What's the best way to filter the output of a H-bridge?

I have a PWM output at +/- 20 V and I want to get a 10V RMS sine wave out at 50hz

Also is there a way to do it without inductors?

  • \$\begingroup\$ electronics.stackexchange.com/questions/109466/… \$\endgroup\$ – Sada93 May 15 '15 at 6:33
  • \$\begingroup\$ Best is a subjective word, and current flow matters. I could do this with no inductors and a circuit called a Bubba Oscillator if it didn't need much current. We need to know everything about what you are attempting to do so that we can help you solve your actual problem. This question is focused on a single part of a partial solution - please edit it to include as much information as possible. Schematics are great, but a thorough discussion of what you've done so far will suffice. \$\endgroup\$ – Sean Boddy May 15 '15 at 7:23
  • \$\begingroup\$ What's the load? What PWM frequency? \$\endgroup\$ – Andy aka May 15 '15 at 8:46
  • \$\begingroup\$ whats the reasoning behind wanting todo it without L? an LC (for a d/dt limiter) or an LCR is the most practical way todo this. \$\endgroup\$ – JonRB May 15 '15 at 14:47

Sure there's a way to do it without inductors providing you are happy about not taking two much power from the H bridge to produce the sinewave. You can use RC low pass filters: -

enter image description here

If your PWM frequency is (say) 10kHz, after filtering with an RC filter having a 100Hz 3dB point you can calculate the attenuation at 20kHz by using the following calculator. I chose R = 160 ohms and C = 10uF.

The cut-off (3dB point) was 100 HZ and, from the graph it appears that 10 kHz is 40 dB down in amplitude.

Of course, knowledge about RC filters (1st order types) informs us that the amplitude rolls away at 20 dB per decade above the 3 dB point. Between 100 Hz and 10 kHz there are two decades of frequency hence a total of 40 dB (about 100:1 attenuation in terms of volts in and volts out).

You can use higher order filters made of RC networks too but you'll not get more than a few watts from this type of filter because of the resistors are in series with the signal.

The same page as the calculator as a host of other filter calculators


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.