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I know about inverters is as follows (correct me if I am wrong) DC source such as battery fed current into transformer by controlled switches such as mosfet or IGBT. This controlled current fed into primary side of step up transformer. While the control circuit modulate the high frequency carrier wave and referance sine wave. By modulating we get gate signals as follow. Pwm waveform

Using fullbridge we converte DC waveform into single sine wave PWM signal. And when fedding it to primary winding of inverter we use LC low pass filter and use filter also on output(I read on research paper but not sure). i explain this circuit(found this picture later)

Ok my question is:

How to filter this PWM sine wave into sine wave? Does LC filter do the job?

This inverter circuit I explain here is seems much simple than commercial inverter. Am I missing anything?

Does output waveform change if load varies?

Correction and suggestions are welcome for question. Thank you.

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    \$\begingroup\$ The output of an H-bridge is low impedance, not high impedance, so the transformer is voltage-fed, not current-fed. Once you have that distinction right, the rest will follow. \$\endgroup\$
    – Neil_UK
    Dec 3, 2017 at 6:10
  • \$\begingroup\$ Also, you seem to assume open loop with a look-up PWM pattern. There is no need for this. You can benefit from a half-wave sine lookup and close the loop by comparing this reference to your output and adjust the duty-cycle accordingly. \$\endgroup\$
    – winny
    Dec 3, 2017 at 8:11

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  1. Yes the LC filter will do the job. Use a choke of 1mH or more, or 2 x 0.5mH, and 1uf or 2 x 2.2uF.

  2. The commercial inverter has the first stage push pull inverter from 12V to 400V, high frequency - small transformer. Then the second stage is the H bridge, no transformer, output filter. It doesn't mean that your circuit is bad, it is just more heavy, but with only one transistor stage so it can even outperform the commercial in terms of efficiency.

  3. Yes of course. You can add a small transformer on output 220/12 for feedback. Let suppose you have a look up table: sine -> PMW, then you can simply multiply the PWM duty ratio proportionally with voltage drop on the output.

Link to italian site for inverter DIY

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  • \$\begingroup\$ You took Value of L and C that has resonant frequency of 5.03 kHz. But in this inverter we have to pass the frequency of 50 Hz. So this value you calculate from any equation or just assume. \$\endgroup\$
    – Ridham
    Dec 4, 2017 at 4:49
  • \$\begingroup\$ @user3785133 Just from the experimental work in that link. But perhaps is better to use a smaller capacitor or to simulate. \$\endgroup\$
    – Marko Buršič
    Dec 4, 2017 at 13:07

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