Plane Splitting - Single winding AC to dual rail on the same board as preamps

Question

So I have asked a similar question and was wondering if I could trouble you guys for a little more advice. I believe I have taken all of the advice from before into account. Ill get right into it.

Above you can see my schematic, where the DC barrel jack sees 18VAC. This is sent through a doubler which is then smoothed and rectified twice. The reason for rectifying twice is to get improved ripple rejection.

There are many other parts to this board that are highly sensitive to any sort of noise, these parts will lie to the right of the power section as per figure 2. I have looked into splitting the ground plane and have come up with the following layout.

So the area on the far left is where the AC supply (18VAC) goes directly into the caps and diodes for rectification. After this, the remainder of the circuit is spead around before the rails emerge at the opposite end.

The ground plane for all of the power supply related compoennts only meets with the signal ground plane near the power inlet (top left of the board, fig 2(b)).

Here is a 3d render of what the components are.

So my question is this, will this actually achieve what I want it to achieve. Will there be no noise from the power section in the sensitive analog components that will be placed all to the right of the power section as per figure 2.

Is there some obvious principal that I have overlooked?

If so, is there any other way you would recommend doing this type of thing? Apologies if this question has offended anyone, I have experience with audio circuit design but have never put an AC power supply on the same PCB as preamps before.

Answer 1

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.

Answer 2

The path highlighted in red on your schematic carries pulsed recritfier currents:

Do not connect your circuit ground along this path! It will be noisy. The correct place to connect supply ground to your analog circuit's main ground plane is the blue circle on the right.

It may be beneficial to add a small capacitor (like 100pF) between the barrel connector's ground and the circuit ground near the barrel connector for EMI control. But if you connect grounds as described in the PCB image in your question, then there will be a ripple voltage on your ground caused by pulsed rectifier currents.

Also the capacitors on your voltage doubler are close to the potentiometer on the left, and the cap's metal cans carry AC voltage, so there may be capacitive coupling between them and the high-impedance pot wiper.

The RCA connector close to the power jack should be the output connector, which is usually low impedance and high level, so less vulnerable to hum coupling.

Vulnerable bits like inputs, especially microphone inputs, and pots should be further away.

Area of loops carrying pulsed currents should also be minimized.