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schematic

simulate this circuit – Schematic created using CircuitLab

I've been using decoupling caps for a while now and I understand their purpose in keeping DC signals clean since capacitors allow high frequency signals through that get passed into ground and don't show up on the capacitor as built up charge in DC form, so they don't disturb the signal on the load.

What I've been wondering for a while is, if a random noise spike is introduced into the system from the power source or from an external source through capacitive coupling, what happens to the charge already stored on the capacitor? If the capacitor is fully charged to 1v as shown in the schematic, and it then instantaneously acts as a short to bypass the high frequency noise into ground, would a short show up on the load? Would the load have 0 volts applied to it in that instant? If this is the case, would the load not be disturbed in operation for that nanosecond in which the capacitor shorts? I'm just having a hard time understanding exactly how the capacitor can bypass charge and act as a short while still maintaining its charge built up from the DC source.

It's been explained to me in school that caps block DC but pass AC, but I feel like that on its own doesn't explain decoupling well enough. Thanks for your time.

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    \$\begingroup\$ Separating DC & AC components of the 1V source is a key concept to understanding the capacitor's bypass function. This concept applies ubiquitously to other circuits - do spend some time getting it straight. \$\endgroup\$ – glen_geek Jan 20 at 15:23
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    \$\begingroup\$ Are you commending OP for asking questions, or insulting them for not already knowing? Your comment can be taken either way. \$\endgroup\$ – WGroleau Jan 20 at 16:33
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    \$\begingroup\$ @peufeu's answer highlights an important fact: The symbols in our circuit diagrams represent ideal components. But sometimes, in order to understand what is really happening, we need to remember that every physical wire actually is both a low-value resistor and a low-value inductor, and every space between two physical wires actually is a low-value capacitor. When you add in a few of the more significant of these so-called "parasitic" components to your diagram, then the simple "bypass capacitor" circuit looks like a low-pass filter. \$\endgroup\$ – Solomon Slow Jan 20 at 23:14
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The capacitor does not "short out", it has charged up to a constant voltage by storing energy as electrical charge, and if something external tries to change the voltage over the capacitor, it means that more or less charge is needed to change the capacitor voltage up or down, and moving charges means is current flowing.

So in short, a capacitor wants to keep the voltage over it constant, and resists any voltage changes by combating it with current. So voltage spikes get attenuated because the capacitor uses energy of the spike to change charge, and the larger the capacitance is, the less the spike can change the capacitor voltage.

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    \$\begingroup\$ thanks, this makes sense. So, because the noise spike is very short lived and doesn't have the ability to supply sustained power into the circuit, the voltage on the capacitor changes very insignificantly as to not disturb the load, right? \$\endgroup\$ – OhmerSimpson Jan 20 at 8:32
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    \$\begingroup\$ @OhmerSimpson Yes, which is precisely the reason the capacitor is there. It makes it take more work to change the voltage so a noise spike with a fixed amount of energy can't change the voltage as much or as quickly. \$\endgroup\$ – David Schwartz Jan 21 at 3:55
  • \$\begingroup\$ @OhmerSimpson and while we are at it, chokes (inductors) work almost the same way in terms of filtering current spikes. They want to keep the current passing through it constant, and they resist any current changes by combating it with voltage. \$\endgroup\$ – quetzalcoatl Jan 21 at 15:24
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    \$\begingroup\$ @OhmerSimpson Like hitting a train with a sledgehammer to see if you can change its speed. Even a pretty big spike can be dissipated with little effect on the train's velocity. \$\endgroup\$ – J... Jan 21 at 19:52
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The fundamental nature of a capacitor is that stored charge = capacitance x voltage: - $$Q = CV$$

We also know that current is the rate of change of charge with time hence, if the formula is differentiated with respect to time we get this: -

$$I = C\cdot\dfrac{dV}{dt}$$

The impact of this formula is that if a current is injected, the capacitor's terminal voltage will ramp up or down depending on whether the current is positive or negative.

If that current were applied to a resistor, there would be a step change in voltage but, for a capacitor, there is a ramping up or down and not a sudden loss of terminal voltage.

Looking at it from a different angle, if a voltage ramp is applied to the capacitor, there will be a constant current flow into and from the capacitor.

Neither of these scenarios imply that the capacitor turns into a short circuit other than to try and counteract changes to its terminal voltage; there can be no sudden fall to zero volts because that would imply an infinite current surge into the capacitor.

Imagine the capacitor like a flywheel rotating at constant speed; any action that attempts to speed-up or slow-down the flywheel requires great torque and that results in only a ramping up or down of speed and not a sudden change of speed to zero.

