STM32 programming and BOOT0 pin The 2019 Stack Overflow Developer Survey Results Are...
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STM32 programming and BOOT0 pin
The 2019 Stack Overflow Developer Survey Results Are InSTM32F102 USB ProgrammingSTM32F303RET's core is always halted after programmingST-Link debugger/programmer failed to find STM32L152 MCU on designed PCBSTM32 prototype unable to connect to PC using ST-Link/V2 SWDSTM32 'Connect under reset' suddenly stopped workingCannot Program a Custom STM32 BoardWhy is this hex file different than the code programmed onto the device?STM32F091 Jump to Bootloader from applicationSTM32 & ST-LINK - SWD connector not workingSTM32F0 - interrupt/breakpoint not working on certain hardware
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I am developing a STM32H7 board which will be programmed via SWD.
I am curious, is there any reason why I should connect BOOT0 pin to GND via resistor and not directly to GND?
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
stm32 bootloader
asked 4 hours ago
ningboningbo
243
$endgroup$
add a comment |
$begingroup$
I am developing a STM32H7 board which will be programmed via SWD.
I am curious, is there any reason why I should connect BOOT0 pin to GND via resistor and not directly to GND?
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
stm32 bootloader
asked 4 hours ago
ningboningbo
243
$endgroup$
$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago
add a comment |
$begingroup$
I am developing a STM32H7 board which will be programmed via SWD.
I am curious, is there any reason why I should connect BOOT0 pin to GND via resistor and not directly to GND?
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
stm32 bootloader
asked 4 hours ago
ningboningbo
243
$endgroup$
I am developing a STM32H7 board which will be programmed via SWD.
I am curious, is there any reason why I should connect BOOT0 pin to GND via resistor and not directly to GND?
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
stm32 bootloader
stm32 bootloader
asked 4 hours ago
ningboningbo
243
asked 4 hours ago
ningboningbo
243
asked 4 hours ago
ningboningbo
243
asked 4 hours ago
ningboningbo
243
asked 4 hours ago
ningboningbo
243
243
$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago
add a comment |
$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago
$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago
$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you're making a board, why not provide for a resistor/jumper to connect Boot0 to +V and GND? Then only populate one of them? Leave your options open. You don't need a resistor. I prefer to use a jumper rather than a resistor.
You can use header shunts as jumpers that can be easily changed but that takes more PCB space, components and more assembly work. Or just place an SMD chip footprint you can either choose to use an SMD resistor or SMD jumper (zero ohm resistor) during assembly. Then you can fiddle with things all you want after-the-fact.
answered 3 hours ago
ToorToor
1,527210
$endgroup$
add a comment |
$begingroup$
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
That's exactly why you might want to use a resistor and not a zero ohm jumper. With a resistor, you can manually override it with a piece of wire to the opposite rail start in bootloader mode once, without having to get out the hot air station and change the resistor.
Granted, if you have the hardware reset line brought out, and an SWD probe that actually drives it, and a suitable SWD software config (both common points of failure - and specifically a failure that may not be noticed in routine use, but only when this kind of recovery fails to work) then that is another way to work around disabled SWD lines.
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you're making a board, why not provide for a resistor/jumper to connect Boot0 to +V and GND? Then only populate one of them? Leave your options open. You don't need a resistor. I prefer to use a jumper rather than a resistor.
You can use header shunts as jumpers that can be easily changed but that takes more PCB space, components and more assembly work. Or just place an SMD chip footprint you can either choose to use an SMD resistor or SMD jumper (zero ohm resistor) during assembly. Then you can fiddle with things all you want after-the-fact.
answered 3 hours ago
ToorToor
1,527210
$endgroup$
add a comment |
$begingroup$
If you're making a board, why not provide for a resistor/jumper to connect Boot0 to +V and GND? Then only populate one of them? Leave your options open. You don't need a resistor. I prefer to use a jumper rather than a resistor.
