If the output is a pulse-width modulated square-wave, then you want a filter that rolls off at somewhat above 50Hz.
The output sinewave will be the same frequency as the input sinewave.
60 Hz cut-off isn't a good idea, too much attenuation at 50 Hz. Another point is that fc involves large and possibly uncomfortable L and C values. On the other hand there should be a certain factor between fc and pwm frequency to get sufficient pwm attenuation. For the intended low pwm frequency, fc of 100 to 200 Hz sounds reasonable.Above 50 Hz ,like 60Hz? I use this equation Fc=1/2*PI*sqrt(LC) in order to dimension a LC filter, so what you tell me is For example : to put Fc=60Hz and get the values of L and C?
Firstly 2Khz is really too low a switching frequency to create a clean 50 Hz sine wave, it becomes much more difficult to filter.
As the switching frequency is x40 a good starting point to set the cutoff frequency would be the square root of 40 or 6.3 times 50 HZ.
So about 315 Hz, which is about 2.5 octaves above 50Hz and 2.5 octaves below 2 Khz. That mid way point is the best compromise between minimum attenuation at 50Hz and maximum attenuation at 2 Khz.
You then need to design a 315 Hz low pass filter for the operating impedance.
It will be difficult with only 2.5 octaves separation.
Adding more filter stages will also increase the losses at 50 Hz.
Much better to switch at 20 Khz.
Frequency difference is x400
Square root is 20
More like 4.5 octaves to play with, and a much simpler filter centred around 1Khz will give much better results.
Firstly 2Khz is really too low a switching frequency to create a clean 50 Hz sine wave, it becomes much more difficult to filter.
As the switching frequency is x40 a good starting point to set the cutoff frequency would be the square root of 40 or 6.3 times 50 HZ.
So about 315 Hz, which is about 2.5 octaves above 50Hz and 2.5 octaves below 2 Khz. That mid way point is the best compromise between minimum attenuation at 50Hz and maximum attenuation at 2 Khz.
You then need to design a 315 Hz low pass filter for the operating impedance.
It will be difficult with only 2.5 octaves separation.
Adding more filter stages will also increase the losses at 50 Hz.
Much better to switch at 20 Khz.
Frequency difference is x400
Square root is 20
More like 4.5 octaves to play with, and a much simpler filter centred around 1Khz will give much better results.
Firstly 2Khz is really too low a switching frequency to create a clean 50 Hz sine wave, it becomes much more difficult to filter.
As the switching frequency is x40 a good starting point to set the cutoff frequency would be the square root of 40 or 6.3 times 50 HZ.
So about 315 Hz, which is about 2.5 octaves above 50Hz and 2.5 octaves below 2 Khz. That mid way point is the best compromise between minimum attenuation at 50Hz and maximum attenuation at 2 Khz.
You then need to design a 315 Hz low pass filter for the operating impedance.
It will be difficult with only 2.5 octaves separation.
Adding more filter stages will also increase the losses at 50 Hz.
Much better to switch at 20 Khz.
Frequency difference is x400
Square root is 20
More like 4.5 octaves to play with, and a much simpler filter centred around 1Khz will give much better results.
I completely agree with your suggestions- particularly about the frequency. I have no idea about the factors or considerations that lead to the original idea for using 2kHz as the switching frequency. Even if you are not interested in creating a 50Hz sine wave.
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If octaves and decibels appear confusing (Yes! I bet), the best way is to take the geometric mean between the two: (50x20000)^0.5=1000.
I do not know why PI filters are not so popular these days but they do a decent job, IMHO. You can wind a simple toroid that can do the job.
Another question , when I use a RLC filter I notice that the sinusoidal wave after filter loss amplitude, how I can reduce this attenuation?
If the filter frequency and the sinusoidal frequency are close enough, then you need to
1. Shift the filter frequency;
2. If that is not possible, make a filter with greater slope (higher order);
3. Make a tuned notch filter.
I will try this tips! So i have used a Fc= 701 hz because i use a geometric mean sqrt(50* 10000)
I use a RL // C, what you suggest?
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