ADC counts Calculation for AD7656


Can any one share me the ADC counts calculation for AD7656?I am following 2s compliment technique. Reference we are using is internal(2.5).The input voltage is varying between +-10V.

With regards,

Letheesh G

For eg:If my input signal is square wave of Vpp=2V what will be  my  AD count?
[edited by: Letheesh at 9:21 AM (GMT -4) on 23 Oct 2020]
  • +1
    •  Analog Employees 
    on Oct 28, 2020 12:41 PM 5 months ago


    The datasheet specifies that the LSB weight (table 9) for a 2.5V reference in the +/-10V range as 305uV (0.305 mV).  This is computed as 

    (4*Vref - -4*Vref)/2^16-1 -> (10V--10V)/2^16-1 =  305uV.

    So for a square wave with Vpp = 2V centered around 0V the positive peak of the square wave will be at 1V which should be at code 1V/305e-6 = 3278d or approximately 1/10th of the fullscale value (0x0CCEh).  The negative peak will of course be at -1V which would be equivalent to -3278d or -1/10th of the fullscale value.  This will result in an output code in two's complement format output code of  0xF332h).

    Hope this helps


  • Thanks Sean for the reply .For better understanding I would like yo ask one more thing .Once we get +1V and _1V at the inputs of ADC, Are we going to subtract  the equivalent ADC counts (3278-(-3278) =6556  then do a 2s complement?

    Can you please clarify ?

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