I am designing a PCB for reading analog signal by the AD7732. I searched a lot to find tips which increase the precision and effective bit resolution. One of these tips is to segregate analog power supply and the Digital one due to switching noise of digital components. I want to Know:

  1. How to design a suitable powering circuits for an ADC. In AD7732 datasheet I did't see any new and different point exclusive for an ADC. It is worth collecting some tips around.

  2. Should I connect analog ground to digital ground? In some datasheets I read they should be connected. If this is true why do we segregate analog and digital power supplies? Doesn't this connection pass noise from one side to other side?

  3. Is two switching power supplies proper to use as analog and digital power supplies?

  • \$\begingroup\$ A simple solution is to use the same noisy digital power supply then filter it with a PI LC filter. There is usually no fast large current variations on the analog part, so it can be filtered without stability issues. \$\endgroup\$
    – TEMLIB
    Sep 12, 2015 at 14:33
  • \$\begingroup\$ By separating analog and digital ground planes, and routing analog signals far from the digital parts, you avoid the return current from the digital circuits pass through the analog part and disturb it. The two ground planes can be connected together on a single point, as a DC reference, but it will behave as an inductor for AC signals and block noise. \$\endgroup\$
    – TEMLIB
    Sep 12, 2015 at 14:37
  • \$\begingroup\$ A very common scheme is to power the digital section from a source voltage through a switching regulator such as converting +9V In to +3.3V Digital. A separate linear regulator is used to power the analog circuits such as +9V In to +5V Analog. Since the analog circuit is usually low current the linear regulator does not have to dissipate much power. \$\endgroup\$ Sep 12, 2015 at 15:44
  • \$\begingroup\$ @TEMLIB what do you mean by single point? I understood making two polygons separately for each ground and then connect them only through a route. yeah? \$\endgroup\$
    – Pana
    Sep 13, 2015 at 6:05

1 Answer 1


My thoughts on the matter:

*Make sure you have proper bypassing on VDD of the ADC. Close to the package as possible. On the other side of the board is usually nice. Combination of both large and small (10u & 0.1u).

*If you can separate the AGND and DGND, you'll see less ground noise from the digital circuits being injecting into the analog signals. You can use an isolator to bring signals between separate ground planes. That may be overkill though.

*Connecting AGND and DGND planes with a ferrite bead can help remove high frequency noise.

*Regarding power, I think either switching regulators or linear regulators are fine, just make sure you have appropriate filtering on the switching regulators and bypass caps on either type of supply.

Some extra links for reading: TI Application Note on grounding in mixed signal. Part 1 Part 2

  • \$\begingroup\$ In application note part 1, what does "Back-to-back Schottky diodes" mean? Is it a configuration of circuit or a type of diode? I am not an electrical engineer, so sorry if trivial. \$\endgroup\$
    – Pana
    Sep 14, 2015 at 6:10
  • \$\begingroup\$ They are referring to 2 components placed in a specific configuration. Schottky diodes are a specific type of diode, they begin conducting current at lower voltages than typical silicon diodes (~0.2 V vs ~0.6 V). It sounds like they are recommending placing 2 Schottkys antiparallel to each other. (annode of D1 + cathode of D2 on one side, cathode of D1, annode of D2, cathode of D1 on the other). Under this configuration, if the potential between the 2 ground planes is ever greater than 0.2 V, one of those diodes will begin conducting and short the ground planes together. \$\endgroup\$
    – Mike
    Sep 14, 2015 at 13:27
  • \$\begingroup\$ I am in agreement with you. In article It isn't clear so I misunderstood and thought that anode of D1 and anode of D2 should be connected. Of course this configuration doesn't function as keeping V < 0.3. I want to thank you by voting up your answer but I can't cause of my low reputation. \$\endgroup\$
    – Pana
    Sep 15, 2015 at 4:59

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