Battery consumption for nrf52840

[preethi16]

Hi There i have a Code handling two events simultaneously, one event is advertisement And the other is scanning which  is done every half an hour. the average current consumption for only advertisement is 75 mA and average current consumption for advertisement plus scanning , happening half an hour once is 150 mA. How should i calculate the battery life in such a case, usually it is batter capacity / average current consumption. But here you have two events are happening

should it be like this result avg current = 75 * (time in hour ) + 140 *(time in hour) - for a period of one hour.  then, battery life = battery consumptiuon/ result avg current.

[Simonr]

Hi Preeti

In order to get the actual average current, you need to measure the full advertising and scanning event as it won't be 75mA during the full advertisement/scanning. Then you have to add this to the full current draw for the rest of the half-hour when the chip is asleep/inactive to find the average current. Then the battery life (in hours) will be the battery's capacity (specified in the datasheet) divided by the avg. current you found.

Best regards,

Simon

[preethi16]

Hi Simon 

Thanks for answering. So it would be advertising in the reset of the period . So the advertising is carried on continuously with an interval of 360ms and scanning is done along with advertising for every half an hour. So only advertising gives me 75 micro Ampere  as avg current and scanning along with advertisement gives me 150 micro AMphere as avg current. So how should i calculate the average current from both to find the battery life.