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Supressing voltage spikes using an LC filter

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Ron H said:
Because the load is a square wave, which has harmonics that extend far beyond the fundamental. If one of the harmonics is at the resonant frequency of the filter, it will ring.

EDIT:
You sucked me in there. :) You read something wrong in my sim. The square wave is at the resonant frequency of the filter.
What I said about harmonics is true, but the amplitude of the 15th harmonic will be 1/15th the amplitude of the fundamental.

Opps sorry... youre right.

Even with the addition to an LC filter, if you design the feedback compensation to have a bandwidth of a frequency much less than the transient load frequency (for example, I design the feedback to have a bandwith of 10k and the load transient is 151k in your example), wouldnt you not see the oscillation like in your example?
 
FusionITR said:
Opps sorry... youre right.

Even with the addition to an LC filter, if you design the feedback compensation to have a bandwidth of a frequency much less than the transient load frequency (for example, I design the feedback to have a bandwith of 10k and the load transient is 151k in your example), wouldnt you not see the oscillation like in your example?
Isn't the filter outside the feedback loop? I don't think you would want to put it inside! Regardless, if your regulator has a bandwidth of 10kHz, its closed-loop output impedance will go up, not down, at frequencies above 10kHz (assuming no passive filter on the output).
 
Ron H said:
Isn't the filter outside the feedback loop? I don't think you would want to put it inside! Regardless, if your regulator has a bandwidth of 10kHz, its closed-loop output impedance will go up, not down, at frequencies above 10kHz (assuming no passive filter on the output).

Yes the filter is outside the feedback loop (as suggested by smps design books and other).

Ok as far as I understand the feedback design, the purpose of limiting the feedback bandwidth to, say, 10khz (if the switching frequency is 50khz) was that any transients inserted into the system much greater than 10khz was that when the transients completed the loop, the gain will be very small (pratically zero) when it shows up back at the input, thus will not oscillate (but still have feedback at lower frequencies).

So even if the LC filter is placed outside the system, wouldnt that not effect it? Please correct me if I am wrong about my analysis, I'm here to learn.
 
FusionITR said:
Yes the filter is outside the feedback loop (as suggested by smps design books and other).

Ok as far as I understand the feedback design, the purpose of limiting the feedback bandwidth to, say, 10khz (if the switching frequency is 50khz) was that any transients inserted into the system much greater than 10khz was that when the transients completed the loop, the gain will be very small (pratically zero) when it shows up back at the input, thus will not oscillate (but still have feedback at lower frequencies).

So even if the LC filter is placed outside the system, wouldnt that not effect it? Please correct me if I am wrong about my analysis, I'm here to learn.
Well, I can't imagine how the response of a filter on the output of your converter could be much affected by the compensation of your converter, but I'm no expert on flyback converters. I'm just an old analog engineer with lots of experience with filters and feedback networks. Some of the other guys here have lots of hands-on experience with DC-DC converters. I hope one of them will jump in here if they have any insight.
 
Ron H said:
This will add a very high resonant impedance between the load and the supply. Do you really think this is a good idea?
The capacitor will stop this being a problem.
 
Last edited:
Sorry, the circuit I had in mind is very slightly different.
 

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But the DC componant goes straight through the inductor, charging the decoupling capacitor which supplys the circuit when L and C are blocking the resonant frequency.
 
Hero999 said:
But the DC componant goes straight through the inductor, charging the decoupling capacitor which supplys the circuit when L and C are blocking the resonant frequency.
Here's a Bode plot of the response of the filter to the load. You can change values around, but the resonant frequency will always be 1/(2*pi*sqrt(L*Ctotal)), where Ctotal is the sum of the two capacitances. The "Q" (bandwidth) will vary depending on load impedance, but ALL of the "signal" generated by the load shows up on VCC at the resonant frequency.
 

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  • LC filter AC Bode.PNG
    LC filter AC Bode.PNG
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Hero999 said:
But the DC componant goes straight through the inductor, charging the decoupling capacitor which supplys the circuit when L and C are blocking the resonant frequency.
Here's a Bode plot of the response of the filter to the load. You can change values around, but the resonant frequency will always be 1/(2*pi*sqrt(L*Ctotal)), where Ctotal is the sum of the two capacitances. The "Q" (bandwidth) will vary depending on load impedance, but ALL of the "signal" generated by the load shows up on VCC at the resonant frequency.
 

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  • LC filter AC sch.PNG
    LC filter AC sch.PNG
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  • LC filter AC Bode.PNG
    LC filter AC Bode.PNG
    18.1 KB · Views: 146
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