# Microcontroller - potential damage to IO pin

I'm revisiting a design I made.

I placed a power loss detection before the polarity protection:

In case the power pins were replaced, the 3v3 zener diode will have forward voltage of ~1V.

will this pin go through R32 to the "unpowered microcontroller" and destroy its IO pin?

The absolute maximum is -0.3V in the mcu:

It's only 100uA limited by R32, but I can't be sure.

Is there such detailed information about the IO pin?

• I don't think I understand your text. You write that there will be "~1V", which is well within -0.3 to +4.0 V. Do you mean -1V? In that case, do you mean to also draw -23 V over the resistors, and reverse the current direction through R32?
– pipe
Commented Apr 9, 2020 at 19:34
• What is the documented impedance on that pin? The pin (and diode) make a voltage divider with R32. Commented Apr 9, 2020 at 22:40
• Also, you want the fuse before the TVS devices, so that a constant overvoltage blows the fuse before the TVSes. Commented Apr 9, 2020 at 22:41
• @pipe yes its -1v but in if you see it from cathode to anode .. yes it goes through R32 to microcontroller io pin .. the pin it self will have clamping diode .. thats why 100uA limited ..^^ Commented Apr 10, 2020 at 14:14
• @JonWatte im not sure i follow up .. there is somthing ~50ohm in the pins to reduce EMI ringing .. the diode will clamp that 1v to 0.3 i think .. Commented Apr 10, 2020 at 14:16

Usually the datasheets, in the absmax section, give an information like "-0.3 V or -10 mA, whichever occurs first", but unfortunately not in this case; therefore a proper answer is not possible, but I can give a bit of insight.

The -0.3V limit is there because for each pin there is a diode from ground to the pin, and from the pin to vdd. With 10 kOhm in series, this is a non issue; I would expect almost any IC to be able to handle at least a few mA in the protection diodes, and here you have a 10 kOhm resistor in series. You can even bump the resistor up to 100 kOhm, if I understand your circuit correctly.

To sum it up, with your series resistor you are not violating the absmax because the diode will conduct and block the voltage around -0.3 V, probably even less. With 100 uA you can sleep safe.

• That's an under-specified aspect of many data sheets. IMHO, it would be more helpful if they were to say something like "The device will tolerate without side-effects beyond power dissipation whatever amount of current flows through the clamp diode at VDD+0.3V, and will likewise tolerate whatever voltage it would reach if 10mA or less is fed into the pin" but some devices may have side effects even when current levels wouldn't damage the device. For example, on one 74HCxx part I used, if an input was driven with 1mA above the rail and the adjacent input was grounded, that 1mA would flow... Commented Apr 9, 2020 at 19:43
• ...out through the adjacent input (which for the circuit in question caused trouble since that input was passively pulled down with a 4.7K resistor). Commented Apr 9, 2020 at 19:44
• Great question and answer also fully agree with @supercat - lazy data sheets are far too common Commented Apr 10, 2020 at 7:27
• Microchip have an appnote for AVR devices where they use the ESD diodes to allow direct connection to the mains through a resistor. ww1.microchip.com/downloads/en/Appnotes/… You're safe. For sure. Commented Apr 11, 2020 at 9:59
• Further, the ATSAM4D datasheet specifies 1mA is acceptable on page 877, section 37.10 but unfortunately, the ATSAM4E has no such section... Commented Apr 11, 2020 at 10:35

According to the datasheet this is indeed a violation of the absolute maximum rating, with the caveat that the current not being specified. It might be okay, but not guaranteed.

I would look into a small power monitor IC which can detect when the input voltage is about to get low and signal your MCU. These are available at a wide variety of ‘trigger’ voltages and can be simpler/cheaper than doing your own comparator circuit.

Just move the voltage detection to after the polarity protection mosfet and your MCU won't see the negative voltage.

Also, for detecting voltage, use a resistor divider circuit that gives you 2.5V or whatever your threshold is at 15V in (so, 56 kOhm to 10 kOhm or so, or even 560k to 100k to save power). Then parallel the zener with the lower leg for protection; far under 3.3V it won't conduct a lot and won't impact your UV detection significantly.

• there are big fat caps there .. i want to detect power failure and write on memory .. thats why i put it before the pmos .. Commented Apr 10, 2020 at 14:23

Insert (add) a diode after R32 for any sensing of voltages this is a standard protection This offer reverse protection to input pins however any I /O pins will have two diodes one to the Vcc in forward bias to take care inputs exceeding the Vcc Other to Gnd to take care of reverse voltages occurring on in put pi

As others have said the current limiting R32 means it will all be ok, but you could move the UVLO tap to the right of the FET and then it would definitely be better. But ...

In your schematic you have a comment about the UVLO circuit being wrong and needing a comparator.

Take a look at shunt regulators such as the tlv431. These three pin devices can be used as comparator with integrated reference and driven straight from pretty much any supply with only a series resistor provided you can protect against reverse polarity.

That will definitely overcurrent the cpu pin.

I just use a schottky diode right on the v+ input to prevent against reverse polarity. FSV10100v might work depending on your current requirements.

You can then spin off a small separate circuit if the supply is reversed to turn on some user indication (led?).

• hi @emBee .. there is already polarity protection PMOS .. i didnt want to add another one .. the power loss detection i wanted it before the PMOS as i want to detect the uvlo before the circuit uses the charges in the capacitors .. Commented Apr 10, 2020 at 14:21

Have a look for injection current in the data sheet. Also fix the divider, set it so the zener doesn't conduct under normal circumstances. Failing this ask their web site. But I agree keep the current into the micro to less than 1-2 mA is usually good enough. An alternative is to use a bas99 as clamp diodes. There was a paper done a long time ago about their use as esd protection. The other part of this is to remember what you are protecting.. micro pin so after the 10k series resistor. Use the 10k as a buffer. Or split the 10k in 2

• this wasn't divider but the zener was working in knee point thats why the current is 5mA .. (16v min - 3.3 / 2200ohm = 5mA) .. any less value than 14v will start making the zener to work below the knee point Commented Apr 11, 2020 at 3:57