What happens to noise after crossing through a bypass capacitor?

Question

I understand that bypass capacitors provide a lower impedance path for oscillating noise. However, capacitors aren't able to dissipate energy (apart from their effective series resistance), so all the noise should get to ground almost without getting affected.
My doubt is what happens to this noise once it reaches ground. My first though was that, as usually the ground is contained in a large ground plane, the noise would break up in every direction, losing its energy in the near surroundings. However, by the current loop theory (https://www.ewh.ieee.org/soc/emcs/acstrial/newsletters/fall08/tips.pdf), the noise should loop back to the initial node below its trace (the path of least impedance). This way, the noise would not be able to fall apart in every direction.

Answer 1

If you have noise, e.g. AC noise in form of a current, riding on DC voltage or current, the capacitor will want to keep the DC voltage over it, and the AC noise current just charges and discharges the capacitor.

As current changes the charge of the capacitor, it means that changing charge on capacitor changes the voltage over the capacitor in proportional to the capacitance.

So the capacitor filters the noise into lower amplitude by charging or discharging, as capacitors try to maintain the voltage it has by sinking or sourcing current.

But the role of a bypass cap is not really to filter noise or shunt it to ground, but to act as a local energy reservoir to provide current quickly to a chip that needs current quickly.

If there is no local bypass cap, and a chip needs current, the inductance of a wire or PCB trace prevents it from getting it immediately, so that is why voltage drops. A bypass cap close to IC is able to give it immediately and even though the peak current may be quite large, the amount of energy is not very large and the voltage of bypass capacitor does not change very much and it is charged back to supply voltage.

So in essence, a bypass current, along with inductance from power supply to capacitor, will essentially filter the spiky current taken by an IC into DC average current, or at least with much less ripple.

Answer 2

Nothing "happens to" the noise, as such. It sounds like you're thinking of the noise as being some type of "thing" that will stick around until something makes it go away, but it's not. Rather, the noise is an event that happens, and we use a bypass capacitor to ensure that the noise has no negative consequences when it happens.

A typical example is that we have a circuit with a digital IC in it, such as a microcontroller.

Digital ICs consume current in a rapidly varying fashion, and this rapid variation of the current is referred to as "noise." It's important to keep in mind that there are always two (or more) pins carrying current in opposite directions; any time that there is 5 mA going into the power pin (or pins), there is also 5 mA coming out of the ground pin (or pins), and whenever there's 15 mA going into the power pin, there is 15 mA coming out of the ground pin.

As a rule of thumb, current almost always flows in loops. If the current through the IC is rapidly alternating between 5 mA and 15 mA, say, and you don't have a bypass capacitor, then that current will alternate in that way throughout the entire portion of the circuit that supplies power to the IC. This rapidly fluctuating current will then produce unwanted voltages due to the resistance and inductance of those traces.

On the other hand, if you do have a bypass capacitor, then the rapidly varying component of the current will go through the capacitor instead of going through all of the power supply circuitry. The resistance and inductance of the bypass capacitor are both very low, so little unwanted voltage is created.

None of the things you've suggested will happen to the noise. It won't travel to the ground plane and break up. It won't keep looping or oscillating. Nothing will "dissipate" the noise.

Now, granted, there is some "noise energy" stored in the capacitance and inductance of the bypass capacitor and the traces connecting to it. Most of this energy will be consumed by the very same IC that created it, and some of it will be dissipated in the resistance of those traces.

Answer 3

Considering the capacitor alone to understand where the noise current and voltage exist does not work.

In the simplest form the capacitor forms part of a series closed path which includes a noise source, a resistance , and an inductance.

The voltage divider rule indicates that the voltage across the capacitor at high frequencies will be small.

The noise will appear across the series inductance and resistance.

The noise is still there just distributed in a way that it has a reduced effect.

The great big ground provides a low impedance path back to the noise source ensuring that the noise voltage across the ground path is zero (almost).

Answer 4

The current loop is formed between whatever is introducing the noise (antenna action of a trace, unwanted higher harmonics from switching, pulses from loads turning on or off, etc) and dissipated as heat and to a lesser extent radiated as RF.

Answer 5

The noise always originates from a source with an internal impedance. Putting a capacitor across means that the resulting noise voltage over the capacitor will be attenuated at the frequencies where the capacitor's impedance can not be considered negligible with respect to the internal impedance of the source.

Answer 6

To answer this question we should remember that, at least for low level signal (as noise usually is), the complex circuit seen by the terminal of a bypass capacitor, say \$C_\textit{by}\$ is equivalent to a real voltage generator (by Thevenin's theorem) or to a real current generator (by Norton's theorem) having the same inner impedance, say \$Z_\textit{eq}\$.

^{simulate this circuit – Schematic created using CircuitLab}

The choice of one or the other of equivalent models is usually based on the values of \$Z_\textit{eq}\$: Thevenin's model is customarily chosen if the impedance is below, say, a hundred ohms while the Norton one is customarily chosen when the impedance is higher.

Said that, we can state that, as long as the lumped parameter hypothesis is acceptable for the circuit under analysis, the energy / power associated to the noise signals \$v_n\$ or \$i_n\$ are dissipated in the real part of the equivalent impedance, i.e. \$\Re Z_\textit{eq}\$, and this implies that the behavior of the noise signal depends strongly on where it is really found the most part of the impedance \$Z_\textit{eq}\$.

1. If the ground plane is nearly ideal, it contributes in a negligible way to the \$\Re Z_\textit{eq}\$, thus the noise signal (be it current or voltage) will flow across the ground plane carrying almost no power/energy an thus being unable to harm other circuits connected to its return path (in short, no coupling with other circuit is possible).
2. If the ground plane is damn bad, it contributes in a significant way to the \$\Re Z_\textit{eq}\$ thus the noise signal will flow across the ground plane carrying a significant power/energy an thus harming other circuits connected to its return path (in short, a really significant coupling with other circuits occurs).

Note. As the lumped parameter hypothesis is not verified, things get worser as vir has described in their answer, since radiation and also other purely propagator behaviours of the noises can occur. Nevertheless, the discriminating condition is again the near ideality or not of the ground plane: if it is almost ideal, the noise will close back to their source without harm, if it is not, the whole behaviour can turn out to be a complete mess.