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I am new to electrical engineering. Please excuse me if I use the wrong terms. (Welcome to correct me tho ;-) )

I am learning about capacitors, and I have been studying a circuit diagram explaining how capacitors can "stand-in" as a power source if the power drops. This article also mentioned that it can help with spikes. Upon researching spikes, I learned that a cap can handle a percentage more voltage than what it is rated for. This is useful in a voltage spike because the cap can "absorb" some of that voltage, decreasing the spike to let's say an IC.

I understand this. BUT. Won't the 5V capacitor now sit at lets say 6 volts? What happens when the power dips and the cap is supplying current to the circuit? Won't it supply that current at 6V instead of 5V because of the previous overvoltage it accepted?

Thanks

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    \$\begingroup\$ Sure. And another part of the circuit is at 4V, and current will flow to make up the difference. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 25 '18 at 12:25
  • \$\begingroup\$ Are you referring to an actual power dip? \$\endgroup\$ – Louis van Tonder Feb 25 '18 at 12:29
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    \$\begingroup\$ Well, yes. A digital circuit is just a pile of capacitors charging and discharging in a pattern. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 25 '18 at 13:03
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Spikes are very short events (otherwise, they would not be formed like a spike), so they do not have much energy and cannot add much charge to the capacitor, so the output voltage does not actually rise by a noticeable amount.

(If you'd connect a 6 V power supply, it would shove as much current as needed into the circuit.)

From an alternate (frequency) point of view, the capacitor has a low impedance at high frequencies, so the high-frequency spike gets shorted to ground.

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The various bypass capacitors will resonate with the various inductors in the circuit. 1uF and 1cm of PCB trace over air (not over a plane), or 1uF and 10nH, will resonate at 0.159 / sqrt(1e-6 * 1e-8)

0.159 / sqrt(1e-14)

0.159 * 10^+7

1.59 MHz.

Its your job to dampen that resonate circuit, to prevent ringing.

I use the formula Rdampen = sqrt(L / C) = 0.1 ohms for 1uF and 1cm PCB trace. Given 0.1 ohms is an awkward value, you may just hope the capacitor has enough ESR to dampen. That is not likely.

VDD ringing is the enemy of high-accuracy circuitry.

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