LC Low pass filter design for SPWM filtering

[godfreyl]

And is the problem solved know ?

No, longcrystal still does not know what values to use or how to calculate them.

For butterwoth , you can see below :
https://en.wikipedia.org/wiki/Butterworth_filter
Or search in goodle .

I know what a Butterworth filter is. Maybe you need to read more about it if you think this formula is correct:

RO=0.5sqrt(L/C)

That formula is not correct for a two pole LCR Butterworth filter.

 

[goldsmith]

That formula is not correct for a two pole LCR Butterworth filter.

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 ?

No, longcrystal still does not know what values to use or how to calculate them.

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 ?

 

[godfreyl]

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 ?

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.

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 ?

You asked me a question. I answered it.

 

[goldsmith]

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.

well if you are thinking that i'm not correct perhaps your mean is professor pressman is not correct too . see blow , please :



Godfreyl , again : try to change your behavior . it is not way for go . why you don't want learn instead of fighting ? i know what will happen if you don't change your behavior . you won't learn . it is the least effect .

 

[godfreyl]

well if you are thinking that i'm not correct perhaps your mean is professor pressman is not correct too . see blow , please :

I don't see the word "Butterworth" on that page.
The picture shows a number of filter responses. None of them show the response of a Butterworth filter.

 

[goldsmith]

Hummm ! is the problem with the word butter worth ? or type of filter ? we are talking regarding filter that author used ? and picture of that book is that filter too . and relations that i wrote ws related to this 2nd order filter ? didn't you understand it ?

 

[godfreyl]

Hummm ! is the problem with the word butter worth ?

Yes. There is an explanation of Butterworth filters here in Wikipedia.
Note:

ωc = cutoff frequency (approximately the -3dB frequency)

If you use R=0.5sqrt(L/C), then response is -6dB at the corner frequency, instead of -3dB.

 

[goldsmith]

Godfreyl ! i thought we are friends and friends should not fight ! we could discuss together without bad words . isn't it ?
So , let me tell you my mean , can you simulate the circuit ? if yes , would you like that i tell you some values to simulate ? if yes thus you can understand what is my meaning , i think you misunderstood my meanings . isn't it ? or perhaps you are angry from me because of each reason that i don't know ? if yes tell me that , please . ;-)

 

[FvM]

Apart from the question, if the terminating resistor for a second order, voltage source driven butterworth filter should be 0.5 or rather 0.7*sqrt(L/C), the more interesting problem is how to make the output filter work over a load impedance range of R0 up to infinity. And how to deal with the fact, that a nonlinear load, e.g. a rectifier power supply with filter capacitor provides no damping at harmonic frequencies, where it represents a pure apparent load.

I tried to give a limited answer in post #32 and #34, depending on the intended PWM frequency attenuation and other parameters, you may want to modify it, e.g. shift the cut off frequency higher, or design it with more damping, although it involves 50 Hz losses, too.

 

[goldsmith]

to make the output filter work over a load impedance range of

Hi dear FvM
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)
and then the result was as i expected and without any problem . ( average of maximum and minimum load , and after that a heavy feed back loop from current and voltage )
Best Regards and Sincerely yours
Goldsmith

 

[godfreyl]

can you simulate the circuit ?

I did simulate it with different values already.

or perhaps you are angry from me because of each reason that i don't know ?

I 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.

the more interesting problem is how to make the output filter work over a load impedance range of R0 up to infinity.

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.

And how to deal with the fact, that a nonlinear load, e.g. a rectifier power supply with filter capacitor.....

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.

 

[goldsmith]

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.

Nice to hear that your aim is teaching but before you , a friend of mine have told me this ! ( long time ago )
I'm not trying to play with names ! but as it appears names are important to you ! you saw that 0.5sqrt(L/C) is correct but you told it is not ! i think now you know thing i told is correct .
And it is good that you have simulated it , so why you are trying to don't accept thing that i told ? however you are seeing that is correct ?

 

[godfreyl]

you saw that 0.5sqrt(L/C) is correct but you told it is not !

No, it is not correct. R = sqrt(0.5*L/C) for a Butterworth filter.

The name is important. The name refers to a particular type of filter, and it is named after the person who invented it.

 

[FvM]

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)

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.

 

[goldsmith]

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.

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 )
Best Regards

 

[LvW]

Hello, I do not want to interfere in this long and interesting discussion - however, I cannot resist to correct the following "definition" :

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.

1.) The corner frequency of a filter can be defined in several ways. It can be defined - and in many cases it is in fact - the -3dB point.
This is normally the case for Bessel, Butterworth and Chebyshev(2) responses - however, in some cases it is, of course, allowed to define another band edge. Why not?
Sometimes this 3dB frequency is even used to describe the band edge for Chebyshev(1) and Cauer responses - although in most cases the ripple factor is used for this definition.

2.) More important: A low pass response is called "Butterworth" response when the magnitude response has a maximally flat characteristic.
Therefore, such a response very often is called "maximally flat" response. This has something to do with the mathematical description of the Butterworth polynominals
but it has absolutely nothing to do with the definition of a corner frequency (which - as indicated above - can be defined at the 1dB, 2dB, 3dB or any other point on the frequency scale).

3.) Another typical characteristic of the 2nd-order Butterworth response is the pole-Q=1/sqrt(2)=0.7071. Higher order Butterworth poles are arranged in the s-plane on a circle that is centered in the origin.

 

[goldsmith]

Hi Dear LvW
Your idea , makes , all things more clear ! and hence thank you for your pretty good , description .
Yours SIncerely and Best Regards
Goldsmith

 

[godfreyl]

1.) The corner frequency of a filter can be defined in several ways.

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.

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.

Yes, agreed. The point I was trying to make is that this is achieved by setting R = sqrt(0.5*L/C).

Thank you for the nice explanation.

 

[LvW]

Thank you for the nice explanation.

To complete the description of a Butterworth response:
It is the only 2nd-order characteristic with a pole frequency that is identical to the 3dB frequency. For all other responses both frequencies are different.

That means: Godfreyls definition as quoted in post #56
"An LCR low pass filter is only called a "Butterworth" filter if the response is -3dB at the corner frequency"
is correct if we replace the term "corner frequency" by "pole frequency".

 

[goldsmith]

Yes, agreed. The point I was trying to make is that this is achieved by setting R = sqrt(0.5*L/C).

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 ?