Can somebody help explain what's happening here?

Discussion in 'Homework Help' started by Bombasa, Apr 4, 2013.

  1. Bombasa

    Thread Starter New Member

    Dec 2, 2012
    My task is to make a graph of the current over C1 and C2 when charging and discharging.

    The picture here represent the charging, but i cant explain what's going on? 0 ampere on both capacitors instantly? Can somebody please explain to me why the circuit behaves as it do?

    And why does it start at 5V and not 10 before dropping?

  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
    The simulation seems to divide the 10V into 2x5V, one for each capacitor, initially. The initial 6A pulse charges the caps equally, they are treated here like some very low value resistors when discharged. Once they are charged the current from the voltage source ceases. After some time C2 will be fully charged and C1 will be fully discharged.

    You didn't indicate what the red curve is, I assume it's the voltage on C1.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Strictly speaking, if the circuit were ideal the initial current into the caps would be of infinite magnitude and of infinitely short duration - which is an impossibility in a physically real system. The observed -6A value at t=0 sec [if that is what the graph is actually indicating] is probably an artifact of the simulation - perhaps due to the specified time step.
  4. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    Yep, it's a perfect "simulation" of why simulators don't work very well on real world circuits... :)

    The 10v source is capable of an infinitely high current with infinitely small voltage sag, and both the caps have infinitely small ESR and inductance, as does all the wiring.
  5. tubeguy

    Well-Known Member

    Nov 3, 2012
    The caps will be perfectly matched in the simulator so I think they both would charge instantly at the same rate, hence the 5volt max.
    With the caps in series in this circuit the mid point could never reach 10 volts.
  6. WBahn


    Mar 31, 2012
    Or, put another way, why a simulation is no more faithful than the fidelity of the model being simulated.

    If you use ideal components, then your models start off with a set of bounds on their fidelity. But if you model the non-ideal properties with some fidelity, you can get amazingly accurate simulations of real world circuits.
  7. Bombasa

    Thread Starter New Member

    Dec 2, 2012
    Thanks for the replies! :D
    So, in a real world circuit with components found in the random university, how would the circuit behave then?
    Is it possible for me to emulate that in LTspice?

    Also, the red line is the voltage across the first node, after C2 and before C1 and R1.
    Last edited: Apr 9, 2013
  8. WBahn


    Mar 31, 2012
    One thing to keep in mind is that your output is graphed on a time scale that doesn't give any insight into what is happening during the initial transient charging event. Even if the 10V supply was only capable of delivering 1mA the capacitors would be fully charged in just over 10ms. So if you were to build this circuit up in the lab and look at it with a scope you would probably see something very similar to what you are seeing here.

    To see the effect of the ideality vs. reality, you need to zoom in to the origin pretty tight. Does the simulation show any ramp at all?

    The biggest thing to do to model the initial transient better is to model the output impedance of the source you are using by putting a fairly low valued resistor in series with it. Another thing you could do would be to model the series resistance and inductance of the capacitors. Of considerably less importance is to model the self leakage of the capacitors by putting a high valued resistance in parallel with each of them.

    You can look at a number of sources for information on rules of thumb to use for the values of these modelling components. You can also get an idea from looking at the response on a scope.