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I'm trying to understand the default voltage tolerance of the boot pin(s) used to power up into the system bootloader of an STM32F3xx. I want to power my whole circuit at one voltage (e.g. 7V), and use this same external voltage to set logic high on the boot pin when performing a (hopefully rare) firmware update over USART1.

AN2602

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  • \$\begingroup\$ Which STM32F3 is it? Different F3 chips have different maximum voltages. \$\endgroup\$ – Justme Feb 2 at 13:59
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From the datasheet, it looks like BOOT0 can tolerate up to 9V, so 7V should be OK.

enter image description here

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  • \$\begingroup\$ Which "whole circuit "? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 at 1:13
  • \$\begingroup\$ I mean as the voltage provided to the BOOT0 pin as well as the voltage provided to a regulator for the MCU and the driver input voltage for a single motor. \$\endgroup\$ – tarabyte Jan 8 at 18:05
  • \$\begingroup\$ @terabite ah ok then 7V is fine. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 at 18:07
  • \$\begingroup\$ but 0x04: voltage range [2.7 V, 3.6 V] and double word write/erase operation is used. In this case it is mandatory to supply 9 V through the VPP pin (refer to the product reference manual for more details about the double-word write procedure). \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 at 18:11
  • \$\begingroup\$ You are looking at an Absolute Maximum, not a recommended condition. Given that you'll want a pull down anyway to keep the state, you could so easily build a voltage divider. Just calculate a series resistor for your available supply voltage that in conjunction with the pulldown will target something around the operating voltage, or at least within the chip's design goal of 5v tolerance for the pins not bounded by Vdd. If you don't want to be prevented from using a lower voltage to drive it, you can put the series resistor in your programming harness. \$\endgroup\$ – Chris Stratton Jan 8 at 19:26

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