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ADS6425

Shahzad Malik97479
Intellectual 295 points
Other Parts Discussed in Thread: ADS6425

I am using the ADS6425 and I have some questions regarding the reference voltage. I am using the internal reference and it is not clear to me how to calculate the LSB size in volts. Normally I would take the reference voltage and divide by the number of bits, in this case 4095 since it’s a 12bit converter. However this part has a "top" reference voltage of 2V and a "bottom" reference voltage of 1V. So do I take 2V/4095 or 1V/4095?

 

Furthermore, table 19 shows some gain settings vs. input voltage. I assumed there is some amplifier that will increase the signal amplitude sent into the converter, and so consequently the maximum signal amplitude applied to the converter has to be reduced when there is a gain applied inside the ADC.... correct? If I use a gain setting, does this have any impact on the VREF or LSB size? Or is the size of my LSB the same regardless of if I set the ADC with a gain or not.

  • 0
    TI__Mastermind 29220 points

    Hi,

    This quetion also came in by way of our Product Information Center and one of our analof field engineers and I answered this in that venue as well, and i would hope that the FAE would get that answer back to you.  But here is what I answered there:

    You would take the default full scale voltage and divide by 4096.  The default full scale voltage (in the absence of selecting any gain) is 2V *differential* peak to peak.  That means each side of the differential signal has a 1V peak to peak swing centered about the common mode voltage of 1.5V.  So each side of the differential signal would have a maximum swing from 0.5V below VCM to 0.5V above VCM.   Thus the reference voltages would be 1.0V on the lower side and 2.0V on the upper side.

     If you choose to use a gain option then you would consider the full scale voltage to be reduced by the appropriate amount.  There is the optional gain setting of 3.5dB (that is gain applied before the ADC circuit I believe), so that would simply reduce the external full scale voltage definition by 3.5dB as measured at the input pins.    There is also the fine gain setting in steps of 0.5dB that would also reduce the external full scale voltage definition as measured at the input pins by the amount of gain chosen.     So as shown in table 19, if you choose 6dB of gain then the full scale voltage becomes 1.0V, and the voltage represented by each increase in the sample code by one lsb would be 1.0/4096 in this example.

     Regards,

    Richard P.

  • 0 in reply to Richard Prentice
    Intellectual 295 points
    I guess I’m still confused. Perhaps it’s because of the differential nature of this ADC. If the reference voltage is 1V on the low side and 2V on the top side… then the range of Vref is 2V-1V=1V, isn’t it? Then therefore the LSB is 1V/4096 (not 2V/4096)? I’m used to thinking about Vref as a voltage referenced to GND that gets divided up by a resistor ladder to determine the size of each bit. I guess I look at this ADC having a resistor ladder between 2V and 1V, thus giving a Vref of 1V…. but maybe this is a flawed way of looking at this ADC.
     
    Also, the response above seems to imply that the size of the LSB varies with gain settings. I can see how applying gain could force you to reduce the magnitude of the applied input voltage, but does it really also scale Vref? Again, maybe I’m over simplifying this ADC but I would think that the LSB size should remain fixed and the gain just scales the input signal to maximize the full scale range.
  • 0 in reply to Shahzad Malik97479
    TI__Mastermind 29220 points

    Hi,

    yes, it is becuase the input full scale voltage definition is *differential* and *peak to peak*.  The max output code would correspond to the IN+ being at 2.0V and the IN- being at 1.0V so the input voltage is +1.0V.   The min output code would correspond to the IN+ being at 1.0V and the IN- being at 2.0V so the input voltage is -1.0V. So the input voltage can swing from -1V to +1V for a peak to peak swing of 2V.    But the swing on each input pin is only 1V but they are differential and need to be th einverse of each other so the net differential change is 2V peak to peak.   So each change in lsb of the output code represents 2/4096 V change in the differential input.

    Now if you chose to use gain, imagine for the moment that the gain stage is in front of the ADC device.  Now if the full scale definition is still 2V adn the amp gain is 6dB, then the full scale definition at the amp is 1V even though the ADC itself is still operating with 2V full scale, and the reference voltages dont change.  And each output code lsb represents still 2/4096 V of change at theADC inputs, but now would only represent 1/4096 V of change at the input to the amp.   Now pulling the amp to be inside the ADC package but before the ADC doesn't change any of these definitions.  I believe the 3.5dB coarse gain option represents exactly this case.

    The fine gain option I believe is done digitally after the ADC, but still the full scale definition at the inputs are reduced by the amount of the fine gain chosen as in the case for analog gain.

    Changing the Vref will change the definition of full scale from 2V diff peak to peak to something else accordingly, but the above analysis of what gain does to *that* new full scale is as we just went through.

    Regards,

    Richard P.