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Achieving the right Cap Size

Discussion in 'Electronic Basics' started by news, Nov 25, 2004.

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  1. news

    news Guest

    Does anyone know of a web site, or a simple way of getting the right caps to
    add up to the number I want.
    Heres what I need.
    I want a .39u Cap I went to the store but they said they don't have it, its
    an odd size. Fine. I know that I can add up Caps in parallel. But I'm not
    that familiar with all the "Common" Sizes. It would be nice if there was a
    web site that I could put in my "Odd" size and it would spew out a couple of
    common combination to get my .39u Cap.
    I down loaded a great Cap Calculator, but now I have to keep trying
    different numbers to add up to .39 and then checking to see if I have those
    caps or if they can be had locally.
    Anybody know of such an animal.

    Regards
     
  2. Steve Evans

    Steve Evans Guest

    4 X 100nF would probably do and theyre easily availble. But getting so
    claose to .39u may not be necessary at all. Whats the circuit? can you
    posta schematic?
     
  3. I have a handy little program that can do that. It's called
    RESCAD.EXE, and is actually for calculating *resistor* values. But if
    you think 'parallel' instead of 'series', and vice versa, it does just
    as well for capacitors too.

    Couldn't recall source, and googling gave some dead ends, but this
    site has it, as ResCad.zip, a 143 KB file.

    http://www.armory.com/~rstevew/Public/Software/

    As an example, if I enter 39 in its Desired Value box, under the
    Series tab I see these results:
    %err R1 R2
    0.00 27R 12R
    1.03- 33R 5R6
    2.56 22R 18R

    That tells me that I could get a 39uF *capacitor* by placing a 27uF
    and 12uF in *parallel*, and that the result would have zero error. Or
    33uF and 5.6uF would give me a value about 1% too low

    Similarly, under the Parallel tab, I get:
    %err R1 R2
    0.70- 47R 220R
    2.10- 56R 120R
    3.79 68R 100R

    That says I can place a 47uF and 220uF capacitor in *series* to get a
    value less than 1% away from the 39uF target.

    The program handles E6, E12, E24 and E96 tolerance ranges; those
    examples above were for E12. Which BTW does include 39uF; try a better
    stocked store next time!

    If you have any trouble, I'll email it to you.
     
  4. tempus fugit

    tempus fugit Guest

    From the link you sent:

    "The value of C1 can be increased to increase the amount of current the
    circuit can supply. With the values shown, the circuit can supply up to
    about 15mA. Remember to increase the size of C2 also. "

    You can no doubt increase the cap to a more commonly available size, like
    0.47uF, without really affecting the circuit too much. This would allow a
    greater amount of current, so you could raise C2 to 330uF or so to help out,
    if it were necessary.
     
  5. news

    news Guest

    This is the circuit that Im trying to do. Im just testing and learning about
    various circuits. Im putting together various circuits for the fun of
    learning.
    http://www.aaroncake.net/circuits/supply5.htm
     
  6. news

    news Guest

    Hey thats a cute trick. Thanks.
     
  7. news

    news Guest

    Thanks. I knew I could increase it somewhat but didn't know by how much. Im
    cautious around this kind of circuit.
    I'm guessing that if I Reduce the Caps I will get less Amperage.
     
  8. Bill Bowden

    Bill Bowden Guest

    www.digikey.com has quite a few .39uF caps.
    I see twenty ceramics on this page:

    http://www.digikey.com/scripts/DkSearch/dksus.dll?Criteria?Ref=36909&Site=US&Cat=30540363

    -Bill
     
  9. Brian

    Brian Guest

    In this case, your best bet would be to go to a 0.47 uf capacitor. There
    is a program that will compute the unknown capacitor size (to be put in
    series), to get the total series capacitance that you want (plus many other
    things). Look at http://www.fncwired.com
     
  10. John Larkin

    John Larkin Guest


    The cap is determining how much current is being dumped into the zener
    shunt regulator. What you need is enough current for the load and a
    bit extra to keep the zener conducting. So the best cap value depends
    on the load current. If you're just experimenting, the size of the cap
    needn't be at all precise.

    If the voltage droops below the zener voltage when your load is
    attached, you need a bigger cap. If the zener gets too hot, you need a
    smaller one.

    John
     
  11. That is a bad circuit to play around with - the DC output can have
    hazardous voltages - there will be 11 volts between the terminals, but
    they may be at full line voltage AC.

    Look for a power supply circuit using a power transformer - much
    safer!
     
  12. John Fields

    John Fields Guest

    ---
    Yes, you will, but for its seeming simplicity, this circuit is
    devilishly complex.

    In this circuit, C1 is being used used like a "lossless resistor",
    with its reactance at the mains frequency determining how much current
    it will allow to flow into the load and into C2. At 60 Hz, the
    reactance of C1 will be:


    1
    Xc = ---------
    2pi f C


    where Xc = the reactance in ohms

    f = the mains frequency in Hz.

