Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitance by energy or time current product?

  1. Jan 4, 2012 #1
    I want a capacitor bank capable of transferring a 350 kJ, 500 ms pulse. Or, rather, a bank capable of supplying 70 A @ 10 kV for 0.5 s. I have located a 2000 μf 10kV capacitor which looks like it might do the job. Unfortunately they are very expensive ($1500 USD) and I can only afford 2 or maybe 3 of them. The problem is I can't seem to figure out whether or not 2 of those capacitors would be enough. The energy method seems to result in twice the value of the time current product method.

    Let C = capacitance in Farads, t = time in seconds, I = current in Amps, V = voltage in Volts, E = energy in Joules, and Q = charge in Coulombs.
    t = 0.5 s
    I = 70 A
    V = 10 kV
    E = 350 kJ
    [tex]
    Q = CV \hspace{1 cm} Q = tI \hspace{1 cm} \rightarrow \hspace(1 cm}tI = CV \hspace{1 cm} \rightarrow \hspace{1 cm} C = \frac{tI}{V}\\
    C = \frac {tI}{V} = \frac{(0.5)(70)}{10000} = 0.0035 = 3500\μF\\
    E = \frac{1}{2}CV^2 \hspace{1 cm} \rightarrow \hspace{1 cm} C=\frac{2E}{V^2}=0.007=7000\μF
    [/tex]

    So which one is correct?

    Edit: Hmm. The latex isn't working. Can anyone see what I did wrong with the latex? I'll write out the equations without the latex.

    Q=CV; I=Q/t; Q=tI; tI=CV; C=tI/V
    C = tI/V = (0.5)(70) / 10000 = 0.0035 = 3.5 mF or less than (2) $1500 capacitors
    E = 0.5CV^2; C = 2E/V^2 = (2)(350000)/10000^2 = 0.007 = 7 mF or (4) $1500 capacitors, which is over my budget

    If the energy equation result is correct then I have a problem. If the time current product over voltage equation result is correct then I don't have a problem.
     
    Last edited: Jan 4, 2012
  2. jcsd
  3. Jan 4, 2012 #2

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    i think answer will show if you work it in your head one millisecond at a time.

    a cap charged to 10 KV will deliver current when connected to a laod
    but as soon as it starts to discharge, its voltage decreases.

    so the first bit of current is delivered at very high voltage

    the next at lower voltage
    the next at lower voltage yet

    so your energy available works out to only half CV^2

    yes you can get all the coulombs out of the cap Q=CV
    but at steadily decreasing joules/coulomb.

    at what voltage will your load stop accepting current? that's the endpoint for your energy transfer. 1/2C * [v1^2-v2^2]
     
  4. Jan 4, 2012 #3

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    @metiman
    jim is making good sense here. Don't leap in with a computer simulation when the basic problem is only two lines of 'proper' Maths! Sledgehammer to crack a wallnut, I think.
     
  5. Jan 4, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Just checking, before you invest heavily. You do realize that there is more involved to producing a heavy pulse than getting enough farads alone? The series resistance, and inductance, of the capacitor impose serious restrictions on the speed at which it can discharge. Instead of one giant cap of 2000 μf you may consider whether performance may be better if you were to parallel 45 caps each of 47 μf, assuming they are obtainable. I'm not talking specifics, but you need to investigate this. Also, a large cap will get hot if you want to charge-discharge it repeatedly and in quick succession. It goes without saying that it would be cheaper to replace a small cap if you manage to destroy it, rather than a single big one. Something about eggs and baskets.... :smile:
     
    Last edited: Jan 4, 2012
  6. Jan 4, 2012 #5
    I get that the potential energy in the capacitor, the voltage across its plates, and the charge stored on its plates all start to decline after t=0, but I still don't understand what is going on here. In both cases you should be able to account for the unused portion at the end of the pulse. In the charge-current equation, perhaps by lowering the initial quantity of stored charge to account for the amount you can't use. In the energy equation, perhaps by increasing your potential energy requirement to account for all the wasted energy. With both equations it looks like you can account for the unused energy. So I still don't see why each equation gives me a different capacitor value, and exactly by a factor of two as well. In the charge-current equation capacitance is inversely proportional to the voltage across the plates. In the energy equation capacitance is inversely proportional to the square of the voltage across the plates, but it also has a much larger value in the numerator. For simplicity sake and for the purpose of comparing the two equations, it might be better to assume that I can make use of the full potential energy of the capacitor all the way down to zero volts and zero joules.

