LSI for IWDG (is resolved)

[veacheslavkris]

Posted on November 21, 2016 at 17:41

Hi, in order to use IWDG should I manualy starting LSI or LSI will start automatically ?

I am going to use CMSIS. Thank you.

[valentin]

Posted on November 21, 2016 at 20:13

You can always check the status of LSI in the RCC register (check ref manual).

And if it's not started where you need it, start it yourself.

[waclawek.jan]

Posted on November 22, 2016 at 11:21

We don't know what STM32 model are you using, but if you look into the respective RM's RCC chapter, chances are that there's ''watchdog clock'' or similar sub-chapter, with a text like this (taken from RM0091):

If the Independent watchdog (IWDG) is started by either hardware option or software access, the LSI oscillator is forced ON and cannot be disabled. I've played a bit with a 'F031 and found that enabling IWDG without prior enabled LSI causes RCC_CSR.LSIRDY being set (without affecting RCC_CSR.LSION), and if RCC_CIR.LSIRDYIE was set previously, RCC_CIR.LSIRDYF gets set too. This is

https://my.st.com/public/STe2ecommunities/mcu/Lists/cortex_mx_stm32/Flat.aspx?RootFolder=https://my.st.com/public/STe2ecommunities/mcu/Lists/cortex_mx_stm32/STM32F0xx%20findings&FolderCTID=0x01200200770978C69A1141439FE559EB459D7580009C4E14902C3CDE46A77F0FFD06506F5B&currentviews=34

behaviour than when ADC autoswitches the HSI14 oscillator...

[veacheslavkris]

Posted on November 22, 2016 at 12:30

Thank you. I think the question now may be closed as resolved.