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I've seen several circuit diagrams having a 0.1 microfarad capacitor connected to the supply voltage.

How do capacitors work to prevent fluctuations in DC source voltage? Why don't they completely block DC and act as a sort of break in the circuit

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    \$\begingroup\$ We need to see the circuit diagram that's causing you problems. A series capacitor will block DC. A parallel capacitor will prevent, or at least, serve to reduce, fluctuations. Show us which you think you have. \$\endgroup\$ – Neil_UK Jul 28 '16 at 17:06
  • \$\begingroup\$ A parallel capacitor. That's what a decoupling capacitor is right? \$\endgroup\$ – LeroyJD Jul 28 '16 at 17:11
  • \$\begingroup\$ This answer of mine can be relevant. \$\endgroup\$ – Lorenzo Donati Jul 28 '16 at 17:59
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The power distribution network (PDN) in a PCB is not ideal. It has non zero resistance and inductance.

Let's say an IC needs a sudden increase on supply current. The resistance and inductance of the PDN could cause that such a sudden increase in current consumption generates a temporary voltage drop (a local drop, close to the IC).

A local capacitor, very close to the IC, has the capability to provide the extra needed current until the PDN can keep up to the pace of change and provide that current by itself, recharging in the process the capacitor.

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  • \$\begingroup\$ Why does the closeness to the IC matter? \$\endgroup\$ – LeroyJD Jul 28 '16 at 17:17
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    \$\begingroup\$ Because the resistance and the inductance of a PDN are proportional to their length. The more length you have, the more (undesired) resistance and inductance you get. \$\endgroup\$ – Claudio Avi Chami Jul 28 '16 at 17:19
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Capacitors "block DC" in the sense that in a steady state DC system the current through a capacitor is zero.

But we aren't typically dealing with a steady state DC system. We are dealing with components that are performing some time-varying task. The power supply demands of those components will in turn be time-varying.

If we had an ideal DC voltage source connected to our components by ideal wires then this variation in current wouldn't matter but we don't have ideal voltage sources and we don't have ideal wires. We have a real power supply with nonzero impedance and we have real wires and PCB traces with resistance and more importantly inductance. The inductance is particulally important it means that the effective impedance of the power network rises with frequency.

So variations in current demand, especially at high frequency, lead to variations in voltage. If those variations in voltage become too large they lead to misbehaviour. For example crosstalk and feedback in analog circuits or glitches in digital circuits.

To reduce the variations in voltage we need to reduce the impedance of the supply to the chip. We can do this by putting a capacitor across it. Depending on the frequency range of the problem different types and sizes of capacitors may be used, but as a rule of thumb for moderate frequencies sticking a 100nF ceramic capacitor across each pair of power/ground pins as close to the chip as practical is normally sufficient.

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The 0.1uF capacitor, is there to avoid high frequency noise. Generally is very useful on digital circuits to prevent microprocessor reset, or false High states on logic inputs. The capacitor will receive energy and will retain it for some time, like a water tank with a little hole is it bottom. If the water supply is constant, the tank will never be empty, but, if you close a little the water supply for a moment, the tank will start to discharge its contents, making the flow change less notorious.

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  • \$\begingroup\$ suppose a source has a potential of 5V and a capacitor is charged to this potential. If for some reason the potential in the circuit rises to 5.5V, then the capacitor would absorb some current to maintain potential yes? But what is the capacitor is full charged? \$\endgroup\$ – LeroyJD Jul 28 '16 at 17:19
  • \$\begingroup\$ In this case, the capacitor could break it self, if it cannot support that potential. Otherwise, if for example, the capacitor is 25V rated, it will charge to the new voltage of 5.5V if it is connected in parallel \$\endgroup\$ – GTRONICK Jul 28 '16 at 17:20
  • \$\begingroup\$ What is the reason for such fluctuations in the source voltage? \$\endgroup\$ – LeroyJD Jul 28 '16 at 17:32
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    \$\begingroup\$ @GTRONICK: You are confusing several things here. Capacitors don't have an RC constant. They have C and maybe a little ESR (equivelant series resistance). You've mixed in input debouncing with a question on power-supply decoupling. The leaking tank analogy is very poor also as self-discharge is not usually a problem. See Claudio's answer. \$\endgroup\$ – Transistor Jul 28 '16 at 18:07
  • \$\begingroup\$ Capacitors do however have a self-resonant frequency. The frequency at which the capacitor's inductance becomes a more significant component of it's behaviour than it's capacitance. \$\endgroup\$ – Peter Green Jul 24 '18 at 17:27

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