LC filter design for DC-AC inverter

Thread Starter

daengg

Joined Mar 8, 2016
8
I have designed a dc-ac inverter. Input is 300V Dc and required output is 240V AC (RMS). I have used PWM with switching frequency of 20kHz. My output frrequency is 50Hz. I need help with the filter design. I want to use 1000Hz as my cut off frequency. But i am not being able to get the correct values for the inductor and capacitor as I am a beginner. I would appreciate your help.
 

Thread Starter

daengg

Joined Mar 8, 2016
8
Show us how you got the values you got. How do you know they are not correct?
Hi! Thank you for replying. I actually used the kvar as 0.5 and then calculated the capacitor value for 13.8nF. Then i calculated the inductor value for the required cutoff frequency. This value came to 0.45H. Then i adjusted it to get the required output of 240V ac (rms). It came to 0.35H. The inductor is very big and the waveform is not pure sin wave. Also not certain of the method.
 

ronv

Joined Nov 12, 2008
3,770
Hi! Thank you for replying. I actually used the kvar as 0.5 and then calculated the capacitor value for 13.8nF. Then i calculated the inductor value for the required cutoff frequency. This value came to 0.45H. Then i adjusted it to get the required output of 240V ac (rms). It came to 0.35H. The inductor is very big and the waveform is not pure sin wave. Also not certain of the method.
We would let you cheat and use this: :D
http://sim.okawa-denshi.jp/en/RLCtool.php
 

Thread Starter

daengg

Joined Mar 8, 2016
8
Sir I tried the simulation but the output is still incorrect and not sinusoidal. I am attaching the screenshot of the schematicUntitled.png file. Please suggest how to fix it.
 

Papabravo

Joined Feb 24, 2006
14,241
Hi! Thank you for replying. I actually used the kvar as 0.5 and then calculated the capacitor value for 13.8nF. Then i calculated the inductor value for the required cutoff frequency. This value came to 0.45H. Then i adjusted it to get the required output of 240V ac (rms). It came to 0.35H. The inductor is very big and the waveform is not pure sin wave. Also not certain of the method.
This is not exactly what I meant by show us how you got the numbers. Your verbal discussion doesn't contain the method or the numbers, and I have no idea what a 'kvar' is.
 

Papabravo

Joined Feb 24, 2006
14,241
KVAr is the reactive power in kilo. The method i used is given in the website link :http://www.electricaltechnology.org/2013/11/how-to-calculate-required-capacitor.html
You meant to say: "KVAr is reactive power in killowatts"
That's great, except in an inverter you're not really concerned with power factor improvement. What you are concerned with is converting a square wave into a sinewave. The problem with that is the required size of the inductor to work at low power line frequencies like 50-60 Hz. Power factor improvement, if it is required at all in your application, comes after your inverter produces something that at least resembles a sinewave. This is the problem with formulas and online calculators -- people use them without having the slightest idea what they are doing. Solve your most important problems first and don't sweat the small stuff and second order effects.
 

Thread Starter

daengg

Joined Mar 8, 2016
8
Yes I agree. In this case can you suggest a source where i can get basic filter design and then how to implement it. I have posted my basic circuit.
 
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