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Limiting current with capacitors

Discussion in 'Electronic Design' started by DillonCo, Feb 4, 2004.

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  1. DillonCo

    DillonCo Guest

    I need to limit current on "wall power" (115V, 60Hz) to 15A. A resistor
    won't work because of the current and resulting voltage drop. An inductor
    won't either simply because I'd need .02H (hard to get at 15A) and I have no
    way of measuring inductance. I therefore decieded to use a capacitor. The
    problem is that I need about 350uF in a non-polar cap at a reasonibly high
    voltage (something I don't have).
    I tried (as a test) connecting a 4.7uF 250V polarized cap in series with a
    diode (to prevent reverseing), but this resulted in the cap exploding after
    a couple seconds. There was no load (a short). I also connected a diode in
    parallel with a similar cap so reverse current would flow through the diode.
    I
    put two of these in series swith the diodes opposing (so current wouldn't
    just drain through them), and connected it again with no load. This time
    the diodes (both!) exploded after a short time. I even tried two .05uF caps
    in parallel, but one of these exploded when the power was turned on
    (probably just caught the wrong part of the sine wave).
    What am I doing wrong, and how can I successfully limit current?

    Thanks.
     
  2. Guy Macon

    Guy Macon Guest

    If your load is drawing too much current at a given voltage,
    you cannot reduce the current without reducing the voltage
    or reducing the load.
    Unless you are willing to reduce the voltage or change the load,
    you can't.

    Tell us what you are plugging in to the wall power and we may
    be able to suggest ways of making it draw less current.
     
  3. DJ

    DJ Guest

    Direct-off-the-line power supplies using series capacitors are possible. They
    are *practical* for currents up to a few mA, not Amps. Maxim have some chips and
    excellent data and application notes.
     
  4. Robert Baer

    Robert Baer Guest

    Capacitors are definitely *not* the "solution", most especially at the
    large 15 amp displacement current.
    There are damn few capacitors that are rated at that high of a
    displacement current, and the prices go way up as the displacement
    current requirements go up.

    The use of an inductor is a good choice, and 20mH at 15A is not too
    difficult to achive with a gapped E-I core and fair-size wire to keep
    the I*R drop down.
    Measuring can be done (with easily available parts) in various ways:
    make a bridge with 1% resistors and a 2% capacitor (the inductor is the
    4th item) and tweak the resistors for the null.
    There are a number of bridges that one could use, including the Owens
    bridge, which is frequency independent.
    Just be advised the bridge exitation frequency must be below the
    inductor resonant frequency for a meaningful result.
    And chokes *are* available with ratings that (may be) inthe ballpark
    of your needs,so that one could buy off the shelf instead of DIY.
     
  5. ID

    ID Guest

    Could you let me know the model of any such chips that Maxim makes? I
    looked at their website and couldn't find anything on line AC->DC
    conversion.

    Thanks,
    -ID
     
  6. Steve

    Steve Guest

    You're messing with lethal voltages and currents when you apparently
    have little knowledge on the subject.

    Did I hear someone say "Darwin awards contender"?

    Tell us WHY you wish to do this and someone might be able to give you
    better advice!
     
  7. DillonCo

    DillonCo Guest

    What am I doing wrong?
    Hardly. I may have little experience as far as induction is concerned, but
    I'm hardly pluging my tongue into the socket. Whenever I test a circuit
    that involves mains power, I stay on the other side of the room and wear
    safety goggles when I turn it on. It wasn't quite necessary in this case,
    but was probably the closest I've come to actually inflicting damage with a
    circuit. I plan on getting my Darwin award far more spectacuarly, thank you
    very much.
    Why is to make a power supply that can drive loads with little resistance.
    Limit the current so nothing explodes, melts or otherwise damaged. The
    current expectation is to power a flyback transformer which has nowhere near
    the inductance or resistance necessary to be powered directly. It's had to
    say what else it might be useful for at this point, but they would be along
    the same lines (low resistance heating element, for instance).

    However, that is largely just to satisfy the curiousity of those wo may
    wonder, because it seems a large inductor is simply the best choice.
     
