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Capacitor as a filter for a rectifier ?

  1. Jan 4, 2008 #1
    in my textbook, they've said that a capacitor is used to filter an input from a full wave rectifier.. So that the output does not fluctuate (much) and hence convert an a/c current to d/c current. But, the textbook isn't clear about how this happens.

    It just says that the time constant for the capacitor needs to be made large, for which the load resistance is increased.

    The textbook diagrams are attached below.. pls have a look

    Attached Files:

  2. jcsd
  3. Jan 4, 2008 #2


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    It will help you understand what is going on if you draw separately diagrams of the following:
    (1) the output voltage from a full wave rectifier alone, given an AC input voltage (2) the output voltage from an RC circuit (capacitor and load resistor in parallel, i.e. a filter stage as you've indicated in your diagram), given a DC input voltage.
  4. Jan 4, 2008 #3
    If you're looking for an intuitive picture, you can use the model where the capacitor acts like a spring. When it is not charged, it is effectively a short circuit, so the current will flow there most easily. Once it gets charged, it "pushes" back, which coincides with the output voltage from the rectifier sagging (with an appropriate choice of capacitance). In other words, the capacitor stores enough charge to "fill in" the troughs in the output current.

    Another way to think of it (which only just occurred to me) is with the good ol' water analogy. If you have a current of water that it pulsing, you could run it through a tall reservoir with an outlet at the bottom. The pressure from the water in the reservoir will compensate for the sags in the input current, giving you a smoother output flow.

    Hope that helped.
  5. Jan 4, 2008 #4


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    The time constant has to be large wrt the period of the input waveform. You can have a simple viewpoint whereby the cap discharges through the load when the output of the rectifier is around zero volts. The cap will then act as a source to the load and will discharge through it. If we however make the time constant very small, then chances are that the cap will discharge significantly through the load causing unwanted ripples in the output. Therefore, the larger the time constant, the longer the cap will take to discharge and the output ripple will be reduced.
  6. Jan 5, 2008 #5

    I think you will find this page very useful in explaining what you need to know.
    http://www.antonine-education.co.uk/physics_a2/options/Module_9/Topic_3/capacitative_smoothing.htm [Broken]

    there are a couple of things to think about as well....

    its always nice if your source has next to zero resistance(impedance Xc) and that when you switch on and off, not too much current is flowing.
    So a capacitance much bigger than your current needs will cause a surge at switch on, which could tax your fused protection and when you switch off, current will flow for a bit after or is kept stored on the cap....just waiting for you to touch it...;)
    Last edited by a moderator: May 3, 2017
  7. Jan 5, 2008 #6
    That's interesting ... it also makes sense if you've ever looked into a high-power device like a power amplifier for a hi-fi system. They typically have monstrous caps in them, and unless they have circuitry to prevent it, you often get a big transient turn-on "thump" when you power them up. They're also know for zapping people who are too quick to go poking around inside them after they've been powered down.
    Last edited by a moderator: May 3, 2017
  8. Jan 5, 2008 #7
    good circuits will have some sort of bleed mechanisms to prevent that....
    you'd hope....;)
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