STM32H750XBH6 What is the resistiance of a GPIO pin when the device is shut down = 0V to STM32 ? The pin is config. to be an output pin( push-pull) when the device has power.

[BHell.1]

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[FBL]

Hello @Björn Erik Hellerstedt​ ,

You can find in datasheet DS12556 Rev 6 of the product section 6.3.15 I/O port characteristics in rev Y, the weak pull-up equivalent resistor is around 40KOhm when VIN=VSS

[eewiz]

The PU and PD resistances are real resistances in series with MOSFET that get turned on to enable the PU or PD current. Since the chip is not powered, the MOSFETs cannot be on so the pin will be floating.

Until you try to pull the pin more than a diode drop above the chips 0 volt, or VSS (i.e. ground) potential.

Then you will run into an internal protection diode connected from the pin to VCC.

As you pull up further you will start to power up the entire chip. If you pull an I/O pin to 3.3 volts for example, you will then find about 2.7 volts on the VCC pins and the chip will start to operate. The maximum injection current is 25mA. If your chip and external circuits connected to the other pins on the chip cause the injected current to exceed 25mA the chip may crowbar, shorting everything out until the source of the injection current is removed. Any CMOS chip powered up backwards through an I/O pin will crowbar at some terminal level of current injected into that pin. The entire chip turns into an SCR that is triggered ON by the injected current. The chip will remain in the ON state SCR mode until the injected current is removed. If the design permits the main power supply to be energized while the chip is in the the crowbar mode, you'll definitely let the smoke out.

All for now

[BHell.1]

Hello eewiz,

I fully agree wíth you.

What is your oppinion to connect gpio on STM32 into a enable pin on a chip, normally it is a op:amp input and the current is approx. 40nA from this pin. I have added a 10K pull down to keep the level to low when SM32 is off and the power chip with enable pin is powered. This means the voltage on the pin will be very low I guess the internal resistor (40K) in STM32 will be in parallel with the 10K resistor?

Thanks

[eewiz]

Hello BHell.1

All correct except for the 40K pull down. The micro's pull down has a MOSFET in series to VSS which has to be OFF because the micro is not powered. Your external 10K * 40nA = 400uV. So long as you hold the micro's pin voltage to less than 300mV all is well. The I/O pin on the micro will be open circuit.

At 300 mV on the I/O pin the leakage current might will exceed 40nA meaning that the 10K resistor may not be needed. If it is not needed in the final result you can always leave it out of the BOM.

All for now

[BHell.1]

Hello eewiz

Thank you. I fully agree.

Have a nice day.