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capacitors in series

Discussion in 'Electronic Basics' started by Ken O, Jul 2, 2006.

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  1. Ken O

    Ken O Guest


    I have two capacitors in series. I want to put 35 volt acrosse ( cause I
    dont have 50v rated) so I want to use 25 v rated. I know its going to make
    50v (and 1/ 1/c1 +1/c2) but they are rated 25v, do I need 50v rating
    capacitors or my two 25v will be fine ? I think it should be ok to use two
    rated 25v but I wanted to make sure. There are a lot of parallel and series
    formula on google, but they dont talk much about the ratings.

  2. Ken O

    Ken O Guest

    it is going to work, sorry for the post.
    One of the capacitor is rated 1000uF and the other 2200uF, I am just
    thinking that one might take more voltage then the other. v=Q/C
    proportionally, so 1000uF will have twice more voltage.

  3. If the capacitors are equal, the applied voltage will split equally
    across them, and you can use two 50V caps in a 35V circuit. If the caps
    aren't equal, the voltage will be higher across the smaller cap. (The
    charge is the same on both caps because they're in series, and voltage
    = charge / capacitance V=Q/C). So if you had two capacitors with a 25V
    rating, and the ratio between their values was greater than 25/10, or
    5/2, the rating of the smaller cap would be exceeded. For example, if
    you used two 25V caps in a 35V circuit, and one was 5uF while the other
    was 2uF, you would end up with more than 25V across the 2uF cap.
  4. Ken O

    Ken O Guest

    I did the circuit like said, I get 35 volt at the bridge, but when I add the
    capacitors, I get 50v (49.8v), I was sure I would be getting 35 volts across

  5. I see you figured it out for yourself. I'd just like to inject a note
    of caution about the electrolytics. It's very common for large caps to
    be 20 or 30% off nominal value, often higher. If the 2200 is that much
    higher and the 1000 isn't, the latter might have end up with more than
    25V on it. It might be alright or not.
  6. For electrolytics as DC filters, the voltage split may depend more on
    the relative leakage current of the two caps than on their capacitance
    values. It's probably safer to use a single capacitor with a voltage
    rating higher than 50V for your power supply.
  7. G. Schindler

    G. Schindler Guest

    Something to consider:
    The tolerance of electrolytic caps is typically very large, maybe even
    +100%, -20% depending on the units in question.

    Second, leakage and ESR, and probably a lot of other things will also
    contribute to voltage drop.

    So, the actual capacitance ratio only controls the AC voltage division
    across the capacitors. If the capacitors are filtering rectified AC and
    thus dropping a significant DC Voltage component then the Capacitance
    ratio cannot be used to predict the (DC) voltage drop across the
    individual capacitors. For this the other effects become dominant. For
    this reason, series connections of electrolytics in power filters is
    probably not advised.

    In situations where it is done by paralleling the caps with resistors
    controls the (DC) voltage division.

    Sorry, but I never went down this road so I don't know the criteria for
    selecting the resistors other than the basic voltage division. I would
    think that the resistor current would have to be selected as a ratio of
    the ripple current but that is only a guess.

    Hope this helps ....
  8. Eeyore

    Eeyore Guest

    You mean you got 35V without the caps ? That's because it was just rectified ac.

    The caps will charge up to the peak of the recified ac's votage which will inded
    be ~ 50V in this cae.

  9. ehsjr

    ehsjr Guest

    Have you considered the ripple voltage? Assuming you
    are using 60 Hz AC, it will be about I*.0083/C where
    I is the current you draw from the supply in amperes,
    and C is the value of your filter capacitor, in farads.
    For example, assume you draw 1/2 amp from your ~50
    volt supply, and have a 2200 uF 63 volt cap. Your
    ripple voltage will be .5*.0083/.0022 or about 1.89
    volts. And rather than using the nominal 2200 uF
    figure, you should look at the the +/- spec on the
    cap and use - spec to compute the lowest value the
    cap might be.

    There's a whole lot more to consider than the above
    when designng a power supply. This site may help:

  10. jasen

    jasen Guest

    are they the same capacitance?

  11. Here is one way to calculate parallel "ballast" or "bleeder" resistors
    for a DC application of 2 series caps:

    First you need to know the approximate worst case leakage of the
    capacitors in question.

    Then take the following worst case conditions:
    - The maximum voltage rating of one of the caps
    - Assuming one cap has maximum leakage and the other has zero leakage

    i.e. The capacitor with the leakage with drag the mid rail voltage away
    from its nominal half rail, creating a greater voltage across the cap
    with no leakage. You don't want to have more than the maximum capacitor
    voltage across the non-leaky cap.

    Once you assume these worse case conditions then the circuit is easy to

    Each resistor is:
    R=(CapLeakRes * (MaxCapVolt-(Vrail/2)) / (Vrail-MaxCapVolt)) * 2

    The leakage will be voltage dependant, but this simple formula is a
    good starting point.

    Dave :)
  12. Alan B

    Alan B Guest

    Assuming you are using a DC voltmeter: you are measuring average voltage.
    assuming you are measuring the output of a rectifier: an unfiltered
    rectifier will have a lower average voltage than a filtered rectifier.

    Now, about ratings. When designing circuits, it is always a good idea to
    use a process called "derating." Derating is the use of components with a
    rating comfortably in excess of the actual running conditions of the
    circuit. In this case, with a filtered DC output of 50V, good derating is
    the use of a capacitor rated at 75-100VDC.
  13. Ken O

    Ken O Guest

    yes you are right, I just calculated that I needed at least 2700uF and
    approximately 75v, but I'll go to 100v to make sure. That means that the
    capacitors I used are know blown up. ..

  14. Guest

    Yes, two 25V caps in series will give 50V operational overall but cap
    total cap will be 1/2.
    I'll also add a 470K resistor across each one to ensure that each one
    share the voltage equally.
    Steve Balstone stephen balstone
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