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AFE for 250MHz 2.5Vpp Signal for ADS9684

Category: Software

Hi,

Preface :

My application requires the single input signal from a optical sensor which is in range of 2.5Vpp to be driven to an ADC through a FDA

The signals are in range of DC-250MHz

I have selected the ADC as AD9684( Only available ADC with Input range of 2.06Vpp Maximum) at 500MSPS

I wish to use the ADC for Both AC and DC coupling so i have to give the option as per requriement.

below is my signal nature 

Q Channle is 2580mVpp and I channel is 2100mVpp

i need to condition this signal to take it to a ADC and make full use of resolution of ADC

1. Do i really require an FDA for the ADC Selected ?

2. If So, Kindly suggest a FDA for driving the ADC and AFE scheme for the input as above

3. The ADC input has range of FSR up to 2.06 but i am confused how my signal will be read by ADC when Vref is set to 1V

above 1V the ADC is saturated but the ADC says 2.06Vas FSR, kindly let me know how to understand this also

4. should i need to shift the DC of signals to make them fall in exact range of ADC FSR ?



ADC Doubts
[edited by: shyam.sunder91 at 2:20 PM (GMT -4) on 1 Jul 2022]
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  • Hi ,

    You'll get the best performance if you drive AD9684 input(s) differentially.

    For driving the AD9684 input(s), you could use a FDA like LTC6419, as shown below to attenuate your signal from a maximum of 2.6Vpp to make sure you don't exceed the ADC maximum FS voltage of 2.06V, as shown below, with a CM voltage of 2.05V per AD9684 input CM voltage specification:

    Here is the LTspice simulation file if you'd like to work with it:
    6036.LTC6419 attenuator FDA ADC Driver EZ 7_2_22.asc

    The analog input requirements of AD9684 call for a CM voltage of 2.05V and a maximum FS input of 2.06V.

    The configuration I've shown above achieves these goals. The 1.0V reference built into the AD9684 is used internally for other purposes.

    Hope this helps.

    Regards,

    Hooman

  • Thankful to your response, i have few basic doubts in high speed adcs and its Vcm

    conventionally i understand that Vref is a ceiling of the ADC Full-scale value

    You meant to say that 2.06 is the Full-scale value which means an input signal of 2.06 would read all 1's in Digital output ? and Vref is nothing more than internal purposes and should be only 1.0V and not more than that.

    what is the input equivalent value of digital 0x0000  ? 0V ?

    what is the range of input then 0-2.06V, where 0x000 is 0V and 0x3FFF is 2.06V ?

    How to read the negative values which are less than 0V ? how to use Vcm as a tool to read negative values

    Can you please expalin with some picture how the ADC 9684 sees input and transaltes to digital equivalen with Vcm played arround, that would help me to do any signal conditioning requried

    you may see back my  signal which is Q Channle is 2580mVpp and I channel is 2100mVpp

    Attenuation would cause low voltage signals which are not shown which are arround 200-250mV to go un noticed, is there any other way arround ?

  • Hi ,

    With regard to AD9684 specs, I'm going by this below from the datasheet:

    But since I'm not an expert on ADC's, I'll try and move your post to the data acquisition forum for better support and in case I've misspoke.

    Some Responses Below:

    what is the input equivalent value of digital 0x0000  ? 0V ?

    Response: digital 0V will be a -1.03V differential input voltage centered at 2.05V CM. 

    You meant to say that 2.06 is the Full-scale value which means an input signal of 2.06 would read all 1's in Digital output ? and Vref is nothing more than internal purposes and should be only 1.0V and not more than that.

    Response: When Vin+ = 2.05V + 2.06/4 = 2.56, and Vin- = 2.05-2.06/4 = 1.54V, you'd register all 1's digital output. On the other hand, all 0's would correspond to Vin+ = 1.54V, and Vin- = 2.57V by looking at the specs above.

    How to read the negative values which are less than 0V ? how to use Vcm as a tool to read negative values

    Response: All 0's = -1.03V diff input with 2.05V CM. All 1's = +1.03V diff input with 2.05V CM.

    you may see back my  signal which is Q Channle is 2580mVpp and I channel is 2100mVpp

    Response: I'd just reduce the LTC6419 driver gain lower for the Q channel than the gain needed for the I channel, if I've understood your question correctly?

    Attenuation would cause low voltage signals which are not shown which are arround 200-250mV to go un noticed, is there any other way arround ?

    Response: As long as the input to the ADC is greater than 1 LSB, the ADC will digitize the analog input accordingly. I don't know what you mean by "it would go un-noticed"! That's another reason I should be moving your query to the ADC forum for better support in case I'm not fully understanding your problem.

    Regards,

    Hooman

  • Hi ,

    I just moved your question to the High Speed ADC forum to make sure the responses I've given are correct and to get better support for your AD9684 specific questions.

    Regards,

    Hooman

  • Hi ,

    Can you please expalin with some picture how the ADC 9684 sees input and transaltes to digital equivalen with Vcm played arround, that would help me to do any signal conditioning requried

    Response: Does this picture drawn with the help from Diff-amp Calculator program help with diff input visualization in seeing the all 0's and all 1's instances in time?

    https://www.analog.com/en/design-center/interactive-design-tools/adi-diffampcalc.html

    Regards,

    Hooman

  • Kindly inform any ADC forum expert to confirm your comments please

  • Hi ,

    I've additionally sent your request (to confirm my responses to your questions) to the Application Eng. in charge of AD9684. Hopefully we'll get a response from them soon.

    Regards,

    Hooman

  • Hi ,

    FYI>>

    I've received feedback from the Application Engineer that the answers I've posted in response to your questions about AD9684 ADC are accurate and you can use that information as-is.

    I wanted to make sure that you get this so that you can have more confidence in my responses.

    Regards,

    Hooman

  • Hi shyam.sunder91,

    All answers posted by Hooman are accurate. ADC inputs are formulated below:

    Vin+ = Vcm +/- (Vdiff / 4)

    Vin- = Vcm -/+ (Vdiff / 4)

    As a supplement, please see the attached articles that are relevant to this conversation.

    raq-issue-93.pdf (analog.com)

    AN-1404 (Rev. 0) (analog.com)

    MT-075: Differential Drivers for High Speed ADCs Overview (analog.com)

    Regards,

    Xavier

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