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Charging / Discharging Capacitors In Series

  1. Jan 24, 2010 #1
    I have 23 450V 1600uf electrolytic capacitors and was wondering if I could use them in a series? If I were to put them in a series, would they act like a single 10,350 volt 72.727 uf capacitor? Or would they melt/fail/leak/explode if I tried to use them that way? Any information is helpful! Thanks!
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 25, 2010 #2
    Place equal valued resistors across each capacitor to equalize the voltage drops. Buy capacitors all from the same lot to mimimize variation in capacitance. Dynamically, the smallest capacitor will develop the greatest voltage. So, no, you don't get the full 23x450 volt V_max. It's reduced by the variation in capacitance. The leakier capacitors will leak their charge on to the less leaky. This is what the resistors are for. They help make each capacitor-resistor pair equally leaky.
  4. Jan 25, 2010 #3
    Would this be correct?


    Attached Files:

  5. Jan 25, 2010 #4
    That would do it, except that you've still quoted the Vmax as 1800V for 4 capacitors. Capacitors can vary wildly in actual value and as a result the voltage partioned between them will vary just as wildly. This is especially true of electrolitics.

    If you're just making one or only a few of this design, you can determine the voltage each capacitor will obtain simply by charging the string to a lower voltage test voltage--say 30 volts and measure the voltage drop across each. The one with the greatest drop will limit the maximum value you can place across the string.

    In your drawing you've placed resistors across each capacitor. But what are their values? Selecting the required value depends upon the leakage current through each capacitor. This is not something easily measured. You're also interested in the leakage current at the maxium temperature in which the capacitors will be operated for your calculations. Capacitors are leakiest when hot. If it's going to get hot you will need a good fudge-factor that will reduce the value of the resistors or test in an oven.

    So look up the leakage current in the component datasheet. You will probably get only one value quoted over operating temperature range.

    Alternately you can discover the value emperically. Warm them to max operating temperature, place a test voltage across the string of capacitors without the resistors. Measure and write down the voltage across each capacitor. Write down the time power was applied. Wait. Measure the voltage again and write it down. Q=CV. Q=current times seconds. I = CV/t.

    We don't really care about the total leakage current, but the maximum difference in leakage between any two capacitors.

    One thing you should be aware of, is that the leakage variation becomes balanced by the resistors because some capacitors develop a greater voltage across them than others. This becomes another reduction in the maximum voltage you can apply across the string.

    Finally, I don't know right off how large the resistors should be whether you obtain data from measurment or from a datasheet. So if you can come up with leakage data, I could calculat your required resistance values, or supply the relevant equations.

    I think you need at least 5 capacitors, btw.
    Last edited: Jan 25, 2010
  6. Jan 25, 2010 #5


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    Working at these voltages and high energy capacitors is dangerous. You really need to understand the risks and proper handling of high voltages before proceeding.
  7. Jan 25, 2010 #6
    BTW, Berkeman who is the gatekeeper of this PF science folder, should show up about now with warnings about high voltage. Don't experiment with it if you don't know what you're doing. Since he hasn't, I'll try to help out. If you are not familiar with the hazards of high voltage, obtain the help of someone that is. If you complete a circuit with your hand wrapped around a hot point, you WILL NOT be able to let go. Your muscles will not respond.

    Berkeman is so much better at this, but it's the best I can do.

    Also, I belatedly notice that you have something you call an NTS. I assume this means Neon Sign Transformer. If this is the case, you cannot charge capacitors directly from a transformer. You will need a rectifier. Additionally it appears to be a 120 VAC to 1800 VAC transformer. If this is the case, the voltage developed across the string of capacitors will be more than 1800 volts. It will be 1800*sqrt(2) volts or about 2500 volts.

    What is the application for this design?
    Last edited: Jan 25, 2010
  8. Jan 25, 2010 #7
    Thanks to all for the warnings, but I am aware of the dangers of high voltage and high current though a friendly reminder never hurts. I also realized my mistake of not adding a rectifier but I had already posted and was too lazy to edit the image. The application for this design was for an experiment that causes water to explode. If you run a voltage of more than about 3KV through water it will allow a large current to flow through. If enough current is passed through, the water essentially detonates. Since my capacitors were only rated for 450 volts, I thought that putting them in a series would allow me to get the voltages I need but I didn't try it because I didn't want to damage my capacitors.

