Guys, in my sketch I used the wdt library and activate it for 8s and clean it throughout the program, but my question is whether in addition to that I have to change the fuse related to wdt as well?
My car broke...what's wrong with it?
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Lockout solenoid. Find the release button near the stick.
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Did you try the key?
See this comprehensive fuse bit tutorial:
https://www.martyncurrey.com/arduino-atmega-328p-fuse-settings/
The question is whether, in addition to programming the watchdog timer in the code, I also need to activate its fuse, in this case it is an Arduino Uno
No, you shouldn't have to modify the fuses.
The fuse is to cause the WDT to be turned on at power-on-reset, which you don't want because the bootloader is going to run first anyway (and also manipulate the WDT.)
As long as you manually enable the WDT in your sketch, the fuse can remain as-is.
1. What do you want to achive once 8s timeout of the WDT occurs?
(1) Do you want that your UNO will be in sleep mode; WDT timeout will wake-up the MCU and the execute WDT interrupt routine first and then will jump at system reset vector?
(2) Do you want that your UNO will be in sleep mode; WDT timeout will wake-up the MCU and then will make a jump at system reset vector? (There is no execution of WDT Interrupt Routine.)
2. You may consult the meaning of the WDT fuse bit from the following Tables:
3. Please, post your sketch.
In fact, I activated wdt in the sketch with a time of 8s and well before that I always clear its counter in the main loop. My question was whether I would need to activate it in the fuse as well, but it seems not.
Please, post your sketch to alow me to understand the purpose of WDT in your project.
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