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Do the capacitors always charge exponentially?

  1. Jul 3, 2012 #1

    Femme_physics

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    The way I see it, the more the plates charge, the harder it is to charge them. Therefor capacitors always charge exponentially and linearly. Is that right?
     
  2. jcsd
  3. Jul 3, 2012 #2

    I like Serena

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    It depends on how the capacitor is charged.

    If the capacitor is charged from a constant voltage source, it will be charged exponentially as you say.

    If the capacitor is charged from a constant current source, it will be charged linearly (as in your recent thread).
    The current source will have to work harder and harder to keep its current constant though. ;)

    If the capacitor is charged in another way, it will yet be different.
     
  4. Jul 3, 2012 #3

    Femme_physics

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    So since we have a transistor mediating between the voltage source and capacitor (like in my last exercise), that means it charges from a constant current source, yes?
     
  5. Jul 3, 2012 #4

    I like Serena

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    Yes.

    It also involves the zener diode that forces the voltage at the base of the BJT (:wink:) to remain constant.
     
  6. Jul 7, 2012 #5
    Capacitors, from every chart I've seen, charge exponentially. Capacitors are reactive and not linear devices, so they charge in a curve, exponential way.

    I read above how they can charge linearly with current. This could be right, but I know nothing about this method.
     
    Last edited: Jul 7, 2012
  7. Jul 8, 2012 #6

    jim hardy

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    remember basic relation for a capacitor: Q = C X V , charge Q = capacitance C X voltage V

    so V = Q / C

    which would make dV/dt = dQ/dt X 1/C

    that should light up the brain cells

    If dQ/dt (current) is constant, so is dV/dt (slope of voltage)

    so - a capacitor charged by a constant current gives a straight line not exponential.

    Observe Voltage would be 1/C X ∫current

    and if current's integral is an exponential, that's what voltage will be.

    is that any help?


    Now draw yourself a simple circuit, battery and resistor and capacitor and switch all in series.
    Initial condition is zero volts on cap, and zero current of course 'cause the switch is open..
    Now close switch.
    Current commences because battery pushes it through resistor
    commencing buildup of voltage on cap
    which subtracts from voltage across resistor
    lowering current
    lowering rate of charge
    so we have a process whose rate depends on its value
    and isn't that the definition of exponential growth?

    So your initial statement could be elegant-ized to "Capacitor in series with just resistance charges exponentially . 'Cause that's how Mother Nature designed e^x. "

    Pardon my simplistic approach. I have to do such thought exercises myself before i can believe the equations because i make so many math mistakes.
    You always ask penetrating questions. I think it belies a very analytical mind.


    old jim
     
  8. Jul 8, 2012 #7
    If you "charge" a capacitor through an inductor, it will charge "sinusoidally", even with a constant voltage source:
    [tex]
    \mathcal{E} - \frac{Q}{C} - L \, \frac{dI}{dt} = 0
    [/tex]
    [tex]
    I = \frac{dQ}{dt}
    [/tex]
    [tex]
    \ddot{Q} + \frac{1}{L C} \, Q = \frac{\mathcal{E}}{L}
    [/tex]
    Define:
    [tex]
    \omega_0 = \frac{1}{\sqrt{L C}}
    [/tex]
    and the solution, with the inital conditions [itex]Q(0) = 0, I(0) = 0[/itex], is:
    [tex]
    Q(t) = \mathcal{E} \, C \, \left[1 - \cos \left( \omega_0 \, t \right) \right]
    [/tex]
     
  9. Jul 8, 2012 #8

    jim hardy

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    PS

    Indeed. Just like squeezing a spring.



    edit late entry:

    indeed the sinewaves are another exponential .

    ei t = cos t + i sin t

    Euler was one heavy thinking dude !
     
    Last edited: Jul 8, 2012
  10. Jul 9, 2012 #9

    sophiecentaur

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    Woa there A constant voltage source has zero source resistance. The capacitor will charge 'instantly' because the voltage across it will suddenly become equal to the supply volts. You must make it clear that a series R is involved if you want to avoid misunderstanding.

    If charged with a 'constant current', then Q=CV rules and the voltage across it will increase linearly with time with no limit.
     
  11. Jul 9, 2012 #10

    I like Serena

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    It's still exponential. :)

    In the extreme theoretical case where the resistance is zero, the exponential function turns into a step response.
     
  12. Jul 9, 2012 #11

    sophiecentaur

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    You said "constant voltage". That has a specific meaning. There are a lot of misunderstanding about this topic and it is essential to be as clear as possible. I try to look at people's posts as if I were a total beginner and see what I could make of them. That's why I am so picky. :smile: They were just that fussy when I was at School and I am eternally grateful.

