2026-05-26 10:31 PM
Hiroshi Yamada
Solved! Go to Solution.
2026-05-26 10:56 PM
The examples in DS11139 (B.2, B.3) suggest 10k. This is obviously low enough to pull down reliably.
Any lower value increases the current without need, but will not hurt besides of that.
2026-05-26 10:56 PM
The examples in DS11139 (B.2, B.3) suggest 10k. This is obviously low enough to pull down reliably.
Any lower value increases the current without need, but will not hurt besides of that.
2026-05-26 11:37 PM
@Hitatachipoo415256 You can also connect the BOOT0 pin directly to GND, but that would no longer allow the pin to be explicitly set to 1. If you use 510Ω, then when setting it to 1 a correspondingly large current would flow, which is probably undesirable and is probably what @mfgkw meant.
Since it is a CMOS input, practically no current flows out of or into BOOT0, so, as also mentioned just now, around 10kΩ is a common value, which also helps to attenuate any small interference picked up (BOOT0 is only queried during a very short time window at power-up anyway).
Regards
/Peter
2026-05-27 9:54 AM - edited 2026-05-27 10:50 AM
@Hitatachipoo415256 wrote:In some cases, lower values such as 510 ohm are used.
Please give a link to such an example
As @mfgkw said, the datasheet is the foundational document.
See also Application Note AN4488, Getting started with STM32F4xxxx MCU hardware development:
2026-05-27 5:50 PM
Hi Andrew,
Best Regards,
2026-05-27 9:48 PM
Using a low value resistor will not increase current draw as the pin is floating. Any value between 0 and 1meg or more will work.
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