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Capacitor in DC current

  1. Oct 6, 2011 #1
    Hi all. Good day to you all. i wanna make one thing clear. In converter circuits, how to trace the polarity of capacitor circuit. Is it following the polarity of capacitor or follow the current flowing in it? I understand that the capacitor is open at dc. So it means it is like a brake part in dc cirucit and the current can't flow through it, is it?. Pls anyone answer for me. Thanks.
  2. jcsd
  3. Oct 6, 2011 #2
    Yes, the current does not flow through the capacitor. Equal and opposite charge builds up on either plate. The polarity of the capacitor depends on its arrangement in the network.
  4. Oct 6, 2011 #3
    Thanks for your reply dim. So if the capacitor connected seires to resistor or other circuit then the current can't go througt them too. One more question, if the capacitor get the open circuit voltage, is there any chance for it to discharge later. How about the inductor?
  5. Oct 7, 2011 #4


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    Hi lumaw. If you post an actual circuit it will let you ask the question more specifically and make it much easier for us to explain.
  6. Oct 7, 2011 #5


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    With DC, current does go through the capacitor, and any circuit it is in series with, but only as a burst of current for a short time when power is first connected (e.g., at switch-on). After that, current is blocked--provided the DC is perfectly constant. When you switch off, current may again go through briefly, in the reverse direction.

    But I agree, you need to show the schematic if you want accurate advice, especially if something will be damaged when you get it wrong.
  7. Oct 7, 2011 #6
    Hi all,
    Please see the attach file. It is positive output LUO converter. at t=0 and let us consider switch is open. At that time, diode will be reverse bias and the current will flow through at first loop only (flow through inducotr L1 only0. But that time the capacitor C will be charged. Is it correct? But how can be -Vc. Pls explain me. Since the capacitor C is opened, the current will not flwo to inductor L2 and Capacitor Co. Any chance Co will be charge at the same time because of open circuit voltage. Pls explain me more detail on these circuit operation. Thanks

    Attached Files:

  8. Oct 7, 2011 #7


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    Ok let me stop you there, you have that the wrong way around. When the switch is closed the diode is reverse biased.

    After that I'm not totally sure what you're asking, but I suspect that you are troubled by the fact that the direction of current flow through the first capacitor appears to be inconsistent with the direction of this capacitors voltage. Is that your query?
    Last edited: Oct 7, 2011
  9. Oct 7, 2011 #8
    The capacitor only behaves as an open circuit at the moment is has become fully charged, meaning t>>5 time constraints. Until that point, the capacitor charges until it develops its "saturation voltage" across its terminals. It is at this point when the capacitor can be considered an open circuit element when analyzing the network.

    Hopefully that helps. If you are supposed to analyze these types of questions, I would suggest reading any chapter out of an introductory circuit analysis book.
  10. Oct 7, 2011 #9


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    Yeah that's true for a DC circuit (or where the switch is assumed to just be thrown once at t=0) but in this type of converter circuit (where the switch is continually operated at a certain duty cycle) then it's not quite so simple.

    There are however some relatively simple principles that you can use to analyses the steady state average capacitor voltages and average inductor currents.
  11. Oct 7, 2011 #10

    jim hardy

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    you have there what looks like a switching regulator or converter, perhaps with step-up.

    look up National's AN556 and digest it.
    It's an intoduction to power suppliues.

    To understand that circuit you will need to "work it in your head"
    s1 cannot remain closed for long else L1 will staurate
    so it's a switching circuit, S must cycle.

    read www.national.com/an/AN/AN-556.pdf[/URL]
    then ask yourself "When S opens, what becomes of the energy that was stored in L1 while S was closed ? "

    old jim
    Last edited by a moderator: Apr 26, 2017
  12. Oct 7, 2011 #11

    jim hardy

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    Aha i finally saw your question in post #6.
    """But that time the capacitor C will be charged. Is it correct? But how can be -Vc. Pls explain me.""

