1000 µF is a quarter of a second.

The capacitance needed to hold a supply droop within a limit during a current burst, plotted against burst duration for several currents. Millisecond bursts need hundreds of microfarads and ten-millisecond bursts need millifarads. — Fig 6 — The fix, sized. A capacitor across the cell supplies the burst and is refilled slowly between bursts, so the cell never sees the peak. The requirement is C ≥ I·t/ΔV, and it is unforgiving: shortening the burst is far cheaper than enlarging the capacitor, because the two trade one-for-one.

Self-discharge, as a current

Self-discharge is quoted as a percentage per month or per year, which cannot be compared with anything else on the schematic. Convert it to a current and the comparison becomes obvious:

CR2032, 1 %/year                 0.27 µA
Li-SOCl₂ AA, 0.08 %/month        2.6  µA
AA alkaline, 0.3 %/month        10.3  µA
NiMH low-self-discharge, 1.5 %/month  41  µA
Li-ion 18650, 2 %/month         82    µA
NiMH standard, 20 %/month      548    µA

Against a 22 µA design, the coin cell’s self-discharge is one per cent of the load and can be ignored. A standard NiMH cell’s is twenty-five times the load — the battery empties itself while the circuit sits there, and no firmware change touches it. This is the entire reason low-self-discharge NiMH exists, and the reason long-life sensors use lithium primaries rather than rechargeables.

Note also the alkaline row: 10 µA is half the example design’s load. A multi-year alkaline product is spending a third of its energy on self-discharge before the circuit draws anything.

Self-discharge expressed as an equivalent continuous current for several chemistries, drawn against the average load current of a low-power design. For the leakiest chemistries the self-discharge alone exceeds the load. — Fig 7 — Self-discharge as a current, which is the only way to compare it with a load. A coin cell losing 1 % a year is 0.27 µA — negligible next to a 22 µA design. A standard NiMH cell losing 20 % a month is 548 µA — twenty-five times the load, which no amount of firmware will out-run.

For rechargeables, the charger takes a cut too

The “usable fraction” of a rechargeable pack is not only the cutoff. TI’s SLUAAR1 (LiFePO4 Design Considerations) tabulates how much capacity a charger’s voltage inaccuracy costs, because a charger that cannot be sure it has reached the target voltage must stop short of it:

charge voltage accuracy    Li-ion loses    LiFePO₄ loses
       ±0.5 %                  3.0 %           1.7 %
       ±1 %                    6.4 %           3.4 %
       ±2 %                   13.2 %           6.8 %
       ±3.5 %                 24.5 %          12.1 %

A discrete charger at ±3.5 % gives up a quarter of a Li-ion pack’s capacity before the load has drawn anything, and SLUAAR1 draws the commercial conclusion: “if your design truly needs a 10Whr battery for your application, and your design doesn’t utilize 15-30% of the battery … you will need to buy a 15-30% bigger battery.”

Bars showing the capacity lost to charge-voltage inaccuracy for lithium ion and lithium iron phosphate cells at four charger accuracies, from half a per cent to three and a half. The losses grow steeply with inaccuracy. — Fig 8 — For a rechargeable, part of the "usable fraction" is decided by the charger. TI's SLUAAR1 tabulates how much capacity is lost to charge-voltage inaccuracy alone: at ±3.5 % a Li-ion pack gives up 24.5 % of its capacity before the load has drawn anything.

Choosing the chemistry is mostly not about capacity

SLUAAR1’s comparison table is the fastest way to see the trade, and the row that decides most designs is the last one.

A comparison table of four cell technologies giving energy density, nominal voltage per cell, cycle life and discharge temperature range, from the application note that publishes it. — Fig 9 — TI SLUAAR1's comparison, which is the fastest way to see why the choice is rarely about capacity alone. Note the discharge temperature ranges: the chemistry that stops first in the cold, Ni-MH at 0 °C, is also the one with the least energy density of the three batteries — and the supercapacitor, which runs the widest range, stores almost nothing.

Temperature is where a runtime estimate meets reality. No cell in that table spans the −40 to +85 °C the rest of the board is usually specified over, and charging is narrower than discharging — a Li-ion pack must not be charged below 0 °C, which is a firmware requirement rather than a note. The CR2032’s own range is −30 to 60 °C.

