PWM frequency and LC filter

[ricperes]

I'm new here, I have some fairly basic doubt. I am creating a PWM signal in MatLab by comparing a sine wave (50Hz) and a triangular wave (2kHz) .... So what is the PWM frequency will output?

After getting the PWM signal filters I want to get a sinusoid, using an LC filter, but do not know what the frequency of cut should I use?

Thanks for help

 

[crutschow]

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.

 

[ricperes]

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.

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?´

thanks

 

[FvM]

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?

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.

You should have noticed that specifying fc doesn't give you distinct L and C numbers. There's one more degree of freedom which defines the filter impedance respectively the quality factor for a known load impedance. A practical criterion to choose the filter may be the 50 Hz voltage drop at full load current, could be e.g. 2 - 5 % of output voltage.

 

[Warpspeed]

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.

 

[ricperes]

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.

thanks for your tips.

So you say me to change switching frequency to 20khz and try this equation Fc=1/2*PI*sqrt(LC) ,with fc=1khz ?

 

[ricperes]

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.

Warpseed, can you show me a reference about how to choose a cutoff frequency please?

 

[c_mitra]

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.

- - - Updated - - -

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.

 

[ricperes]

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.

- - - Updated - - -

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?

 

[c_mitra]

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.

 

[ricperes]

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?

 

[ricperes]

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?

I made a RC filter with R=200ohm and C= 2.2uF, I want to filter this PWM in the figure with 10k frequency, and get a Sinusoide with 50hz, but in this case I loss amplitude how i can do this???

 

[FvM]

No idea what's the circuit belonging to this waveforms.

 

[ricperes]

No idea what's the circuit belonging to this waveforms.

The PWM is created by comparing a Sin 50HZ with a tringular 10Khz

 

[Warpspeed]

An LC filter will be much better than an RC filter as it will have far less attenuation at 50 Hz.

The problem will be ringing at the filter cut off frequency when there is little or no load on the output.
That ringing can be reduced by shunting the L with a resistor, but its not going to be a perfect cure.