Connect with us

Capacitor charging problem

Discussion in 'Electronic Design' started by john, Nov 27, 2007.

Scroll to continue with content
  1. john

    john Guest


    A constant current source is in series with a capacitor ( 10volts,
    0.5uF) and a resistor ( load) of 100kilo ohms.
    What happens if

    1. A constant current source charges up the capacitor upto its maximum
    voltage and keep charging it for a long time. The constant current
    source has following specs

    a. compliance voltage +/- 15 volts
    b. current range 0 to +/- 600uA (ac)
    c. biploar
    d. dc leakage current of 100nA

    2. The capacitor will show high impedance to the current source but
    get charged by the DC leakage current constantly. Will the AC flow
    through the capacitor anymore?

  2. Eeyore

    Eeyore Guest

    Homework was it ?


  3. Hello John,

    The output stage of your current source will run
    into saturation beyond the compliance range.
    This will result in distortion and less amplitude
    on your 100kOhm load.

    Best regards,
  4. No, it will initially show a short!
    That is why it starts at zero volt across the capacitor.
    ..6mA in 100kOhm makes 60V, you only have 15V, you example makes no sense.

    If you omit the resistor for a moment, then the voltage across the capacitor is
    The charge Q = C * U = I * t

    So U = (T * t) / C
    C in Farad, I in ampere, and t in seconds.
    The voltage across the capacitor will then rise LINEAR with time, until it reaches
    the 15V (theoretically).
    However, with the resistor in parallel, the max voltage will be limited by I * Rp, or
    the 15V, whatever is lower.
    In this case 15V is lower then 60, so 15V
    And the charge current in the capacitor is no longer linear, as
    as teh voltatge rises, the current through the resistor is increasing,
    substract that from the current in the capacitor.

    The capacitor will charge like this:
    see under 'Time domain considerations'.
    There is no longer a linear rise in voltage.
  5. john

    john Guest


    The leakage current will charge the capacitor upto its maximum
    voltage , because the leakage current does not have any other place to
    go it will keep charging the capacitor. Now, my questions are

    1. will the current source shut down?

    2. for longer times, the charge will break the dielectric and the
    leakage current will increase?

    Please advice!
  6. 1) provide current diagram
    2) what leakage current
    3) what ARE you trying to drive, so what IS your load.
    4) what frequency are you talkinga bout when you said 'AC' in a previous posting.
    5) why do you drive the load, or try to drive the load with a current osurce.


  7. Eeyore

    Eeyore Guest

    That'll depend on the design of the current source but most just continue
    delivering the rated current up to theit maximum voltage compliance.

    Not as long as the capacitor's voltage rating isn't exceeded.

  8. legg

    legg Guest

    If you're looking for something strange to happen, you have only one
    real practical unknown - the leakage current of the capacitor. This
    will only be a factor if the leakage current develops a signifigant
    voltage across a 100K resistor.

    When you're considering practical components, ratings and breakdown,
    you'll be dissapointed to know that there's no guarantee that anything
    will or will not happen outside paper ratings, unless published
    otherwise or practically demonstrated by somebody with the time money
    and need to do so.

    Leakage in electrolytics can increase with shelf age. There's a
    typical method recommended for 'reforming' higher voltage parts

  9. YD

    YD Guest

    The whole thing is probably unanswerable from a theoretical point of
    view. He states that the CS has a compliance of 15 V but the capacitor
    has a 10 V rating. So once the cap's charged beyond 10 V it may die
    with a bang or a whimper and possibly just take it like a hero. Most
    likely just start leaking, had an odd one like that the other day. A
    leaky cap making very low frequency noise, it would charge up for a
    bit, then drop a few volts, charge up again and so on at an average
    period of a couple of seconds, very irregular both in period and
    peaks. It was below rated voltage so probably just old age that got

    - YD.
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