I'm using buffer SN74AVCA164245 and I feed one of the I/O with 20V and using voltage divider, I expected to see 2.34V on the input of the Buffer and to see 'High(1.8V)' on the output of the buffer, and I can't understand where is gone ~0.4V! I read every line on the datasheet and can't understand it. why I read 1.98v instead of 2.34?
EDIT: Description of DIR,OE: 
Your VccB is 1.8 volts.
Input protection devices are beginning to switch on because the pin voltage exceeds the VCC on that side of the buffer device. So current is starting to flow into the device pin, reducing the voltage.
Because it is coming from a fairly high impedance source, not much current will be flowing : causing a 0.36v voltage drop from a 5879 ohm (50000 in parallel with 6650) impedance source. So thats 61 microamperes going into the chip pin.
You don't state it but it's a design error.
You cannot feed more than 1.8V into an input pin that is powered from the 1.8V supply side, and it stated in the datasheet.
So what you see is normal in the case you have, it is not unexpected but expected.
Input pins must always be within the supply voltage or current flows through pin to supply.
