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How to test ADC’s CMRR and PSRR?

Nicole Han
Intellectual 2230 points
Other Parts Discussed in Thread: ADCPRO, ADS1248

 For example ADS1248, I tied AINP and AINN together to AN0 , and AN0 to Vc1, read the voltage output Vo1 on ADCPro; then tied AN0 to Vc2, read the voltage output Vo2 on ADCPro. Then got CMRR as bellows.


The test CMRR of ADS1248 is better than datasheet  typical value.

PGA=1

Data Rate=5SPS

voltage change

 

Vcm

output(code)

CMRR(dB)

1.6034V

-1996.17

-1.607

-1975.73

5.12uV

115

PGA=8

Data Rate=5SPS

voltage change

 

Vcm

output(code)

CMRR(dB)

0.1796V

-1848.08

-0.2021V

-1853.77

0.174uV

127dB

 

I just want to know whether this test method of CMRR is right? Thank you.

  • 0
    TI__Guru** 113765 points

    Hi Nicole,

    Your method appears to be correct, but I didn't actually verify your data.  You basically are making a DC CMRR test with a single input channel shorted internally.  Normally this test is run using an external short so that the entire path of a normal configuration is used.  You could repeat your test using a number of different data points throughout the input range to see how well the results track.  One thing to consider here is calibration of the external metering to make sure what is measured is accurate.

    Another method that could be used is an AC common mode signal (within the bandwidth of the measurement ) where you can then analyze the output using a FFT.  In the case of PSRR you would modulate the AC signal on the DC power input. You would most likely need a faster data rate to use this method.

    Best regards,

    Bob B

  • 0 in reply to Bob Benjamin
    TI__Intellectual 2230 points

    Hi, Bob,

    Thank you very much. The reason why I can not use DC common voltage to test CMRR is ADS1248' offset voltgage is not constant, right?

    I use a AC common voltage, 2Hz, 1.2V to see the FTT analysis, while I can not see the 2Hz signal. I think maybe ADS1248's CMRR is too high, so the offset comes from AC common voltage is drown by noise, right? see below

    But when Vcm amplitude is 1.8V and 0.8V,the 2Hz siganl can be seen only a little higher than noise.

     

    Best Regards,

    Nicole

  • 0 in reply to Nicole Han
    TI__Guru** 113765 points

    Hi Nicole,

    You can use the DC test as the numbers given in the datasheet Electrical Characteristics table are for DC at a gain of 1 and also at a gain of 32.  However you must also consider the noise within the measurement process.  Also, your measurement result is only as good as the performance of the test equipment used (such as the stability of the voltage source and accuracy of the voltmeter).

    Using the AC test, you must also consider the affects of the digital filter in the measurement result.  The AC testing that is done is with respect to power line frequencies, which is the normal mode rejection characteristics.  The table of results for these tests is shown in Table 9 of the datasheet.  I will state again that your measurement results are only as good as the test equipment used.  You may see added noise and distortion as a result of the signal generator used.  You probably want to take a lot more data points, such as 8-16k instead of 512.  Also, you cannot use the SNR numbers from the FFT as the SNR would relate to the differential input and not the common mode input.

    In either case, the device is specified in the datasheet and is tested as meeting a minimum CMRR of 80 dB, and typically 90 dB at a gain of 1 with DC measurement.  This could be better depending on the device tested, but will not be worse throughout the operating temperature range of the ADS1248.  So I would not discount the original numbers that you had calculated, although it does seem better than what I would expect.  You might want to retest the DC measurement and go within 90% of full-scale at the DC measurement points for the input.  90% would be about +/-1.84V for a 2.048 reference and a gain of 1.

    Best regards,

    Bob B

  • 0 in reply to Bob Benjamin
    TI__Intellectual 2230 points

    Bob,

    Thank you for answering my question. I got your meaning.