How many Farad is enough in the bridge rectifier ?

How many Farad is enough in the bridge rectifier ?

I see some people use over 10000uF in tube pre-amp  ](*,)  [-X
(PS : that tube pre-amp only charge 20W RMS power from AC mains)

So, how many Farad (uF) is enough ?

How to calculate?

KK.HO

See below :

o Minimum input voltage on top of the transformer: this is the peak    rectified minimum AC input voltage minus the peak-peak ripple voltage

See the simple equation

..... just a simple equation made up by a combination of alphabets, numericals and symbols that should have been used on earth by someone.....

KK,

I assume:
f line means freq of the AC mains, which is 50Hz in HK;
P o-max is the max power required in the pre;
V DC-min is the VDC that the loading needs;

but what is V AC-min??

f line means freq of the AC mains, which is 50Hz in HK;
Correct

P o-max is the max power required in the pre;
Correct

V DC-min is the VDC that the loading needs;
Min operating voltage of the device.

but what is V AC-min??[/quote]
Your transformer should have diff on no-load and full load, under the full load, the AC output from the transformer will drop.
Or you can use 5-10% on it.

Basically, I am using high power DC supply as source to the device, and monitoring the current using Current probe with OSC. (not digit reading type, it cannot show the transient response!!!)

Therefore, I can easily to find the "min" capacitor value that for the device.

KK.

Here is another formula:

Smoothing capacitor for 10% ripple, C =  5 × Io/ (Vs × f)

C  = smoothing capacitance in farads (F)
Io  = output current from the supply in amps (A)
Vs = supply voltage in volts (V), this is the peak value of the unsmoothed DC
f    = frequency of the AC supply in hertz (Hz), 50Hz in the UK

For typical preamp +/-15V application, regulated from A/C 21V, 0.5A output.
C=5x1/(21x50) = 2380uF

For power amp +/- 32V application, supplies from A/C 32V, 2A output
C= 5x2/(45x50) = 6250uF

For both application, 10uF is overkilled already :