• State Resolved
  • Locked Locked
  • Replies 4 replies
  • Answers 1 answer
  • Subscribers 46 subscribers
  • Views 556 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.

Question on ADS1246 Input Signal Range

chen shi
Prodigy 155 points
Other Parts Discussed in Thread: ADS1246

In my application, ADS1246 needs to sample a single ended signal from 0.2 to 4.8V, besides a differential signal.

But discription about input signal range and common voltage input range is not very clear in the datasheet of ADS1246.

With the datasheet, common voltage input range Vcmi is restricted from (AVss+0.1+VinPGA/2) to (AVDD-0.1-VinPGA/2), where Vin, as I understand, equals to AinP-AinN.

In the case that, AVDD=5V, AVss=0 (single supply), AinN=2.5V, PGA=1, Vref=2.5V:

When AinP=4.8V, Vin=AinP-AinN=2.3V, Vcmi should be in the range of 1.25V to 3.75V, and since VinN=2.5V, Vcmi is the common component of AinP and AinN, Vcmi=2.5V.

When AinP=0.2V, Vin=AinP-AinN=-2.3V, Vcmi range is from -1.05V to 6.05V, Vcmi, as the common component of AinP and AinN, is 0.2V.

As the result, when AinP is from 0.2V to 4.8V, Vcmi always satisfies the restriction in datasheet, which means I can use ADS1246 to operate AinP voltage from 0.2 to 4.8V when AinN=2.5V.

But, I cannot confirm this, since Vcmi range of -1.05V to 6.05V seems to be some wrong.

I want to realize about the Vcmi range of  "(AVss+0.1+VinPGA/2) to (AVDD-0.1-VinPGA/2)" :

1. Does Vin equals to AinP-AinN, even when AinP<AinN.

2. Does Vcmi equals to the lower one of AinP and AinN, with the reference of GND.

3. Shall I use (AVDD-AVSS)/2=2.5V as the reference to calculate Vcmi.

 

Thanks a lot.

  • 0
    TI__Guru* 97455 points

    Chen Shi,

    I think the datasheet has a difficult wording. The best way to think about this is as the input range, and not the common-mode input range.

    The input range is AVSS+0.1V to AVDD-0.1V. AINP and AINN should be between these voltages for the input range.

    In the datasheet we say that the input common-mode range should be (AVSS+0.1+VIN*PGA/2) to (AVDD-0.1-VIN*PGA/2). This attempts to say that the inputs should stay within the range of AVSS+0.1V to AVDD-0.1V, but that any input signal will be limited to the full scale range set by the PGA.

    Note that the common-mode input is the average between AINP and AINN and that AINP can be lower than AINN. In your example, if AINP=0.2V, and AINN is 2.5V, the common-mode input is 1.35V.

    So in answer to your questions:

    1. VIN does equal AINP-AINN, even when AINP<AINN. Note that this should be scaled by PGA gain.

    2. The common-mode input is the average of AINP and AINN or Vcmi=(AINP+AINN)/2.

    3. I'm not sure what you're asking about in this question, but let me say that 2.5V is an appropriate reference voltage. Depending on what you are measuring, if you need a common voltage to measure different channels, 2.5V is also appropriate for your AINN. It is set in the middle of your common-mode range and can certainly be used to measure against other AINP inputs.

    Joseph Wu

  • 0 in reply to Joseph Wu
    Prodigy 155 points

    Joseph Wu

    Thank you very much for your explanation.

    But there are still some questions about my application.

    In your reply,

     "This attempts to say that the inputs should stay within the range of AVSS+0.1V to AVDD-0.1V, but that any input signal will be limited to the full scale range set by the PGA. "

    How can I deduce this conclusion from the equation given by the datasheet.

    When AINp=4.8V, AINn=2.5V, Vcmi=3.65V, and the Vcmi range is from 1.35V to 3.65V, it is easy to understand.

    But, when AINp=0.2V, AINn=2.5V, Vcmi=1.35V, the Vcmi range is still from -1.05V to 6.05V, but it seems of no sense, so the effective limitation is from 0.1V to 4.9V?

    In other words, in my application, any voltage of  AINp from 0.2V to 4.8V can be correctly sampled into code range of -2.3V to 2.3V, if PGA=1 and AINn=2.5V?

     

    Thanks.

  • 0 in reply to chen shi
    TI__Guru* 97455 points

    Chen Shi,

    Part of the confusion comes from the fact that the equation should be the absolute value of Vin. I'm sorry about that I should have realized this before.

    Just to make sure, lets look at this using your examples. I'll need to change it to explain it better though.

    First the common mode input (Vcmi) is AINP+AINN/2 or the average of the two inputs.

    Now, lets say AVSS=0, AVDD=5, PGA=1, and VIN=AINP-AINN=2.3V. In this case, the common-mode input range is:

    From the bottom: (AVSS+0.1+|VIN|*PGA/2) = 1.25V
    to at the top; (AVDD-0.1-|VIN|*PGA/2) = 3.75V

    In your example, AINP=4.8V, AINN=2.5V so that VIN=2.3V, and Vcmi=3.65V. The 3.65V is between 1.25V and 3.75V so this is good.

    In the second example AINP=0.2V and AINN=2.5V. This makes VIN=-2.3V, with Vcmi=1.35V. However in the calculation for common-mode range, it is done with the absolute value of VIN.

    From the bottom: (AVSS+0.1+|VIN|*PGA/2) = 1.25V
    to at the top; (AVDD-0.1-|VIN|*PGA/2) = 3.75V

    Which is exactly the same as before. Since 1.35V is between 1.25V and 3.75V so this is also good.

    If Vin=0, then the input can go all the way from AVSS+0.1V to AVDD-0.1V. If Vin is larger, then the common-mode input range is restricted.

    I would also note, that the input range is +/-VREF. If VREF=2.5V, then you will only be able to sample inputs as large as 2.5V. The maximum will be AINP-AINN=2.5V and the minimum will be AINP-AINN=-2.5V. Input signals beyond this will overrange the output data to the maximum 7FFFFFh or minimum 800000h codes.

    Joseph Wu

  • 0 in reply to Joseph Wu
    Prodigy 155 points

    Thank you Mr. Joseph Wu.

    Thank you very much.

    The key to understand the Vcmi range is the absolute value of Vin, but not the raw Vin with sign.

    I have ever supposed it is the absolute value, but I was not very sure, since no absolute value symbols appears in the equation.

    I make so bold as to suggest the datasheet editor to verify this part to avoid possible confusion from users.

    Thank you again.