2022-12-15 04:43 PM
Hi team,
I'm developing a PCB that implements STM32MP151AAA3.
I think VBAT is a backup power supply for RTC, so I'm planning to supply power to VBAT from a button battery.
Will the RTC be reset when this button battery is changed?
If it is so, is there any way to avoid resetting, such as connecting a capacitor on the VBAT input terminal?
Best regards,
Solved! Go to Solution.
2022-12-19 05:12 AM
Hi @tatsuya
VBAT (without RETRAM enabled) is guaranteed down to 1.2V (as per datasheet).
if you change battery when it is 1.8V, you have at least 0.6V headroom.
With roughly 1uA, if you compute C=I*t/V, you end up to something in the range of 100uF.
You might use a smaller capacitor value if you ask user to change battery at an higher threshold (e.g. 2.5V seems a more sensible value for a 3V lithium battery) and/or if you allows less time for battery replacement.
Beware that depending on the capacitor size and dielectric technology, 100uF might exhibit a self discharge current which has to be taken into account (especially at elevated temperatures).
Take also into account the real capacitor value (including tolerances and voltage derating if applicable to the capacitor used). It is somewhat surprising when you dig into spec of capacitors.
Regards.
In order to give better visibility on the answered topics, please click on 'Select as Best' on the reply which solved your issue or answered your question. See also 'Best Answers'
2022-12-15 11:43 PM
As soon as the battery is removed or runs empty and no supply voltage is applied, the backup domain and thus also the RTC will lose all information. One way to bridge such interruptions of VBAT is indeed a capacitor in parallel with the battery, the size of which depends on the time to be bridged.
Regards
/Peter
2022-12-18 11:02 PM
Hello Peter-san
Thank you for the reply.
> One way to bridge such interruptions of VBAT is indeed a capacitor in parallel with the battery, the size of which depends on the time to be bridged.
When battery output is 1.8V and it needs to be replaced.
In case it takes 1 minutes to replace the battery, how much is the capacity of the capacitor?
(Is 100uF enough? Under the condition shown in the attached figure for example)
Best Regards,
2022-12-19 05:12 AM
Hi @tatsuya
VBAT (without RETRAM enabled) is guaranteed down to 1.2V (as per datasheet).
if you change battery when it is 1.8V, you have at least 0.6V headroom.
With roughly 1uA, if you compute C=I*t/V, you end up to something in the range of 100uF.
You might use a smaller capacitor value if you ask user to change battery at an higher threshold (e.g. 2.5V seems a more sensible value for a 3V lithium battery) and/or if you allows less time for battery replacement.
Beware that depending on the capacitor size and dielectric technology, 100uF might exhibit a self discharge current which has to be taken into account (especially at elevated temperatures).
Take also into account the real capacitor value (including tolerances and voltage derating if applicable to the capacitor used). It is somewhat surprising when you dig into spec of capacitors.
Regards.
In order to give better visibility on the answered topics, please click on 'Select as Best' on the reply which solved your issue or answered your question. See also 'Best Answers'