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Filters and equations

Discussion in 'Electrical Engineering' started by Mark or Sue, Jan 15, 2004.

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  1. Mark or Sue

    Mark or Sue Guest

    I want to make some X-10 filters, and my equations aren't coming up with the values I commonly see
    for X-10 filters. Assuming the following are correct, what are the units:

    R =FL, R=1/FC (first order filters)
    F=sqrt(1/LC) (resonant filter)

    I assume C in farads, L is in henrys, and F is in hertz.

    My answers sometimes are off by a factor of about 6. Should there be a 2pi as part of F?

    The frequencies in question are 60Hz to block, and 120KHz to pass. Noise filters will be the
    opposite. I'm first trying to make a bandpass "signal coupler" that passes 120 KHz with easy to find
    component values. I'll use a cap and inductor in series. Suggestions are 0.1uF and18uH, but those
    don't fit the equation all that well. What am I calculating wrong?

    Thanks.
     
  2. Guest

    I *believe* the coupler is a capacitor only - no L.
    The reactance is found by Xc = 1/2piFC where Xc is the
    capacitive reactance in ohms, F is the frequency in
    Hertz, and C is the capacitance in Farads.

    Anyway, your equations are off by 2pi, as you stated:
    Xc (you call it R) = 1/2piFC
    Xl (you call it R) = 2piFL
    Your third equation is off by more than 2pi - your
    sqrt is misplaced. The equation is F = 1/(2pi*sqrt(LC))

    If I did my math right, you could parallel a
    ..1, a .05, a .02 and a .003 to get .173 uf
    and series a 1000 uH to get ~12100 hz resonant
    freq. You'd have about 15 mA through the inductor.
    But it's early in the AM - check my math!
    And I'll say it again - I *believe* that a coupler
    for X10 is just a cap. Check on that before
    going further.
     
  3. Mark or Sue

    Mark or Sue Guest

    Thank you very much, the numbers are working out now. I've always heard of just the 0.1uF cap for an
    X-10 coupler, as this is a simple high pass filter. The reactance equations you provided seem to
    correlate with the leakage current I've also heard for a single capacitor filter. You must have made
    a mistake though, because the equation seems to work out that the LC product should equal about 1.8
    when using uF and uH. Your 1000uH coil is way too big -- needs to be more like 10 uH.

    I found an X10 site with info for how to make your own coupler, and speculation that this is what
    was inside the Leviton phase coupler. It said use an 18 uH inductor in series with a 0.1uF cap, and
    put a fuse in line for good measure too. This would connect across a double pole circuit breaker.

    How does one calculate the reactance of a bandpass filter (the L and C in series with a resonant F
    of 120 KHz)? I believe the idea of adding the inductor is to lower the overall impedance at 120 KHz
    while also blocking any other noise above 120 KHz. Using that inductor by itself would seem to have
    an excessive impedance at 120 KHz, thereby defeating the purpose. When calculating currents through
    a resonant filter at various frequencies, do you look at each component individually, or does
    something special occur that gives a nearly 0 impedance at the resonant frequency and much higher at
    all others? Intuitively, it seems like any L and C that multiply to 1.8 would not be a universal
    solution. I would expect there to be some optimal L or C.
     
  4. Guest


    You are absolutely right! I just finished checking my math
    before I opened your post. I was doing it bleary-eyed last
    night and I was two decimal places off. The inductor would
    turn out to be 10 uH, not 1000 uH! But with the inductor
    much smaller, the approach of paralleling caps to come to
    resonance can be re-investigated, as there are many small
    inductor values available. The new computation yields
    a .1 uf cap and a 17.3 uH inductor. You can get a 15 uH
    or an 18 uH off the shelf. With an 18 uH, the net impedance
    is about .5 ohms at 121 KHz, which is consistent with the
    information you posted below.
    Slight word change - it should be "how does one calculate
    the *impedance* of a bandpass filter".

    Let me digress a moment, before the math. At resonance, the
    reactance in the capacitor (Xc) and the reactance in the
    inductor (Xl) cancel each other out. You can imagine it
    this way: when one of them wants to "give away" energy
    (in the form of electrical current), the other wants to
    "take" exactly the same amount of energy the other one
    wants to give. So the two components "satisfiy" each other,
    and those components don't impede the rest of the circuit.
    When they are not able to "satisfy" one another, they
    do impede the rest of the circuit.

    Ok, at resonance, they "satisfy" one another, so there is
    no impedance to the rest of the circuit. Mathematically,
    Z (impedance) = 0 at resonance for series LC. Also, in
    general (not just at resonance), and ignoring resistance,
    Z = absolute (Xl - Xc). Absolute simply means ignore the
    sign of the result. So if Xl was 1 and Xc was 1.5, the
    result absolute(Xl-Xc) would be .5

    The formula is Z = Xl - Xc and the result is signless.


    Another problem: inductor resistance. It is so small that the
    inductor would burn out. With no C, you put a 10 uH (or any
    value, for that matter) L across the phases. Say the inductor
    is rated to carry ~200 mA. What limits the current? With a
    series LC across phases, the capacitor limits the current. The
    resistance of the inductor is usually less than 1 ohm. With no
    C, the L is exposed to 240 volts at 60 Hz, and its resistance is
    usually less than 1 ohm. Call it 1 ohm to make the math easy.
    Its reactance at 60 Hz is 2piFL or 6.28*60*.00001, which equals
    ~ .0037 ohms. The total impedance is Z = Xl + R, so
    Z = .0037 + 1 At 1.0037 ohms, there would be almost 240 amps
    drawn, and the inductor would burn out immediately.
    Well, it's sort of special. As frequency changes, the reactance
    of one goes down and the reactance of the other goes up. The
    reactances always change in opposite directions with changes
    of frequency. It acts like a see-saw, where there is only one
    point of perfect balance.

    You look at the components individually to see how each
    will handle/affect the current that the *overall* circuit
    will impose on the component. Maybe an example, using
    C = .1 uF and L = 18uh will help.

    At 60 Hz, Xc = ~26526 ohms, and Xl = ~.0067 ohms so
    impedance is ~26526 ohms.
    At 240 volts, the current drawn will be ~240/26526 or
    about .009 amps. The 240 volt supply is an enormous
    transformer (typically) on a pole outside your house.
    It has a very low internal impedance, and can provide
    a huge amount of current, but the capacitor limits the
    current through the series LC to about .009 amps.

    At 120 KHz, Xc = Xl so impedance is 0, implying
    infinite current. However, the source is the
    X-10 transmitter, and it has a high internal
    impedance, so it limits the current to a very low
    value.
     
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