You can use header shunts as jumpers that can be easily changed but that takes more PCB space, components and more assembly work. Or just place an SMD chip footprint you can either choose to use an SMD resistor or SMD jumper (zero ohm resistor) during assembly. Then you can fiddle with things all you want after-the-fact.
answered 3 hours ago
ToorToor
1,527210
$endgroup$
add a comment |
$begingroup$
If you're making a board, why not provide for a resistor/jumper to connect Boot0 to +V and GND? Then only populate one of them? Leave your options open. You don't need a resistor. I prefer to use a jumper rather than a resistor.
You can use header shunts as jumpers that can be easily changed but that takes more PCB space, components and more assembly work. Or just place an SMD chip footprint you can either choose to use an SMD resistor or SMD jumper (zero ohm resistor) during assembly. Then you can fiddle with things all you want after-the-fact.
answered 3 hours ago
ToorToor
1,527210
$endgroup$
If you're making a board, why not provide for a resistor/jumper to connect Boot0 to +V and GND? Then only populate one of them? Leave your options open. You don't need a resistor. I prefer to use a jumper rather than a resistor.
You can use header shunts as jumpers that can be easily changed but that takes more PCB space, components and more assembly work. Or just place an SMD chip footprint you can either choose to use an SMD resistor or SMD jumper (zero ohm resistor) during assembly. Then you can fiddle with things all you want after-the-fact.
answered 3 hours ago
ToorToor
1,527210
edited 3 hours ago
answered 3 hours ago
ToorToor
1,527210
answered 3 hours ago
ToorToor
1,527210
answered 3 hours ago
ToorToor
1,527210
1,527210
add a comment |
add a comment |
$begingroup$
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
That's exactly why you might want to use a resistor and not a zero ohm jumper. With a resistor, you can manually override it with a piece of wire to the opposite rail start in bootloader mode once, without having to get out the hot air station and change the resistor.
Granted, if you have the hardware reset line brought out, and an SWD probe that actually drives it, and a suitable SWD software config (both common points of failure - and specifically a failure that may not be noticed in routine use, but only when this kind of recovery fails to work) then that is another way to work around disabled SWD lines.
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
$endgroup$
add a comment |
$begingroup$
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
That's exactly why you might want to use a resistor and not a zero ohm jumper. With a resistor, you can manually override it with a piece of wire to the opposite rail start in bootloader mode once, without having to get out the hot air station and change the resistor.
Granted, if you have the hardware reset line brought out, and an SWD probe that actually drives it, and a suitable SWD software config (both common points of failure - and specifically a failure that may not be noticed in routine use, but only when this kind of recovery fails to work) then that is another way to work around disabled SWD lines.
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
$endgroup$
add a comment |
$begingroup$
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
That's exactly why you might want to use a resistor and not a zero ohm jumper. With a resistor, you can manually override it with a piece of wire to the opposite rail start in bootloader mode once, without having to get out the hot air station and change the resistor.
Granted, if you have the hardware reset line brought out, and an SWD probe that actually drives it, and a suitable SWD software config (both common points of failure - and specifically a failure that may not be noticed in routine use, but only when this kind of recovery fails to work) then that is another way to work around disabled SWD lines.
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
$endgroup$
I won't be using the bootloader but can it happen that I accidentally disable the debug port from software and then can't program the MCU via SWD anymore and need to use bootloader mode by putting BOOT0 high?
That's exactly why you might want to use a resistor and not a zero ohm jumper. With a resistor, you can manually override it with a piece of wire to the opposite rail start in bootloader mode once, without having to get out the hot air station and change the resistor.
Granted, if you have the hardware reset line brought out, and an SWD probe that actually drives it, and a suitable SWD software config (both common points of failure - and specifically a failure that may not be noticed in routine use, but only when this kind of recovery fails to work) then that is another way to work around disabled SWD lines.
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
answered 3 hours ago
Chris StrattonChris Stratton
23.2k22865
23.2k22865
add a comment |
add a comment |
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$begingroup$
if you have a resistor to ground, then you can change the state of the pin by connecting it to Vcc ... if the pin is connected to ground directly, then you have to disconnect it before it can be pulled high
$endgroup$
– jsotola
41 mins ago