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The Arduino Uno Rev3 has an OpAmp (TI LMV358IDGKR datasheet) where the non-inverting input ("CMP") is connected via a voltage divider to Vin. The OpAmp's supply is connected to 5V.

part of schematic

On the Arduino Uno Rev3 product page (click) they specify the recommended and limit ratings for the boards input voltage:

input voltage

Looking at the LMV358IDGKR absolute maximum ratings i am wondering why having (in the case of say input voltage > 11.4V) a voltage above 5.7V at the OpAmp's input ("CMP") isn't a problem? For my understanding the OpAmp will blow up at say 15V input voltage?!

(Well voltage at barrel jack might be ok up to roughly 12.5V because there is a diode between power jack and VIN, but above?)

absolute maximum ratings

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2 Answers 2

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The "input" in question is the power supply voltage. That Arduino uses an SPX1117 regulator.

enter image description here

Other considerations are thermal (on the high end) and the typical vs. worst-case dropout voltage of the regulator (on the low end).


Not important to this question, but for those using clones, note that the popular AMS1117 has a lower rated absolute maximum input voltage (15V).


As far as your op-amp circuit the Absolute Maximum input voltage to the divider would be 11.4V. The LMV358 is a bipolar technology part and does not have diodes from inputs to Vcc, so care is called for. You can add a diode if you need to.

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  • \$\begingroup\$ @glen_geek I think that says there are 6 analog input pins. \$\endgroup\$ Mar 4, 2021 at 15:21
  • \$\begingroup\$ So applying a voltage to the Arduino Uno Rev3's DC jack that's above (11.4V+ Vf) could at least possibly damage the OpAmp, right? (Vf... forward voltage of input diode) \$\endgroup\$ Mar 4, 2021 at 15:32
  • \$\begingroup\$ More than 6.6V won't do anything. If you add the diode (preferably a Schottky such as 1N5819) from pin 3 to +5V you can exceed 11.4V by a bit, but at some point (depending on the total load on the regulator) too much current will go into the 5V supply and it will rise out of regulation (not good). You could also add a 4.7V zener from pin 3 to ground which wouldn't have that problem. Or just don't put so much voltage on the input (or add series resistor to the input, adding to the 10K on the board). \$\endgroup\$ Mar 4, 2021 at 15:37
  • \$\begingroup\$ "More than 6.6V won't do anything." What do you mean by this? Where does this 6.6V value come from? "or add series resistor to the input, adding to the 10K on the board" guess you mean in series with RN1A so the voltage across RN2A (rexpectively voltage at CMP) is only a third of VIN \$\endgroup\$ Mar 4, 2021 at 16:01
  • \$\begingroup\$ It's comparing Vin/2 to 3.3V so applying more than 6.6V doesn't do anything (well, until it breaks something). I see a divider with two 10K resistors so the chip sees Vin/2. \$\endgroup\$ Mar 4, 2021 at 16:03
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If there is an internal protection diode to Vcc in the Op. Amp., due to the 10k resistor, when the diode conducts, the input voltage will not be above the maximum rating.

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Edit: the 20V input with the 10k voltage divider form the equivalent voltage source above, limiting the current that goes to the Op. Amp.

Page 16 of the datasheet:

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Edit: I was considering the issue unsolved, but I'd like to add two extra points:

  1. In another TI datasheet (TL084*), I found an explicit note on input protection:

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  1. I actually measured VIN and CMP with an Arduino Uno Rev3 and the measured values do not indicate the presence of a clamping diode without internal current limitation (the slope of the curve does not change after CMP > 5.7V) or maybe the protection diodes are already gone... Note: I always powered this board from USB.

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  • \$\begingroup\$ I have doubts that represents the actual circuit since they allow 5.7V to be applied to the inputs with zero Vcc. Maybe there's a resistor, maybe something else. \$\endgroup\$ Mar 4, 2021 at 15:40
  • \$\begingroup\$ @SpehroPefhany. Could it be that the "absolute maximum rating" is for an ideal voltage source? If so, we could go above the 5.7 with limited current. The circuit in question certainly does. \$\endgroup\$
    – devnull
    Mar 4, 2021 at 16:14
  • \$\begingroup\$ Yes, usually it is. But if Vcc was 0V and there's really just a diode there, it would be vaporized. That's why CMOS circuits are usually specified as Vcc+0.3, so when Vcc is 0V the input can't go higher than 300mV (ideal voltage source and worst-case hot temperature). \$\endgroup\$ Mar 4, 2021 at 16:16
  • \$\begingroup\$ I see. Probably ESD protection implies a series resistor, or something like that. \$\endgroup\$
    – devnull
    Mar 4, 2021 at 16:18
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    \$\begingroup\$ Found this TI forum entry which looks helpful in this matter link. If i got it right it is not enough to just limit the input current with a series resistor if permanent overvoltage (> 5.7V) is applied to an input. But it is a sufficient solution for short time events (like ESD...). Can someone read that forum entry as well and confirm or refute my interpretation? Thanks \$\endgroup\$ Mar 8, 2021 at 17:42

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