How do I separate ground planes on a PCB with two different voltages?

Question

I am using a four-layer PCB with the following stack up: top signal, ground, 3.3V VCC, and bottom signal.

All of my components require a 3.3V input except for this sensitive optical sensor (MAX30102) that needs both 1.8V and 3.3V inputs. In the MAX30102 datasheet, the 1.8V needs to run to a separate ground from the rest of the system.

How should I go about separating the ground planes? I am using two linear voltage regulators in parallel and have attached an image of my design.

Pad 12 is the ground for the 1.8V input and pad 4 is the ground for the 3.3V input on the MAX30102. How should I connect pad 12 to ground to minimize noise? Furthermore, how do I tie both 1.8V and 3.3V grounds together?

I'm guessing that I make a separate ground plane underneath anything that connects to 1.8V and then connect that to the ground of the 1.8V linear voltage regulator.

Or maybe, I have to connect the 1.8V ground to the main ground source at only a single point with a track? Currently, the second layer ground is tied to all of the other components powered by 3.3V.

Please let me know, I'm a student.

Answer 1

Simple answer is you don't split planes unless you know what you are doing and why you are doing it.

Separate the ground planes and route signals over the the gap in a split ground planes at your own peril.

The lowest inductance path for return currents is on the ground plane directly under the signal trace. This forms the smallest possible 3D loop.

But the return current from such traces cannot cross the gap in the split plane so will flow around the gap instead making a big fat loop.

You do not want to split a plane without justification because you then can only route traces over the bridge between split planes or else you end up with the aforementioned big fat loop.

Instead, what you do is just use a single unsplit ground plane but partition components to different areas on the board so ground currents of the noisy parts do not flow through the ground plane under the sensitive parts.

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The only cases I know where you might split the plane is you need really, really low noise but can't partition sensitive section of the PCB far away enough.

Current flowing on a plane kind of spills out and smears to the sides of the linear path it is taking. Split plane can stop the spillage from leaking onto areas of the plane that are under other components because you cannot space them far enough in your partitioning with a single unsplit plane. But it's still contentious whether this is actually needed.

Images taken from: http://www.hottconsultants.com/pdf_files/june2001pcd_mixedsignal.pdf

Answer 2

Following up on the excellent answer of DKNguyen, you can best localize the Ground currents of that 1.8volt super-delicate sensor ---- by using a LOCAL BATTERY.

A local battery is a resistor PLUS large capacitor, acting as a LOW PASS FILTER for trash from the outside, and providing almost all surge currents needed by your sensor from that LARGE CAPACITOR.

You place the LARGE CAPACITOR across the sensor's GND and VDD pins. Bring in the 1.8 volts through a 10 ohm or 33 ohm or 100 ohm or 330 ohm resistor. Place that series resistor right against the LARGE CAPACITOR.

This placement of the LARGE CAPACITOR and the series resistor is a fine isolator of trash coupled onto your 1.8v regulated voltage; these LDOs are poor rejectors of high frequency trash, such as MCU spikes or switch_reg spikes; adding the LARG CAPACITOR and series resistor provides a guaranteed low-pass-filter of high-speed trash and spikes and switch-reg ripple.

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The datasheet shows up to 20mA from the 1.8v supply. So I'd make the resistor be 3.3 or 4.7 or 5.6 ohms, and use 470uF or 1,000uF for the LARGE CAPACITOR.

With 4.7 ohms and 1,000uF, the time constant is 4.7 milliSeconds, or about 90Hertz F3dB; this provides only 1 dB attenuation to 60Hz ripple, though the LDO should attenuate that frequency.

However, this R+C will greatly reduce any high frequency trash that the LDO is unable to attenuate. All LDOs have internal servo (regulation) feedback loops, and the LDO allocates a small amount of current for the transistors in that loop. Small amounts of current tell us the speed, of controlling the large internal (on_chip) linear dissipation transistor, will be slow.

Additionally, that large transistor has large junctions and large parasitic capacitances that provide a direct path from Vin_raw to Vout_clean; thus our LOCAL BATTERY is essential to correct the high_frequency weaknesses of LDOs.

Answer 3

Actually reading that datasheet seems to indicate that you are overestimating the sensitivity of the MAX30102 to noise by a lot. There are Arduino compatible eval boards plugged into breadboards that work fine; it's actually remarkably insensitive to noise.

LED currents max out at 200mA and flow between pins 10 and 4--a standard 1.0oz plane won't even notice this drop. 1.8V current is limited to 20mA. Move that SCL line so it actually goes out the right and move the SDA line away from pin 4 and things should "just work".

Common grounds should be more than sufficient for this design.

However, do pay attention to the fact that pin 10 and pin 4 can be carrying 200mA, you probably need to tie them to the GND/3.3V planes with a couple of vias, not just a single via.

Answer 4

I agree - Don't cut planes

What I'd do is connect the 1.8 regulator ground to your IC pin 12 with a nice fat trace. Then connect pin 4 and pin 12 to the ground plane with vias right at the pins. So the two grounds meet at only one place (pin 12's via). The idea is prevent currents from either voltage domain from mixing with the other. If you have only 1 place they connect, they won't be able to mix. Physics will drive the return currents back to their sources. This is a take on a 'star ground' scheme, very very common in audio to prevent unwanted noise.

https://resources.altium.com/p/how-to-use-a-star-point-for-analog-ground-digital-ground-connection#:~:text=A%20star%20ground%20is%20a,as%20to%20eliminate%20ground%20loops.&text=In%20the%20same%20way%2C%20your,only%20at%20the%20star%20point.