    Here's what i think:

    While switch is closed:
    Left hand side of C is at voltage of Vin.
    Current is established in L1

    When switch opens:
    Current in L1 must be maintained. That's what inductors do.
    So voltage at top of L1 reverses polarity to maintain current flow. That's what inductors do.
    Note that the polarity reversal forward biases diode D.
    So the capacitor C becomes charged + on right, - on left
    with right side at about -0.6 volt. Left side is highly negative.

    switch closes
    left side of capacitor is immediately raised up to Vin which is positive
    right side will rise an equal number of volts
    which reverse biases diode
    and leaves right side of capacitor more positive than Vin,,,
    and begins to re-establish current in L1.

    when switch opens again cycle repeats.

    so you have a "boost" regulator, so long as the switch cycles.
    It will be the designer's responsibility to control the ON and OFF times of switch to keep the regulator "boosting", ie always allowing time for L1's current to build back up and never allowing capacitor to completely discharge.

    L2 and CO form a filter to smooth out the pulsations from voltage at node of diode/C/L2 before presenting it to load resistor.

    i think there's a better deescription in that National AN.

    old jim
  13. Oct 9, 2011 #12
    Jim thanks for your answer. This is exactly what i want to ask. Yeah your explanatoin is very clear. So it means the capcitor polarity will be positive on the right hand sided during unit step response (switch close) and the polarity will be reverse when switch open ( natural response). Is it right? so during switch close it will be charged by Vin and during switch open it will discharge and transfer charge to Co, am i right? Jim. I just post one more today regarding for some problem during shore power supply for the ships. Pls take a look for me and please advice your opinion.
  14. Oct 10, 2011 #13

    jim hardy

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    "" So it means the capcitor polarity will be positive on the right hand sided during unit step response (switch close) and the polarity will be reverse when switch open ( natural response). Is it right? """

    well, nearly.

    be aware that the circuit's operation is cyclic not just a one time switch closure.
    the link to circuit has disappeared from my screen so i'll abswer from my memory of the circuit.

    forget about unit step: this circuit is best understood the old fashioned way, using your imagination to push charge around it.

    Start with switch open, all currents and voltages are zero.
    Capacitor has no charge so there's no voltage across it.

    Now close switch. Hold your mind at that instant.
    Capacitor cannnot accept charge any so voltage across it stays zero.
    Reason it cannot accept any charge is there's no way for charge to get out of it.
    Charge could enter from left if there were a way out - but look--
    ---charge can't go down through doide, that's backward
    ---charge can't go to right toward load resistor because inductor won't allow it (remember we are locked at one instant of time)
    so both sides of capacitor are lifted to Vin.

    Now let your mind unfreeze time,,,,
    current begins to rise in both of the inductors..
    in left inductor it increases and just draws current from the supply
    in right inductor it begins carrying charge toward load and begins slowly discharging our capacitor
    so cap will get slightly reverse charged on first switch closure.
    left side is Vin, right side is a teeny bit less.

    Note top of first inductor(on left) is at +Vin

    now open switch----
    immediately current stops flowing out of supply
    but left inductor won't let its current stop immediately, remember it's an inductor
    so its polarity reverses to push the same amount of current it was pulling an instant ago.
    that means left side of capacitor is yanked from +Vin to negative something substantial,,
    ,, this next step is tricky repeat it until it works for you
    ---right side of capacitor tries to follow left side but it can't go negative because of diode -- dode passes current holding cap's right side near zero
    so the inductor on left pushes current into capacitor and that leaves capacitor charged substabtially negative on left and positive on right(even zero is positive enough).
    we have replaced a small badkward(+ on left) charge on capacitor with a large one in proper direction, + on right
    now close switch
    left side of cap is yanked back up to +Vin
    Cap's right side jumps up same number of volts leaving it lots higher than Vin
    and current flows from cap into second inductor towards load but at higher voltage than Vin

    and the cycle repeats until somebody stops it.

    they didnt show the circuitry that contols timing of switch
    and they didn't give values for inductors and capacitor
    some designer must size those parts and select timing for the load he wants to handle.

    Such a circuit would be real handy for lighting LED's that require 3 volts from a 1.5 volt battery
    or charging an 18 volt computer battery from a car cigarette lighter plug

    but i think he point of the exercise was to get you workng circuts in your head
    a must for any electronics guy's "bag of tricks"

    practice until you can make this one go

    then if you decide on a Vin, Vout and current you could sart sizing parts and timing.

    old jim
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