I use an Arduino Due which is rated 3.3V as a kind of waveform oscilloscope to display signals on my PC. As I knew I had to scale my signals to 3.3V to avoid damaging the analog pins.

Here is the warning on the Arduino Due specifications page:

Warning: Unlike most Arduino boards, the Arduino Due board runs at 3.3V. The maximum voltage that the I/O pins can tolerate is 3.3V. Applying voltages higher than 3.3V to any I/O pin could damage the board.

Some time ago I used a voltage divider to do this but lately not sure for what reason I used just a 20k resistor.

The Arduino Due analog pin seems fine, I have used it to measure 4.5V (as measured by the multimeter) for quite some time, so can someone clarify to me why the pin or the board is not damaged, what difference makes a voltage with very small current ( 0.265mA in this case ). Does the voltage measured by the ADC on the Due reach the microcontroller? (I suppose yes as the ADC is part of it)


MCU pins have clamp diodes on the pins to protect against static discharge (ESD). They work by clamping the pin voltage to the power supply rails. However, they can only handle so much current. If whatever you are driving the pin with itself only has limited current capability then it might bog down and be unable to drive enough current through these diodes to damage them.

One method that is sometimes used is to add in a series resistor to limit the current that flows through the diodes. It plays the same role as the output impedance in the schematic below. It drops the excess voltage across the resistor rather than the nearly constant forward voltage drop of the diodes which in turn limits the current that flows through the diodes. This lets the pin safely clamp much higher voltages. I don't recommend it though...I always add external clamp diodes rather than rely on the MCU's internal ones for such a thing.


simulate this circuit – Schematic created using CircuitLab


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