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ADS1281EVM input divided by 2?

Yannick Niane
Prodigy 90 points
Other Parts Discussed in Thread: ADCPRO, ADS1281, OPA1632

Hi All,

I'm using an ADS1281EVM on MBBO Motherboard with ADCPro. I am bit confused on how the diffrential input is converted.

  I have done the following test  :  Ain + = 3.28V and Ain - = 1.19V . So i was expecting  to read Vout = 3.28V - 1.19V = 2.09V. Instead i read 1.1V, half the input.

  So i measured with Ain + and Ain - directly on the ADS1281 pins and it turned out that (Ain+) -  (Ain-) = 1.1 V.  I have done other tests by changing Ain+ and Ain- , and the input measured directly on ADS1281 pins is always half the input measured on header J6 of  ADS1281EVM. Any ideas why?

Thank you

Yannick

  • 0
    TI__Mastermind 44680 points

    Hi Yannick,

    What is the voltage at the VOCM and VOUT-  pins of the OPA1263? Are you providing external +/-10V supplies to the MMB0 to power the OPA1632 buffers? See figures 7 and 8 in the ADS1281EVM User's Guide. Also, check jumper "J13" on the MMB0 to see if it is set like figure 5 in the User's Guide.

    The buffer could be trying to output a negative voltage and saturating at the negative rail - which would explain the gain of 0.5.

    Regards,
    Chris

  • 0 in reply to Christopher Hall
    Prodigy 90 points

    HI Christopher,

    Thanks for your help.

    1- VCOM= 2.5V.  If i apply Ain+ = 3.28V and Ain- = 1.66V  I have Vout+ = 2.92V and Vout- = 2.06V 

    2- Yes i have external +-10V supplies to the MMB0

    3-J13 is set according to figure 5 in User's guide

    it still doesn't explain the  0.5 gain

    Yannick

  • 0 in reply to Yannick Niane
    TI__Mastermind 44680 points

    Hi Yannick,

    What are you using for your input sources? Might they have a high source impedance that is creating a resistor divider? You could try removing the input resistors, R12 and R13, on the ADS1281EVM.

    I tested this out on my ADS1281EVM here in my lab and could not replicate your results. Here are my measurements will a DMM:

    AIN+ (V) AIN- (V) VCOM (V) VOUT+ (V) VOUT- (V) (VOUT+) - (VOUT-)  (V)
    3.25 1.17 2.49 3.53 1.44 2.09
    3.25 1.64 2.49 3.30 1.68 1.62
    3.26 1.18 0 1.09 -0.98 2.70
    3.26 1.65 0 0.86 -0.74

    1.60

    As you can see, I tested both of your input test conditions that you mentioned and I also tested them for both bipolar and unipolar ADC supplies (by grounding the GPIO3 pin on the serial header), which changes the VCOM voltage.

    Regards,
    Chris

  • 0 in reply to Christopher Hall
    Prodigy 90 points

    Christopher,

    I'm using the 3.3v  digital supply from J8 and i've made a basic voltage divider to generate lower voltages.

    I was not able to proceed, U13 burned up and i think other components might have been damaged also. I've ordered a new one, it will be a good thing to start with a new kit and compare with your results.

    However a little remark on the the "ADS1281EVM and ADS1281EVM-PDK User's Guide" , in table 2 page 4 , it says that if GPIO3 is High then we're in bipolar mode(+/-2.5v) and when GPIO3 is low we're in unipolar mode. However U10 and U13 are enabled and +-2.5V generated  when GPIO3 is low since the signal is inverted by U8 . So it seems to be the contrary.

    Thank you for precious your help

    Regards

    Yannick

     

  • 0 in reply to Yannick Niane
    TI__Mastermind 44680 points

    Hi Yannick,

    You're correct. The ADS1281EVM pulls GPIO3 high by default for single-supply operation. I'll make a note to correct the User's Guide.

    Take a look at the ADS1281 schematic at the end of the User's Guide. The feedback circuitry around the OPA1632 (and R1) lower the input impedance of the buffer. I think this is placing a load on your voltage divider and causing the unexpected output voltage.

    Regards,
    Chris

  • 0 in reply to Christopher Hall
    Prodigy 90 points

    Chris,

     I've received the new  ADS1281EVM and was able to reproduce the same tests and this time i have the good results, the output is not divided by two.

    So the kit i was using at the first place  was certainly defective.

    Thanks for your help

    Yannick