  8. I read in sci.electronics.design that DillonCo <>
    You appear to say that you are putting all these circuits directly
    across the mains supply. Of course stuff will explode.
     
  9. Tim Dicus

    Tim Dicus Guest

    Hi John,

    I missed the original post, so I am going to respond here.

    To DillonCo:

    I think what John means is that a device that has any type of impedance with a voltage applied will cause heat.

    In your case, the 4.7uf cap across the 115vac 60Hz supply has a capacitive impedance of about 500 ohms. With the voltage drop of
    115v, that would put the current at about 1/4 amp. That would be about 26va (watts kinda). That is gonna torch just about any device
    you put in there to take its place. It does not matter if the impedance is inductive, capacitive, or resistive.

    The 350uf cap will go into orbit (parts of it anyway) if you put it across the mains! The correct use is to connect the diode and
    cap in series across the power supply and take the power from across (parallel to) the cap, not in series with it.

    The only exception that may work is an incandescent light bulb. Maybe a 25 watt bulb in series should limit the current to about 1/4
    amp. Be really careful with this. If you become part of the circuit, the bulb will not allow enough current through to blow the
    power panel breaker, but enough to blow yours. May I suggest using a GFI circuit (like in your bathroom) on the mains power supply
    when playing with this. Please wear eye protection, like safety glasses, and be really careful...

    Tim
     
  10. N. Thornton

    N. Thornton Guest

    So you dont know what youre doing. Using caps on mains is a recipe for
    electrocution, mainly after the power is switched off. Seriously,
    someone gave you good advice, either learn about mains and safety or
    dont play with it.

    These are not great choices. I used a heater as a series load for
    suspect compressors, and bulbs for lower power items. A large inductor
    will step your mains up to voltages that can jump through insulation
    and turn you into instant toast.

    What youre trying to do wont work, not with any passable amount of
    safety. Learn the dangers and how to work round them or quit while
    youre ahead.


    Regards, NT

    PS Darwin awardees never planned to get Darwinated.
     
  11. dated Wed, 04 Feb 2004 00:42:48 GMT,
    DillonCo, <> says...


    YOU ARE A TERRORIST.
    YOU ARE A TERRORIST.

    YOU ARE A TERRORIST.
     
  12. dated Thu, 5 Feb 2004 11:43:48 +0000,
    HE IS A TERRORIST.
     
  13. Its OK on neutral most of the time.
    LOL, now you design THAT bad;-)?

    Better keep doing it that way!
     
  14. Steve

    Steve Guest

    Fair enough, glad you take it semi-seriously. It's just that when I
    read about you blowing up capacitors by connecting the across the
    mains supply... 'nuf said. If you had the knowledge and had
    confidence in your design you wouldn't feel the need to stand on the
    other side of the room with goggles on!!

    When you say "power supply" do you mean AC at mains voltage (but
    current limited)? That is gonna be hard to achieve.

    Why?! You don't back up any of these statements with fact! A large
    inductor across a mains supply is even MORE dangerous than
    non-mains-rated caps. And it's much worse than blowing aluminium cans
    across the room!

    Understand that an inductor resists a CHANGE in current flow. This
    means that the inductor tries to drive the back EMF to infinity as the
    magnetic field collapses. Across mains it does this every cycle (60
    times a second!)

    Translation; it will self destruct trying to do its job unless
    properly "contained", and in the process it could create incredibly
    high voltages that may arc.

    Tesla coils are basically "large inductors" Go read up about them,
    look at some pictures of them working, note the humungous arcs leaping
    huge distances from the top of the device... read about Teslas ideas
    for creating weapons from his designs... read about the bird that flew
    across the beam of one of Teslas inventions and was instantly
    obliterated (not even a pile of feathers remained!) Read about the
    theoretical link between a Tesla experiment and the Tunguska
    explosion. Read about the Tunguska explosion and understand that
    large inductors are BLOODY DANGEROUS!

