Connect with us

Uncharged capacitor is a short circuit?

Discussion in 'Electronic Basics' started by Ron J, Jan 9, 2006.

Scroll to continue with content
  1. Ron J

    Ron J Guest

    Hey all,

    I'm curious why an uncharged capacitor can be viewed as a short circuit
    by, say, an amplifier? I thought an uncharged cap would be more like an
    open circuit.

  2. The relationship between capacitor voltage and capacitor current is
    I=C*(dv/dt), where I is in amperes, C is in farads and dv/dt is in
    volts per second.

    When you first connect a capacitor to a stiff voltage source (one that
    puts out as much current is necessary to maintain the voltage) the
    source tries to alter the voltage of the capacitor very quickly. This
    is a big dv/dt. Si it takes a big current to do that. Short circuits
    also draw big currents. The difference is that as the cap voltage
    approaches the source voltage, the dv/dt goes down and approaches
    zero, and so does the current. A short circuit just keeps sucking
  3. Don Bruder

    Don Bruder Guest

    Only until the first juice hits it, and only when talking about DC. (AC
    basically sees a cap as a dead short - lots of "special rules" about
    that for various frequencies of AC versus values of capacitor, but
    that's the basic idea, anyway) At first, there is practically no
    resistance. Like a short. Charge flows freely into it - For a moment.
    Once it takes on its first "droplet" of charge, it becomes more like a
    gradually increasing-value resistor until it's fully charged, where it
    behaves more or less like an open to DC. Trade a load for the charge
    source at this point, and it behaves like a current source until the
    plates equalize (full discharge) then it's back to being like a short
    until the next dose of charge hits it.
  4. because even though there is an open circuit between the plates of a cap
    there is a force across them. This force is felt on the other side when an
    electron moves onto one place. In essence it pushes off an electron on the
    other plate when it arrives... the same kinda stuff happens in a wire where
    one electron feels the repulsive force of another... in this case though
    there is a definite force but there is a "gap" that prevents the electron
    from continuing.

    Thing of blowing up a water rocket with water. It requires a certain force
    to push the air in and initally is pretty easy to do... the more
    air(electrons) you put in the harder it gets until eventually you cannot
    push any more in(or it explodes) and the forces are counter balanced. The
    same thing happens in a cap except in this case we can think that we have a
    reservoir of air connected to the pump(so there is a limited amount of air)
    and when we pull the air from the reservoir and push it into the rocket it
    is likened to a current flowing(the air being analogous to the electrons)...
    if you put the reservoir and the water rocket close to each other you can
    envision that there is a "hypothetical" current flowing between them where
    it seems that the air is flowing continously through(ofcourse we know it
    isn't but it sorta looks like that from a distance).


    (need to view this with a fixed width font)

    |-----| | |-------|
    | Air ==|== Water |
    |-----| |-------|

    === Water || Air ====
    // |------||-----| \\
    || ||
    || ||
    \\ //

    So as we pump it first is very easy but gets harder and harder. Also we feel
    as if the air leaves the Air chamber and enters the water and we might even
    suspect that some air leaves the water chamber and goes across to the Air
    chamber(maybe through a little valve betwen them). This is not what happens
    but if we didn't not know they were not connected and we started to pump we
    would see a air current flowing from one around the "circuit".... but we
    would notice it would get harder and harder to pump. We could hold the air
    that we pumped into the water chamber but as soon as we released the pump
    all the air would rush back into the air chamber(until it equalized).

    A capacitor is very similar except that it uses conductive
    plates(reservoirs) that hold electrons. When the electrons arrive on one
    side of the plate they "push" off electrons on the other side and it
    "simulates" a current(even though there isn't any "physical" current. The
    more electrons that arrive makes it harder and harder to push more
    on(because the electrons start to repel each other(ones on the same plate)

    What you have to understand is that just because there is a "gap"(an open
    circuit) in a capacitor that it doesn't stop electrons from flowing...
    Electrons can easily flow through a gap(say a vaccumn) if they have enough
    energy. Since capacitors are made of conducting material electrons move
    freely on on or off its plates... if, say, one of the plates were an
    insulator then it would not conduct electricity.

    You can also think that a battery pulls an electron off one side one
    plate(making it +) and puts it on the other plate(making it -).... this
    causes a potential difference across the cap and the more electrons that the
    batter does this two the stronger the ptotential difference is... if you
    have N electrons that we could hypothesize that the potential would be
    proportial to N(which it theoretically is)... but the battery can only push
    so "hard" and eventually it's voltage would be "countered" by the voltage
    across the cap.

    Hope that helps some,
  5. Pooh Bear

    Pooh Bear Guest

    It can ? How so ?
    Actually it's more like a capacitor. Capacitors are 'reactive' not
    resistive btw.

  6. Guest

    NO, a typical capacitor has allmost NO resistance, this resistance is
    CONSTANT irrespective of the state of charge. When a current flows
    through a capacitor it developes in internal electric field (( current
    x time ) / capacitance) which is a voltage which opposes the external
    voltage. So eventually the internal voltage will balance out the
    external voltage and so the current will be zero BUT notice the circuit
    resistance is still the same.
  7. Guest

    Guest Guest

    : Hey all,

    : I'm curious why an uncharged capacitor can be viewed as a short circuit
    : by, say, an amplifier? I thought an uncharged cap would be more like an
    : open circuit.

    : Thanks!


    An alternate way to look at this is that the capacitor has an
    impedance (similar to resistance) equal to 1/(2*j*pi*f*C) where j =
    sqrt(-1), f = The frequency of the source, and C = the Capacitance. The
    magnitude of the impedance is called the reactance and is simply equal to
    1/(2*pi*f*C). Even if you don't understand everything that I have just
    said, what you should take away from it is that the impedance of the
    capacitor is inversely proportional to the frequency -- i.e. the higher
    the frequency, the lower the impedance (think resistance.) Note that if
    you plug f = 0 (i.e. DC) into that, the capacitor has infinite impedance
    and acts like an open circuit.

    Under some circumstances, the capacitor can look like an open
    circuit -- namely when the frequency is very high. Steps and impulses
    contain high frequency components, and thus, at the onset of the
    application of a step or an impulse to a circuit containing a capacitor,
    one might think of the capacitor as an open circuit (to calculate max.
    current draw or something like that.)

    If you are talking about a capacitive load on an amplifier that
    has a continuous time input signal, then the capacitor neither behaves as
    a short or as an open circuit, but something in between. It has an
    impedance given by the equation above, and, for a single frequency, could
    be modeled as a resistive load with a value of the capacitive reactance.
    Certain amplifier-containing circuits cause their input/output to
    transition in steps. Switched-capacitor circuits are an example.
    Therefore, once again, it is sometimes useful to consider the capacitor as
    a short circuit, because at the beginning of one of these steps, the
    current required to charge/discharge the capacitor is a maximum, and this
    is almost certainly important for the design of the amplifier.

    That's probably more information that you really wanted, but
    hopefully it's helpful.


    P.S. Those who said that capacitors don't have resistance, etc, are, of
    course correct, but if your are only interested in the magnitude of the
    impedance, then you can simply treat a capacitor as a resistor with a
    resistance given by the formula I posted waaaaaaay above. If you care
    about phase, then things are not as simple.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day