Electronic – Why have a 0.1 uF capacitor in parallel with 2500 uF in a power supply filter

To answer in your question about the “PC power supply”. Briefly the switching mode ps does not have a good transient response in load changes and also they are generates lot of EMI. So you can not use in ,lets say, inductance dominated loads or in high end audio applications. It is common practice to banking (parallelize) capacitors, since this allows to sum their capacity as well as reduce their series resistance. So we can handle bigger currents with less dissipation and heating, since ripple current per microfarad is often increased by reducing capacitor sizes. Regarding max ripple current, do not forget to include the "ripple current multiplier" given at the operating temperature, an d also keep in mind that temperature of the capacitor core is hotter 3~5 oC /watt of ripple power compared with the case temperature. Usual heat dissipation factor is 0,00093W/oC/cm^2. So in a 4700μF/35V capacitor, a 10oC temperature rise limits the current to 2A In case of capacitors banking, main task is to keep them in low temperature (i.e do not place close to high heat generated or dissipated components, or use forced cooling methods) Bonding each cap even with hot glue to a cooler base helps allot. Also protect capacitors from high frequencies (i.e using of very fast and exceptionally soft recovery power diodes, or RC snubers) since elcap does not respond well in hf and ESR decrease more than twice at resonance. The type of capacitors that you have (temperature rating as well as geometrical size) plays important role to the dissipation factor (or tanδ) stated in data sheets (there is also a temperature and frequency derating). For banking we often use large diameter and short length capacitors. My experience limits to 20x1000μF/35V bank, but your is 50x4700μF.... Assume that your capacitors are uniform, or if you mix them keep in mind that each capacitor have a different discharging time. As already stated above, rectifier bridge plus heatsink sizing should carefully design because on this VA rates, the equivalent series resistance of transformer it is very low and now comparable to the total capacitor ESR. So it is very possible to need an external power resistor.

Regarding the comparison cost of the two: do you have price of a good torroidal transformer 2500VA that you need?

RMS versus peak: -

RMS is the equivalent DC voltage that would dissipate the same heat in a resistor when the AC voltage is applied. You can do the math\$^1\$ if you want but, for a sinewave, the RMS is 70.71% of the peak OR the peak is \$\sqrt2\$ higher than the RMS.

When ever we talk about AC voltages (without specifying peak or peak-to-peak) we imply RMS hence, a 6V RMS signal will have a peak of 8.49 volts (nearly 9 volts). After passing through a rectifier diode this might be more like 8 volts DC on the smoothing capacitor.

However, transformers unloaded will run at a slightly higher voltage on the output so expect a volt more (or so) on the output.

\$^1\$ Take a 10V peak sinewave. Square the waveform - this produces a double-frequency sinewave between 0V (lowest peak) and 100V (highest peak). Take the Mean - 50V then take the Square Root - 7.071 volts i.e. 7.071 volts is the RMS value of a sinewave having a peak of 10V.

R = root, M = mean, S = square = root of the mean of the squared signal.