Use of Capacitors

Discussion in 'General Electronics Chat' started by dalydir, Sep 7, 2012.

  1. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    This is my first time posting and I hope this is rather simple.

    Here is my example:

    Take a 220pf capacitor, of 20kv.

    Now, from what I've read that means something along the lines of, that the capacitor's potential (voltage) will increase at a rate of 1 volt a second when 220 trillionths of an amp is applied to it.

    Here's the issue I haven't been able to resolve:

    Does this mean it would take 20k seconds to fully charge aforementioned capacitor?

    All measurements I've read regarding charging capacitors do NOT reference the voltage of electricity being applied to it. I'm not looking to charge this thing with extreme amperages (I want the result to be exceedingly low amperage, as this is a pretty high and potentially dangerous voltage).

    Does the voltage applied to the capacitor hold significance? Or is it just the amperage? And what if the amperage is like .05 amps (much greater than this one is rated for in farads)? What difference do these things make?

    Why?
    This is an experiment and has nothing to do with an existing circuit and is also my attempt at understanding capacitors better than what I've been able to ascertain on the internet. I'm not necessarily going to DO this or use a similar capacitor to what I've described, as I need to understand this better first. Its not that I don't entirely understand the equations that I find, but I'm unable to find full examples of the equations in actual numbers and explanations of variances (variances of amperages and or voltages of the input voltage).

    Thank you for any help
     
  2. fila

    Member

    Feb 14, 2011
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    Time to charge a capacitor depends on the time constant of the circuit. It is usually in the form of

    τ = R * C

    where τ is the time constant, R is resistance and C is capacitance. In theory you can charge the capacitor instantly by directly placing a voltage supply across it's leads (don't do this).

    If you know the capacitance and voltage across the capacitor you can calculate the charge

    Q = C * U.

    So if you put 20 kV across your 220 pF you will get

    Q = 4.4 μC.

    That is maximum charge you can accumulate with your capacitor. Higher values will create electric fields that will damage the capacitor.
     
  3. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    Thank you, I think I'm starting to understand it better.

    But what I'm curious about is something like this:

    20kv 220pF capacitor, lets say I apply 1.5 volts at .05 amps.

    What would be the dynamics of THAT? Such as time to charge it and would such a variation damage it? Or what if I altered the volts and or amps, how would I take into account such changes and its effect upon the capacitor?

    And when you say "if you know the capacitance and voltage across the capacitor" are you referring to what the capacitor is designed to do or are you referring to what I have actually charged it with?
     
  4. ramancini8

    Member

    Jul 18, 2012
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    I=C(DV/Dt) You know DV and C, so the equation reduces to I=k/Dt. Do not place the full voltage across the capacitor I will be huge and may destroy the capacitor.
     
  5. fila

    Member

    Feb 14, 2011
    64
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    You should read Vol. 1 -DC, chapter 13 of the e-book http://www.allaboutcircuits.com/vol_1/index.html

    Post a circuit schematic. The behavior depends on the arrangement of circuit elements. But if you really want some calculations...

    So I = 0.05 A through the capacitor all the time.

    Generally I = C(dV/dt). So if I is constant as we assumed and initial voltage across the capacitor is zero you get

    I = C(dV/dt)
    I*dt = C*dV
    ∫(I*dt) = ∫(C*dV)
    I*t = C * V

    For V = 20 kV we get t = 88 μs. So it will take 88 μs to charge up to 20 kV. Sounds scary. I will have to double check this. :)
     
  6. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    Can you please define each variable you have specified, I want to ensure I'm understanding what your equations are referencing. I've seen so many people use so many variables with different meanings I don't want to misunderstand it.
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    The voltage and the current are not independent, that is, you cannot arbitrarily choose both. You choose one and the other will set itself depending on the circuit configuration. In fact, the other parameter is not a constant and will change over time.

    Let's take some examples, and I will use simple numbers and units in order to keep the math simple. Suppose you have a constant 1A source charging a 1F capacitor that was completely discharged. The voltage will rise equally and steadily with time.
    After 1 second, you would have supplied 1 coulomb and the voltage across the capacitor would have risen to 1V.

    Now let's look at the other scenario. Suppose you had a 1V source and you connected it to a 1F capacitor that was completely discharged. The current would be infinite because there is nothing to impede the flow of current. In practice, the initial current would be large until about 1 coulomb was transferred at which point the voltage across the capacitor would be 1V.

    So let us put a 1Ω resistor to impede the current and slow things down. Initially the current would be 1V/1Ω = 1A and the voltage would be 0V. Over time the current will gradually diminish from 1A to 0A and the voltage across the capacitor will gradually rise from 0V to 1V.

    How long would it take to charge the capacitor to 1V? In theory, it would take forever because the current is gradually diminishing to zero. The voltage does not rise equally with time. It starts off rising quickly and then starts to slow down. If you were to draw the graph of the voltage over time, it would look like the opposite of an exponential function (or the inverse of a slow decay).
     
    Last edited: Sep 8, 2012
  8. fila

    Member

    Feb 14, 2011
    64
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    Nice explanation from MrChips.

