Reduce Power Consumption

[Mohsen]

hi everybody.

Please help me on this

I use nRF51822 as iBeacon with ble_app_beacon in Nordic DK in below state:

• only advertise in iBeacon mode
• interval = 100 ms
• DC/DC disable
• All Peripherals are disabled like GPIO

My source code is here:

int main(void)
{
ble_stack_init();
advertising_init();
advertising_start();

for (;; )
{
    power_manage();
}

}

As you see, I didn't set power.(it means its 0 dBm) and none of peripherals are Enabled. in this case I use multimeter for measuring current. I use multimeter in uA AC (true RMS) position. in this case, I see 280 uA on multimeter display. now with my calculation, it endure 1 month approximately with CR2030 coin cell battery. is may way true or false. how can I reduce my battery life. because 1 month is very low for my application. I notice that after Programming, I reset my Board and disconnect debuger

my main.c:

main.c

thank you

[Sigurd]

In order to reduce the power consumption, you can try to:

• Increase the advertising interval
• Enable DCDC

There is a tradeoff between the discovery time and the advertising interval, try to do experiments to determine the slowest advertising rate that still provides a fast enough discovery time.

[Mohsen]

thank you Sigurd for your response.

another question. did you see my source code and explanations? are they corrects? could somethings be incorrect? is 280 uA logical? i thought in this case my current be about 20 to 50 uA.

nordic recommend to disable DC/DC in some versions of nRF51822. (i have one of that versions)

thank you.

[Sigurd]

I didn't find anything incorrect. 280 uA seems normal with only 100ms advertising interval. Did you try to increase the advertising interval ?

[Mohsen]

No. I don't want to increase. But I expected the current consumption be lower than this. thank you

[Sigurd]

Is there any reason you won't increase it? From 100ms to 200ms, you will get ~50% reduced power consumption.

[Mohsen]

because I think It may has bad effect on user experience. although as I saw, with reducing interval to 50%, Power will reduce to 33% due to a steady constant current.

thank you