I generally work in a purely digital world, however I'm working on a design at the moment that requires a separate power supply for analogue and digital components. What's the best practice from providing separate analogue and digital VDD from a single power source, such as a battery?

I assume that I'd just take the raw battery, have separate conditioning/power management for the analogue and digital supplies and then adequately decouple?

I'm asking this question as I've perused the internet but can't seem to find any good resources on this subject. I know that resources must exist but my Google skills obviously aren't strong enough to find them - any pointers to some relevant literature will be greatly appreciated.

EDIT: I should add that the analogue end of this circuit is for the precise measurement of a very low power signal.

  • \$\begingroup\$ Are you pretty much just asking for 2 separate power rails from a single supply? \$\endgroup\$
    – MCG
    Apr 26, 2019 at 9:33
  • \$\begingroup\$ I'm looking for the best way to prevent a potentially noisy digital power rail affecting a separate analogue component power rail and how to absolutely ensure that a spike in power draw on DVdd has minimal affect on AVdd - AVdd is solely being used as a supply to an ADC's Vref through a voltage reference. \$\endgroup\$
    – amitchone
    Apr 26, 2019 at 9:37
  • 2
    \$\begingroup\$ I would use a separate reference device, but the implementation at layout is critical. See electronics.stackexchange.com/questions/185306/… as an example \$\endgroup\$ Apr 26, 2019 at 9:48
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    \$\begingroup\$ Don't forget to partition your analog and digital components so digital return currents don't flow under the analog components in the ground plane. Remember currents spread out when flowing in a plane. Remember to minimize current loops and return paths, especially digital ones. You can filter the Vdd and use that to supply all your analog but that is not ideal. Also google "switching reference planes" if you have more than one plane layer (i.e. power+gnd plane counts as two planes, two gnd planes count as two planes) \$\endgroup\$
    – DKNguyen
    Apr 26, 2019 at 14:49
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    \$\begingroup\$ Regarding current spreading out, look up stripline and assymetric stripline current distributions. This reduces the distance a return current spreads out in the plane under the signal trace so you don't have to space things as far. In addition, placing opposing gnd/pwr vias next to each other reduces via inductance due to field cancellation, even on little chip caps if you put them next to pads without a trace on the SIDE of the chip rather than the component ends so they are closer together. More vias around the pad helps too. Everything I said is in Henry Ott's book BTW in more detail. \$\endgroup\$
    – DKNguyen
    Apr 26, 2019 at 15:09

1 Answer 1


Since linear regulators have ZERO ability to reject high-frequency spikes and tone-bursts, you must provide the high-frequency attenuation. And, to avoid re-contamination by the ground currents, you must use Ground Plane. [And keep magnetic-fields as some distance, discussed at end of answer.] Like this, as shown below.

The 100 ohm and 100uF are 0.01 second time constant, or 16Hz F3dB. The 1uH (inductor, not Ferrite Bead) and 100uF produce a 16KHz F3dB, except the resistor dampens any resonant benefit. Thus the inductor has a benefit only above 10 or 20MHz, to further reduce high frequency spikes and ringing.

There are TWO such networks. Notice the emphasis on NOT SHARING VIAS, on keeping the VIAS well separated; centimeters apart.

You want to create a CLEAN VREF. This "cleanliness" is expensive to produce (components and PCB Area), and needing shields to sustain the cleanliness.


simulate this circuit – Schematic created using CircuitLab

Copper foil of PCBs will not provide magnetic shielding, below 4MHz. And above 40 (Forty) MHz, you'll get about 25dB attenuation; above 400 MHz, expect 80dB.

How bad can magnetic fields degrade your "precise measurement of a very low power signal"? here is math:

Vinduce = 2e-7Henry/meter * Area/Distance * dI/dT

Assume a trash Transmitter (TX) long straight wire 1 centimeter away from this LowPassFilter's output. Assume current in that wire is one amp, turning off/on in 0.1 uS, or 10^+7 amps per second. And assume your LPF's clean output is 2mm (1/12 inch) above the Ground plane, parallel to the TX wire for 0.1 meters. Lets insert the numbers

Vinduce = 2e-7 * (2mm * 0.1 meter)/1cm * 10^+7 amp/sec

Vinduce = 2e-7 * 0.1meter * 2mm/10mm * 1e+7

Vinduce =2*2/10 * 1e-7 * 1e+7 * 0.1 = 0.04 volts

Your "clean" voltage reference is not clean.

You must keep black-brick battery chargers, and other power supplies, far away.

If you want to shield, use thin steel sheets.


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