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I don't "get" capacitor usage if it blocks current

Discussion in 'General Electronics Discussion' started by NuLED, Jun 22, 2013.

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  1. NuLED

    NuLED

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    Jan 7, 2012
    So back to my somewhat inanely basic questions about basic electronics...

    It seems my [theoretical] understanding of capacitors is wrong, based on my [applied] experimentation lately.

    If a capacitor in a DC circuit is getting charged up, it seems that loads (e.g., resistor) in series "receive" less and less voltage/current as the voltage across the capacitor's dielectric increases. Until there is almost negligible current flow (?). I wired an LED in series and it got dimmer and dimmer.

    I thought the capacitors were supposed to charge up to the input voltage, and then "release" it at that level, and release their charge when the input voltage drops (like a bucket or balloon of stored "water" when the faucet turns low or off).

    Now it seems it just blocks everything like a flood dam?

    I am a fair bit confused on this. I understand in AC that it is almost transparent to the circuit since the polarity changes so much that no charge accumulates across the dielectric.

    But what about DC?
     
  2. duke37

    duke37

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    771
    Jan 9, 2011
    Think of a capacitor as a rubber diaphragm in a water pipe. If the pressure goes up and down, the diaphragm will flex and pressure will go up and down on the other side. (AC)

    If the pressure rises on the input and stays there, the diaphragm will flex and water will come out the other side but will eventually stop. (DC)

    If the pressure (voltage) gets too high the diaphragm will burst and there will be no limit.
     
  3. eKretz

    eKretz

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    Apr 8, 2013
    Right, and if the pressure drops on the supply side for a short time, the flex in the rubber will provide a little cushion so the pressure in the system reduces slowly rather than abruptly dropping.

    Your original post had it right where you said it charges up to input voltage level then stops, but if the input voltage drops, then it is released back out.
     
    Last edited: Jun 23, 2013
  4. NuLED

    NuLED

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    Jan 7, 2012
    OK, so I think I am starting to get it. If the voltage across the capacitor has MORE potential difference than the input voltage, then it will re-establish equilibrium to match the "status quo" - but I don't understand the part where current STOPS (or slows) when the input voltage matches the capacitor. What good is it if it acts like a huge resistor at that point?
     
  5. NuLED

    NuLED

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    Jan 7, 2012
    The only two instances I can think of are 1) putting it in parallel as a sort of stop-gap against a sudden voltage drop and 2) smoothing fluctuating DC current as a buffer, but in both cases the input voltage must not cause the capacitor to be fully loaded and become a huge resistor. So that's what I don't quite get, if it is to be used in series.
     
  6. NuLED

    NuLED

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    Jan 7, 2012
    Duke - for this bit, when you say "burst" do you mean it is like a reverse-biased diode that FAILS due to overload, or do you mean it is like a Zener Diode that will burst to ALLOW (by design) current flow? (that is, ensuring a sufficient voltage level and nothing below that)
     
  7. eKretz

    eKretz

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    Apr 8, 2013
    In your instance of number 2, the cap would be connected one leg to circuit and one leg to ground, so as the input dropped, the cap would discharge into the circuit. Another example where it can be used is wired in parallel with a load, like say an LED. The LED and it's load resistor wired in parallel with a cap and resistor will allow the LED to start softly and grow brighter as the cap is charged. Once the cap is fully charged, it blocks the parallel flow of current and the LED runs at full brightness.

    And in Duke's example, that would be a failure mode, they aren't designed to do that.
     
  8. NuLED

    NuLED

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    Jan 7, 2012
    ok thanks - I am going to read more about it (am going through several beginner books actually). I have found no single author really covers the ground comprehensively, and all of them make certain assumptions. Your collective advice here is helping tremendously in covering those assumption gaps so, thanks again. Hopefully in a matter of months I will be somewhat competent. One can hope! :)
     
  9. eKretz

    eKretz

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    Apr 8, 2013
    I'm still pretty new to a lot of this stuff too; I am doing the same as you, learning as I go.
     
  10. NuLED

    NuLED

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    Jan 7, 2012
    Talk about SYNCHRONICITY! Ha!

    Posted yesterday!

     
  11. eKretz

    eKretz

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    Apr 8, 2013
    Lol, that's hilarious!
     
  12. BobK

    BobK

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    Jan 5, 2010
    Perhaps this will help you understand your original question better.

    cap.JPG

    In the circuit on the right, does current flow through the resistor? The answer depends on the voltage of the two batteries. If the are the different, current flows, if they are the same none flows.

    Now look at the circuit on the left (the one you were asking about)

    If the capacitor has no charge on it, it acts like a battery with 0V. So if the battery has, say 10V current will flow through the resistor, charging the capacitor. After a while, the capacitor has charged to 5V. Now only half the current flows, because the voltage across the resistor is down to half. And eventually it reaches 10V. Now it is like the circuit on the left when the batteries are equal. No current flows.

    Bob
     
  13. duke37

    duke37

    5,364
    771
    Jan 9, 2011
    In the water analogy when the diaphragm bursts, it will not work again and needs to be replaced. In the capacitor case, when it fails it may just go short circuit or, if using high voltage and power, may explode. I always use safety glasses when using over 100V.

    Do not use a capacitor in excess of its voltage rating.

    Consider a valve (tube) amplifier, the anode of one valve may be at 200V and will fluctuate about this value when a signal is present. The grid of the following valve needs to be at 0V and will need to fluctuate about this with the signal. A capacitor between the two with 200V across it will pass the signal but block the DC.

    In the case of a power supply, the capacitor is connected across the supply so that the AC component is shorted to ground, leaving a more constant DC component for the output.
     
  14. NuLED

    NuLED

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    Jan 7, 2012
    Wow, Bob thanks! Yours was a very clear explanation!

    I had actually finally understood it the way you explain it, and when you did explain it, it confirms my understanding.
     
  15. NuLED

    NuLED

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    Jan 7, 2012
    Thanks Dave
     
  16. eKretz

    eKretz

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    Apr 8, 2013
    Ah yes, here's another good example of very common DC capacitor use, forgot about that one.
     
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