Capacitors blocking DC

Discussion in 'General Electronics Chat' started by Voltboy, Nov 6, 2007.

  1. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
    I was reading about capacitors and I read that capacitors block DC current, but I dont get the point "For circuits with a constant (DC) voltage source and consisting of only resistors and capacitors, the voltage across the capacitor cannot exceed the voltage of the source. Thus, an equilibrium is reached where the voltage across the capacitor is constant and the current through the capacitor is zero"
    I really dont get the point of why it block DC current, can someone please explain me what Wikipedia and every other tutorial is trying to say?
    I know the question is kinda stupid, dont blame me for not being smart (or for being dumb).
  2. Salgat

    Active Member

    Dec 23, 2006
    Applications of capacitors include smoothing out a rectified voltage, or blocking DC and passing AC. The reason why capacitors block DC is because they are simply two separated plates that charge(a gap between them prevents current from passing from one plate to the other). You can think of them as electric sponges. Electricity will be sucked into the two plates but since the plates are separated, once the two plates are fully charged they won't accept anymore current(and thus begin to block any further DC). By putting this electric sponge in parallel with a circuit, any dips and spikes in voltage will result in the capacitor either outputting or absorbing those drops and increases in voltage in order to compensate. The reason why AC passes is due to how capacitors can absorb and give off current to an extent, while in DC the capacitors just absorb until they reach their full capacity and reject any more current.
  3. cheddy

    Active Member

    Oct 19, 2007
    On topic, but tangent to the discussion. The other day I tested this out to see for myself. I connected a 470ohm resistor to a 100uf capacitor to a LED around a 9 volt and it was interesting the LED started at full brightness and over the course of a few seconds dimmed to near completely black.
  4. thingmaker3

    Retired Moderator

    May 16, 2005
    A capacitor is not quite an open circuit. The conductors are close, very close, but they are not closed. So a capacitor can pass AC, if the frequency is high enough, but not DC.

    Capacitive reactance is inversely proportional to frequency. So, at zero Hz, capacitive reactance is just about infinite. Same as an open circuit.
  5. chesart1

    Senior Member

    Jan 23, 2006
    First and foremost, the question is not dumb and does not cast a negative reflection on you. Perhaps the best way to explain this is via the mathematical formula for the current through a capacitor at any instant in time.

    The current through the capacitor at any instant in time is referred to as the instantaneous current.

    The instantaneous current through a capacitor is equal to the capacitance in farads times the change in voltage across the capacitor.
    Using this formula, it is easy to determine that when the voltage is not changing, the current is zero.

    An electric field exists between the plates of a capacitor. I have not seen a good explanation of what happens in the electric field when a voltage is applied across a capacitor.
  6. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    For practical purposes, a capacitor will block DC once it is charged. As the LED experiment above shows, initially the capacitor is a short until it charges to the source voltage.

    This is one of the areas where inexperienced engineers make a mistake by not accounting for that initial short, and their circuit can act weird. A good example is audio "pop" when turning on an amplifier or switching between modes such as CD to radio. Although there can be other reasons as well, it is often caused by the charging or discharging of coupling caps within the circuit.
  7. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
    Thanks guys. Now i got the idea.
    And cheddy, I did that experiment and saw the effects.
    Thanks for helping
  8. Papabravo


    Feb 24, 2006
    One application for blocking capacitors is coupling the output of one amplifier stage to the next. If you don't use a blocking capacitor a DC offset in the output will affect the biasing of the following stage. Generally speaking it's a bad idea to let this happen. It is quite common, especially, in tube amplifiers to have multiple voltages running around.
  9. guitardenver

    Active Member

    Jan 24, 2009

    Some reason this stuff never works for me. Could you provide a link to a diagram. Is the capacitor in parallel with the light or is it one series circuit?

    Is there an experiment I can build easy to help me under stand why you would use a capacitor in a dc circuit? I'm one confused guy.
  10. Wendy


    Mar 24, 2008
    Look at the time date stamp, the original poster is long gone.
  11. hobbyist

    Distinguished Member

    Aug 10, 2008
    The capacitor is in a series circuit with the led and resistor.
    That's why the led shines bright initially then grows dim.

    But if you put the capacitor across the led and resistor, then there will be a short delay then the led will start out dim and grow brighter.

    However you need to put a large enough capacitor and a resistor in series with it, across the led and it's series resistor to see the change.

    The reason the led starts bright then dims in the series circuit, is the capacitor is chraging through the led.

    In the parrallel circuit the capacitor is in essence acting as a short across the led, but as it begins to charge up it begins to act like a variable resistance across the led wich starts out low and increases in resistance across the led therefore more current flows through the led as the resistance of the capacitor increases due to it charging.

    This resistance is actually called REACTANCE, because it is not a physical resistor, but acts and reacts like one hence the name reactance.

    It is given by the formula: Xc= 1/ (2*pi.*F*C) or Xc= 1/(6.28*F*C)

    So when the frequency (F) is at zero (direct current), than reactance aproachs infinity.