No, longcrystal still does not know what values to use or how to calculate them.And is the problem solved know ?
I know what a Butterworth filter is. Maybe you need to read more about it if you think this formula is correct:For butterwoth , you can see below :
https://en.wikipedia.org/wiki/Butterworth_filter
Or search in goodle .
That formula is not correct for a two pole LCR Butterworth filter.RO=0.5sqrt(L/C)
I thought you have read at least one book regarding this issue . see professor pressman's book . filter sections ( and compensation ) . he told this formula too . and this formula is in many books too . didn't you try ?That formula is not correct for a two pole LCR Butterworth filter.
If he/she tell me that he couldn't calculate the elements with those formulas , thus i can help him / her . but i don't think that you are a lawyer for each people ? are you ?No, longcrystal still does not know what values to use or how to calculate them.
By definition a Butterworth filter is -3dB at the corner frequency. Thus R = sqrt(0.5*L/C)I thought you have read at least one book regarding this issue . see professor pressman's book . filter sections ( and compensation ) . he told this formula too . and this formula is in many books too . didn't you try ?
You asked me a question. I answered it.If he/she tell me that he couldn't calculate the elements with those formulas , thus i can help him / her . but i don't think that you are a lawyer for each people ? are you ?
well if you are thinking that i'm not correct perhaps your mean is professor pressman is not correct too . see blow , please :By definition a Butterworth filter is -3dB at the corner frequency. Thus R = sqrt(0.5*L/C)
You said R=0.5sqrt(L/C), which is not correct.
I don't see the word "Butterworth" on that page.well if you are thinking that i'm not correct perhaps your mean is professor pressman is not correct too . see blow , please :
Yes. There is an explanation of Butterworth filters here in Wikipedia.Hummm ! is the problem with the word butter worth ?
If you use R=0.5sqrt(L/C), then response is -6dB at the corner frequency, instead of -3dB.ωc = cutoff frequency (approximately the -3dB frequency)
Hi dear FvMto make the output filter work over a load impedance range of
I did simulate it with different values already.can you simulate the circuit ?
I am not angry. I am just trying to teach one simple thing:or perhaps you are angry from me because of each reason that i don't know ?
Agreed. A damped filter, as you showed in post 34 is one way to deal with it. The alternative, as pointed out by Goldsmith, is to use feedback to control the response.the more interesting problem is how to make the output filter work over a load impedance range of R0 up to infinity.
The current drawn by a load like that will have high levels of harmonics at frequencies well above 50Hz, so it is important for the inverter to have a low output impedance even at those frequencies. I guess that's one good reason to make the inductor smaller rather than larger in value.And how to deal with the fact, that a nonlinear load, e.g. a rectifier power supply with filter capacitor.....
Nice to hear that your aim is teaching but before you , a friend of mine have told me this ! ( long time ago )am not angry. I am just trying to teach one simple thing:
An LCR low pass filter is only called a "Butterworth" filter if the response is -3dB at the corner frequency. Any other LCR low pass filter is not a Butterworth filter.
No, it is not correct. R = sqrt(0.5*L/C) for a Butterworth filter.you saw that 0.5sqrt(L/C) is correct but you told it is not !
I didn't particularly mean the effect of load variation on output voltage. It will be present, also without any filter, so output voltage stabilization isn't a bad idea in any case. I meaned that the filter will only act as "butterworth" under a certain load condition (e.g. full load), and have much higher Q with lower load. You won't see filter related voltage variation at the fundamental wave, because it's sufficient below the cut-off frequency, that's the optimistic estimate of post #32. But it emerges if the resonance is excited, e.g. by pulsed currents.I removed this problem with DC feed back from out put . i have rectified a sample of out put and then i have affected it ( after filtering ) to the first stage ( stage which converts low voltage to the 311 volts)
Yes i agree , but an old friend of mine , some years ago , told me , if we predict minimum and maximum value of load and then take average from it , and then design our filter according to that , the problem will be lower , ( as i have tested it too , the problem was lower ) ( it is the best condition , i think )I didn't particularly mean the effect of load variation on output voltage. It will be present, also without any filter, so output voltage stabilization isn't a bad idea in any case. I meaned that the filter will only act as "butterworth" under a certain load condition (e.g. full load), and have much higher Q with lower load.
Sorry, I was not clear. I meant Fc = 1 /(2*Pi*sqrt(LC)). At that frequency the response of a 2nd-order Butterworth filter is -3dB.1.) The corner frequency of a filter can be defined in several ways.
Yes, agreed. The point I was trying to make is that this is achieved by setting R = sqrt(0.5*L/C).2.) More important: A low pass response is called "Butterworth" response when the magnitude response has a maximally flat characteristic.
3.) Another typical characteristic of the 2nd-order Butterworth response is the pole-Q=1/sqrt(2)=0.7071.
Thank you for the nice explanation.
Hi again ! how ever this is not for a butter worth condition , ( 0.5sqrt(L/C) ) , but it is best situation for , a filter , ( for this aim ) aren't you agree with this ?Yes, agreed. The point I was trying to make is that this is achieved by setting R = sqrt(0.5*L/C).
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