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ADS42LB69: input voltage range

Depeng Man
Intellectual 2030 points
Part Number: ADS42LB69


Hi Team，

what will be the maximum input range supported by the ADC?

what will be the digital value for 2Vpp and 2.5Vpp？

Best Regards

James Man

  • 0
    TI__Intellectual 2030 points

    In any case, the customer can calculate the LSB in voltage as Vpp / 2^16. In this sentence it is mentioned 3fff, what is the analog input value it is representing 2Vpp or 2.5Vpp?

    but while calculating LSB we are getting 30.5x10e-6, so the output vale for maximum peak is 0.5 V

  • 0 in reply to Depeng Man
    TI__Mastermind 24180 points

    Hi Depeng,

    The LSB in voltage can be calculated as shown in the original post. The customer will need to look at the digital gain they are using to find which analog input range is active. By default (0dB gain), the input range is 2Vpp. The VCM for the ADS42LB69 is a 1.9V common-mode.

    Input range (Vpp) / 2 + VCM = voltage corresponding to maximum supported input

    Input range (Vpp) / 2 - VCM = voltage corresponding to minimum supported input

    For example, if using a 2.5Vpp input range, the maximum ADC code will occur at (2.5Vpp / 2) + 1.9V = 3.15V.

    Thanks, Chase

  • 0 in reply to Chase W
    TI__Mastermind 24180 points

    Hi James, does this make sense? Does the customer need any further guidance on this? Let me know.

    Thanks, Chase

  • 0 in reply to Chase W
    TI__Intellectual 2030 points

    Hello ,

    Thank you for your quick reply.

    I've conveyed it to my customer.

    Have a good day .

    James

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi Chase W

    In case we are giving adc input as1Vpp sine wave of frequency 20MHz  , we get digital code as 13294, how to decode this to analog voltage while we setting digital gain is 2VPP,

    ADC voltage calculation 1LSB = 2vpp/65536 Digital code Digital code * 1 LSB 
    1 30.05*10^-6 13294 0.405700684
    1.1 30.05*10^-6 14101 0.430328369
    1.2 30.05*10^-6 15722 0.479797363
    1.3 30.05*10^-6 17300 0.527954102
    1.4 30.05*10^-6 18741 0.571929932
    1.5 30.05*10^-6 19333 0.589996338
    1.6 30.05*10^-6 20059 0.6121521
    1.7 30.05*10^-6 22215 0.677947998
    1.8 30.05*10^-6 23761 0.725128174
    1.9 30.05*10^-6 24395 0.744476318
    2 30.05*10^-6 26810 0.81817627
    2.1 30.05*10^-6 28088 0.857177734
    2.2 30.05*10^-6 29524 0.901000977
    2.3 30.05*10^-6 30856 0.941650391
    2.4 30.05*10^-6 30746 0.938293457
    2.5 30.05*10^-6 32767 0.999969482

    Please do the needful

    Regards

    Sabitha M

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hi Sabitha,

    It seems like your table above is not incorporating the common mode voltage. Please find attached a spreadsheet showing how to do conversion between analog input voltage and the digital code. The green fields are informative only and should not be written to. I hope this helps!

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi sir,

    As per above formulae we get , Min voltage is -0.65

    min Voltage Formulae is (VPP/2)-VCM=[(2.5vpp/2)-1.9]

                                                                  = -0.65V 

    Max Voltage Formulae is (VPP/2)+VCM=(2.5VPP/2)+1.9

                                                                    =+3.15V

    But you have shared an attachment , the min Voltage Formulae is VCM-(VPP/2)  & Max Voltage Formulae is VCM+(VPP/2)

    Please share the correct formulae to find min & max Input voltage 

    ADC output voltage formulae= Min voltage + [1LSB*Digital Code]

    why we are not using Max voltage to find ADC Output Voltage

    Thanks & Regards

    Sabitha M

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hello Sabitha,

    The correct formula is VCM +- Vpp/2. I've just noticed the error in my previous message. Sorry for this confusion! 

    You can use either the min or max to calculate the ADC input voltage. There is no ADC output voltage, just digital code.

    Volts = VCM-Vpp/2 + 1LSB*Digital code

    Volts = VCM+Vpp/2 - 1LSB*(2^16‐Digital code)

    It's a bit easier to calculate using the minimum voltage.

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points
    ADC INPUT VOLTAGE DIGITAL CODE
    1 12378
    1.1 13766
    1.2 14345
    1.3 16268
    1.4 17886
    1.5 18860
    1.6 19949
    1.7 21033
    1.8 22827
    1.9 23907
    2 25070
    2.1 26464
    2.2 27986
    2.3 28965
    2.4 30614
    2.5 31738

    Hi Chase W,

                  I herewith attached the table for various input voltage which I am generated using FG of sine wave with 23.5 MHz frequency,

    and the corresponding digital code of the ADC are mentioned which where the digital code which I get from the peak of the sine wave from the ILA using the Xilinx Vivado tool.

    And I am setting the Digital gain as 2.5Vpp, Could u please explain me the calculation for the above table. Because I am not getting 1Vpp corresponding value when multiplying the 1LSB with the  Digital code when I am using the formula you mentioned for calculating the Voltage. 

    Thank you,

    Regards, Sabitha

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hi Sabitha,

    What are the units for your table? Vpp? Volts? The spreadsheet that I sent a few days ago performs the calculations correctly. If the received ADC code is not as expected, then the recorded input voltage is not correct. Are you measuring the input voltage using a differential probe of sufficient (>10x) bandwidth directly at the ADC's input pins? Or is this at the signal generator? What signal generator are you using? Can you send me the ILA data in CSV form so I can check it in our software?