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    \$\begingroup\$ Thanks, this makes more sense. I was thinking of an extreme in which the capacitor is suddenly a short circuit. So what this entails is that the capacitor attempts to discharge/charge based on the noise, but since the signal is such a fast transient the charge on the cap doesn't change enough to disturb the load, right? \$\endgroup\$ – OhmerSimpson Jan 20 at 8:31
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    \$\begingroup\$ Load disturbances are inevitable if the surge or transient is sufficient but for moderate stuff and a load that is capable handling minor disturbances then there should be no problem. I think you should move away from thinking a capacitor is suddenly a short circuit unless you think of it as a short circuit to transient effects only. The underlying DC conditions on the capacitor are not suddenly changed (unlike a resistor where an injection of current results in an instant change in resistor voltage). \$\endgroup\$ – Andy aka Jan 20 at 8:44
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The capacitor does not become an actual short circuit.

When we say that a capacitor is a short circuit at high frequencies, we are talking about the impedance of a capacitor when a sine wave voltage is applied to it. The impedance is the voltage divided by the current (similar to resistance).

$$ Z_C = \frac{V}I = \frac{1}{2\pi jfC} $$

It's an imaginary number, which means the current and voltage are out of phase. Apart from that, impedance is very similar to resistance.

Let's say you have a 1uF capacitor and a 1V (rms), 10kHz sine wave. The math says the impedance is 15.9 imaginary ohms. Therefore the current is \$\frac{1V}{15.9\Omega} = 0.063A\$. Now calculate the current at 100kHz. The current is \$0.63A\$. Now try 10MHz. The current is \$63A\$. Lots of current!

A real short circuit would allow infinity amps to flow, of course. As we increase the frequency, the impedance gets closer to 0 and the current gets closer to infinity. That's what "capacitors are like short circuits at high frequencies" means. A high frequency voltage signal will cause a large current to flow. (Or, equivalently, a high frequency current signal will only create a small voltage)

So what happens if you have a signal consisting of several frequencies? Capacitors are called linear elements, which means that we can analyze each frequency independently. If you have a DC voltage (0 Hz), and a high frequency noise voltage, then the DC signal will not create any current through the capacitor (the impedance is infinite), but the high frequency noise voltage will create a large high frequency noise current through the capacitor which acts to cancel out the noise voltage.

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if a random noise spike is introduced into the system from the power source or from an external source through capacitive coupling, what happens to the charge already stored on the capacitor?

It becomes easier to understand if you model the noisy power supply as a DC + AC noise voltage source having an output impedance "Z":

enter image description here

This makes a simple voltage divider between the power supply impedance Z and the capacitor impedance \$ Z_c = \frac{1}{2 j \pi f C } \$.

If C is a real capacitor, it will have a bit of series resistance (ESR) and series inductance (ESL) too.

If we model the noise source with a bit of series impedance, say some wire inductance and resistance, and the cap as having ESR and ESL, this is again a simple voltage divider and you can calculate it using the usual voltage divider equations (using complex impedances). Top graph is Vout/Vin ratio, and bottom graph is impedance of capacitor (red) and noise source (green):

enter image description here

An ideal capacitor has infinite impedance at DC and zero impedance at infinitely high frequency so you can say it "blocks DC and shorts AC". But in real life its impedance will never be zero at "AC" since frequency is not infinite. It will be \$ Z_c = \frac{1}{2 j \pi f C } \$ plus its ESR, plus \$ 2 j \pi L f \$ impedance for its inductance, which causes total impedance to rise at high frequency.

what happens to the charge already stored on the capacitor?

You can calculate voltage ripple on the cap using the voltage divider formula, and knowing q=Cv you know what happens to the charge. It moves around according to current flowing in the cap.

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The capacitor acts as short for the noise not for the signal, thus keeping the total signal value at a constant voltage which is the intent of the power supply.
If there is a noise or interference signal signal coupled to supply at frequency \$\omega\$, then total supply voltage becomes: $$v_{in}(t) = V_O + A_nsin(\omega t)$$ Here, \$A_n\$ is the interference voltage amplitude.
After passing through the capacitor (assuming sufficiently large capacitor) becomes: $$v_{out}(t) \approx V_O + \frac{A_{n}}{\omega CR_{in}}sin(\omega t) \approx V_O + 0$$ for large decoupling capacitor, where the 0 is the short circuit at high frequencies.
Hope it is clear.

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Capacitors and inductors used for decoupling are called filters for a good reason.

Compare a coffee filter: Unless you apply way too much pressure (that would be the case with a capacitor if you grossly exceed its voltage rating!) or let the filter rot and break (some capacitors age!), it shorts water and dissolved coffee (AC, Actual Coffee) into the coffee pot without that having any bearing on how much coffee grounds (DC) are passed through.

BTW, decoupling capacitors do not only short out high frequency noise - they SUPPLY it when the dynamic current demand of the load asks for it. A sudden current demand is taken from the decoupling capacitor instead of the power supply (since the longer wires to the power supply act as an inductor), making the RF current flow rather local (this is key to it not messing with other circuitry, not messing with the load itself(!!) and not being radiated as noise!) to the circuit formed by decoupling capacitor and load. "Messing with the load itself" happens when the inductance of the power wiring causes a voltage drop upon sudden current demand.

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