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Safety of 300uA pull to 12V on IO pin before MCU startup
Psi:
I have a circuit with 12V going to an ADC pin of ATMega809 through a 39k resistor (part of batt Vsense divider)
The MCU is powered from a 7805 output from that 12V source too.
When the power switch is closed the 7805 will take some time to startup but the ADC input gets 12V through the 39k instantly. So for a very short time the GPIO maybe above VCC/AVCC and the protection diodes will try to
shunt that 300uA to VCC.
Is there any risk of damage to the MCU from this 300uA?
I would have expected it to be totally fine due to only being 300uA and only for a very short time until the 7805 starts. I also have 100nF across the ADC input so the cap has to charge first.
Although I guess if you cycle the power quickly the cap will already be charged.
But I'm trying to track down an issue where the MCU is getting damaged and develops an internal short on AVCC to GND. So I just thought I'd confirm that my understanding is correct and that 300uA isn't going to do shit so i need to look elsewhere for the issue.
Thanks.
Psi:
My justification that it's fine is the pic below from datasheet.
Which seems to suggest the inputs will be fine as long as current is limited to 1mA into the port.
ataradov:
I'm not familiar with ATmega behaviour, but this is the exact text and limiting values used in the SAM D21, for example. And there your use would allowed and perfectly within the spec.
Psi:
I'm somewhat worried the AVCC pin might be a little more sensitive on this MCU vs preveious ATMegas.
It's one of the newer MCUs Microchip has released since they bought ATMEL and some of the peripherals are quite different register wise compared to an older ATMega, so I do wonder what else has changed inside the chip.
hm.. is it a good idea to use a zener or TVS to clamp AVCC under VCC?
I don't actually need the full 5V on AVCC since the ADC reference being used is the internal 2.5V ref.
ataradov:
How exactly is your capacitor connected? Is it right on the pin with no further limiting? Would not that make it discharge into the pin with the current limited only by the ESR when power disappears?
But also, if it is a resistive divider, then the pin would still only have the divided voltage no matter what. So, it will never see 12 V.
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