I am going to measure VCC for my project to secure my entire circuit. VCC voltage is too high for the ADC pin of the microcontroller, so I am using a voltage divider circuit.

I don't know the purpose of the highlighted part of the circuit. Could somebody explain the purpose of the highlighted part of the circuit?

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  • \$\begingroup\$ Highlighted? Just note that your R3 is "perhaps" too high (depending on type MCU used). To be checked. \$\endgroup\$
    – Antonio51
    Sep 7, 2022 at 8:32
  • \$\begingroup\$ When i say highlighted, I was talking about squared part that contains two diodes. By the way, i am using STM32G4. \$\endgroup\$
    – Clyasenth
    Sep 7, 2022 at 8:40
  • \$\begingroup\$ Keep in mind your ADC has an input impedance. That'll be effectively a resistor to ground (not shown in your schematic) i.e. if you don't account for it, your voltage readings will be wrong. The datasheet should clearly tell you this value. \$\endgroup\$
    – Kyle B
    Sep 7, 2022 at 9:05
  • 1
    \$\begingroup\$ @KyleB ADC input has impedance but it is not a DC load, it is an AC load. The capacitance is low enough AC impedance for making ADC conversion, and as long as ADC conversions are not made too often to let the capacitor charge, there is no problem. \$\endgroup\$
    – Justme
    Sep 7, 2022 at 10:06

1 Answer 1


You have chosen to take the circuit for a general purpose input, where fault conditions, electrostatic discharge, or abuse could put very high voltages on the input. Then the D1/D2 pair clamp the voltage to the ADC at no more than a diode drop outside the voltage rails, to protect it in these conditions.

In this case, where the input is hard-wired to VCC, D1/D2 may be superfluous. If VCC goes above the 36 V or so needed to take the ADC voltage above 3.3 V, then it may have already killed other components it's driving.


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