Why might one of these decoupling capacitor schematics also include an inductor and the other not?

Question

I am looking at a third-party schematic (centered around a small IC) and it includes a selection of power schematics. They connect between a general 3.3 V supply and various other users of 3.3 V.

In the image, you can see the schematic on the left takes in 3.3VD and then feeds it directly back out to VCC3IO, with decoupling capacitors connecting to ground.

The schematic on the right looks similar, but it includes an inductor in line between 3.3VD and VCC33A. What feature might the inductor offer here?

On a similar note, they actually have four different sub-schematics on this one board, each connecting 3.3VD to other places with a set of decoupling capacitors. Am I correct in assuming that this is to meet the needs of each power rail individually, rather than trying to calculate/guess how to group them into one monolithic set of capacitors?

Answer 1

Some parts of the circuit are usually more sensitive to power supply noise than others (especially high frequency noise), and some parts of the circuit may also inject different amounts and frequencies of noise into the power supply.

For example, on a microcontroller, the core and I/O will generally make a lot of noise by drawing pulsed current with all the processing and switching of I/O lines going on, while the ADC makes little noise but it is sensitive to power supply noise. So they have different power pins. Likewise for a sensor, these usually like a clean power supply (you can check PSRR vs frequency in the datasheet).

The usual way to solve this without resorting to one LDO per circuit block is to partition power into several local "islands", each with its own decoupling capacitors, and isolate them from each other at high frequency with ferrite beads. This puts a CLC filter in the way of noise from the aggressor to the victim.

So from the point of view of the sensitive circuit, it will filter the power supply, but it will also work the other way around: it will prevent the noisy circuit from polluting the whole board's power supply. It works both ways because the ferrite bead increases impedance at high frequency, which forms a current divider with the capacitors. HF current follows the path of least impedance, which is the closest capacitors to the load drawing that current, and hopefully this current will ignore the path of highest impedance, which is the ferrite bead. This prevents the noisy circuit from drawing power from the decoupling capacitors of the sensitive circuit, which would increase noise where you don't want it.

Another important feature is that when the noisy chip is forced to draw current only from its local decoupling capacitors, and not from other capacitors a bit further away, then return currents through your ground plane stay in a tight loop around the chip and its decoupling capacitor. This means the ferrite bead in the power supply rail of the noisy chip also makes the ground plane cleaner. That one is not obvious, but it is very useful.

Note that the 0 Ω resistor may mean they thought they might need a ferrite bead or a resistor to make a filter, but it didn't end up being necessary, so instead of redoing the board they substituted it with the cheaper 0 Ω resistor.

Ferrite beads, being inductors, will create resonances at low-ish frequencies with the decoupling caps. These can be damped with a resistor in series, or with a large value capacitance with some ESR, like the 33 µF capacitor.

Answer 2

The inductor with capacitors filters noise and ripple so that the power to thermal sensor is clean so it can measure temperature accurately.

The other set of capacitors is for generic digital IO which does not need an inductor for filtering, it can work with more noisy and ripply supply voltage. But without the inductor the digital IO switching can cause ripple and noise back to the supply voltage, which might be the reason the temperature sensor supply is filtered to begin with.

Although it reads "NM" under the inductor, which might mean "Not Mounted", so the inductor may not be present and there is an alternate route to power the sensor.

The bypass capacitors are local to chips to be effective, because there needs to be local energy storage that is not far away as the more there is wiring the less effective the bypass caps are. So one set of bypass caps somewhere won't work if the chips need local bypass.

Answer 3

The presence of the inductor or ferrite bead depends how sensitive your load is to noise and how noisy your source is.

The reason you might see four "identical" sub schematics is is because of that little series resistor which is grouping each chunk off into its own section. It is a power jumper for debugging (or filtering) purposes.

Answer 4

The most important part of any filtering circuit is the actual placement of the elements on the board.

The capacitors are drawn together in order not to mess with the interesting part of the schematics.

In reality, these capacitors are placed each one closest to its corresponding noisy element (e.g. a digital integrated circuit or other switching element).

If one wants to imagine (or simulate) what happens in the filtering circuit, the important part of it are the intrinsic ("parasitic") resistances and inductances of the printed circuit board tracks. It is them who creates the voltage noise in the power rail in the first place, it is them who creates and dampens the unwanted resonances with the capacitors in question.

In short, both pictures in the Q show nothing more than the bill of materials. The real circuit that one deals with has twice the elements and their values depend on the actual PCB layout.