    I think the reason I am getting different answers is because I am asking different questions. I need to understand how the questions are different. So I am going to try to look at each question in greater depth. One equation relates capacitance directly to stored charge and indirectly to a constant current for a given amount of time based on the idea that a given charge Q on the plates can only sustain a particular current until the charge from the plates has been used up. The other equation relates capacitance to the total potential energy stored by a capacitor and the voltage across its plates. With both equations a larger voltage allows for a smaller capacitance. In the energy equation a larger potential energy requirement results in a larger capacitance and this potential energy is derived based on the input power of the load and the length of time that power is required. In the charge-current equation a larger time with a fixed current or larger current with a fixed time requires a larger capacitance.

    I was interpreting tI = CV to mean that a capacitor of of 3.5 mF would be capable of discharging 70 amps of current for 0.5 seconds from a peak plate voltage of 10 kV. But maybe it just means that that is what would happen if I could somehow keep the plate voltage at 10 kV during the entire discharge, which is almost doable with a pulse forming network at the expense of a high bandwidth square wave pulse shape. The more I think about this equation the more I think that I am somehow misusing it and it probably has something to do with the continually decreasing values of V and Q. Capacitance is just a simple ratio between Q and V. Once the charge starts to move you can substitute tI for Q but that doesn't necessarily give you a prediction about how I changes over time as Q and V decrease. Still, theoretically, you could place the capacitor in a pulse forming network and keep the voltage drop to a minimum throughout most of the pulse. Especially with such a relatively long pulse. If you assume the presence of sufficient inductance to keep V from dropping until the end of the pulse then it seems like this equation should be valid because V would be more or less constant for the network as a whole.

    But can you really substitute tI for Q? Both equations represent definitions. The first for capacitance and the second for current. So clearly each is true individually and no matter what kind of crazy electrical machine you design current is always going to be flow of charge per unit time and capacitance is always going to be stored charge per unit voltage. Defining charge as time current in amp seconds does seem kind of strange, but as long as you have some kind of moving charge it would seem to be valid. Before the charge starts to move you don't really have a current, but as soon as the plates start giving up their charge you do have a (transient) current. So it does seem like you should be able to substitute tI for Q after t=0. Still I can't help thinking somehow that these equations were just not meant to be used this way.

    It's clearly true that for a given stored charge on the plates that t and I will be inversely proportional. The greater the discharge time the smaller the possible current and the greater the current the smaller the possible discharge time. I think it is that relationship combined with the fact that capacitance for a given charge Q on the plates is inversely proportional to the voltage across the plates due to the definition of capacitance as Q/V which gives the first equation its meaning in terms of my application. But does this inverse relationship of t and I with respect to Q account for the fact that V, and therefore I, is not going to be a constant? I don't think it does account for that. The relationship between tI and Q is too simple. For this reason I am thinking that the other equation may be better for a continually changing voltage. But if you assume a constant voltage due to a PFN then this equation should work just as well as the other.

    It's true that V is inversely proportional to C for a given value of Q and that t and I will always be inversely proportional for a given value of Q, but does it necessarily follow from those relationships that I can determine I for a given t or t for a given I? As Jim pointed out, after t=0, V is continuously dropping. As V drops so does Q. As Q drops so does the tI product. I suppose you could re-evaluate Q for each change in V that you want to examine using Q=CV. I guess Q would sort of be analogous to potential energy in the other equation. In the other equation what you really want to know is not the total potential energy, but the total potential energy that you can actually use. In this equation you want to know the total charge that you are actually going to use. The equation relates capacitance either to the product of discharge time and current produced from the discharging capacitor and the decreasing voltage across the plates or to the product of charge time and the current produced by the voltage source during the charge process with increasing voltage across the plates. Presumably these two scenarios are equivalent and the time t to charge is the same as the time t to discharge. The equation essentially predicts the charge/discharge behavior of a capacitor in terms of the quantity of charge on the plates, the magnitude of the electric field across them, and the time it takes for the plates to lose their charge when connected to a load. Electric potential energy is not explicitly mentioned, but it is there nevertheless, hidden in the voltage, which is just potential energy per unit charge in joules per coulomb.

    The other equation relates capacitance directly to the electric potential energy stored in the capacitor to the voltage across the plates. It also derives from Q=CV but not quite so directly and is analogous to the potential energy equation in mechanics. Because it is derived so directly I would tend to trust it more, but it's not giving me an answer that I like, and I still don't really understand why that answer is so different from the other equation.

    Ultimately what I want to know is how long will the capacitor be able to supply a significant amount of current and voltage. Both equations seem to provide an answer. Both seem to be valid, although the charge-current equation seems to be better for a static voltage. In both cases you can account for the diminishing voltage from the capacitor by just evaluating the equation at multiple voltages. I wish I were still an EE major. These are exactly the sort of questions I used to love torturing my poor professors with.
     
