So let's recap.
You have an AC voltage. The basic idea is to convert it to a DC output using a bridge rectifier (or four independent diodes), then use a capacitor to smooth out what comes out from the bridge rectifier, and then you can use a voltage regulator to get exactly 5v or whatever you want.
Whenever you rectify the AC to DC, the energy goes through two diodes, so there's always going to be a loss equal to 2 times the voltage drop on an individual diode in the bridge rectifier.
For low currents (let's say 100-300 mA), you could use some Schottky diodes which can have a voltage drop as little as 0.1-0.3v. Bridge rectifiers made for high currents can have a drop of about 1-1.2v per diode.
Then, depending on the capacitor size, the amount of voltage you will get will vary. A simple formula that approximates how much capacitance you need is : C = Current / ( 2 x AC Frequency x Vripple) where Vripple is how much you're willing to let the voltage sag.
Then, you have all kinds of linear regulators, or switching regulators... some are better than others. Linear regulators typically need to have the input voltage higher than the output voltage by some amount, in the case of a classic chip like 7805 you're looking at about 2v.
Other classic regulators like 1117 only need about 0.8-1v to work properly, but they can only handle a lower amount of current, in the case of 1117 it's about 1A compared to 1.5-2A for the 7805.
There are also other linear regulators like (for example)
LT1085 / LT1084 / LT1083 which can handle up to 3A / 5A / 7.5A with just 1V drop but when you go above around 2-3A, it's not a good idea to use linear regulators because of the power dissipated in the regulator which would force you to use a big heatsink.
With linear regulators, everything between the input voltage and output voltage is converted to heat and must go somewhere... for example if you have 8v at input and want 5v at the output at 2A, that means the regulator will produce (8v - 5v) x 2A = 6w of heat
So let's look at an example... you have a 6.3v AC transformer and you want 5v @ 0.5A .
This means the peak DC output before the losses in the bridge rectifier will be equal to 6.3v x 1.414 = 8.9v DC.
Now, I could use a bridge rectifier but considering it's only 9v peak and 0.5A, I'm looking at some diodes like the classic
1n5817-1n5819 and I'll go with the middle one, 1n5817 which can handle up to 20v and 1A and have a typical voltage drop of about 450mV at 1A (scroll down in datasheet, value is there).
The voltage drop is important.. but note it's specified at 1A and it's "typical" so depending on your luck, the actual drop may be a bit higher, or a bit lower, at 1A. As our maximum current is only 0.5A, we're obviously not going to get close to 450mV drop @ 1A, so the voltage drop on the diode will be smaller than that figure of 450mV. But just to keep things simple, let's say the voltage drop per diode is 500mV.
With this in mind, we now know that the peak DC output after the bridge rectifier made out of 4 x 1n5817 diodes will be 6.3v x 1.414 - 2 x 0.5v = 7.9v
Now you need to pick a capacitor to smooth out this dc output... the more capacitance, the most time the output voltage will be closer to that peak of 7.9v .. BUT there's no point using a very large (and expensive) capacitor to keep the voltage high, if the linear regulator only needs a bit of voltage above the 5v it needs to output. So you have to pick your linear regulator, determine the minimum input voltage that regulator needs to output 5v properly, and then choose a smoothing capacitor accordingly.
In this case, as I have a maximum of 5v @ 0.5A, I can use a
LM1117 linear regulator, which is designed to output up to 0.8A of current with only about 1.25v drop at 0.5A (it's all in those tables in the datasheet).
So, the regulator will need 5v + 1.25v = 6.25v at minimum to output 5v properly, and we have a dc voltage with a peak of about 7.9v. To be very safe, let's choose the smoothing capacitor to make sure there's always at least 6.5v DC to the linear regulator, which means we can afford a ripple voltage of 7.9v - 6.5v = 1.4v
Now we can use that formula C = Current / ( 2 x AC Frequency x Vripple) = 0.5 A / ( 2 x 60 Hz x 1.4v) = 0.5 / 168 = 0.002976 Farads or 2976 uF. The next standard value would be 3300uF, so a 3300uF 16v rated or better would be perfect.
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For 5v and 10A, if you really want linear regulators I suppose you could use two LT1083 regulators in parallel (see page 14 in datasheet), but you'd still need to feel them with at least 6.5v (due to the dropout voltage of 1.5v) and you'd need a bridge rectifier capable of at least 10A which means the voltage drop inside the bridge rectifier will be much larger) and also you'll need much much larger smoothing capacitors.
So the best idea would be to use a transformer designed for more than 6.3v AC, for example let's say 9v AC.
In that case, you'd have a peak dc voltage of about 9v x 1.414 = 12.7v , minus about 1.2v x 2 = 2.4v in the bridge rectifier , so about 10v. Now determine capacitance size to make sure there's always 6.5v even at 10a... so C = 10A / [ 2 x 60Hz x (10v - 6.5v) ] = 10 / 120x3.5 = 0.02380 Farads or 23800 uF so something bigger than that would work, like 27000 uF or 2 x 15000 uF for example.