10 years =~ 87650 hours.
1 uA drain will require 87.75 mAh in 10 years.
With som shelf life degradation that's close enough to
= 10 mAh / uA / year or
= 100 mAh / uA / 10 years
So your cited 163 mAh battery will supply 1.63 uA mean.
Pushing technology, size and luck may get you to say 5 uA mean.
There are 86400 seconds/day.
There are 1440 minutes/day.
You will find that eg alarm use is much restricted in the allowable use to get 10 years. If 1 uA of the drain is for alarm use then you get 24 uA.hr/day or 86400 uA.seconds or 86 mA.seconds. That's about 240 mW seconds at 3 V. Or say 5 x 50 mW x 1 second burst/day.
An LED can provide ample lighting at 1 mA. Use it 5 times/day x 1 second = 5 mA.sec = 5000 uA.sec or "only" 5000/86400 = 0.06 uA mean drain. Increase as desired and allowed.
Can you run a time keeping IC on say 1 uA?
Probably yes.
So overall it all falls in the area of "notionally possible if really really really clever and careful".
Casio can be expected to be quite clever by now.
Note that if any sort of energy harvesting is being used then all bets are on. Harvesting a uA or few sounds doable.
REAL WORLD EXAMPLE:
There are many others.
In September 2012 user Hli commented:
An EFM32, which is an ARM Cortex M3 MCU, can run on about 1.45µA while
driving a LCD (550nA for the LCD, and 900nA for running the RTC and
keeping its RAM). So a chip keeping only time should be capable to run
on much less than that
The link he then provided is now broken, so:
EFM32 "Gecko" family are M0+, M3, M4 ARm Cortex microcontrollers from Silabs
Silabs EFM32 search
Wonder Gecko
EFM32™ Wonder Gecko 32-bit ARM® Cortex®-M4 Microcontroller
Silicon Labs’ EFM32™ Wonder Gecko 32-bit microcontroller (MCU) family includes 60 devices based on the ARM® Cortex®-M4 core, which provides a full DSP instruction set and includes a hardware FPU for faster computation performance.
Wonder Gecko MCUs feature up to 256 kB of flash memory, 32 kB of RAM and CPU speeds up to 48 MHz. The MCUs incorporate highly differentiated Gecko technology to minimize energy consumption, including a flexible range of standby and sleep modes, intelligent peripherals that allow designers to implement many functions without CPU wake-up and ultra-low standby current. With the lowest active and standby power consumption, the Wonder Gecko is the world's most energy friendly Cortex-M4 MCU.
Other xxx-Gecko variants M0+, M3, M4
Digikey listings of "Gecko" - legion
Lowest cost in 100's with LCD EFM32TG822F32-QFP48T$US2.03/100 Digikey
Lowest power useful mode with RTC running - EM2 - deep sleep
In EM2 the high frequency oscillator is turned off, but with the 32.768 kHz
oscillator running, selected low energy peripherals (LCD, RTC, LETIMER,
PCNT, LEUART, I
2C, LESENSE, OPAMP, WDOG and ACMP) are still
available. This gives a high degree of autonomous operation with a current
consumption as low as 1.0 µA with RTC enabled. Power-on Reset, Brown-out
Detection and full RAM and CPU retention is also included.
EM1 - sleep
In EM1, the CPU is sleeping and the power consumption is only 51 µA/MHz.
All peripherals, including DMA, PRS and memory system, are still available
EM0 - running
In EM0, the CPU is running and consuming as little as 150 µA/MHz, when
running code from flash. All peripherals can be active.
So running in EM0 for 1 ms/s adds 0.15 uA to the EM2 standby load.
Overall, operating in EM2 at around 1 uA mean plus EM0 as required would allow
the 10 years / 163 mAh example target to be met.
___________________________________
Energy harvesting:
Vibration and motion may well be possible energy sources.
A silicon solar PV/solar panel seems viable.
Very roughly power available is 150 Watts/m^2 at 1 sun = 100,000 lux.
A 10mm x 10mm "panel" at 10 lux at those ratings would provide ~=
150 Watt x (0.01m x 0.01m) x 10lux/100000lux = 15 microWatt.
10 lux is dim roomlight - at the level where colour fades into monochrome. Dim!
If that level of sensitivity can be maintained at such low light levels (as it quite possibly can with other 'chemistries') the light powering looks viable.
The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.
ΔV = I × Δt / C
Solving for C gives:
C = I × Δt / ΔV
Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:
C = 25 mA × 25 ms / 0.1 V = 6.25 mF
The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.
For the second example, the numbers work out to:
C = 50 mA × 100 ms / 0.1 V = 50 mF
Average current = 50 mA * 100 ms / 1.0 s = 5 mA.
Best Answer
You can only rely on the (Amp Hour) the battery has. Or for small batteries the mAh (milli amp hour). Anyway all these figures are estimates anyway. The lifetimes depends e.g. on temperature.
I don't know what you need the LED for but you can get LEDs which are a lot more efficient: 1 mA would give you 20x as long.
As K H said: you are wasting energy in a series resistor: Try to find a switching current source. A simple PWM circuit with an inductor comes to mind.