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5V operation with internal pull-up resistors (STM32F205)

Question asked by schneider.patric.001 on Apr 12, 2012
Latest reply on Apr 13, 2012 by schneider.patric.001
I'm currently developing a "plug-in adapter" using the STM32F205 that is to be used in a variety of legacy designs. These old designs all rely on 5V components and at least some of them require bidirectional communication over an open-drain line.

Because I cannot foresee in which applications the adapter will be used and how the other components will be configured, I would like to use an open-drain output stage for all I/O pins and also enable the internal pull-up resistors (external resistors are not an option because they won't fit onto the adapter).

How does a configuration like this work in a 5V environment?
From the data sheet (Doc ID 15818 Rev 7) I was able to gather the following information:
- Injected current (Iinj) on 5V-tolerant pin: -5/+0 mA (Ch. 5.2, Table 8)
----- Positive injection (Vin > Vdd) is not possible on these pins. Iinj must never be exceeded (Footnote 3 referring to 5V-tolerant pins)
- Maximum input voltage (Vin) on 5V-tolerant pin: Vdd + 4.0V (Ch. 5.2, table 7)
----- VIN maximum value must always be respected. Refer to Table 8 for the values of the maximum allowed injected current (Footnote 2, referring to 5V-tolerant pins)

How am I to understand these values? Vin would be in the permitted range [0V...3,3V+4V]. According to Table 8, positive current injection is not possible, so the output configuration should not matter as long as the input voltage stays below 7,7V.
However, wouldn't there still be a current going through the pull-up resistor (thereby violating the allowed injected current)?

Can you help me understand these specifications? Would my application be possible under these circumstances?