I was going through this article for learning about ADC error analysis.

In Example 1: Applying TUE and Monitoring a Power Supply you can see the below thing:

enter image description here

May I know from where that loss equation (log2(4.13) = 2.05) came? I mean it's derivation.

2.05 bits of accuracy: that ADC in the example is 12-bit. Does 2.05 bits accuracy mean ENOB is 9.95 bits?

  • \$\begingroup\$ \$log_2(x) = \dfrac{log_{10}(x)}{log_{10}(2)}\$ \$\endgroup\$
    – Andy aka
    Apr 16 at 10:04
  • \$\begingroup\$ from where that loss equation came \$\endgroup\$
    – Confused
    Apr 16 at 10:07
  • 1
    \$\begingroup\$ Did you ask this question on a TI site 7 years ago: e2e.ti.com/support/data-converters-group/data-converters/f/… <-- I believe the answer is in there i.e. loss of resolution (an alternative way of describing effective resolution \$\endgroup\$
    – Andy aka
    Apr 16 at 10:22
  • \$\begingroup\$ That question was asked by me only,7 years ago ,I forgot it. Thanks for finding it out. But this is different I believe. \$\endgroup\$
    – Confused
    Apr 16 at 10:29
  • 1
    \$\begingroup\$ No, it's related. The effective resolution and the ideal resolution can be subtracted to find the loss of resolution. I'm not trying to be cryptic; I have tried a little to find the proper derivation but ran out of time. \$\endgroup\$
    – Andy aka
    Apr 16 at 10:33

1 Answer 1


May I know from where that loss equation(log2(4.13) = 2.05) came

Lets try this approach. If I asked what \$log_2(256)\$ is, the answer you would get is 8.

  • 256 is equivalent to the number of LSbs that you get from an ideal 8 bit ADC.

  • In other words \$2^8 = 256\$ is just like when using regular logs: \$10^3 = 1,000\$

In your question it uses \$log_2(x)\$ to find the equivalent of the error (or loss) in bits. It starts with LSbs and converts them to equivalent ADC bits.

So, in that example of a 12 bit ADC that you linked, the loss of bits is seen as 2.05 making the effective resolution of the ADC drop from 12 to 9.95.

Does 2.05 bits accuracy mean ENOB is 9.95 bits?

  • You have a 2.05 loss of bits and not a 2.05 bits accuracy
  • No, it's not ENOB; that's a bit different
  • ENOB requires an analysis of the ADC output when the input is a sinewave

Maybe this document (Understanding Noise, ENOB, and Effective Resolution in Analog-to-Digital Converters) might help.


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