And what happens at the cold end is not primarily a capacity loss. The handbook attributes it to resistance:

Cold temperatures cause the electrochemical reactions that take place within the battery to slow down and will reduce ion mobility in the electrolyte. … For example, a wireless garage door sensor could stop functioning in the cold of winter due to an excessive voltage drop.

That is the pulse-ceiling mechanism again, with the internal resistance raised by temperature instead of by age. A design with enough buffer capacitance to survive an aged cell usually survives a cold one for the same reason.

Operating temperature ranges drawn as horizontal bars for several cell technologies, against a marked industrial range. Several chemistries do not span it. — Fig 10 — Published operating ranges, taken from the Energizer CR2032 datasheet and SLUAAR1's table. The handbook is blunt about what happens outside them: "a wireless garage door sensor could stop functioning in the cold of winter due to an excessive voltage drop" — not because the capacity has gone, but because the internal resistance has risen.

One more effect, if the cell has been in a drawer

Long storage grows a passivation layer on the lithium anode. The handbook describes it as a trade the chemistry makes deliberately — “This layer reduces the rate of self-discharge of the battery by slowing the reaction between the lithium metal and the electrolyte” — with a side effect on first use:

The passivation layer due to long term storage can contribute to a slightly higher initial internal resistance and subsequently an increased voltage drop when the battery is first put into use. Once a load is placed on the battery, the passivation layer will become thinner, and internal resistance typically returns to normal.

So a device that fails its first transmission after sitting in a warehouse, and then works, is not faulty. It is worth knowing before spending a week on it, and worth designing around if the first thing the product does is a radio burst.

Putting the derating together

The factors that apply to the worked design are the ones this article has already met, and the handbook’s own curves put sizes on them. Its pulse-effects figure shows a 2032 under 75 Ω pulses — about 30 mA, a little larger than the design’s 20 mA bursts — meeting the 2.0 V cutoff at about 190 mAh, 81 % of the nameplate: that is what unbuffered bursts on an ageing cell look like. Its temperature figure shows the same cell at 0 °C delivering about 150 mAh to 2.0 V under a 1 mA drain, against about 228 mAh at 21 °C, a factor of 0.66. Self-discharge at about 1 % a year over the runtime that is left costs well under 1 %.

Two factors that are often listed do not apply here, and it is worth saying why. The nameplate is already quoted to 2.0 V, so a 2.0 V cutoff removes nothing from it — the handbook’s warning is about cutoffs above 2 V. And a 22 µA average is below the 190 µA rating drain, so the drain-rate correction, if anything, runs slightly the other way.

The two that do apply multiply: 235 mAh becomes about 125 mAh, a factor of 1.9, and 1.22 years becomes 0.65. They are not independent — both are the internal resistance at work — so treating them as a product is the pessimistic reading, which is the one to design to.

A waterfall from the nameplate capacity down to what the worked design can use, with each step labelled by the effect that removed it and the handbook curve it was read from: unbuffered pulses reaching the cutoff early, the cold end of the range, and self-discharge over the remaining runtime. — Fig 11 — The same 235 mAh cell, stepped down by the effects the worked design actually meets, each sized from the handbook's own curves at the handbook's test conditions. Unbuffered pulses and cold are both the internal resistance at work, so multiplying them is the pessimistic reading — but 125 mAh is the number to design against, not 235.

The estimate that survives a field trial

Fig 12 — The order that catches the failures. Most estimates do step 1 and step 4 and skip the two in the middle, which is exactly where the missing factor of two lives.

Measure the sleep current on the real board, not from the MCU datasheet. Pull-ups, a regulator’s quiescent current, a floating input on a CMOS pin and flux residue between a high-impedance node and a rail all land here, and they are the usual reason an estimate is out by a factor of five.
Check the peak separately from the average. They fail independently: one is a voltage question about internal resistance, the other a capacity question. A cell can pass either and fail the other.
Look up what drain the capacity was rated at and how far your average is from it.
Buffer the bursts if the peak is more than a few milliamps from a coin cell, and size the capacitor from $I t/\Delta V$ with the cell’s aged resistance in mind, not its fresh one.
Derate for the cold end of the range the product will actually see, which is rarely 21 °C.
Then halve it and see. If the halved number is still acceptable, the design has margin; if only the unhalved one is, it does not.

Work the average and the runtime out in the battery runtime calculator, which takes the two-state profile, the usable fraction and the self-discharge, and reports how much of the total drain is self-discharge — because when that share is large, the answer is a smaller battery of a better chemistry rather than a bigger one of the same.