nRF54L10 with watchdog

[omt]

When I turn on watchod current consmuption rasises about 2uA ( it is  about 50% more than without watchdog) - I want to get lowest power consumption while waiting for semaphore.

Is it possible that simple watchgod uses so much current ?

Initialziation looks like this (30sec reload):

int WatchdogInit(void)

{

int wdt_channel_id;

int err;

    WdtDev = DEVICE_DT_GET(DT_ALIAS(watchdog0));

    struct wdt_timeout_cfg wdt_config = {

        .window.min = 0U,

        .window.max = WDT_TIMEOUT_MS,

        .flags = WDT_FLAG_RESET_SOC,  

        .callback = NULL

    };

   wdt_channel_id = wdt_install_timeout(WdtDev, &wdt_config);

   wdt_setup(WdtDev, WDT_OPT_PAUSE_HALTED_BY_DBG); 

   CLR_WDT();

   return wdt_channel_id;

}

I have an external 32 kHz crystal. And current consumption without watchodg is 4.3uA

[AHaug]

Hi,

This is as I understand it expected. The hardware based watchdog will cause some increased consumption.

4.3 µA idle without WDT is reasonable for System ON idle with LFXO running and RAM retained. +2 µA for the WDT is the cost of an always-running watchdog on this part. You already have the external 32 kHz crystal running (for GRTC), so the WDT is not adding an oscillator, it is just the counter/comparator logic in the low-power domain.

The current increase you see matches the current increase from active WDT: https://docs.nordicsemi.com/r/bundle/ps_nrf54l15/page/chapters/current_consumption/doc/current_consumption.html-unique_1647170005 

Kind regards,
Andreas