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi sir,

    ADC input frequency is 23.5MHz 

    ADC sampling frequency is 188MHz

    For test case, I had set the digital gain as 2.5vpp through register configuration for both channel(A&B) and gave 2vpp as an input voltage for ADC.

    The ILA sampling depth is 4k. The ADC_CHANNELA_DATA(signed decimal format)  column D in the excel sheet is the digital code which I am getting for respective input voltage(2vpp).

    Kindly find the attachment below,

    iladata_25_2vpp.csv

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi sir,

    ADC input frequency is 23.5MHz 

    ADC sampling frequency is 188MHz

    For test case, I had set the digital gain as 2.5vpp through register configuration for both channel(A&B) and gave 2vpp as an input voltage for ADC.

    The ILA sampling depth is 4k. The ADC_CHANNELA_DATA(signed decimal format)  column D in the excel sheet is the digital code which I am getting for respective input voltage(2vpp).

    Kindly find the attachment below,

    4186.iladata_25_2vpp.csv

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi sir,

    ADC input frequency is 23.5MHz 

    ADC sampling frequency is 188MHz

    For test case, I had set the digital gain as 2.5vpp through register configuration for both channel(A&B) and gave 2vpp as an input voltage for ADC.

    The ILA sampling depth is 4k. The ADC_CHANNELA_DATA(signed decimal format)  column D in the excel sheet is the digital code which I am getting for respective input voltage(2vpp).

    Kindly find the attachment below,

    6518.iladata_25_2vpp.csv

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hi Sabitha, thanks for providing the data! I will review this next week.

    However, I do notice from your description that the input frequency is an even integer multiple, 8, of the sample rate, meaning the exact same 8 points along the input sinusoid will be sampled each period, not ideal. If you can adjust the input frequency slightly to an arbitrary input frequency (such as 23.4790251 MHz) then the sinusoid should be sampled in a more random pattern and the samples should be spread out along the sinusoid. I can provide more information on this next week.

    Regards, Chase  

  • 0 in reply to Chase W
    TI__Mastermind 24180 points

    Hello Sabitha,

    As I suspected, the samples are the exact same points along the sinusoid, which is not so good. I am still curious how you are finding the input voltage, as well as the quality of the input source. If you change the input range from 2.5Vpp to 2Vpp, will the codes decrease by 25%?

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi sir,   

            As you said I have changed the input voltage range from 2.5vpp to 2vpp, the digital code has decreased  by 25%. The ADC_CHANNELA_DATA(signed decimal format)  column D in the excel sheet is the digital code which i am getting for both the input voltage and I have attached the CSV file for your reference below. kindly refer the document.

    FILE REFERENCE:

    iladata_25_2.5vpp.csv -> ADC digital gain 2.5vpp  and input voltage range is 2.5vpp.

    iladata_25_2vpp.csv -> ADC digital gain 2.5vpp and input voltage range is 2vpp.

    QUERY:

    How to calculate the SNR value for ADC? Does TEXAS INSTRUMENTS provide any tool for plotting the SNR value?

    iladata_25_2.5vpp.csv

    5282.iladata_25_2vpp.csv

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hello Sabitha,

    The only tool we have for this is HSDC Pro. This is our software suite for our evaluation modules (EVMs). You can download HSDC Pro at this link: https://www.ti.com/tool/DATACONVERTERPRO-SW

    You will have to import the data as CSV by placing channel 1 data in column 1 and channel 2 data into column 2. (see my attachment below as an example containing 2^16 samples).

    I think the "unexpected" output codes are mainly due to front end loss and/or equipment. Can you share with me a screenshot of the front end circuitry?

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi Sir,

    Please refer the front end circuit which i have attached below. Incase if any mistakes in my circuit kindly suggest me the proper front end circuit for ADC. 

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hello Sabitha,

    The front end circuit looks fine. The WBC1-1TL balun seems to have an attenuation of around 0.8dB per transformer. This will mean the front end losses will be around 1.6dB at minimum, equating to roughly 400mVpp loss (assuming a perfect 50 ohm).

    In the last set of data you sent (testing the 2.5Vpp input while ADC is set for 2.5Vpp input range), the maximum difference between codes is 52698. This equates to utilizing roughly 80.4% (52698/65536) of the available input range. However, this is assuming there is no losses in the front end. If you consider the 400mVpp loss into the 2Vpp applied signal as a result of the qty. 2 baluns, then the normalized input voltage is actually about 2.1Vpp (around 84% utilization). This leads to an expected ADC code of around 55050 (2.1Vpp/2.5Vpp*65536). This is fairly close to the measured values in the previous post. All which you are seeing if the front end loss inherent to the front end design, which is normal. This just means you will need to increase the amplitude of your source by ~400mV in order to reach closer to the ADC fullscale.

    Regards, Chase

  • 0 in reply to Chase W
    Prodigy 130 points

    Hi Sir,

        In our application the ADC input range should be 0dBm to -40dBm.But our front end circuit is currently accepting input range of -22dBm(Digital gain is set to 1Vpp).Is there any possible way to increase the input range to -40dBm by changing the circuit or register configuration? 

  • 0 in reply to SABITHA MUTHUVIJAYAN
    TI__Mastermind 24180 points

    Hi Sabitha,

    I'm confused what is meant by the input range is currently accepting -22dBm. Is the input -22dBm and there is nothing showing in the ADC sampled data? The minimum LSB is 15.25uVpp (digital gain set for 1Vpp) which is adequate to capture/detect a very low input, such as near -90dBm. 

    Regards, Chase