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I've found multiple sources (eg. Microchip, Analog) saying something like "An ADC with DNL error less than ±1 LSB guarantees no missing code."

If the DNL is calculated only as a width of the "step" and not with relation to the ideal linear transfer fcn... won't having several wider codes cause a disappearance of some other one?

Unless they only meant to write that DNL > -1 means no output code has 0 width?

Edit: added links. After reading more into the details in the Analog's description,

The key is having -1LSB as the low limit.

It seems like my understanding was correct and that just Microchip worded it poorly.

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  • \$\begingroup\$ Please link one of the papers you refer to as multiple sources. \$\endgroup\$
    – Andy aka
    Commented Jan 18 at 19:02
  • \$\begingroup\$ The wording is correct, since each step is independent, the missing code can come from >+1 or <-1, so ±1 LSB. \$\endgroup\$
    – StainlessSteelRat
    Commented Jan 18 at 23:45
  • \$\begingroup\$ @StainlessSteelRat yes, but <+1 does not guarantee no missing codes, only >-1 does \$\endgroup\$
    – herhor67
    Commented Jan 20 at 11:21

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It does guarantee no missing codes because if you run a linear sweep, you won't see any missing codes because they have spaced the voltage bins close enough that you will convert every value and every value is within 1 of the other value, meaning if you do a sweep and subtract every value from the next one there won't be any missing values.

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  • \$\begingroup\$ Sorry, I cant understand anything you wrote. Feels like you went a bit off topic, explaining the effect of DNL or checking of missing codes. I was just asking about the mathematical reasoning, because I wrongly focused on the positive bound instead of negative one, as specified in Analog's doc. \$\endgroup\$
    – herhor67
    Commented Jan 19 at 9:56

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