Bklaric,
Since the internal circuit is not disclosed, this is a general theory and my imagination.
The reason why the AD conversion value is smaller than the input voltage may be due to the effect of the voltage divider circuit.
There is a sample-and-hold circuit at the input of the AD conversion circuit, which samples and holds the input voltage in a small capacitor and converts the voltage of the held capacitor to AD. If there is no voltage divider circuit, the input voltage is sampled and held with a small time constant, and the input voltage and the capacitor voltage almost match.
However, if the resistance of the voltage divider circuit is as large as 220k, the capacitor voltage rises slowly with a fairly large time constant and is held before it fully reaches the input voltage. As a result, a value lower than the input voltage is AD converted.
To check this, try 2k2 instead of 220k, and the difference will be smaller. Also, if you connect 100nF to the input pin, the difference will be smaller because the capacitor will sample and hold without passing through 220k.
In any case, if you are using an accurate measuring instrument, I would recommend adopting its value to correct the AD conversion result.