0
\$\begingroup\$

This is a full wave bridge rectifier

enter image description here

I have read that if you put a capacitor parallel to the load resistor instead of DC pulsed signal we get a more smooth DC signal.

enter image description here

However I am unable to understand the working principle and I have read many websites and couldn't figure it out.

Any help? I know that this has to do with discharging and charging of the capacitor.

Any help?

\$\endgroup\$
1
\$\begingroup\$

The input AC waveform charges the capacitor when it has higher voltage than what the capacitor has so the capacitor voltage goes up from the charging current.

When AC waveform has lower amplitude than what the capacitor has, the capacitor voltage drops because the current to load discharges the capacitor.

\$\endgroup\$
3
  • \$\begingroup\$ But what doesn't let the current of the capacitor flow in the circuit not just the full wave bridge rectifier? \$\endgroup\$ – Maddy Wells Jul 5 '20 at 12:10
  • \$\begingroup\$ I'm sorry but I really don't follow you. The capacitor will be in parallel with the load. The diode bridge will only allow current to flow from the AC side to the capacitor and load when the AC side voltage is higher, and the diode bridge will block current from flowing from capacitor to AC side when AC voltage is less than capacitor voltage. \$\endgroup\$ – Justme Jul 5 '20 at 12:18
  • \$\begingroup\$ Ah OK nvm my bad. \$\endgroup\$ – Maddy Wells Jul 5 '20 at 12:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.