    A step function is not just a very fast exponential function; no time is involved for the transition.
     
  13. Jul 9, 2012 #12

    NascentOxygen

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    Only if the transistor is arranged to operate as a constant current source. The transistor circuit could be designed to provide some other characteristic, perhaps something non-linear, so you mustn't assume it to be a constant current source without analysing it.
     
  14. Jul 12, 2012 #13

    Femme_physics

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    Hmm. How would I know if the transistor acts as a constant current source or not?
     
  15. Jul 12, 2012 #14

    sophiecentaur

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    You would need to look at the circuit or add one that you had designed. What is the exact context of all this? Do you not believe the theory?
    If all you want is to verify the theory by looking at waveforms on a scope then use a low voltage (low output resistance source) and a moderately high series resistance or a high voltage and a very high resistance (effectively a current source) to show the extremes. Don't ask what would constitute low or high because it would depend on the capacitor value. Basically, all you are doing is using a shortish time constant for one - so you get a recognisable exponential curve because the volts on the capacitor will go from zero to almost the supply volts - and a very long time constant for the other, which will give you a curve that looks like a straight diagonal line whilst the capacitor volts are low. Of course, when using the high voltage supply, you would need to cut it off when the capacitor volts get too high. If you actually want a good constant current supply then there are tons of circuits available.
     
  16. Jul 12, 2012 #15

    NascentOxygen

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    You derive equations for its voltages and currents, to establish which variables influence the current charging the capacitor. Just like you did in that other memorable thread. :smile:

    If the equation for the charging current involves terms which are fixed and constant (so that means it cannot include a term related to the capacitor's voltage, since that voltage changes as the capacitor charges) then it will be constant current charging.

    You don't need to painstakingly analyze every circuit at each encounter. Having thoroughly examined each once or twice, the essential details will remain with you. Honest. :wink:
     
  17. Jul 12, 2012 #16

    Femme_physics

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    Hmm... ok then, I think I understand. But as far as the constant voltage issue from before... I am in basic electronics and we're only working with ideal voltage sources. I'm not sure how can a capacitor charge "instantly" if everything in nature takes time. Even if it's 0.000000000000000000000000000000001 nanosecs
     
  18. Jul 12, 2012 #17

    NascentOxygen

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    Well, if you connect an ideal voltage source across an ideal capacitor you have one embarrassing situation. :eek: It's best to think in terms of elements approaching the ideal, and describe the situation accordingly. Leave the ideal for the idealists. :wink:
     
  19. Jul 12, 2012 #18
    It does not charge instantly if you connect it to an ideal current source. It charges linearly and indefinitely in time, according to:
    [tex]
    Q = I \, t
    [/tex]

    EDIT:
    As for an ideal voltage source, it ought to charge instantaneously according to:
    [tex]
    Q = C \, V
    [/tex]
    But, the electric field inside the capacitor will change as a Heaviside step funciton:
    [tex]
    E = \frac{V}{d} \, \theta(t)
    [/tex]
    and the displacement current density is:
    [tex]
    J_{\mathrm{disp.}} = \frac{\partial D}{\partial t} = \epsilon_0 \, \frac{\partial E}{\partial t} = \frac{\epsilon_{0} \, V}{d} \, \delta(t)
    [/tex]
    This delta-like displacement current will cause an impule in the magnetic field, which, in turn will cause an induced emf that will oppose the sudden rise of the electric current (Lenz Law).

    So, if you do not neglect the displacement current, you will have a finite charge time.
     
    Last edited: Jul 12, 2012
  20. Jul 12, 2012 #19

    jim hardy

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    indeed we wind up with division by zero
    so we evaluate its limit instead
    that was the fun of freshman calculus...

    Dick has brought up self inductance of the space between the plates, i think,,,, but my math isn't that good.

    i do know a very fast current pulse has a tremendous magnetic field. In the power plant we noticed effects of lightning stroke on nearby loops of instrument system ground wires.

    old jim
     
  21. Jul 12, 2012 #20

    I like Serena

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    If you charge a capacitor through a smaller and smaller resistor, you'll get something like this.

    attachment.php?attachmentid=49057&stc=1&d=1342125120.jpg

    The curve comes closer and closer to a so called step response.

    When we consider ideal wires, an ideal capacitor, an ideal constant voltage source, and an ideal switch, the graph will ideally be a step response.

    But you're right, that's not realistic!
    It's just an ideal and abstract approximation of reality.

    (There! I included a smiley!)
    .
     

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