The way I understand that is its not going to pass voltage through to Vdd until the input voltage reaches Vdd plus the Vf of the diode.

So what happens when Vdd is 3.3Vdc and the input is 5Vdc? Would that not pass 5-(3.3+.6) = 1.1Vdc additional on to Vdd? Sorry if thats a dumb question, but I'm trying to understand how that protects the over voltage.

That's why you need the series resistor.

 

You only need two diodes if you want to also protect against negative transients, which is not the case here.

The 1n914 is an one-cent part (Taiwanese) More like .5 cents for Chinese made. In the real world you would use two diodes for the sake of completeness.

 

The 1n914 is an one-cent part (Taiwanese) More like .5 cents for Chinese made. In the real world you would use two diodes for the sake of completeness.

Well, I am in the real world and I would not add the extra diode due to the extra capacitance, and one more part to fail.

 

Well, I am in the real world and I would not add the extra diode due to the extra capacitance, and one more part to fail.

Paper, scissors, rock. If you win then no diode

 

I found in this document, that maximum current for protection diodes of Cmos -Ic is 20mA.
So a few kohm series resistor in input should be safe.
"Designing With Logic" - Texas Instruments
https://www.google.fi/url?sa=t&sour...vOhY6kXYj9cxWklxQ&sig2=g68MRExfZgVjttcjrJLnUg

 

Paper, scissors, rock. If you win then no diode

Well, that's one novel way to make an engineering decision.