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PSoC™ 5, 3 & 1 Forum Discussions

Tim
Level 2
First solution authored 10 replies posted 10 sign-ins
Level 2

I've attached two really good docs that somewhat answer my question but not completely.    The third .png doc shows an INL20 of +-32.   Surely this must be a typo?   What is the INL of the PSoC 5 when the 20 bit AtoD is used?

Same question for the enob on the PSoC 5?   What is it?  I realize it depends on a lot of things as the app note explains but most data sheets contain the enob info.  It specced DNL and INL, power supply rejection ratio, and other parameters but no mention of enob.

 

Thanks.

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1 Solution
DennisS_46
Employee
50 sign-ins 100 replies posted 25 solutions authored
Employee

Tim:
Expanding a little on the comment from my co-worker Vasanth . . .
The 20 bit Delta-Sigma converter is a native 16 bit D-S vastly oversampled and averaged. This oversampling and averaging reduces the noise and suppresses DNL errors, but it does not change the fundamental INL. The INL has a characteristic curve:

DennisS_46_0-1659022203639.png

Note that at 16 bits it is well under +/- a single bit of INL. The 20-bit version will follow the same curve, but multiplied by 16. The typical peak value will be less than +/- 16 counts and the device will handily meet the 32 bit spec. A comment on the shape of the curve. Note that at the ends, the value increases considerably. This is a function of the decimator averaging, it is not an analog circuit "problem" per se.
Noise is reduced by averaging, but since noise is power it takes n^2 samples to reduce the noise by a factor of n.  This is simple physics, not circuit design or post-sampling digital averaging effects. You can expect a 20 bit converter of this type to be at least 16 times slower than a 16 bit. Averaging even more will reduce noise, but will not change the INL.
Finally, the INL of the 20 bit converter is 32/(2^20) = 32ppm or 0.0032% deviation from straight line. Example: 500 lb weigh scale: resolution = 500/2^20 = 0.000477 lbs; 32 count DNL = 0.0153 lbs. I don't track myself that closely. On the other hand, if you are measuring satellite parts, it's probably not good enough. Whether this is good enough for your system is your determination to be made. 
If you have further questions, send me your phone number in a private email and I'll be happy to share all that I know about the topic.
---- Dennis Seguine, PSoC Applications Engineer

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2 Replies
Vasanth
Moderator
Moderator 250 sign-ins 500 solutions authored First question asked
Moderator

Hi Tim,

The INL20 value is 32 bits as shown in the datasheet. Meanwhile the ENOB will depend on your system. You can follow the suggested method for your system and calculate the ENOB.

Best Regards,
Vasanth

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DennisS_46
Employee
50 sign-ins 100 replies posted 25 solutions authored
Employee

Tim:
Expanding a little on the comment from my co-worker Vasanth . . .
The 20 bit Delta-Sigma converter is a native 16 bit D-S vastly oversampled and averaged. This oversampling and averaging reduces the noise and suppresses DNL errors, but it does not change the fundamental INL. The INL has a characteristic curve:

DennisS_46_0-1659022203639.png

Note that at 16 bits it is well under +/- a single bit of INL. The 20-bit version will follow the same curve, but multiplied by 16. The typical peak value will be less than +/- 16 counts and the device will handily meet the 32 bit spec. A comment on the shape of the curve. Note that at the ends, the value increases considerably. This is a function of the decimator averaging, it is not an analog circuit "problem" per se.
Noise is reduced by averaging, but since noise is power it takes n^2 samples to reduce the noise by a factor of n.  This is simple physics, not circuit design or post-sampling digital averaging effects. You can expect a 20 bit converter of this type to be at least 16 times slower than a 16 bit. Averaging even more will reduce noise, but will not change the INL.
Finally, the INL of the 20 bit converter is 32/(2^20) = 32ppm or 0.0032% deviation from straight line. Example: 500 lb weigh scale: resolution = 500/2^20 = 0.000477 lbs; 32 count DNL = 0.0153 lbs. I don't track myself that closely. On the other hand, if you are measuring satellite parts, it's probably not good enough. Whether this is good enough for your system is your determination to be made. 
If you have further questions, send me your phone number in a private email and I'll be happy to share all that I know about the topic.
---- Dennis Seguine, PSoC Applications Engineer