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ADS1259 without buffer at input

Other Parts Discussed in Thread: ADS1259

Hi,

I am using ADS1259.  This is switched capacitor type ADC. I want to use ADC without buffer. Can I use it? if yes, what factors i need to consider.

Input is single ended 4-20mA across 130 ohm.

Can you help?

Regards,

Vinayak Aghor

  • Vinayak,

    You can use the ADS1259 single ended without a buffer, but you will introduce an error.  The significance of the error is relative to the impedance involved.  If the driving impedance is small, then you won't see much of an issue. 

    Essentially you have a current divider.  The parallel impedance will lower the voltage drop across the 130 ohm resistor.  If the differential impedance is approximately 120k, you will lower the shunt resistor about 15m ohm, which is about 0.1%.  This will change slightly over temperature, but may be well within your measurement tolerance.  You may also find that you can calibrate out this error as it should be fairly linear.

    Best regards,

    Bob B

  • Hi Bob,

    Thanks for answer.. However, so as not introduce noise into system we want to use 2nd order LPF as well. We want to drive the ADC directly through it. This will require 100K and 0.22uF components. However this will not satisfy the settling time criteria of 0.5 us. (given FLCK= 7.3728MHz, Serial CLK = 1MHz, sampling rate = 50 SPS & reference of 3V and power supply of +5V and 0V)

    How to satisfy the settling time criteria?

    Regards,

    Vinayak

  • Hi Vinayak,

    You've added a significant new issue from your first post.  You have now gone from a current divider to a voltage divider.  For my first response all that was needed was a simple RC anti-aliasing filter (approximately 50 ohms and 10nF.)  If the desire is to create a low frequency filter, and if you add large resistance values you move from a simple current divider to a voltage divider.  Also settling time will be very large.  With a 24 bit converter your settling time will be related to resolving to 1/2 LSB.  If you have a large time constant this may be problematic if your input signal changes often and quickly.

    There are trade-offs.  What level of noise can you accept and what resolution is required?  If you need the low frequency filtering, you will need a buffer.

    Best regards,

    Bob B

  • HI Bob,

    I had tested the 10nF and 47 ohm resistance filter, but i did not get expected result. I have also tested RC LPF (100K & 0.22uF) connected before ADC inputs..Still did not get expected results. I am using cut off frequency of 7 Hz. Input direct current is 4-20mA. Still I need buffer?   Also I did expect 10nF and 47 ohm filter could work for ADC, but it did not. Have you any thoughts on probable cause?

    Regards,

    Vinayak Aghor

  • Hi Vinayak,

    If you use a series resistance of 100k, you will need a buffer.  When you say you did not get the expected results, what do you mean?  What were the test conditions?  What did you expect and what did you get?  I can try to duplicate your setup, but I need more information.

    Best regards,

    Bob B

  • HI Bob,

    With 47 ohm & 10nF cap filter I got ADC count as: (here i have shown first 16 bits from MSB)

    4mA --  0x0C07

    12mA – 0xE0C0

     20mA – 0xF424

    I expect ADC count as

    4mA --  0x2CCA

    12mA – 0x8E89

    20mA – 0xE826

    Test conditions : I am giving 4-20mA current across 130 ohm resistor and then giving it to RC filter & then giving it to ADC input. Even if I use RC LPF filter of 100K and 0.22uF then connect it to ADC inputs, i did not get expected results.

    With 47 ohm and 10nF filter, time constant of 0.47us which is around 1/ (Fmod/2) = 0.5us. Input needs to settle within 0.5us.

    what should be the cause or remedy to get expected result?

    Regards,

    Vinayak Aghor




  • Vinayak,

    The data from the ADS1259 is binary two's complement.  I think there is something wrong with your calculations and perhaps read back of the data.  You should only be seeing codes from 0x000000 to 0x7FFFFF, unless you are adjusting the gain calibration register.  Can you show me how you are making your calculations?

    Your settling time is relative to the analog filter settling to 1/2 LSB size.  This is the time that it takes for a step change voltage to respond at the output of the filter to within 1/2 LSB value. LSB size is full-scale range divided by 2^24 bits.  This will be many time constants, so you can see that if you increase the time constant it will take a very long time for the analog input to settle.

    Outside of a calculation error,  you may not be seeing the circuit settle.  Are you taking multiple readings at each current level, or are you taking single point measurements?

    Earlier today I took some measurements using slightly different resistance (200 ohms) and current levels from 4 to 10mA, and had measurements very close to the expected and quite linear.  There was some gain error as expected.  I will try to capture some data and send you my results.

    Best regards,

    Bob B

  • 6724.calc.xlsx

    Hi Bob, I have inserted calculations in file. I can see that gain error of 2 is present in the counts. Please let me know if there is wrong in procedure. in file 1 table calculates ideal count and second table calculates current from count. we are considering first 16 bits of ideal count.

    you can change current and ADC count that you got in sheet, and compare results.

    For settling time, you suggest , around 350ns can be ideal settling time (=3V / 2^24). Is it correct?

    i am taking multiple readings however ENOB that is considered is 16, we do not need to average the ADC count.

    Also i have kept some margin of 3mA more. So 0x7FFFFF will be at 23.07mA

    Regards,

    Vinayak

  • Vinayak,

    Settling time will be approximately 12 time constants for 16 bits and about 17 time constants for 24 bits for analog settling.  I'm sorry but I've been rather busy today, so I haven't had time to fully analyze the spreadsheet but it appears that your initial calculations are correct.

    It also appears that the calibration constant in the FSC register is forcing a gain of 2.  I will have to check into this a little further on my end to fully understand what might be happening.  Are you doing any offset or gain calibration? 

    Best regards,

    Bob B

     

  • Hi Bob,

    Kindly tell us how you calculated 12 time constants figure. should i figure out 12 times less than sampling time for internal capacitor of 8pF+2pF?

    it comes out as 1.1us / 12.  1.1us comes from 1/Fmod.I will choose external RC filer of time constant of 90ns. is this you are suggesting?

    I do not want to calibrate the system. I am jest reading raw data.

    Regards,

    Vinayak

  • Hi Bob,

    when I connect 130 ohm resistor to ADC input with 10nF cap across ADC inputs the I get expected ADC counts with some acceptable offset and gain error.

    Here i did not use anti-alias filter before ADC inputs. This is the real problem.

    Regards,

    Vinayak Aghor

  • Hi Vinayak,

    This shows that there was some settling error.  I understand now how you were calculating settling.  You are only considering the internal capacitance of the ADS1259.  What dominates is the external capacitance.  The RC time constant of the analog filter prior to input will be much larger than the switched capacitor input stage.  Both should be used in the calculation, but as the external capacitance is much, much larger than the internal we ignore that capacitance to keep the calculations simple.

    The number of time constants necessary for proper analog settling will be related to the number of resolution bits and the exponential charge (or discharge) of the capacitance.  The simplified formula is: (N+1) ln(2), where N is the resolution of the converter.  The time constant tau is equal to the input resistance of the anti-aliasing filter and the capacitance of the filter.

    For 47 ohms and 10nF the time constant is 470ns.  To settle to the 24 bit level it will need to be a little over 8 us.  If you are to buffer the input with 100k and 10nF, for example, you will still need to have the analog circuit settle.  In this case you would need to wait a little over 17 ms.  If you increase the capacitance, you will have to wait even longer.

    Best regards,

    Bob B