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Why are resistors required on the boot pins?

flyer31
Senior
Posted on January 17, 2012 at 10:46

Hi,

on the pins BOOT0, and also BOOT1 if used, the demo boards as STM32F4Discovery or others always use a resistor in line with the pin. The value varies quite much - in the AN3320 reference design, they use 10kOhm, in the Discovery-board, they use 510Ohm to GND and 10kOhm to VDD (510 ohm, as the 10kOhm always is present - so 510 ohms should pull BOOT0 down to 510/10000*3V3 < 0.2V).

In the Datasheet or in the app notes I do not find any explanation for this. Anybody knows, why to use resistors here?

2 REPLIES 2
rosarium
Associate II
Posted on January 17, 2012 at 18:18

This is to make the user more flexible, whether he wants to boot the program from system memory or flash. Either of the one resistor you can mount as per your requirement.

emalund
Associate III
Posted on January 17, 2012 at 18:37

<i>The value varies quite much</i>

pulling a CMOS input pin permanently high or low is where the expression ''grab bag resistor'' comes into play.

You can pull a CMOS input pin high or low with anything from 0 to several M ohm.

You tend to grab in the middle, a very low value will make it go poof if you are wrong, a very high value will make it noise suspectible.

Erik