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AD9208-3000EBZ common voltage & signal level

Question asked by TERRY123 on Apr 12, 2018
Latest reply on Apr 12, 2018 by UmeshJ

Hello, I am using AD9208-3000EBZ with internal reference mode.


I have some questions. I need help....

Above, this schematic is AD9208-3000ebz's A channel input.


1. I set the board configuration as an internal reference mode & AC coupling(VCM buffer is connected input of ADC).

   when i do not input sampling frequency, the offset of VIP_A node is 0.5V approximately.

   when i input the sampling frequency(3000MHz),  the offset of VIP_A node is 1.34V.

    Is the VCM buffer's output affected by sampling frequency, isn't it? Then why ?



2. And also when i change 0x1A4C(Buffer Control, adjusting amount of VCM buffer's current), the offset of VIP_A node was changed. (1.3V~1.4V)

i want to know the exact calculation of ADC input common voltage(ADC input reference voltage). in the internal reference mode, the internal reference voltage of ADC = 0.5V. please explain this in detail ... ^^




3. AD9208 datasheet said : Use Register 0x1910 to change the internal reference voltage. Changing the internal
reference voltage results in a change in the input full-scale voltage.

But, when I change the 0x1910(1.13Vp-p ~ 2.04 Vp-p), i could not feel the offset voltage difference of VIP_A node  between them. please explain about this...




4. I think the most of existing ADCs, input Full-scale Vp-p follows ADC's input reference voltage(X2) when using the internal reference mode.

   if the ADC's input reference voltage(input offset) were 1V, the input full-scale range(Vp-p) would be 2V.

  But, is the AD9208 different, isn't it? 

 My understand is right???


5. Setting Bit 2 of 0x1908 as 1(for DC coupling) is switch off between VCM and ADC input.

Then, the offset of VIP_A node should be 0V in ad9208-3000EBZ?

Is that right? 


Thanks for helping me!!