    At very least, you should consider using a mains isolation transformer
    and an earth leakage detector. Potentially, neither of these will
    save you from being a Darwin awardee... but it will be a more
    entertaining read when you get extra credit for ignoring warnings and
    killing yourself DESPITE the recommended safety measures.

    niftydog
     
  15. Jim Thompson

    Jim Thompson Guest

    On 5 Feb 2004 15:37:47 -0800, (Steve) wrote:

    [snip]
    [snip]

    What do you think folks... "Dummy of the Month" or "Dummy of the Year"
    Award ?:)

    ...Jim Thompson
     
  16. DillonCo

    DillonCo Guest

    [snip heat calculations]
    I've always been curious, what's the difference between VA and W?
    I'd say so! I'm supprised it lasted that long!
    I was under the impression that non-resistive impediance created no heat...
    Or is that just when a load is connected?
    Actually, I tried this, but as I said, my diodes exploded. Perhaps I'm
    misunderstanding.

    [snip incandesant bulb recommendation]
    GFI's a good idea, even though I ensure the power is off before I do
    anything. Already use safety glasses. Never hurts to be paranoid about
    something potentially deadly though ;).
     
  17. Tim Dicus

    Tim Dicus Guest

    VA is the AC version of watts.
    VA = watts * 1.414
    watts = VA * .707
    All impedance creates heat when voltage is applied. If it creates a voltage drop while carrying current, it creates heat. It is
    usually so little voltage and current you don't see or feel the heat, but it is there none the less.
    Maybe I wasn't clear. The cap will go to "transistor heaven" if you connect it across the mains with no diode.

    The diodes...were they like 1N4001? 1N4002? 1N4003? If so, I will bet you exceeded the PIV rating. If the cap positive terminal
    (diode cathode) charges to 150v on the positive half of the AC cycle, then the diode anode goes to -150v on the negative half of the
    cycle, the voltage differential is 300 volts. Say "Goodbye, diode! Hello, fire and smoke!" Use 1N4004 instead (400PIV).
    It is better to be safe than sorry. I have been shocked badly by circuits I "knew" were switched off (and some were!). Watch those
    caps. They have teeth! Even when the power supply is removed, the larger values can store a lethal charge.

    I have almost been blinded by exploding caps that I was "certain" were polarized correctly. Doesn't happen very often anymore, but
    it only takes once. That stuff that sprays out is hot and sticky, but I guess you know that by now...

    Tim
     
  18. DillonCo

    DillonCo Guest

    Ah, Ok.
    voltage drop while carrying current, it creates heat. It is
    it is there none the less.

    Two questions:
    Will the inductor (wire) or the core get hot?
    If I use an inductor to limit the current on a transformer with little
    resistance or inductance, will the inductor still get hot?
    connect it across the mains with no diode.
    [snip diode statement]

    Well, I used a MR814 which has a reverse breakdown of 400V. If this isn't
    what you mean by PIV, I don't know (don't have the data sheet handy, and
    it's not on the 'net). They have a max current rateing of 1A, BTW.
    I can relate. Voltage multiplier: 10nF, 1kV. Not much, but ... Suprise!
    They hold charge considerbly longer than expected.
    Actually no. I was on the other side of the room (I heard electrolytics can
    explode viciously) when it steamed. I just pulled the power and left the
    room ;). Didn't know it was sticky though. Hot + Sticky = :(.
     
  19. (snip)

    VA tells you nothing about the phase relationship between the volts
    and the amps. It is just the product of RMS volts times RMS amps.

    Watts are the average power flow. If the volts and amps are in phase,
    then the instantaneous product of volts times amps (instantaneous
    power flow) is always positive (negative volts times negative amps or
    positive volts times positive amps, always power in the positive
    direction from source to load). But if the voltage and current are 90
    degrees phase shifted, there are times when the voltage and current
    have opposite signs, so those times (two quarters of each cycle) have
    power going the other way, back to the source and an average power
    flow of zero. Inductive or capacitive loads do not use up energy, but
    just borrow and return it twice a cycle. Energy is just sloshing back
    and forth. But devices like power lines and transformers still have
    to handle that current. So for those devices, VA is a better
    representation of what they have to deal with than watts.
     
  20. I read in sci.electronics.design that Jim Thompson
    Century.
     
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