Does decoupling capacitors try to charge the fluctuating source back?

Question

In common examples of decoupling capacitors, the scenario is like this: When the voltage fluctuates below or above its ideal value, capacitor either stores or releases energy to compensate the fluctuation in power so that the load gets a nice, flat voltage.

^{simulate this circuit – Schematic created using CircuitLab}

Now, when I read the circuit form left to right (first looking at the source, then the cap parallel to it, then the load parallel to it) this explanation sounds right because it seems like power is delivered from left to right and capacitor behaves like a buffer in between.

But when I think about it, if cap is at 1V and power source suddenly drops to 0.95 volts, there is a short circuit between 1V and 0.95V so a relatively high current must flow from the cap to the voltage source. This not only tries to charge the source back but also wastes most of the power on that little short between 1V and 0.95V.

So, how should I think so that this circuit will make sense to me?

Answer 1

Generally speaking, it isn't the source that fluctuates, it's the load. One common example is digital circuits that draw short-term spikes of current when their outputs switch. And the best way to think about it is that there is some significant impedance between the source and the decoupling cap, especially at higher frequencies.

^{simulate this circuit – Schematic created using CircuitLab}

So while the DC signal flows from the source to the load, the high-frequency AC noise imposed by the load flows mostly through the capacitor, both charging and discharging it. It "decouples" the noise from flowing through the power supply impedance, shorting it to ground instead. The goal is to keep the green noise loop as short as possible.

The peak noise current is never greater than the DC current, so there is never a net flow "backwards" through the power supply.