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INA236: ENOB calculation

Part Number: INA236

Tool/software:

Hi Sir,

I am trying to check ENOB table as follows, when test conversion time 140us, sample 1 time and read ADC shut voltage 1000 times, 

I short IN+ and IN- together, so the ideal register result should be 0x01= 0x0000(ideal case) 

however, register result of max was 0x000A, min was FFF7, 

could you help me to figure out if my test result is correct? 

or how to measure or calculate ENOB of CSA?

thanks for your help!

  • Hi,

    Based on your measurements, the peak to peak noise is 19*2.5uV=47.5uV (0x000A=10, 0xFFF7=-9, note two's complement, more details in datasheet). This translates to RMS noise of 47.5uV/6.6=7.2uV. To calculate the ENOB:

    ENOB = log2 [81.92mV*2/(7.2uV × √12)]=12.7

    Regards

    Guang

  • Hi Zhou Sir 

    Thanks for yor reply, 

    I would like to know is the formula of "ENOB" you provide was same meaning of "Noise-Free ENOB"(table was Noise-Free ENOB)?

    I am not sure if we consider noise free condition, was the formula same?

    and I know the table was typical value of Noise-Free ENOB, so how much difference is reasonable,

    (table of ENOB was 13.1, and measurement result was 12.7) 

    BTW, could you help me to know where the formula of "√12" come from? is it constant value of ENOB? or it related to the ADC bit.

    Appreciate your help.  

  • Hi,

    I’m using this equation for the calculation:

    ENOB = log2 [full-scale input voltage range/(ADC RMS noise × √12)].

    Regards

    Guang

  • Hi Zhou Sir, 

    Thanks for you formula of ENOB.

    But I have a question regarding the INA236 datasheet. The table in the datasheet shows the noise-free ENOB, but it seems like it should be labeled as "ENOB" only, not noise-free resolution. Is this a typo in the datasheet?

    Thank you for your clarification.

     

  • Hi Liu,

    I don’t think it was a typo, but I agree it is a bit redundant to include the word noise-free given how the parament is calculated.

    Regards

    Guang