Filters and equations

values I commonly see

filters will be the

KHz with easy to find

and18uH, but those

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.

values I commonly see

as part of F?

filters will be the

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

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.

-- Mark Kent, WA

just the 0.1uF cap for an

you provided seem to

filter. You must have made

should equal about 1.8

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.

speculation that this is what

with a 0.1uF cap, and

pole circuit breaker.

series with a resonant F

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.

itself would seem to have

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.

individually, or does

frequency and much higher at

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.