    Here is a youtube video
    http://www.youtube.com/watch?v=IvFVu7Jxa2I

    First graph that is shown is the charge. But because the voltage is proportional to charge you can view it as a voltage graph. The other graph shows current. As the capacitor charges up it builds voltage up and counter-acts the voltage supply EMF thus decreasing the current. New charge that wants to come to the capacitor plates has a hard time overcoming the force of the charge that is already on the plates.
     
  9. fila

    Member

    Feb 14, 2011
    64
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    C - capacitance
    I - current
    dt - small time interval
    dV - voltage change
    ∫ - integral symbol ( I don't know if you know how to do integrals? )
     
  10. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    Thank you guys. I am reading that chapter now. And MrChips, thank you as well. You've given me some things to think about that I had not considered. Though I was starting to see what you're saying (using a resistor).

    Thus far, the electricity text book is just a review for me, but I'm sure that'll help still.

    I think my problem, atm, is inductive vs deductive reasoning, though I'm not sure. (Sadly, I think this is a common issue when I'm trying to understand things in forums).

    Lets say this:

    I hook up a 850mah battery, 1.5 volts to 20 - 20kv 220pF capacitors that are in parallel.

    So I would have 1.5V going to each of these 20kv 220pF capacitors with a current of approximately (assuming resistance in the wire is negligible), .0425Amps.

    I know that one must consider amps and volts, but in so many examples I'm reading, one or the other appears to be left out of the equation. Or is the best way to look at it to simply look at the wattage? (Volts * Amps = Wattage)? When someone excludes Amps in the equation for a capacitor, does that mean the amps being SUPPLIED to the cap is irrelevant? If someone leaves out Volts in the equation, does that mean AMPS being SUPPLIED to the cap is irrelevant?

    EDIT: When I say "irrelevant" I mean both in time to charge, but ALSO, in whether the cap can handle it without destroying itself.

    My goal is to somehow make (preferably) millions of volts come out of my circuit at a VERY LOW amperage (don't want myself or anyone else to get hurt). Using capacitors seems to be the easiest/cheapest way to achieve this. Like converting a normal 1.5V 850mah AA/AAA battery into millions of volts at a VERY LOW amperage.
     
    Last edited: Sep 7, 2012
  11. fila

    Member

    Feb 14, 2011
    64
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    I see what is confusing you. Read this

    http://forum.allaboutcircuits.com/showthread.php?t=69757

    The 850 mAh doesn't stand for current. That is not current. That is stored charge inside the battery. Depending on the circuit and the wiring the circuit elements are going to consume some amount of current.

    Say you know all of the currents in circuit. You sum them all up

    Total current = I = Ʃi

    where i is one of the currents and you get for example

    I = 50 mA.

    If you divide 850 mAh with 50 mA you get 17 hours. Your battery will last 17 hours. For some other circuit configuration I can be 100 mA. Now this time the battery will last 8.5 hours.

    As for your million volts. 1.5 V can't create million volts with a capacitor. You can't even create 1.6 V. You would have to use transformers to bring the voltage up. My advice to you - don't mess with high voltages unless you know what you are doing!

    EDIT: I see now what you want. You want to charge up the 20 kV capacitors and arrange them in some way to get million volts. Right?
     
    Last edited: Sep 7, 2012
  12. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    Regarding your edit, Yes, that is correct Fila. That was how I was looking around to doing it. What are your thoughts on that (the transformer idea aside)?

    And yes, I don't intend to "play" with high voltages without several safety measures taken (one of which being to keep a distance from the charged device once I've turned it on), but is also why I want VERY LOW amperage. I've heard stories of electricians and the like working with VERY high voltages with VERY high amperages, having tons of equipment to protect themselves etc, and, at least several feet away from a power source/wire/etc, having the electricity jump through the air and knock them back and leading them to be happy to even be alive (I haven't heard many of people dying from such arcs without physically contacting things, though I'm sure those stories simply have not reached me yet).

    Your battery info has reminded me; you're right on that. I was just kind of using that as the basic start of what I wanted to do. Though, admittedly, I have obviously forgotten that the circuit itself will determine the amperage coming from the battery over time.

    I am not set in using a 1.5v AAA battery with 850mah either, btw. Its just an example that I can wrap my brain around to extrapolate on other things.
     
  13. fila

    Member

    Feb 14, 2011
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    Well because you can get yourself killed (do you know about Murphy's law?) when playing around with high voltages I would advise you to think again about what you are doing. The idea sounds very interesting but I can't give you any advice or suggestion because I don't have any experience with high voltages and I think it would violate the Terms of service of this site. You are making this thread kinda dangerous talking about millions of volts. Right? :)
     
    Last edited: Sep 7, 2012
  14. dalydir

    Thread Starter New Member

    Sep 7, 2012
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    Well at extremely low amperages, it is my understanding that such "dangers" would not exist. Either that or they would not exist in a greater amount than wrapping one's arm around an (obviously) grounded copper water line, while lying on a floor and touching a live 120v wire....I did that, wouldn't recommend it, and didn't realize the circuit had the wire live when it was not in use (in fact I was quite sure it was NOT live, though it wasn't my design)...didn't kill me or harm me, though it gave me a jolt I have no intention of recreating. Also been shocked by a dysfunctional gas cooktop before, the ignitors (which I've read are easily many thousands of volts DC but low amperage). It was dysfunctional in that the arc ignitors on it were perpetually ready to discharge and should NOT have discharged when I moved my hand near them. Admittedly, it hurt REALLY bad, though briefly, and was a different hit than the aforementioned one.

