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Capacitor Dirac impulse

  1. Oct 1, 2011 #1
    Let

    [itex] u(t) = \begin{Bmatrix}
    1, & t \geq 0 \\ 0, & t<0 \end{Bmatrix}[/itex]

    and lets have a simple circuit. Solo capacitor, connected to a DC voltage U0, a switch S exists.

    For purposes of this problem, I can mark the voltage across the capacitor as Vc(t)

    Vc(t)=u(t)*U0

    Current through the capacitor is

    [itex] i(t)=C \frac{du_{c}(t)}{dt}=CU_{0} \frac{du(t)}{dt}=CU_{0}\delta(t)[/itex]

    As we know that the derivative of the Heaviside step function is Dirac Impulse function.

    From this we can conclude that there is a initial spike in current when you turn the switch on.

    Does this mean, that at first, very first moment you turn the switch on, capacitor acts as short circuit?


    Second part:

    When I saw graphs of currents of capacitors in RC circuits connected to a DC voltage, I saw that the graph starts at V/R and falls of with exponential law.

    But how can current be immediately U/R, isn't there a transient state, due to RC self-inductance, to V/R, then exponential fall?
     
  2. jcsd
  3. Oct 1, 2011 #2

    AlephZero

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    Science Advisor
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    Both your examples are talking about ideal components. They only correspond approximately to "real life".

    In the first case, yes in theory the capacitor "instantly" charges so there is a charge of Q = CV on the plates. You can think of that as "an infinitely large current for an infinitely short length of time" if you want, but doing that probably doesn't lead anywhere useful.

    Real capacitors and real power supplies have internal resistance and self inductance, so in practice nothing "infinite" happens.

    The second example is basically the same. An ideal CR circult doesn't have any inductance. That's why the current starts at V/R and then decays exponentially.

    These "idealized" models are good enough for many purposes. For eaxmple if you are using a CR circuit to debouce a mechanical switch, it would probably have a time constant of the order of 0.01s to 0.1s. If the self-inductance of the cap and the physical layout of the circut creates a self-resonant frequency of say 100 MHz, there isn't much reason to be bothered about that.

    Of course if you are designing a CR circult that is meant to operate with a time constant of 10ns not 10ms, then you do have to bother about such things.
     
  4. Oct 1, 2011 #3
    I understand. So to take in self-inductance, current would rise to V/R in very very very short time, because self-inductance is low, with exponential law, and then fall exponentially?

    And you got me interested with this debounce mechanical switch. What is that?
     
  5. Oct 1, 2011 #4
    Microprocessors typically evaluate digital/analog inputs at kilohertz speeds. If you press a button with your finger and look at the voltage signal on a scope you might see the signal oscillate between the low and high thresholds a few times. Such behavior might cause a program to take action (do some sort of calculation, initiate a process, etc.) when it's not supposed to. Therefore, as Alephzero pointed out, you would incorporate an RC circuit (either in the analog domain or in the code) to ignore the fast frequency transitions that are physically meaningless (how many times per second can you flip a switch?)

    Regarding the original question:

    When you connect two voltage sources with no series element that can store energy in a magnetic field, you will get an infinite current. For example, if you take two caps with the same capacitance 1F and 1F, precharge one to 1V, and connect them, you will see that the final voltage will be 1/2V on each.

    However, the initial energy is 1/2*C*V^2 = 1/2J but the final energy is 2 * 1/2 * 1 *1/2^2 = 1/4J. This is called a "two-capacitor paradox" and it cannot be explained through basic circuit analysis tools. The lost energy is radiated.

    When you take a large power supply (with a few mF output capacitance), set the output to 100V+ and connect it to a ultracapacitor, you will hear an audible click and your USB computer mouse can disconnect for a second (the system generates a lot of EMI).

    Hope this helps,

    SunnyBoy
     
  6. Oct 2, 2011 #5

    Thank you, I understand what you are saying. I guess I will learn this in more detail in upcoming years.
     
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