The reset circuit as shown in the figure below: The article says that there is a problem with this circuit. Change 0Ω to 1KΩ to get it. The reason is that the external reset IC drive type is CMOS push-pull drive. I don’t understand it~~
Ask me why the resistance should be changed to 1KΩ?
William Posted on June 30, 2020
If the stm32 is reset, the NRST pin will output a short low level. If the external reset IC is a push-pull output, it should normally output a high level. When the middle string has a 0 ohm resistor, is it not equivalent to a short circuit? The 1K resistor on the string is used to limit the current. At the same time, the voltage division between the 1K resistor and the 10K resistor can also ensure the reliable reset of the STM32 during external reset.
Kate Posted on June 30, 2020
The guess should be the voltage division at NRST. When the reset is released, its level cannot be pulled high. After increasing the resistance of this resistor, the voltage division can meet the requirements.
Camden Posted on June 30, 2020
When the reset IC output is high, and the resistance is 0Ω, NRST is almost directly connected to the power supply. At this time, the short reset pulse of the STM32 MCU watchdog will not make NRST low (logically, There has been a risk phenomenon).
When the reset IC output is high, the resistance is 1KΩ, that is, a 1KΩ resistor is pulled on the NRST signal line. At this time, the watchdog can be reset normally.
