LF filter value calculation

Hi
sorry can i post a new topic about:
how can i calculate the value of LC filter on the output of full bridge dc-ac converter?
Vdc = 24v and switching frequency is about 800Hz.
inverter load is RL that R = 100 ohm and L = 1mH ( for example)

i want to remove frequencies above 1.5 KHz. so is it right that i choose L=100mH and C = 0.1uF ?
i must buy power inductor? with how much power rating?
is there any formula for calculating the value of LC filter?

an second question is how can calculate the power rate of my load? ( R= 100 ohm and L= 1mH)
what kind of R i must buy? 1W? 2W? or ....
i really dont know how can i found power rating .
please help
thanks

From what I have seen in simulations, it works okay when you choose L and C so they yield the same impedance as your resistive load.

Use formulas for inductive and capacitive reactance, below. (X is impedance.)

X_L = 2 Pi f L

X_C = 1 / ( 2 Pi f C )

By varying the values, you can shape the waveform. You can alter (a) its amplitude, and (b) similarity to a sinewave.

A simulator is invaluable to let you examine the output.

Do you plan to turn a square wave into a sinewave?
If so then I get different results than the ones you quote. I get 20 mH and 2 uF.

Hi,
thanks BradtheRad for reply,
dc-ac converter is controlled by a controller. i am trying to use sigma delta modulator in controller to push the noise outside the base band. ( assume base band is about 2KHz and sampling frequency of sigma delta is 4KHz), so i need LC filter on the inverter output to filter this high frequency noise that is pushed away by modulator.
the final goal is to be able feeding R-L load by sine wave that its harmonic contents reduced. it's why we use LC filter.
i want to filter frequencies above 1.5KHz. but in a paper i saw that we must choose LC such that the inductor ripple is 20% peak to peak output current and is calculated based in this formula:

L = Vdc / (4*fsw*delta_i) ======> delta_ i is current ripple value

if R=10 ohm and L = 1 mH and fsw=800Hz and Vdc =24v then : p-p output current is about 4.8A (2*( 24v/10 ohm)) , so if we choose L = 100mH , the current ripple is about 75mA. i thinks this is reasonable ripple for this case.
then choosing C= 0.1uF . so cut off frequency is about 1.5KHz and it is appeared that everything is good.
but i don't know is this thought right?
if i don't choose right forum for this topic please help.

thanks

From your description I gather you wish to turn a square wave into a sine wave, by using a second order LC filter.

Here is my simulation.

**broken link removed**

Notice the waveform coming from the secondary. Square wave, 270VAC amplitude. However the load receives a sinewave with a peak amplitude of 330V. The LC filter does not simply filter out high frequencies, but re-shapes the waveform.

Component values need to be custom adjusted to achieve this. You can find optimum efficiency by experiment. The Ampere waveform drawn from the supply is of interest. Notice it is not a square wave, even though the supply voltage is a square wave. It is influenced by action at the secondary side of the transformer.

As it turned out I now have lower LC values compared to my post #2.

Hi BradtheRad,
thanks for your help, but i can't open the attachment.

My attachment expired. (This happens sometimes. I guess it even happens to forum moderators.)