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I have simulated a H-bridge PWM temperature controller for Peltier module. To smooth the PWM output I have added a LC low pass filter, the cutoff frequency of the low pass filter is 734 Hz. I am giving a PWM of duty cycle 80% and frequency of 25 kHz. How can I minimize the initial current spike of the output, as this spike can damage the Peltier module (TEC1-12715)?

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    \$\begingroup\$ Actually, 734 Hz is the resonant frequency of the LC filter, and by the looks of it, that ringing happens approximately at the resonant frequency of the system (you have two LC filters), as that is the response of a LC filter to a step function. You can easily damp the ringing, by calculating a suitable damping resistor value over the inductor. \$\endgroup\$
    – Justme
    Aug 5 at 6:40
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    \$\begingroup\$ Well, you have made yourself a nice 14 dB gain peak at about 420 Hz. If you try and account for some series resistance in the inductors, it may help simulation. \$\endgroup\$
    – jonk
    Aug 5 at 6:40
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    \$\begingroup\$ Those two are two completely separate things. You determine the cutoff frequency based on how much ripple you want or tolerate at PWM frequency. Then you can select components randomly and damp it separately or you can design the LC filter values to match your load resistance to begin with so you get a non-ringing/overdamped DC step response. Your current (single-ended) filter has a Q factor of 5.8 and damping ratio of 0.085 so it rings. You could at least try to do some research yourself how to calculate RLC filters, they are pretty basic stuff. \$\endgroup\$
    – Justme
    Aug 5 at 7:06
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    \$\begingroup\$ thanks for advice @jonk \$\endgroup\$ Aug 5 at 7:38
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    \$\begingroup\$ Also the problem is not the duty cycle but the rate of change of duty cycle. If you change immediately from any duty cycle to another, that's a step function and the RLC will ring. If you have a slow rate of sweep from one duty cycle to another, like go from 0% to 100% linearly in 10ms or 100ms, there will be very little ringing. \$\endgroup\$
    – Justme
    Aug 5 at 7:48
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With the values roughly shown you have a very high-Q low pass filter: -

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Image taken from this interactive calculator.

May I suggest you make amendments to your values: -

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  • \$\begingroup\$ thank you for advice @Andyaka \$\endgroup\$ Aug 5 at 7:21
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    \$\begingroup\$ @powermachines Just don't forget that you have two inductors and two capacitors in your circuit (aside from the parallel cap across the load.) So the values computed here are a little different than they would be in your circuit case. Nothing much really changes, though, about the reality Andy is discussing, except some of the quantities involved. I took into account the entire resonant system in earlier computations. \$\endgroup\$
    – jonk
    Aug 5 at 7:46
  • \$\begingroup\$ @jonk a fair point. If you split the resistor into two equal halves and assume the centre node is 0 volts, you can analyse as a single inductor and 2 capacitors of 470 uF with a load of 1.35 ohms. Resonant frequency is now 519 Hz and Q is 4.14 but, still quite peaky hence, a modification of the values is still required. \$\endgroup\$
    – Andy aka
    Aug 5 at 8:21
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    \$\begingroup\$ @Andyaka Exactly! Nothing important changes about the points being made. Just a few nudges of values, is all. \$\endgroup\$
    – jonk
    Aug 5 at 8:31

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