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Associate
August 22, 2026
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BOOT0 pull-down resistor value for STM32F469 and STM32F412

  • August 22, 2026
  • 5 replies
  • 24 views
Hello, 
We are using STM32F469 and STM32F412 in our design. 
Regarding the BOOT0 pin, we understand that a pull-down resistor is required to ensure boot from Flash. However, we could not find a clearly specified resistor value in the datasheet or application notes. 
In many reference designs, a 10k ohm resistor is used. In some cases, lower values such as 510 ohm are used. 
Could you please clarify the design criteria for selecting the BOOT0 pull-down resistor value? 
Is there a recommended range?
What factors should be considered, for example noise or leakage current?
Why are lower values such as 510 ohm sometimes used? 
We would like to have a clear technical justification for our design.

 

Thank you.

Hiroshi Yamada

Best answer by mfgkw

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.

5 replies

Andrew Neil
Super User
August 22, 2026

@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 AN4488Getting started with STM32F4xxxx MCU hardware development:

faa06f9c-7a46-98bb-7741-65dee30126fc.png

 

A complex system that works is invariably found to have evolved from a simple system that worked.A complex system designed from scratch never works and cannot be patched up to make it work.
Associate
August 22, 2026

Hi Andrew,

Thank you for your response.
This is the EVK with the 510? resistor mounted as shown below.
The board name is STM32F469I-DISCO.

Best Regards,

Peter BENSCH
ST Technical Moderator
August 22, 2026

@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 510?, 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 10k? 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

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TDK
August 22, 2026

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.

"If you feel a post has answered your question, please click ""Accept as Solution""."
mfgkwBest answer
Associate II
August 22, 2026

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.