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There are LC filters put at the output of the inverters to filter and produce a pure sine wave. There is no clearly calculated and well explained document about this topic.

The first picture shows a full-bridge schematic and the output waveform of the inverter.
The second picture shows the filter used in the circuit and calculations.
These two pictures belong to the same document as shown below.

My questions are as follows :

  1. If the output of a full-bridge inverter is designed for 50 Hz output, what is the reason of calculating the cut-off frequency of the filter as 18 kHz in the document ?

  2. L and C are calculated as 72 mH and 0.004 μF in the document; how? For example, I should select L first so that I can calculate C or vice-versa, but how to select the value of L in the first step, and based on what?

  3. The document only uses a general well-known formula for LC, but is it that easy? what are the other considerations that needs to be calculated or taken into account?

enter image description here

Source: Design and Real-Time Implementation of SPWM based Inverter , https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=8338637

enter image description here

Source: Design and Real-Time Implementation of SPWM based Inverter, https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=8338637

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  • \$\begingroup\$ 1. The switching frequency might be high enough so 18 kHz could be sufficient, and the output waveform might be "close enough" to a sinewave or the distortion might be acceptable for the application. The article might have a clue inside. 2. There might also be a clue in the article about, for example, a relation between L and C e.g. why L is that big and C is that small. 3. Losses e.g. L's DC resistance and core loss. \$\endgroup\$ Mar 11 at 8:08
  • \$\begingroup\$ Hello Sir, if the 18 kHz is for pwm frequency that means the mosfets are being drived by 18 kHz, if so how am I gonna end up with 50 Hz output ? @RohatKılıç \$\endgroup\$
    – Mhan
    Mar 11 at 8:25
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    \$\begingroup\$ 72m and 0.04u doesn't bring a cut-off of 18k. The cut-off frequency, if I'm not wrong, is ~3 kHz for these values. So the attenuation at 18 kHz could be high, as shown in Antonio's answer below. \$\endgroup\$ Mar 11 at 8:29
  • \$\begingroup\$ Hello sir, thank you for your nice explanations but ı still couldn't understand even if the cut-off is 3 kHz how the output after the filter is reduced to 50 Hz ? because pwm frequency is so high and cut-off frequency is high as well . @RohatKılıç \$\endgroup\$
    – Mhan
    Mar 11 at 8:52
  • \$\begingroup\$ how the output after the filter is reduced to 50 Hz ? that is what PWM does. PWM is one of the few methods to adjust power. Duty-cycle (ratio of pulse duration to the period) adjusts the "average", so if you take the average of each switching cycle (1/18k = 55.5 us) you'll see that the duty-cycle varies, which basically means a different average value. So the filter gives you these average values (with some ripple of course). If you join all the average values together you'll get a sinewave (distorted but still looks like one). \$\endgroup\$ Mar 11 at 9:53

1 Answer 1

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The goal is don't pass the PWM frequency (i.e. 18 kHz).
So the "cut-off" frequency is 18 kHz @-40dB ...
Note also that the filter should be a "floating" device.

enter image description here

And here is the result of filtering PWM (+100V, -100V, modulated 100 %).
You can see that "harmonics" (and others ...) are below 0.5 V, so less than 1% distortion.

enter image description here

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  • \$\begingroup\$ Hello sir, if the 18 kHz is for pwm frequency that means the mosfets are being drived by 18 kHz, if so how am I gonna end up with 50 Hz output ? @Antonio51 \$\endgroup\$
    – Mhan
    Mar 11 at 8:23
  • \$\begingroup\$ Yes. And as I don't have access to the link on ieeexplore... , ... I can't say more about. \$\endgroup\$
    – Antonio51
    Mar 11 at 8:30

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