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Most of the times attached picture is used to depict the use of Capacitor as filter in the rectifier circuit. I guess, i understand the broad concept but I am struggling with the fact that charging happens for a very little time (as per the picture) and discharging happens for alomost one half cycle. The time constant for charging and discharging is same. How can charging for a small amount allows it to discharge for such a long time.?

enter image description here

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  • \$\begingroup\$ I see that you don't accept any answers as solution in any of the questions posted by you. That's not a good way of using this site. Accept an answer as solution once it solves your problem. \$\endgroup\$
    – Mitu Raj
    May 26, 2021 at 4:25

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The time constant for charging and discharging is same. How can charging for a small amount allows it to discharge for such a long time.

No, they are not the same. The capacitor charges from the rectified voltage source, only through the voltage source internal resistance and the diodes. When the input voltage drops below the capacitor voltage, the diode opens and the capacitor discharges through the load resistance, which is much higher (so the current is much smaller).

As an example, check this simple half-wave rectifier:

enter image description here

R1 represents the voltage source internal resistance and R2 is the load (the circuit powered by the rectified voltage).

enter image description here

After the transient and with a fixed load, the charge lost by the capacitor while diode is opened equals the one quickly recovered when the diode conducts, hence the higher peak current.

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The charging and discharging happen through two separate paths, controlled by the rectifier. The charging path has a very low impedance compared to the discharge path (i.e., different time constants), which means that a larger current can flow during the shorter charging time.

High current × short time = low current × long time, so the long-term average charge on the capacitor remains constant.

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In this case the capacitor charges faster than it discharges because the input current to the circuit is greater than the output current.

Generally a capacitor charges at about the same rate as it discharges.

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    \$\begingroup\$ "Generally a capacitor charges at about the same rate as it discharges" is plain wrong. The rate depends on the time constant of the charge/discharge path, that, generally, is whatever it is. \$\endgroup\$
    – Vladimir Cravero
    May 22, 2021 at 6:48
  • \$\begingroup\$ Technically that's true but for most practical application, including the scenario in the OP here, the capacitor itself is NOT the reason for the difference that is being asked about. For me, answering a elementary question does not call for a treatise on capacitor theory. Please feel free to post your own answer including all the details of what's going on here if you think that might help the OP's understanding of what is going on here. \$\endgroup\$
    – jwh20
    May 23, 2021 at 10:53
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    \$\begingroup\$ I do not think that you need a "treatise on capacitor theory" to spell out that the rate of charge depends on the time constant. Moreover, you also have causality wrong in the first part of your answer. (peak) input current is higher than (peak) output current because the rate of charge/discharge is different, because the time constants are different. Average current, instead, is the same. I won't add an answer since there already is a very good one - and my downvote stands. \$\endgroup\$
    – Vladimir Cravero
    May 23, 2021 at 18:54

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