[MOVED] VDD and VDDA microcontroller

[Alireza770717]

Hello everybody
I have an STM32F030F4 microcontroller that has two VDD and VDDA pins that I can use a jumper to turn on and off the 3.3V power supply. The GND pin is also connected directly to the microcontroller. I use an LM317 voltage regulator to supply the necessary voltage to the microcontroller, which provides 3.3V.

The problem I have is that without connecting the 3.3V jumper to the microcontroller, I have a voltage of 1V on the VDD and VDDA pins, which changes slightly when I change the voltage with potentiometer. I have not been able to find the cause of this, and on the other side of the jumper I have 3.3V, which is used to power the microcontroller when the jumper is connected.

In addition, GND is connected to the microcontroller and the necessary decoupling capacitors are in place. I have checked all traces and there are no short circuits.

 

[KlausST]

Hi,

Show your schematic.

I don´t know why you have jumpers on VDD and VDDA. Does it make sense at all?

What "potentiometer" are you talking about?

The problem I have is that without connecting the 3.3V jumper to the microcontroller

Simply don´t do this.

Klaus

BTW:
You are posting in ANALOG / IC_DESIGN section... I don´t see that this is an analog problem, nor do I see that you want to design your own IC.

 

[danadakk]

You are meeting the 3'rd constraint ?

And

Regards, Dana.

 

[betwixt]

As a guess, the potentiometer is feeding a voltage in when the jumpers are not fitted and the pin protection diodes are conducting it through to the VDD/VDDA lines.
If that is the case, there is a risk to the MCU of reverse voltage damage. The supply to the potentiometer should also be disconnected -or- get its supply from the MCU side of the jumpers so it disconnects at the same time.

Brian.

 

[Alireza770717]

Hi,

Show your schematic.

I don´t know why you have jumpers on VDD and VDDA. Does it make sense at all?

What "potentiometer" are you talking about?


Simply don´t do this.

Klaus

BTW:
You are posting in ANALOG / IC_DESIGN section... I don´t see that this is an analog problem, nor do I see that you want to design your own IC.

--- Updated Apr 16, 2024 ---

As a guess, the potentiometer is feeding a voltage in when the jumpers are not fitted and the pin protection diodes are conducting it through to the VDD/VDDA lines.
If that is the case, there is a risk to the MCU of reverse voltage damage. The supply to the potentiometer should also be disconnected -or- get its supply from the MCU side of the jumpers so it disconnects at the same time.

Brian.

Schematic at your service

 

[KlausST]

Schematic at your service

But no question answered: Why? Should we guess?

Looking at the schematic I have to agree with Brian: You will probably destroy the microcontroller.

Klaus

 

[Alireza770717]

But no question answered: Why? Should we guess?

Looking at the schematic I have to agree with Brian: You will probably destroy the microcontroller.

Klaus

Looking at the schematic, why am I getting a voltage of 1V on pin number 1 when the jumper (P4) is on the two VDDA and VDD pins?

And you said that I would destroy the microcontroller. What is the reason for this and what is the mistake in the schematic design?

 

[KlausST]

Hi,

I guess I would annoy you you if a ask a third time for the same information ... so it´s better for both of us if I leave this thread.

Klaus

 

[betwixt]

Assuming the potentiometer is on P2, as explained before the reason is there are two diode junctions on most MCU pins. The diode is a 'feature' of the silicon design but for practical purposes it looks like a diode between ground (VSS) and input and a diode between input and VDD. The diodes are normally not conducting because the input has to be more positive than ground and lower than VDD, they are there to provide a degree of protection if the voltage applied goes outside that range by conducting it to ground or the VDD line. What you are doing is removing VDD so it falls lower than the voltage your potentiometer is providing, the input pin has a higher voltage than VDD so the 'top' protection diode on "BOOT0" is conducting and carrying the slider voltage to the ICs internal VDD line.

The easiest fix is to source the 3V3 to the potentiometer and R4 from the MCU side of the jumper so they are all turned off at the same time. You should also check that any other pins on the MCU are not affected in the same way by whatever they connect to.

Note that the protection diodes are not intended to be used this way and are not rated to carry full MCU current, they are there for protection against transient voltages, not ones present for significant length of time.

Brian.

 

[dick_freebird]

Open supply pins will let the ESD network on the chip,
conduct forward from an input or output through the
"up diode" of the signal pin into whatever supply rail
pertains. Some input or output is probably "left hot"
around 1.8V.

 

[Alireza770717]

Open supply pins will let the ESD network on the chip,
conduct forward from an input or output through the
"up diode" of the signal pin into whatever supply rail
pertains. Some input or output is probably "left hot"
around 1.8V.

So, according to your explanation, this is a normal occurrence?

--- Updated Apr 17, 2024 ---

Assuming the potentiometer is on P2, as explained before the reason is there are two diode junctions on most MCU pins. The diode is a 'feature' of the silicon design but for practical purposes it looks like a diode between ground (VSS) and input and a diode between input and VDD. The diodes are normally not conducting because the input has to be more positive than ground and lower than VDD, they are there to provide a degree of protection if the voltage applied goes outside that range by conducting it to ground or the VDD line. What you are doing is removing VDD so it falls lower than the voltage your potentiometer is providing, the input pin has a higher voltage than VDD so the 'top' protection diode on "BOOT0" is conducting and carrying the slider voltage to the ICs internal VDD line.

The easiest fix is to source the 3V3 to the potentiometer and R4 from the MCU side of the jumper so they are all turned off at the same time. You should also check that any other pins on the MCU are not affected in the same way by whatever they connect to.

Note that the protection diodes are not intended to be used this way and are not rated to carry full MCU current, they are there for protection against transient voltages, not ones present for significant length of time.

Brian.

thank you

So, you're suggesting I get a jumper pin for R4?

 

[FvM]

So, you're suggesting I get a jumper pin for R4?

Makes no sense. You have more pins driving VDD through substrate diodes while P4 is removed. The good news is that this doesn't kill the processor as long as the "injected current" maximum rating is kept, 5 mA per pin, 25 mA in total. Problem is however that it's no possible to completely remove VDD while any input pin is driven high. Therefore P4 has no practical use, except perhaps for current measurement.

 

[betwixt]

So, you're suggesting I get a jumper pin for R4?

No, I'm suggesting the 3V3 to R4 and pin 3 of R2 is connected to the MCU side of jumper P4 so when the jumper is removed it also removes 3V3 from the potentiometer and NRST pin. In other words, disconnect 3V3 from R4/P2 and connect that point to the top of C4/C5/C6/C7 instead.

Brian.

 

[Alireza770717]

Makes no sense. You have more pins driving VDD through substrate diodes while P4 is removed. The good news is that this doesn't kill the processor as long as the "injected current" maximum rating is kept, 5 mA per pin, 25 mA in total. Problem is however that it's no possible to completely remove VDD while any input pin is driven high. Therefore P4 has no practical use, except perhaps for current measurement.

Alright, I understand the problem now. Thank you for your response and for your time.

--- Updated Apr 18, 2024 ---

No, I'm suggesting the 3V3 to R4 and pin 3 of R2 is connected to the MCU side of jumper P4 so when the jumper is removed it also removes 3V3 from the potentiometer and NRST pin. In other words, disconnect 3V3 from R4/P2 and connect that point to the top of C4/C5/C6/C7 instead.

Brian.

Very good, thank you for the appropriate solution you gave. I solved the problem. Thank you.