I want to use a zener diode to clamp the input voltage. Input voltage is 12V with a +-4V noise. I want to know that if the noise in input voltage can affect zener voltage or not? If I have noise in input, will I have noise in zener too? Is this true?
In AC, the usual circuit having a resistor R and a zener in series forms a voltage divider with the diode's dynamic resistance.
Check zener datasheet for Rd, "Dynamic resistance" or "Differential resistance" of the zener (it depends on current).
Example: BZX84C-6V2 at 5mA Rd=6 to 10 ohms
Therefore, output voltage variation will be input voltage variation multiplied by Rd/(R+Rd).
The PSRR is not usually that good with zeners, unless a current source is used.
Here's an input voltage DC sweep. Top graph is voltage on zener. Middle graph is zener dynamic resistance. Bottom graph is PSRR in dB.
Whether this matters or not depends on what you want to do.
Yes, you will have some noise left, even if the Zener removes most of it.
Zeners only have the rated voltage over them at rated current. So with a simple resistor-Zener circuit, varying the input voltage will cause the Zener current to vary too, which varies the output voltage.