Originally Posted by sniper007

What does this mean? Sorry, but my English is not very well...

hi,
It means the voltage on the capacitor will only fall by a small amount.

A simple numerical example would be:
Assume the peak of the positive half wave cycle input charges the capacitor to 20Vdc

As the cycle falls to zero, the capacitor has to supply the current to the load
and the voltage on the capacitor decays exponentially.

The capacitor has to continue supplying current to the load until the next positive half cycle input voltage
exceeds the voltage on the capacitor [allow for diode Vdrop].

When the positive half cycle exceeds the voltage on the capacitor the diode will conduct
and the capacitor will be recharged to say 20Vdc

The diode has to conduct a high current in a short time in order to recharge the capacitor. Look at my diagram

Another rough example, suppose the capacitor supplies a current of 1amp to the load for about 20mSec [50Hz mains]

If the voltage on the capacitor has only fallen by say 5V during the 20mSec discharge time, means that the diode has to top up [replace the capacitor charge] in about 2 mSec, so it has to conduct a 10amp current for 2mSec.

These are rough numbers for explanation only, the actual values will depend upon your circuit.

Do you follow OK.?

OK, now understand

Thanks you very much

Another question

Excuse me, if it nonsense but nevertheless...


Is there when signal cross 2 diodes (bridge rectifier) for first half period present any crossover distortion like this:

hi,
As diodes have a nonlinear forward voltage/current characteristic distortion will occur.

You may know that a 1N4001 starts forward conduction at about 0.65V.

The rise in current after conduction starts as the applied voltage increases is not linear.

Is this what you mean.?

Yes, tnx