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Explorer
January 25, 2026
Question

Can I pull-down boot0 with 10kOhm instead of having 3 pin jumper

  • January 25, 2026
  • 4 replies
  • 299 views

STM32f103
In the design guide it recommends to have something switching between vcc and gnd with 10kOhm series resistor, and i need my design to be grounded by default so i may not solder the header even, is it okay if i left the design like fig 1 instead of fig 2, and would it cause issues if i needed to pull it high?

4 replies

gbm
Principal
January 25, 2026

No issue. You may pull it down with anything up to 100k and provide some contacts for connecting it to Vdd for bootloader activation.

You may short BOOT0 to ground directly with a PCB trace if you don't need to invoke the bootloader via hardware.

My STM32 stuff on github - compact USB device stack and more: https://github.com/gbm-ii/gbmUSBdevice
Peter BENSCH
Technical Moderator
January 25, 2026

The pull-up resistor has the advantage of allowing you to set the pin in question to a different level directly at the pin without creating a short-cut. If you are absolutely certain that you never want to switch the pin, you can also connect it directly to a fixed potential (e.g. GND for the BOOT pins) without a pull-up resistor. You can also omit the pull-up resistor if you use a switch/jumper.

Regards
/Peter

gregstm
Senior II
January 25, 2026

To minimise board space, I've always just used a single pull down resistor (no series resistor) with a test point/contact  (close to the micro) I can use to pull Boot0 high if required. I think it is a safer design to always have a pull down resistor. I personally would never connect Boot0 directly to ground, sometimes you need to pull Boot0 high to revive a stuck chip.

TDK
Super User
January 25, 2026

If you don't reassign the SWD pins, and that's how you program, there's little need to be able to pull BOOT0 high.

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