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How to calculate the ENOB of a SHA?

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pikky

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Some papers(all of them used the bipolar process) said that the ENOB of a SHA is obtained by this formula:
ENOB=(|THD|-1.76)/6.02
is that right?
Or can we use this fomula in CMOS process?
Why don't use SNDR instead?
Where can I find the original explanation?


Thanks you.

pikky
 

1. We use SNDRef to define ENOB. SNDRef is the "measured" effective resilution of a converter in dB, which includes all the non-ideality effects of a practical converter. ENOB= (SNDRef-1.76)/6.02.
2. This formula is derived using math only and does not depend on process
 

yxo, thank you for your reply.
I know we use the tested SNDR in the definition of the ENOB of the AD. That's why I presented this question.

Added after 5 minutes:

《A bipolar 2-GSample s track-and-hold amplifier(THA) in 0.35um SiGe technology》
this paper presented the exact formula I mentioned.
You may refer to it if you have time.

Will anyone give me some useful infromation?
 

can't see the file?

I upload again. Hope this time you'll see it.
 

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