    I'm not listing my "qualifications" of what some might refer to as "stupidity", but rather that I have inadvertently had experiences whereby I am certainly aware of considerations I should be making with regards to this situation and that, in consideration of that, all the more reason why I want VERY LOW amperage (probably referable to as "pico amps" or something like that).

    If you think giving me any information would result in a violation of the TOS, I understand. I have read them over and don't believe that it would on your part or my part. If I wanted to do something particularly dangerous I'd get some silverware and stick into an electrical outlet, though, I doubt that would kill me, its not something I want to experience either. When thinking things through, I'm actually quite paranoid about being shocked and usually take more than the necessary precautions to avoid such.

    EDIT: I'm not looking for advice "on" high voltage. Just advice on how to create it, such as I've been saying; using a small 1.5v 850mah battery and converting that small amount of wattage into around 1+ million volts. I'm talking about around 1 watt of electricity. The human brain itself is theorized to be only around 20 watts, and the jolts I've been hit with certainly exceed that wattage.
     
    Last edited: Sep 7, 2012
  15. vk6zgo

    Active Member

    Jul 21, 2012
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    If you charge a capacitor from a 1.5 volt battery,all that capacitor will ever have across its terminals is 1.5 volts.

    You could leave it there for a week,then disconnect the battery & measure the voltage across the terminals,& it would still be 1.5 volts.

    You suggested charging 20 capacitors.
    If you did so,then connected them in series so that all the 1.5 volts on them added.all you would get is 30volts!

    The voltage rating of a capacitor is simply the highest voltage that the cap can be used safely at---it is NOT some kind of magic magnification figure which will increase the voltage at the terminals above the original charging voltage.

    It is many years since I had occasion to use the formula Q = C * V.,
    but the "V" ("U" in your case) is the applied voltage,not the capacitor voltage rating.

    For instance, a 1uF 25v capacitor charged by a 1.5 volt battery contains the same amount of energy as a 1uF 25000v capacitor charged from a 1.5 volt battery.

    If it was different,it would make it extremely difficult to choose capacitors!:D
     
    shortbus likes this.
  16. fila

    Member

    Feb 14, 2011
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    OK. Well vk6zgo made a point by saying

    If you are using a 1.5 V battery you can charge any type of capacitor with just 1.5 V. That is all the voltage you can put on it. In order to reach 1 million volts you would have to put 666 667 charged capacitor in series. That isn't practical.
     
  17. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    So far it seems you are talking about the amperage of the charging circuit. Apart from the fact that you need to at least overcome the leakage current in the capacitor, which for such voltages would not be very small, there is the other side of the problem. Once you get a significant charge stored in a capacitor, it can dump said charge in a very short time leading to current of many hundred amps.
    So you need not only small charging current, but you also need to make sure that the discharge current cannot be deadly as well, for example with a series resistor bank.

    I still don´t understand why do you want such high voltage, it seems you want to have high voltage for the sake of having high voltage. Do you realise how big isolation distance you need to keep 1MV from arcing? In dry air you need at least 1m of distance to other parts around the charged conductor, humans should be kept at least 5 meters away.

    I suggest you look on youtube at photonvids´channel and see what a few kV can do, not to say a few MV.
     
  18. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    And once you price 20kV caps you will want to find a new hobby. :) If you just want high volts to make a spark, look into making a Van de Graaff generator - http://www.google.com/search?q=van+...ox&ie=UTF-8&oe=UTF-8&sourceid=ie7&rlz=1I7GWYE Or Wimshust machine - http://www.google.com/search?q=wims...ox&ie=UTF-8&oe=UTF-8&sourceid=ie7&rlz=1I7GWYE

    Both of these can be bought or home made. The Van de Graaff was even used to make high voltages during the research of splitting the atom. And from personal experience they can be made and are fun to play with.
     
  19. JMac3108

    Active Member

    Aug 16, 2010
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    Dalydir,

    One comment first ... you do understand that the voltage rating on a capacitor is only a measure of the maximum voltage it can have applied across it. Right? It has nothing to do with how fast it charges or anything like that. For example, if I'm designing a 12V circuit and I need a filter cap, I need to use a capacitor with a voltage rating greater than 12V. If I used a 6.3V rated capacitor is would fail.

    As far as your other questions, I think you are confused about the basics of charging a capacitor. The answer to your question depends on how the capacitor is charged. It could be charged through a resistor from a constant voltage source. Or it could be charged with a constant current source. Or a voltage could be directly applied to it. In each case the results are different and you have to apply the equations that others have been giving you.
     
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