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I never quite understood the finer details of why a diode is required for turning capacitors into "smoothing" capacitors instead of just low pass filters. The simulations I have done in multisim just seemed to add into the confusion. I realize that capacitors just adds a phase shift to the current and does nothing to smooth out AC voltages by themselves unless we add a diode in series with the AC signal. However the question comes when we try "smoothing" DC voltage with added thermal noise.

below is a circuit with diode D1:

enter image description here

and here is the oscilloscope plot :

enter image description here

However when I remove the diode D1 :

enter image description here

the plot becomes :

enter image description here

Now the biggest issues comes when I change the thermal noise frequency to 100hz, now both design without or without the diode are able to perfectly smooth out the noise. The 100hz is chosen at random and not a fixed threshold.

The R and C forms a basic low pass filter with cutoff at around 98Khz and are modeled as ideal components. My question is why is the diode required for the capacitor to perform its "smoothing" operations and why at 100hz it doesn't matter if the diode is there anymore, also why a diode is needed in general in series with an AC signal for capacitor to smooth out AC into DC signals. Finally does that mean i should add a diode to LDO or buck regulators if I want capacitors to "smooth" out ripples instead of bypassing high-frequency components? Thanks

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    \$\begingroup\$ Tip: "Hz" or hertz. 'k' for kilo. 'K' for kelvin. See SI units. \$\endgroup\$
    – Transistor
    Commented Oct 25, 2020 at 12:45

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A 1 kΩ thermal noise with a bandwidth of 10 MHz at 27 °C produces an RMS value of 12.87 μV. At 100 Hz, the value is 0.04 μV i.e. over 300 times smaller. I can't explain your scope noise pictures but the numbers speak for themselves.

also why a diode is needed in general in series with an AC signal for capacitor to smooth out AC into DC signals

The diode is acting as a peak detector. Once the capacitor is charged up by the voltage peak, the subsequent falling voltage at the diode's anode reverse biases the diode and the capacitor remains charged to that peak.

Finally does that mean i should add a diode to LDO or buck regulators

No, you shouldn't.

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    \$\begingroup\$ You can also use HTML entities &Omega;, &mu;, &deg;, &times;, etc. as well as <sup>...</sup> and <sub>...</sub> in the posts (but they don't work in the comments). \$\endgroup\$
    – Transistor
    Commented Oct 25, 2020 at 12:48
  • \$\begingroup\$ I can never remember this!!! but i'll edit. \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2020 at 12:48
  • \$\begingroup\$ does this mean that full-bridge rectifiers with diodes are all essentially peak detectors in nature? \$\endgroup\$
    – attle
    Commented Oct 25, 2020 at 13:02
  • \$\begingroup\$ @attle they sure are. \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2020 at 13:02

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