AD7768 : 8 CHANNEL ADC CAN SUPPORT SINGLE END - BIPOLAR INPUT.

Hi All ,

I have an requirement to convert the 8 channel analog input (Single ended - bipolar) signal with the range of 10Vpp, The waveform will be sin, ramp, square,and Sweep.

Hence I am planing to use AD7768. In this I need a clarification of following points:

1. I have 8 channel single ended bipolar signal inputs (10Vpp, +/- 5 ), So I will connect this signal to the correspoding port of (AIN0+ to AIN7). My question is, May I Ground it the remaining ports of AIN-0 to Ain7- ?

2. My input signal should support (+5V to -5V , 10Vpp), so in REF1- and REF2- may I connect to Ground or -2.5V should give ?

3. If I ground the REF1- and REF2- , The AD7786 support the input signal swing 10Vpp?

4. Even though My Signal is single ended bipolar , Is there I want to make this signal to converted into differential bipolar (True Biploar signal)?

Please clarify the points as soon we are going to finalise the design stage.

  • 0
    •  Analog Employees 
    on Aug 15, 2018 10:10 AM

    Hi Sidiq,

    You can use a fully differential amplifier like the LTC6363 or ADA4940 to perform this function. This will allow you to use the full range of the AD7768. Please note you will need to use a 5V reference such as ADR4550. The picture included below is from the LTC6363 datasheet which you will find here. In this type of configuration both AIN+ and AIN- are connected to the LTC6363 outputs. REF1- and REF2- can be connected to ground since the common mode output from the AD7768 can be used to set the common mode of the LTC6363 to 2.5V.

    Regards

    Niall

  • Hi Niall,

    Thanks for your inputs,

    As we are using 8 Channel Inputs, Each channel inputs I should convert single end to differential end bipolar signal using LTC6363. 

    With your design reference discrete components occupy more spaces. 

    My board dimension will be 4.5 cm circular dia.

    I have another query,

    1 ) If I will attenuate the signal range from 10Vpp to 5Vpp (That is AIN input signal range  +2.5 to -2.5V) signal range.

    How i should connect the power inputs (AVDDx , AVSSx) and reference inputs (REF+x and REF-x)?

    2) If I give differential Input signal (-2V at AIN1+ ; +2V to AIN1- ) what will be the output data? can you give me the conversion formula ? what REF+x and REF-x I should give ?

    Thanks and Regards,

    Mohamed Sidiq.H

  • 0
    •  Analog Employees 
    on Aug 16, 2018 2:05 PM

    Hi Mohamed,

    You will need an ADC driver to achieve the same level of performance using the AD7768 as in the datasheet. So a 4.5cm board will be a challenge as you will need the ADC, reference, reference buffer and 8 x ADC drivers.

    1. If I will attenuate the signal range from 10Vpp to 5Vpp (That is AIN input signal range  +2.5 to -2.5V) signal range.

      How i should connect the power inputs (AVDDx , AVSSx) and reference inputs (REF+x and REF-x)? AVDDx will need to be at +2.5V referenced to DGND and AVSSx will be set to -2.5V referenced to DGND. Thus there will be 5V between AVDDx and AVSSx. All analog circuitry will then need to be referenced to AVSSx such as the driver amplifier rails and the reference. The 5V reference will be +2.5V above DGND on the REF+x pin. REF-x will be set to -2.5V when referenced to DGND (you can connect directly to AVSS. 

    2. If I give differential Input signal (-2V at AIN1+ ; +2V to AIN1- ) what will be the output data? can you give me the conversion formula ? what REF+x and REF-x I should give ? The ADC will see (AIN1+ minus AIN1-) = -4V. The output from the AD7768 is in two's compliment format. So -4V corresponds to 1110 0110 0110 0110 0110 0110 (0xE66666) with a 5V reference. A simple way to decode this is to convert to straight binary and use code 8,388,608 as mid-scale, code 0 as 0V and code 16,777,215 as +10V. You can then work in straight binary so the -5V to +5V input range becomes 0 to 10V. The picture below is taken from the AD7768 datasheet and explains the Twos Complement coding.

    Regards

    Niall

  • 0
    •  Analog Employees 
    on Aug 21, 2018 11:01 AM

    Hi Mohamed,

    Please see my answers to your questions below:

    1. Is this LDO Current is enough for AD7768 Supply? ADP7118 is used on the AD7768 customer evaluation board and so it is a proven combination. The ADP7183 is needed to create the analog ground plane at -2.5V and should not need to supply large amounts of current to the ADC. 
    2. The LTC6363 can perform both single-ended to differential conversion and attenuation using the circuit I included above. Therefore this one part can be used to replace the two parts above and also achieve better performance. I would recommend using this instead of AD826 and ADA4922, especially since you mentioned board space was of concern. 

    Another factor you should also consider is simplification of the 8th order low pass filter. A 4th-order filter will suffice if the ADC is operating in the optimum configuration. The neat thing about the AD7768 versus a traditional SAR ADC is that the digital filter will take care of the anti-aliasing for you. The potential aliasing points which you need to protect against are shown in the image below and depend on the speed of the modulator. If you run the modulator at a faster rate the unprotected regions shown below can be pushed much higher than the 30kHz bandwidth you require and the complexity of the antialiasing filter can be reduced. 

    Regards

    Niall

  • Hi NiallM,

    Thank you very much for your inputs !!!

    I am going with AVDDx, REF+x = +2.5V  using  ADP7118 ; and AVSSx , REF-x = -2.5V using  ADP7183ACPZN2.5-R7

    Is this LDO Current is enough for AD7768 Supply ?

    I have an rearranged architecture of my 8 Channel Receiver card.

    Before the architecture is :

     

    Re-arranged architecture :

    Can you give me your clarification of following parts.

    1.        AD826: I am going to attenuate my signal ranges from 10Vpp to 5Vpp (+/-5V to +/-2.5V ). By selecting RF /Rin with inverted output (Example:  RF = 1K ; Rin 2K = -0.5 Gain). This Output is fed to the input of another channel of  AD826. Hence I will get +/-2.5V attenuated signal.

            Is this right way of attenuate my signal range ? or do you have any recommended part for this design?

    1.         ADA4922-1 This attenuated signal is fed to the ADA4922-1 to convert the single ended bipolar signal to   differential bipolar signal and fed to the ADC (AD7768) input.

    This single ended to differential end conversion is there right part to do the conversion instead of you recommended LTC6363.

     

    Please confirm may I go ahead with this part ?

     

    Thanks and Regards,

    Mohamed Sidiq.H