Can any one tell the reason for the current through the smoothing capacitor to be pulsating in the full bridge rectifier.
I have attached the screenshots of the circuit and the wave form obtained in Multisim.This is the circuit what  have used an i need to find the capacitor current

These are the wave forms i got for the capacitor voltage and current

  • \$\begingroup\$ Because the voltage is rippling, and \$I_C = C\frac{{\rm d}v}{{\rm d}t}\$. \$\endgroup\$
    – The Photon
    Sep 15, 2017 at 3:51
  • 2
    \$\begingroup\$ Diodes act like a voltage comparator switch between input and stored voltage so Ic * t charge area must equal Iavg * t discharge area and is limited by I=delta V/(cap ESR+ diode ESR + source R) so peak/avg current is inverse to % ripple \$\endgroup\$ Sep 15, 2017 at 3:54
  • \$\begingroup\$ Did you expect something different? \$\endgroup\$
    – Dave Tweed
    Sep 15, 2017 at 4:14
  • \$\begingroup\$ @Prudhvi Ask a good specific question and you'll get better answers. \$\endgroup\$
    – Voltage Spike
    Sep 15, 2017 at 4:58
  • \$\begingroup\$ @Dave Tweed, let me know if you have different explanation \$\endgroup\$
    – Prudhvi
    Sep 15, 2017 at 17:29

2 Answers 2


The diode will only conduct current to the capacitor when the cyclic voltage from the transformer exceeds the voltage remaining on the capacitor. At other times the diode will block and therefore there will be only a leakage flow of current (very small).

Given that most folk want a DC output that is fairly smooth (i.e. the capacitor is large) the voltage on the capacitor causes the diode to be reverse biased most of the time and the diode will only allow current when the cyclic voltage from the transformer is at or near the peak of it's waveform. Thus, the diode ceases to conduct once the voltage from the transformer starts to reduce.


The transformer can only supply current to the load while the transformer voltage is above the capacitor voltage. At all other times the capacitor supplies current to the load.

Since all power ultimately has to come from the transformer and the diodes only conduct part of the time then the current pulses will be many times that of the average load current.

enter image description here

Figure 1. The half-wave rectifier version of the waveforms. The green current waveform has been superimposed on the original graph.

See my answer to Interpreting the ripple curve of a half wave rectifier which discuses the same question for a half-wave rectifier.

  • \$\begingroup\$ This question comes up often ;) But you forgot transformer inductance, which changes the point diodes stop conducting a little bit... \$\endgroup\$
    – bobflux
    Sep 15, 2017 at 14:16

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