  7. Jan 4, 2012 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    There was one instance where you used a round bracket instead of a curly one.
     
    Last edited: Jan 5, 2012
  8. Jan 4, 2012 #7
    Thanks for the advice. I will consider those issues as well, but for now I am just trying to get an order of magnitude estimate for the amount of capacitance needed. I don't think the inductance of the capacitor will be too much of a problem because, at 500 ms, my pulse length is so long. It will still have to be measured and taken into account, but I don't really see it as a limitation. As for the ESR, I intend to measure that as well and take it into account. The danger there is that my current (70 amps) might dissipate too much power and overheat (or melt) the plates. One of the advantages of using such a high voltage is that the resistive power losses are much reduced for a given conductor size. I'd like to go to an even higher voltage if I could. Actually my original plan was to go with 15 kV. I'm only using the value of 10 kV due to the commercial capacitors I found. My duty cycle will only be something like 11% at the most. My target is a half second pulse about every 4 seconds. So it will have some time to cool. The enormous General Atomics capacitors I'm looking at have sturdy metal cases with castor oil inside and look like they could be water cooled if they got too hot.

    The capacitors I'm looking at are so expensive that I am seriously considering the idea of making my own out of copper sheeting and mica sheet dielectric. I was thinking of something along the lines of attaching copper sheeting to an entire wall of my house, gluing mica sheeting to that and then gluing another layer of copper to that. If I made my own capacitors I would consider going with a really high voltage. Maybe as much as 100 kV. I'd have to look at different mica thickness availability and its dielectric strength as well before deciding on an optimum voltage with safety factors added in for the inconsistencies of amateur construction. Copper will be expensive and comparatively heavy, but it has a lower resistance resulting in lower ESR and it has a lower thermal resistance making it easier to cool. You can also attach leads to the plates by just soldering them.

    If the numbers from the energy equation are valid then it looks like I'm either going to have to make my own capacitors or attempt to draw 700 kW pulses directly from the power company transformer. That would be nearly 3000 amps at 240 volts. So I think learning how to make my own mega-capacitors is the better option. Oh, as far as a bunch of smaller capacitors, I don't have anything against the idea, but it just doesn't work in terms of cost. It would cost more than the entire budget for the project. As far as I can tell those expensive giant capacitors are actually the best deal.
     
  9. Jan 5, 2012 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    Are you going to run into radio frequency interference with these pulses? You might tangle with a govt dept if you are found to be transmitting noise over wide bandwidths.

    I seem to recall a tv doco describing Tesla's UK trans-Atlantic CW station. Amidst his antenna farm he built a barn the height of 3 storeys to house his huge capacitors, copper plates with air gaps, he had them hanging from the rafters. Or something like that. (This will give our history buffs the lead to jump in and correct me, and tell it like it really was.)
     
  10. Jan 5, 2012 #9
    Well if the local government sees that I am creating a lot of RFI presumably they will let me know (and maybe fine me or something), and then I can fix the problem. RFI is tricky. I don't know how to calculate the unintentional creation of radio waves. Many pulsed power applications use very short pulses, often in the nanosecond range, in order to get very high powers but with very wide bandwidths. I'm using long pulses with narrow bandwidth. A 0.5 second pulse only has an inherent bandwidth of 2 Hz. So maybe that will help keep down any RFI to a narrow spectrum. I could also try to enclose all my equipment inside a Faraday cage if necessary.

    That's interesting about Tesla. I'll have to do some reading up on him. He sounds like an interesting guy. I love the idea of 3 story high hanging copper capacitors. Luckily I don't think I'll need quite so much capacitance for my application. It's funny to think of the size of your house as a limitation on your capacitance. I like the idea of a house where the inside and outside walls and ceilings and floors consist entirely of copper-mica-copper sandwich material. That would also give you a nice Faraday cage. As a flooring material I wonder if the mica would crush. Ground mica could probably be mixed up as a kind of thin cement in that case.
     