    Phrak, thanks for the great answers, but I am somewhat unclear about a few things. The first thing I am unsure about is the relationship between maximum voltage and leakage current. I think what your saying is that when the capacitors are charging, some leak more than others, which means that some get to their max. voltage before others reducing the amount of energy that can be stored. So to handle this problem we must balance the leakage so all of the capacitors charge up at the same rate. This seems to make sense but I just want to make sure. The second I didn't understand was what you meant by measure the voltage drop when you said, "If you're just making one or only a few of this design, you can determine the voltage each capacitor will obtain simply by charging the string to a lower voltage test voltage--say 30 volts and measure the voltage drop across each. The one with the greatest drop will limit the maximum value you can place across the string." Do you mean put them in series, charge them up to voltage x, and then measure the voltage across the terminals of each? This may seem like a stupid question, but I still have a lot to learn.
  9. Jan 25, 2010 #8
    OK. You are making just one thing, and you intend to discharge it almost as soon as it is charged. These are good things to know. It makes things simpler.

    Leakage current will overvoltage the strongest capacitor in a series over time. This is so because the leakage from the rest will continue to charge the one that leaks least. This is something that happens over time. Since you want to apparently discharge within a few minutes, it may not be a problem. But it still bothers me. Look up the leakage current for your part numbers in an internet search.

    If your capacitors are labeled with a WVDC (it means maximum working voltage maintained at the labeled valued") derate them 50 volts. If they say 450 WVDC don't take them over 400 volts.

    Next, your capacitors should inform you of the actual capacitance you can expect for any one given capacitor. It should say something on the case such as "72 uFards +50%, -10%." What does it say on the case?
    Last edited: Jan 25, 2010
  10. Jan 25, 2010 #9


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    Okay, I'll bite. Why do you want to make water explode? You part of the Mythbusters team? An applicant?
  11. Jan 27, 2010 #10
    Hello everyone,
    I wasn't able to get to a computer yesterday so I wasn't able to respond but I'm on one now, so I'm responding.

    I cannot seem to find any information on my capacitors anywhere. I've attatched some pictures of one and here's the text since its hard to read:

    160 W/S 450VDC
    65C 84-11

    I've tried entering all of the numbers along with the word mallory as well as mallory capacitor but there doesn't seem to be any documentation. The information I had came from the product description on ebay where I bought them. Also thanks for explaining the leakage current and why its important because I now understand.

    I have no affiliations with Mythbusters and do not plan to in the near future. I came across this experiment a little while after I bought my capacitors and decided it would be fun to try. Don't worry, I promise I won't hurt myself or anyone else.

    Attached Files:

    Last edited: Jan 27, 2010
  12. Jan 27, 2010 #11
    I can't determine the capacitance from those numbers. How do you know they are 1600 uF?

    Duracap manufactures Mallory EAF type capacitors. I would drop them an email with the case marking information and ask them if they can send you information on the capacitance value and tolerance. It can't hurt.
    http://www.duracap.com/photoflash_service_capacitors/photoflash_service_capacitors.pdf" [Broken]
    Last edited by a moderator: May 4, 2017
  13. Jan 28, 2010 #12
    Hey Phrak, to answer your question, I bought them from ebay and the capactiance and voltage was in the product description. Also, Great idea about sending them an email and I'll try doing that tonight. Thank you for all of the information and help so far!
    Last edited by a moderator: May 4, 2017
  14. Jan 29, 2010 #13
    I'm suspicious of the quoted 1600 uF value you were quoted. So I scrolled around the internet, but came up empty.

    The labeling as you've given it,

    160 W/S 450VDC
    65C 84-11

    seems to say you have a 450 volt capacitors and they are guaranteed to deliver energy to a flash bulb at the rate of 160 Watts per second and are serviceable up to 65 degrees centigrade. I would expect something in the part number, 602-11264-01, to suggest 1600, like "162", but it's not there.

    Ebay--the buyer beware. Neither the distributors Digikey nor Mouser lists 602-11264-01 in their catalogs.

    If you want test the claim, you can use a charging resistor, a known DC voltage, a clock with a second hand, and an DVM to determine the capacitance.
    Last edited: Jan 29, 2010
  15. Jan 30, 2010 #14


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    "deliver energy to a flash bulb at the rate of 160 Watts per second "
    A Watt is one Joule per second - could that be 160W was meant?
    Watts per second would be the units for rate of increase in power with time.
  16. Jan 30, 2010 #15
    Opps. that doesn't make any sense does it?
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