Plane Splitting - Single winding AC to dual rail on the same board as preamps
lets compute a magnetic-field interference, across 5 centimeters separation between the power region and your "highly sensitive" region.

We will compute an "induced voltage", using some reasonable assumptions. If the computed voltage is much smaller than your measurement budget, then you should be OK. Otherwise you get to put on your engineering hat.

We need the vulnerable-region's "area", and the "distance" from the aggressor to the victim (also known as the Transmitter and the Receiver), and we need the SlewRate of the current.

We'll assume the "area" is 1cm by 5cm. We'll assume the "distance" is 5cm.

Now for the slewrate of the current. (by the way, I don't have any idea how big or small the induced voltage will be).

Assume 1.0 amp PEAK current into the voltage-doubler, as the large caps are recharged.

How fast will the diodes turn on? Assume 25 volts peak from the transformer. The zero-crossing (not what we want) slewrate is 25 * 2*pi*60Hz == 25 * 377 == ~~ 10,000 volts per second. Now assume the dV/dT near the peak, where your diodes turn on, is 10X slower, at only 1,000 volts/second. The diode turnon occurs over about 100 milliVolts; we'll assume 0.026 volts ( K * T / Q) which gives e^1 or 2.718 factor increase.

How fast will the diodes turn on? 0.026 / 1,000 == 26 microSeconds. Thus in 26 uS we are estimated the current rises from ZERO to 1 amp. Our current has slewrate of 1amp/26uS, or about 40,000 amps/second.

We have the numbers for the assumptions. Are we having fun? How big will the induced voltage be, in your "sensitive regions"? We are about to compute that.

Use Vinduce = [ MU0 * MUr * ReceiverLoopArea / ( 2 * pi * Distance) ] * dI/dT

where we ignore a natural-log factor, for easy observation of causality.

With MU0 being 4 * pi * 1e-7, and MUr being 1 (for air, copper, FR-4), we have

Vinduce = [ 2e-7 * Area/Distance ] * dI/dT

Is this exciting? What if the result is 1 millivolt? 100 milliVolts? or 13 microVolts? or 45 nanoVolts? at what level of interference will you take action? by introducing some shields between Power region and Sensitive region? and these must be STEEL shields, because at these slow edges (the 60Hz) the Skin Effect in PCB copper foil will not attenuate the magnetic fields. OK enough asides about shielding. Lets run the math.

V = [ 2e-7 * Area/Distance ] * dI/dT

Vinduce = [ 2e-7 * (1cm * 5cm)/5cm ] * 40,000 amps/second

Vinduce === getting nervous yet? == 2e-7 * 1cm * 40,000

Vinduce = 2e(-7-3) * 4e+4 = 8e(-7-3+4) = 8e-6 = 8 microVolts.

...................... 8 microVolts .......................

Do you have any amplifiers, with gain of 100X? this will produce 8*100 = 800uV or 0.8 millivolt of evil-sounding 60Hz "singing". Not a hum, but an irksome narrow-edged singing.