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In many examples for DZ snubber zener voltage is calculated from maximum MOSFET rating and some arbitrary choosen voltage leakage (usually a percentage of Vin). They don't even use leakage inductance (available in some transformer datasheets).

It seems rather wrong, I would expect it to depend on Lp and Ip somehow. Or maybe, they use something close to maximum FET rating because it indeed provides best performance?

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  • \$\begingroup\$ I suggest you get a good book by Keith Billings to answer all your questions\?/ \$\endgroup\$ – Sunnyskyguy EE75 May 29 '18 at 2:37
  • \$\begingroup\$ This one? Thanks! \$\endgroup\$ – Maple May 29 '18 at 3:18
  • \$\begingroup\$ yes, Keith Billings is a Chartered Electronic Engineer who has specialized in switchmode power supply design and manufacture for 45 years. He owns the consulting company DKB Power Inc. Mr. Billings was lead author on Switching Power Supply Design, Third Edition, filling in for the late Abraham Pressman, and he frequently presents Pressman’s four day course on power supply design and his own one day course on magnetics. books.google.ca/… \$\endgroup\$ – Sunnyskyguy EE75 May 29 '18 at 3:55
  • \$\begingroup\$ e.g. powerelectronics.com/engineering-essentials/… \$\endgroup\$ – Sunnyskyguy EE75 May 29 '18 at 4:12
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Using your link and a specific circuit in the data sheet might help: -

enter image description here

The snubber zener is rated at 24 volts and the absolute maximum SW node chip voltage (from the DS) is 65 volts. Hence, with Vin at maximum (36 volts), the voltage that appears on SW is clamped to 36 volts + 24 volts + a bit more for luck + forward voltage of the schottky. I would say this will be no more than 62 volts and is inside the limit of 65 volts imposed by the data sheet.

It seems rather wrong, I would expect it to depend on Lp and Ip somehow.

Designers know that there may be 5% leakage inductance (worst case) and that there will be a back emf that needs to be quenched but knowing what the leakage inductance is is irrelevant to designing the snubber clamp voltage.

Sure there may be a peak current of (say) 1 amp in the primary when it gets open circuited and this (with a leakage inductance of 5% x 9 uH) is an energy storage of 225 nano joules that needs to be burnt by the snubber. It needs to burn it maybe 500,000 times per second and this results in a power dissipation of 112 mW.

So we do need to understand the leakage but only to determine the power of the snubber.

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  • \$\begingroup\$ Thanks. Just one more, to make sure I understand this. Zener voltage selected by your calculations (and by design example in the datasheet) keeps max Vsw close to RATED max of the switch (save safety margin). What I wanted to know, is: wouldn't it be beneficial if it is kept MUCH LESS than that, i.e. discharge coil faster. The power dissipation will still be the same. \$\endgroup\$ – Maple May 29 '18 at 18:50
  • \$\begingroup\$ Aha, no.... the higher the voltage that it clamps, the quicker the energy is dissipated. Simple reason; power is volts x amps. Amps are dictated by the charging phase and the flyback phase leakage inductance voltage is the only controllable entity hence, more volts equals quicker discharge time. Or put another way, if a pure inductor carrying one amp we’re shorted out that current would flow for all time because there is no energy dissipation. \$\endgroup\$ – Andy aka May 29 '18 at 21:07

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