LC filter increases ripple ?? - Page 1
Forum section: electronics / general

author: hedley
author_rank: Contributor
timestamp: January 21, 2017, 08:30:15 am
content: Any help in understanding this will be appreciated

author: hedley
author_rank: Contributor
timestamp: January 21, 2017, 09:37:03 am
content: as a sanity check I wired up the filter with a signal generator as the input and a scope on the output . I thought there will be little attenuation of the signal p/p before the cutoff frequnless I am using the wrong formula or something else really dumb I expected the cutoff to be 15,915.49 HZ . You can see from the attached that the attenuation starts well before that ??Normally with these problems there is something really basic that has been overlooked .

author: hedley
author_rank: Contributor
timestamp: January 21, 2017, 10:26:57 am
content: tried running on LTspice which confirms what I expect to see but measurements dont tie up .Conclusion - it must be something to do with the measurement / scope / ground / probe

author: Kleinstein
author_rank: Super Contributor
timestamp: January 21, 2017, 10:38:09 am
content: The simple second order LC filter as shown has a high Q (especially the simulation). Without extra damping (e.g. series or parallel resistor at the inductor, load) there will be an increased amplitude in the transition region. Often the load attached to the output will provide some of the damping and thus should be included. Also the source impedance can be important - otherwise there would be no sense in the capacitor on the input side.

author: hedley
author_rank: Contributor
timestamp: January 21, 2017, 12:15:23 pm
content: thanks for input on the ltspice . The original prob remains in that on the actual circuit measured on the scope there is attenuation even at 1.5 khz

author: bktemp
author_rank: Super Contributor
timestamp: January 21, 2017, 12:57:40 pm
content: Your signal generator has an output impedance of probably 50ohms. Because of this rather high series resistance at the output you have also created a RC lowpass filter. If you do the math, you get a corner frequency of less than 1kHz.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 21, 2017, 01:32:44 pm
content: Your L & C values are the problem.Formulas work OK with silly values,but in the real world you don't just need the capacitive & inductive reactances to be similar at the 3dB cutoff point,they need to be reasonably high values of Reactance at that frequency.Your chosen components have Reactances (at 15kHz) of :-XL1=0.9425 Ohms &XC1=1.061 Ohms,respectively,which are comparable to circuit resistance.If you are measuring the output across C2,as the value ofXC1falls with increasing frequency & that ofXL1increases,the Outputwillrise above the "design" frequency ,but it will not be evident due to the low values of both Reactances.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 21, 2017, 10:54:13 pm
content: There is no loss indicated in that circuit.Likely the LTSpice defaults are providing it for you (which is something a simulator should NEVER do, but alas, they did). If you didn't set these properties to zero, the actual circuit isn't what's shown here.A filter is only meaningful when there is resistance in the circuit: it is resistance that does thework(the physical meaning of "work", i.e., energy and power), whether it's the source, or load, or both.Tim

author: IanB
author_rank: Super Contributor
timestamp: January 21, 2017, 11:03:04 pm
content: Quote from: T3sl4co1l on January 21, 2017, 10:54:13 pmBy way of clarification, if I connect an ideal voltage source to an LC filter constructed of ideal capacitors and inductors, will I not see an appropriately filtered voltage waveform on the output side (assuming an ideal measurement instrument)?Or is it the case that while the above may be true, the presence of non-ideal source and load impedances will disturb the operation of the filter and lead to changed results?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 21, 2017, 11:49:41 pm
content: Quote from: IanB on January 21, 2017, 11:03:04 pmThe asymptotic response will still be there (for a low-pass: gain of 1 at 0Hz, gain of 0 at \$\infty \, \textrm{Hz}\$), but you don't have any control over the transition band response. Power flow is undefined, because there's no resistance for it to flow into (or not)!In the simple case of a two pole LC filter (as above), it's a resonant circuit, with one unlimited peak. At frequencies besides resonance, the source current, and capacitor voltage, are finite, but they rise towards infinity as you approach resonance exactly. This is simply because the series impedance goes to zero, so the current (and thus source VAs) go to infinity at exactly resonance.You certainly can't tune such a filter for a nice smooth Butterworth response, or something like that. There must be some notion of real power flowing, from (and back to) the source, and to a load. And it turns out, there is only one component which exhibits this mathematical property: resistance.It's a very deep notion -- that resistance is everywhere, and necessary -- and cannot be emphasized enough. This connects with many other fields, like polynomial equations and factorization, complex analysis, and causality.Tim

author: NiHaoMike
author_rank: Super Contributor
timestamp: January 22, 2017, 01:44:58 am
content: You just discovered why noise filters use lossy inductors. One suggested way to make an inductor more lossy is to add a parallel resistor.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 22, 2017, 05:11:17 am
content: The above comments are valid forXc=nOhms,&XL=nOhms when the Reactances are practical values,but whennapproaches zero,the thing stops looking like an LC network.At audio frequencies,a 10uH inductor will look like a short circuit,so the circuit will be C2 in parallel with C1.bktemphas the correct explanation for the practical test ,where the 50 Ohm generator output impedance is the R component & C1 parallel with C2 is the "C" component.