I am modelling a PWM switching whose model is shown here below, where the Pulse generator represents my previous PWM stage. enter image description here

So far, I am experiencing some weird behaviour in the inductor, I see how it does not exhibits a symmetrical behaviour in its current, as in show below when showing the input to the MOSFET and the output after the filter

enter image description here

I wonder why does this happen, is it owed to how it was modelled? Any ideas?

Thanks in advance,


Focusing in the inductor current ,I_L, ** it is shown that there is not symmetry in the current in the Y axe (A) coloured in red**

enter image description here

  • 2
    \$\begingroup\$ Can you explain what it is that you consider to be wrong? \$\endgroup\$
    – Andy aka
    Dec 8, 2016 at 11:39
  • \$\begingroup\$ Why would it be symetrical? The current from the primary during the "charging"/up-slope phase is determined by the input voltage and the "discharge"/down-slope is determined by the output voltage, both for a given inductance value. \$\endgroup\$
    – winny
    Dec 8, 2016 at 12:21
  • \$\begingroup\$ @Andyaka What I would expect is that the Vout averaged slope is the same as the one of Vin while discharging \$\endgroup\$
    – ndarkness
    Dec 8, 2016 at 12:28
  • \$\begingroup\$ @winny shouldn't the peak of Vout start decreasing at the same time at the one of Vin? \$\endgroup\$
    – ndarkness
    Dec 8, 2016 at 12:34
  • 1
    \$\begingroup\$ The apparent slope difference is caused by your input and output waveforms having different scales. Output voltage should be normalized from 14V (maximum output voltage range) not the peak-to-peak voltage of the output waveform (which is reduced due to low-pass filtering). \$\endgroup\$ Dec 8, 2016 at 13:17

1 Answer 1


Here's a quick simulation of a classic LR low pass filter fed with an asymmetrical sawtooth wave form: -

enter image description here

Sawtooth rise time is 20 us and fall time is 80 us. Period is 100 us. Note that I just used numbers that were conveneient and I wasn't trying to exactly simulate your circuit. The result is typical of an LR LPF.

This looks just like what you are describing so, if you are still not convinced, please expain what facet of the waveform in your question is not described here.

  • \$\begingroup\$ if you compare your outcome signals with my first figure, you can see that I have a different slope in Vout averaged and Vin averaged while discharging, whereas you have the same slope while discharging. That's wat doesn't convince me ;) \$\endgroup\$
    – ndarkness
    Dec 8, 2016 at 12:59
  • \$\begingroup\$ The key concept here is that the inductor current is proportional to the integral of its applied terminal voltage. Thinking about it the other way around, the voltage is the derivative (slope) of the current. The slope of the current does indeed start to reduce as soon as the voltage starts reducing, but it's still rising, so the actual peak of the current is delayed until the inductor terminal voltage passes through zero. \$\endgroup\$
    – Dave Tweed
    Dec 8, 2016 at 13:04
  • \$\begingroup\$ @ndarkness I don't fully understand where Vin (averaged) comes from or how you calculated it. Maybe you should give details about this if it is some numerical average through some form of digital filtering? \$\endgroup\$
    – Andy aka
    Dec 8, 2016 at 13:06
  • \$\begingroup\$ @ndarkness I don't see Vin (averaged) as a measurement node on your schematic so this adds to the lack of clarity in your question. \$\endgroup\$
    – Andy aka
    Dec 8, 2016 at 13:13
  • \$\begingroup\$ Ok I will do it in a different way, I have replicated @Andyaka circuit so it looks like this where the output signals are these and as you see the voltage in the inductor seems to be different accross its terminals \$\endgroup\$
    – ndarkness
    Dec 8, 2016 at 13:41

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