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Here is what I read from Ridley Engineering http://www.ridleyengineering.com/images/phocadownload/12_%20flyback_snubber_design.pdf :

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I do not get why if we increase the clamp voltage, ie, if we reduce the capacitor and increase the resistance, we reduce the dissipation as the energy to dissipate is always the same and depends only on the leakage inductance of the transformer. Nevertheless i could understand that if we reduce the time during which the inductance release its energy, the transition between the on state and the off state will be faster and then there is less losses. But this has nothing to do with power loss into the snubber. Could someone explain it to me ?

Thank you very much !

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    \$\begingroup\$ If you reduce the capacitor value then clamp voltage rises but, if you reduce the resistance, clamp voltage falls. I don't think you meant that in your final paragraph did you? Did you mean increase the resistance and capacitance? If you reduce capacitance, you would naturally increase resistance to maintain the same RC time. \$\endgroup\$
    – Andy aka
    Commented Aug 4, 2020 at 12:08
  • \$\begingroup\$ You re right ! This is my bad ;) \$\endgroup\$
    – Jess
    Commented Aug 4, 2020 at 12:36
  • \$\begingroup\$ It may be an error of interpretation from my side . I will edit the question \$\endgroup\$
    – Jess
    Commented Aug 4, 2020 at 12:40
  • \$\begingroup\$ I highlight the setences :) \$\endgroup\$
    – Jess
    Commented Aug 4, 2020 at 12:44

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I don't think that the assertion is theoretically true.

The snubber (aka diode-RC clamp) power dissipation will be the same if you halve the capacitance. The power dissipation is produced by the resistor because it needs to "reset" the voltage acquired by the clamp capacitor after it received leakage inductance energy during flyback. It has a certain length of time to reset that voltage hence RC must be fixed.

So, if you half the capacitance, you must double the resistance.

Given this well-known formula for capacitor energy (W): -

$$W = \dfrac{1}{2}\cdot CV^2$$

You can re-arrange for voltage: -

$$V = \sqrt{\dfrac{2\cdot W}{C}}$$

So, if you halved the capacitance, voltage would rise by \$\sqrt2\$.

The power dissipated by the resistor would be related to the peak voltage thus: -

$$P = \dfrac{V_{PK}^2}{R}$$

So if the capacitance halved, R would double and \$V_{PK}\$ would increase by \$\sqrt2\$: -

$$P = \dfrac{\sqrt2^2\cdot V_{PK}^2}{2R} = \dfrac{V_{PK}^2}{R}$$

In other words no change in dissipation.

There may be other nuances I've not considered of course but for a simple circuit, I believe the power dissipated by the clamp is unchanged when C changes.

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  • \$\begingroup\$ I do not know if it is correct ! But I totally agree with your demonstration ! \$\endgroup\$
    – Jess
    Commented Aug 4, 2020 at 14:20
  • \$\begingroup\$ I found a demonstration which assumes that the voltage across the capacitor is in average constant but it continues to write that the power dissipated into the snubber depends on the resistance of the clamp. I will take a look closer ! If you are interested in : how2power.com/pdf_view.php?url=/newsletters/1511/articles/…. It begins page 7 ;) \$\endgroup\$
    – Jess
    Commented Aug 6, 2020 at 7:59

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