    C = the capacitance in farads


    So, for 0.39µF we have:


    1 1
    Xc = --------- = --------------------------- ~ 6800 ohms
    2pi f C 2 * 3.14 * 60Hz * 3.9E-7F


    Now, if we forget about C2 for a moment and look at the circuit like
    this:

    +-----+
    MAINS>---[C1]--|~ +|----+
    | | |
    | | [RL]
    | | |
    MAINS>---------|~ -|----+
    +-----+
    FWB

    all the bridge is doing is rectifying the mains, so we have,
    essentially:


    120VRMS>-----+-----E1
    |
    [6800R}
    |
    +-----E2
    |
    [RL]
    |
    120VRMS>-----+-----0V


    Now, since the current in a series circuit is the same throughout the
    circuit and, since


    E = IR


    you might think that if the reactance of the capacitor is equal to
    6800 ohms and has the same current flowing in it that RL does, if RL
    is equal to 6800 ohms it will have the same the same voltage dropped
    across it, but it won't.

    If you understand everything so far, and you're still interested, I'll
    continue...
     
  13. Brian

    Brian Guest

    I agree with you, this is not a circuit to play with unless you really
    know what you are doing. Two other things it should have (if one still wants
    to use this circuit), is a fuse (in case C1 should short out) and a
    polarized plug (so that the "-" side is always near earth ground).
    Brian
     
  14. Clarence

    Clarence Guest


    ..39u is a standard size, and most distributors stock the value. Digikey has
    many listed in several different styles.
     
  15. This supply should be handled with care. Also, you should put a small
    (.25A) fuse and a 1/4 W 10 ohm resistor in series with the 0.39uF cap.
    The cap should also be rated for AC. There are big square caps, called
    X1 caps, that are made for use with AC circuits. Everything should be
    protected from being touched somehow, perhaps with a small enclosure.

    If you are going to use this for other projects, I would suggest instead
    using a wall-wart to supply the DC. You can buy them pretty cheap, and
    they are going to be much safer. This supply would only be appropriate
    for a device that could never be touched by users.

    If you decide to do this, make sure that the high side (the side with
    the cap) is connected to 'Hot' aka 'Line', and the low side (the return)
    is connected to 'Neutral'. The way this stuff works is that the
    'neutral' wire (which is supposed to be white in the wall) is attached
    to ground when it comes into your house. 'Line' (which any color wire
    other than white or green or bare in the wall), carries the big voltage
    differential with ground. This is US 110V.

    <http://www.epanorama.net/documents/groundloop/electrical_wiring.html>

    The outlet is supposed to have Line, the dangerous one, coming out the
    small rectangular hole, and Neutral coming out the big rectangular hole.
    The roundish hole should be connected to ground, which is an alternate
    path to ground (usually a green or bare wire in the wall). You should
    make sure that your plug is wired properly rather than just assuming it.
    Take an AC voltmeter, and test the voltage between the round hole and
    the small hole. It should be near 115VAC. The voltage between the round
    hole and the big rectangular hole should be less than 5 volts.

    If you get this backwards, there will be a low resistance path between
    Line and your 'ground' in your circuit, the point marked - on the
    diagram. Thus, if you touch ground (which you might be likely to do),
    it's like sticking your finger into the socket. You can get a nasty shock.

    Using an isolation transformer (as you will with a 'wall wart', or the
    audio transformer suggested by the web page) will protect you from a
    shock by 'floating' the output near ground.

    Remember, when working with AC, always keep one hand in your pocket.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  16. News

    News Guest

    Thanks all for the advice. I would definetely put a fuse in. I do now in all
    my AC projects. In part yes I am learning things. In part I think that this
    circuit would be handy if I wanted to build something self contained that
    needed low DC voltage. Plugging into a wall would be so much easier. I mean
    if you build something but then have to plug an AC/DC adapter into it, this
    increases the cumbersomeness of it all. Albeit It is the safest way to go.
    Regards
     
  17. News

    News Guest

    Well not completely but go on ..
     
  18. John Fields

    John Fields Guest

     
  19. News

    News Guest

    Well John,
    Im not going to presume that I know as much as individuals who have most
    likely gone to college for this topic or those electrical engineers in my
    company that every once and a while deign to speak to me in a condescending
    manor when I ask then a question. Which sometimes only encourages me to go
    on and sometimes discourages me..

    But Im not sure what you mean by the numbers 3.9-E-7 .
    Is that the .39 Capacitor It could only be based on your equation but what
    is the -E-7

    I found a great little site that had a nice little definition of things and
    caclulator see the bottome of ..
    http://www.eatel.net/~amptech/elecdisc/reactnce.htm
     
  20. John Fields

    John Fields Guest

    ---
    The level of your education and the relationships which exist between
    you and your co workers are of no concern to me. What I'm interested
    in doing, here, is helping out where I can when people ask for help.

    In your case, the power supply you're interested in building can kill
    you if you don't know what you're doing and, since you're obviously a
    novice, I elected to stop where I did to make sure that you understand
    what I've given you so far.

    If you have a problem with that, just let me know and I'll be happy to
    spend my time doing something else.
    ---
    ---
    0.39µF = 3.9 * 10^-7 farads, and 3.9E-7F is a commonly accepted way of
    writing a condensed version of 3.9 * 10^-7 farads, where 3.9 is the
    mantissa, E indicates that an exponent to the base 10 will follow, -7
    is the exponent, and F is the abbreviation for farad.
    ---
     
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