The STM32F103VBT6 resets with a button. After releasing the button, it takes three or four seconds for the microcontroller to execute the program.

把手给我丶

STM32F103VBT6 resets by pressing the button. After releasing the button, it takes three or four seconds for the MCU to execute the program. During these three or four seconds, the voltage of the JNTRST (PB4) pin is 2.3V. After removing R13, the voltage of the pin is 3.3V, but other IOs are all normal low levels, and the chip without code also has only JNTRST (PB4) high level. JNTRST (PB4) is used for ordinary IO ports, connected to IO_LOCK_5, and the hardware is pulled down, as shown in the figure below.

Question 1: After releasing the reset button, the program will be executed three or four seconds later. How can I make the program execute immediately after releasing it?

Question 2: During these three or four seconds, JNTRST (PB0) is at a high level. How can we make it the same as other IOs?

maychang

"STM32F103VBT6 push button reset"

I can't find where the button is in the picture.

It would be best if you could post the complete electrical schematic.

maychang

JNTRST says PB4 before, but then says JNTRST is PB0. It also says "JNTRST (PB4) is used for ordinary IO port, connected to IO_LOCK_5".

把手给我丶

maychang posted on 2019-11-30 14:26 JNTRST is said to be PB4 at the beginning, and then it is said that JNTRST is PB0. It also said that "JNTRST (PB4) is used for ordinary IO port, connected to IO_LOCK_5" ...

The last part is wrong. It is PB4

把手给我丶

maychang posted on 2019-11-30 14:24 "STM32F103VBT6 button reset" I can't find where the button is in the picture. It's better to post the complete electrical schematic.

把手给我丶

https://en.eeworld.com/bbs/forum.php?mod=attachment&aid=NDQ3ODE0fGVjNTc5NmY2YWY5M2I4M2Q3Yzk3NmQ5OTk1ZDE4NzI2fDE3MzE2MzE3OTQ%3D&request=yes&_f=.pdf Schematic diagram PDF file

maychang

Give me your hand. Posted on 2019-12-1 21:47

This diagram is a very normal push-button reset circuit, nothing special about it.

But what is the relationship between this picture and the first picture? I can't tell.

huo_hu

3~4 seconds is definitely not right. JNTRST is the debugger reset pin. This may be the problem. Your program has already run. Initialize and disable JTAG mode first, then configure JNTRST.

caizhiwei

Maybe it is waiting for the internal clock to stabilize? Check the clock initialization, Flash initialization, and while(...) waiting related codes.

ddllxxrr

The original poster's circuit only has a pull-down function, so when it reaches a high level, there is no direct resistor connected to it, which means that it takes a long time to reach a high level. In this way, the reset is always pulled.

把手给我丶

maychang posted on 2019-12-2 08:20 This picture is a normal button reset circuit, nothing special. But what is the relationship between this picture and the first picture? I can't tell.

I posted the download link of the schematic on the sixth floor.

把手给我丶

ddllxxrr posted on 2019-12-4 11:29 The OP's circuit only has pull-down function, so when it reaches a high level, there is no direct resistor connected to it, which means that it takes a long time to reach a high level. In this way, the reset is always blocked...

The reset pin NRST is pulled up. When it is reset, it is measured that the reset pin level changes almost immediately when pressed and released, which is normal.

ddllxxrr

Give me your hand, published on 2019-12-4 14:06 The reset pin NRST is pulled up. I tested it during reset. When pressing and releasing, the reset pin level changes almost immediately, which is normal.

That means there is something wrong with your program, not the hardware.