Surge Current on full wave rectifier

[saviourm]

I read from the Malvino that when a fullwave bridge rectifier (with filter capacitor and load resistor) "before the power is turned On the filter capacitor is uncharged. At an instant the power is applied, the uncharged capacitor looks like a short. Therefore, the initial charging current may be quite large. in worse case, the circuit may be energized at an instant when line voltage is at max. this means V (peak) is across the secondary winding with the filter capacitor uncharged".
My question why the filter capacitor is uncharged even there is Vpeak across it?

Thanks
SM

 

[KerimF]

A capacitor is said uncharged when its voltage on its terminals is zero.
When the transformer primary is connected to mains, the equivalent impedance (internal resistance and leakage inductance of the transformer that are seen from the secondary terminals) limits the current that charges the capacitor. The initial charging current is relatively high, and as you said, it also depends on the initial instantaneous voltage applied on the transformer at t=0.
Also the dynamic resistance of the diode bridge being in series with the secondary coil contributes in limiting the charging current. But, practically speaking, the diode resistance is usually negligible in comparison to the internal impedance of the transformer.

 

[saviourm]

A capacitor is said uncharged when its voltage on its terminals is zero.
When the transformer primary is connected to mains, the equivalent impedance (internal resistance and leakage inductance of the transformer that are seen from the secondary terminals) limits the current that charges the capacitor. The initial charging current is relatively high, and as you said, it also depends on the initial instantaneous voltage applied on the transformer at t=0.
Also the dynamic resistance of the diode bridge being in series with the secondary coil contributes in limiting the charging current. But, practically speaking, the diode resistance is usually negligible in comparison to the internal impedance of the transformer.

First on switch 'On'' Let say during the first positive half cycle across the secondary of the transformer and the bridge ,the capacitor I think its charging but the text say that in the worst case the capacitor is uncharged, why? even there is V peak across it.
Thank for your attention
sm

 

[albbg]

I think the text said the capacitor is uncharged as worst case because you can switch the power supply OFF and ON againg after a delay no long enough to discharge the capacitor (it depends from time and load). In this case during the first cycle the capacitor will not be fully uncharged.
If you wait enough, before to turn the supply ON after having switched it OFF, the capacitor will be uncharged.
Considering then the capacitor fully uncharged the very worse case happens if you switch the power supply on when the main supply has reached its maximum voltage (its probabilistic matter), that means the output of the transformer is at its maximum output (neglecting inductive effects of the windings). All this voltage will suddenly appear across the capacitor through a quite small resistance (due to windings and diodes).
However this problem is present in both single and full-wave rectifiers. In full wave rectifier the probability to swicth the power supply ON when the main voltage is maximum is doubled because it rectify both negative and positive peaks.

 

[FvM]

the text say that in the worst case the capacitor is uncharged, why? even there is V peak across it.

It's just a problem of reading. The text doesn't say that. The capacitor is first "uncharged" (zero voltage across it) and then gets charged in very short time.