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.
Best Answer
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.