  11. Jan 5, 2012 #10

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    OOhhhh you're sooo close to the answer . all the necessary thoughts are there in #5. the light will come on soon.

    bravo for contemplating such things and cross-checking equations. that's how one develops an accurate intuitive "feel" for how the world works.


    capacitor : mechanical analogy to a spring

    force of a spring is F = K X d, K = spring constant and d = distance it's compressed.

    let it extend to its full length and its force will be zero. it's relaxed.

    energy in the spring is 1/2 ( k X d^2 ) - shape of that equation look familiar? it's the result of an integral is reason it's got a ^2 of course.

    energy you can extract from a spring is difference between its stored energies at starting compression and ending compression.
    if you let it relax completely, ie ending compression = zero, you get all the energy.

    cap is same

    it starts at 10KV and ends at some other voltage.
    if ending voltage is zero you get the full 1/2 X ( C X 10kv^2) energy
    and the remaining charge is zero.

    if it discharges to only 2 kv
    there's 20% of the charge left in it
    and you extracted 96% of the energy.
    i think. please check my arithmetic.
    i've been making a lot of dumb mistakes lately

    does (10kv^2 - 2kv^2 ) / 10kv^2 = 96% ??

    note how little energy those lowest voltage coulombs carry.
    just like compressing a spring - first little bit is real easy.


    wow you are talking about REAL high energy here. be mighty careful for that cap could take an arm off....
     
  12. Jan 6, 2012 #11
    I do get the spring analogy. If the load cannot make use of the last 20% of the energy then that energy is wasted then that has to be accounted for with a larger capacitor. That isn't what I am confused about. I am confused about why the two equations give different answers when I assume that the load can make use of 100% of the stored energy from the capacitor. The charge-current equation models the loss of charge on the plates and the resulting current. The energy equation models the creation or loss of potential energy resulting from the plate charge. The equations seem to indicate that the amount of capacitance needed to store the necessary energy is twice the amount of capacitance needed just to supply the required current for the required time and that is very weird. For now I'm just going to have to assume the energy equation is the more accurate model and plan to design and build my own capacitors.
     
  13. Jan 6, 2012 #12

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    yep your thinking is straight.

    charge isn't energy.

    charge is amp-seconds, which are named coulombs.
    energy is watt-seconds, which are named joules.

    thought experiment::
    first amp-second might be at 10 kv.
    power is volts X amps
    so first amp-second carried 10KW-seconds of energy into your load.

    Next amp-second might be at only 9kv
    so it carries only 9kw-seconds of energy

    and so on

    the decreasing energy per amp-second is why you only get half.


    you haven't said whether you've had a freshman calculus course
    if so you recognize the integral

    if not, here's a mental process i use that APPROXIMATES integral (it's mathematically impure but we all crawled before we walked)
    if you do that little exercise above 1 millisecond at a time, 500 times for your half-second
    using calculated amp-milliseconds and volts for your capacitor and load
    and added up the resulting watt-seconds
    you'd get a result close to that energy equation 1/2 C X V^2

    but the amp-seconds would add up to C X V

    the smaller you make your time interval the closer you are to the integral, which is mathematically pure.

    too tedious for me on my sliderule
    but if you do excel or Basic it'd be fun to let the computer chew on it a microsecond at a time...


    wow i'm talking way too much.
    over and out.
     
    Last edited: Jan 6, 2012
  14. Jan 6, 2012 #13

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Using the right value of series inductor and appropriate switching, it is possible to use almost 100% of the stored energy, btw.
     
  15. Jan 6, 2012 #14
    I took at least two years (4 semesters) of calculus back when I was an EE major in college, but that was over 2 decades ago and it seems I have forgotten pretty much everything. I am in the process right now of going back and reviewing, starting with derivatives. All I can really recall is that derivatives have something to do with tangents or slopes and rate of change and integrals have something to do with finding the area of an irregular shape. Use it or lose it I guess.:blushing:

    I did notice that in my physics book (Young 8th Ed.) 1/2CV^2 is derived from Q=CV through an integral. That was just one of the steps involved. So basically the energy equation accounts for the continuous drop in voltage as the charge on the plates equalizes through the load. The charge-current equation doesn't account for that and is really only good for one infinitesimally small slice of time during the charge transfer. In which case it seems like the charge-current equation is actually kind of deceptive and perhaps even useless. Is there any scenario where it would provide a useful result? I feel like I really am starting to get it now. Thanks for explaining.

    What if the voltage from the capacitor were held high via a pulse forming network so that the V(t) output were squared off? The first problem I can think of with this is that the simple Q=CV capacitor charge equation would probably no longer be valid. You would then have to account for the behavior of a network with inductors in it as well. If you had the right equations for that network and it was capable of producing a perfect square wave, then maybe you wouldn't need the integral because the voltage would remain constant throughout the pulse.
     
  16. Jan 6, 2012 #15

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    You can't change Q=CV. That's fundamental but you can meter out the energy in a capacitor by other means than just a resistor (switched mode) so that the power out is constant over a period of time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Capacitance by energy or time current product?
Loading...