• State Resolved
  • Locked Locked
  • Replies 2 replies
  • Subscribers 46 subscribers
  • Views 3191 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

ADC Dynamic Range and Input Voltage-Loss in Enob Calculation

Hari T O0
Genius 3820 points


Hi,

The attached file contains the information regarding Loss in ENOB when there is a huge difference exists between reference voltage and input voltage.

Finally the author obtains loss in Enob as .4bits. The last derivation is not clear for me. Can anyone please tell how the ".4bits is obtained.

Regards

Hari

  • 0
    TI__Guru* 98725 points
    Hari,


    You can find the difference of 0.4 bits through a calculation from the resulting resolution.

    With the the number of transitions at 65536, this is the equivalent of 16 bits of resolution. It can be calculated from:

    log(65536)/log(2) = 4.816/0.301 = 16 bits of resolution.

    This is apparent because 2^16 = 65536.

    Going to the reduced resolution of 49652, you can calculate it in the same way:

    log(49652)/log(2) = 4.6959/0.301 = 15.60 bits of resolution.

    Here, we've lost 0.40 bits of resolution. Calculating the resolution from the number of bits, 2^15.60 = 49652. ENOB is a measure of the resolution in base 2.


    Joseph Wu
  • 0 in reply to Joseph Wu
    Genius 3820 points
    Thanks Joseph..