Estimation of battery life while only advertising.

[Jitendra]

Hi there ..

I have question regarding estimation of a coin cell battery life while advertising on nrf52840 PDK. I followed here https://devzone.nordicsemi.com/b/blog/posts/nrf52-online-power-profiler to compute the average current consumption of BLE advertising event. 

Here:e.g.

"

BLE_charge = 4.92 ms * 2.85 mA = 14.022 uC
BLE_avg = 14.022 uC / (500 ms + 5 ms) = 27.77 uA
TOT_avg = 27.77 uA + 2 uA * (505 ms -  4.92 ms)/505 ms = 29.8 uA

" 

Q: TOT_avg is average current consumption of one adertising BLE event? as with parameter "APP_ADV_DURATION"  one can set the timeout of advertising.So for whole "APP_ADV_DURATION " do i have to multiply " TOT_avg* number of advertising BLE event(as one can count from radio notification)" to get total current consumption during the "APP_ADV_DURATION "?

Thank you.

[run_ar]

Yes, the total average is for one advertisement interval. So you can multiply this by the number of advertising events to get the total power consumption for advertising.

Please note that the power calculator is for the nRF52832 so it might deviate from the actual consumption for the nRF52840. I would suggest that you by a power profiler kit to measure the power consumption for the nRF52840.

[Jitendra]

Hi Run_ar

Thank you for yor reply. So following the above formula, I calculated 24.5 uA for one BLE advertising event on nrf52840 pdk running power profiling application(0dBm, 100msec ADV_interval, 1sec APP_ADV_DURATION ). For battery life(220mAh) computation, if wake up the system for  APP_ADV_DURATION, and during APP_ADV_DURATION  10 BLE advertising events occur and I'm waking system by button 10 times a day. Then battery life would be in hours:

battery life(hours) = 220mAh/(24.5 uA*10*10) = 89 hours ??

Please correct me if I'm wrong.

Thank you 

[run_ar]

Actually I think you are mixing units in this case. When converting from coulomb to µA the time unit was removed. So 24.5µA doesn't have a time unit. If you draw this amount of power for 1 second you will get a consumption of 24.5µA/s. Sending 10 advertisements 10 times per day means you are drawing 245µAs per day (it takes ~1 second to send 10 advertisements at 100ms adv int * 10 times per day.).

So the calculation should look like this: 220 mAh / (0.0245 mA/s) = ~8980 h of continuous advertising. Adjusting to advertising for only 10 seconds per day means you can "advertise forever" (neglecting sleep current and battery discharge current).

Basically if this is your use case you can neglect the advertisement power consumption and calculate the battery life time based on sleep current and the battery discharge current.

[Jitendra]

Hi Run_ar..

I think you ignore the 24,5uA current is average over the one advertising interval, which means each advertising interval consumes 24,5 uA current. So 245uA/sec and 2450uAs per day (24,5uA*10*10)??

Thank you

[run_ar]

you are drawing 24.5µA for 0.1 seconds for each advertisement. Doesn't that mean you are using 2.45µAs per advertisement. Multiply this with 10 advertisement events per second and you are using an amount of energy equal to 24.5µAs per second?