LPF Chebychev Q re: input/output capacitors

for 20 dB Return Loss and the other params...

Chebyshev filter Low-pass filter Equal-ripple pass band Bandedge loss = 0.043648 dB. Upper passband edge frequency = 165.000000 MHz Specified stopband type Multiplicity of zero at infinity = 13 Overall filter degree = 13 Input termination = 50.000000 ohm Output termination = 50.000000 ohm Requested termination ratio = 1.0000000D+00 Nearest available termination ratio = 1.0000000D+00

01 +---R---+ 50.000000 ohm 03 +---C---+ 20.021157 pF 04 | L 71.431026 nH 05 +---C---+ 38.881642 pF 06 | L 82.438483 nH 07 +---C---+ 40.916227 pF 08 | L 84.246015 nH 09 +---C---+ 41.264976 pF 10 | L 84.246015 nH 11 +---C---+ 40.916227 pF 12 | L 82.438483 nH 13 +---C---+ 38.881642 pF 14 | L 71.431026 nH 15 +---C---+ 20.021157 pF 17 +---R---+ 50.000000 ohm
Reply to
Simon S Aysdie
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Simon, you are way ahead of me, your previous two posts had different values for the same filter, and neither matched the values that are in my filter. My caps were ?, 30, 36, 40, 36, 30, ?, the two end ones I don't know.

I need to pull one of them and check it, I rewound and padded the caps in the filter for a lower frequency and it swings wild in the passband, so I think I may have used the wrong values. Thanks, Mikek

Reply to
amdx

You have the 'ideal' problem. That is, a data point that works and you'd like to move slightly over to a new data point! Do that!

The filter is approximately 165MHz, and you want to move to 90MHz

Therefore take each inductor you have times 165MHz/90MHz, 1.83, and every cap times the same! and the filter will move from 165MHz down to 90MHz with just about identical characteristics, but with a cutoff of 90MHz.

Reply to
RobertMacy

They are not "the same." The first has a return loss of 9.6 dB (0.5 dB rip ple) and the second a return loss of 20 dB (0.044 dB ripple). I gave you b oth these Cheby "extremes" so you could compare them to your values, suspec ting your's is somewhere "in there." In fact, your filter may have had eve n less ripple than 0.044 dB. Or it may not be "pure" cheby. If you want t o reverse engineer what you believe the original synthesis is, be my guest, but I mean... reverse engineer a Cheby? I would not bother.

I am not exactly sure why your values are what they are. They are probably COTS "standards" and so are only approximations to the ideal synthesis val ues. That is probably at least part of the explanation of *approximate* ma tch to pure synthesis values. You can't tell what the original goal of des ign ripple, exact cutoff, impedance, and so forth are. (You can't tell us "your" original synthesis.) COTS parts are almost always an approximate ma tch to the ideal synthesis values; you should not expect them to match "per fectly."

I don't know what "padded the caps" means.

Just re-synthesize for the frequency and ripple/return loss you want. It i sn't exotic -- it is a Cheby (all-pole) low-pass or similar. It is about a s basic as it gets, aside from the order being a bit high.

Reply to
Simon S Aysdie

I didn't remove the original caps,I just added additional capacitance. If the original was 30 pf and I needed 56pf, I just add a parallel 26pf.

I used the values a got from this calculator.

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I used 13 elements, 110MHz, 0.1db ripple and 50 ohm.

It is about as basic as it gets, aside from the order being a bit high.

That may be, but the inductors are small and I have an imperfect way to measure them. I'm sure putting them in the enclosure modifies the inductance to some extent. I have no way to characterize the attenuation. I'm trying to look at the input impedance variation, for what it's worth. Thanks, Mikek

Reply to
amdx

So what are your specs? (What do you really need?) That is the real question.

You have a 13th order filter. What if 7 order is fine?

They aren't that small. Resonate them with precision caps, if you must.

Start with inductors designed with the following program, and bend them until the filter looks good. (Use precision caps.)

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Klappenberger used to work for K&L.

Well that sucks. Anyway, if the filter is near loss-less (which it should be if using Hi-Q parts), then if the signal is not reflected, then it is mostly transmitted (the Feldtkeller equation and conservation of energy).

But it would be best to know the "go in" versus "go out" more directly.

Reply to
Simon S Aysdie

It's not a big deal, just making sure I don't generate any harmonics from a 87.5 Mhz transmitter. It is 13 element because that was the filter I had in my junkbox to start the project. The pictures are on alt.binaries.schematics.electronic

That's what I did, but my maximum generator frequency is 50MHz.

then it is mostly transmitted (the Feldtkeller equation and conservation of energy).

Yep, maybe someday. I grabbed the program. Thanks, Mikek

Reply to
amdx

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