I'm using an nrf52 and an OPR5005 reflective optical sensor which takes roughly 20mA. In the design I have only 30mmx30mm space, so only suitable battery is cr2032 or cr2477 so far. The question is how far I can go without switching the sensor on and off? (sensor should be always open)
EDIT : Here is my final schematic file after answers below.
I decide to go with nrf24l01 and for mcu I put a stm8l also I add a load switch to on/off the sensor
Typical power consumptions for IC's:
nrf24l01 : 600nA at powerdown, 0.12mA at TX (it will only transmit from eeprom and it will do it only once when I power up device)
stm8l : 3.7uA at sleep, 0.5mA at run mode (it will wake every 25ms and will enable sensor and record the data to eeprom and will sleep again. It will take almost 5ms estimated, including settling times)
OPR5005 : 15-20mA in run mode, 12nA at sleep (leakage from load switch)
Best Answer
A human eye blink is almost always longer than 50 ms. If you're only counting eye blinks, you only need a few measurements during each blink—one measurement every 15 ms would be plenty. The datasheet for the OPR5005 shows typical rise and fall times of below 100 μs as easily achievable, so having the sensor on for 300 μs before each measurement should be plenty to allow it to settle.
Therefore, the sensor need only be on with a duty cycle of (300 μs / 15 ms) = 2%, which would reduce its average current consumption to (20 mA · 2%) = 400 μA.
If this was the only load on a CR2477 battery (capacity = 1 Ah), it would allow for a battery life of (1 Ah / 400 μA) = 104 days. In practice, other loads, such as the nRF52, will reduce this substantially.