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the equation is used to calculate the power of RCD, I know this power is from leakage inductor the inductor energy can use 1/2Li^2 to express, but I don't know how to get this document equation.

Can someone give me some help?

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  • \$\begingroup\$ Simplify it. The *F bit converts the energy dissipated by the RC in the off part of the cycle to power. Then, the *delta t convers the power of the terms preceding it to an energy dissipated per cycle and that leaves you with half of Vclamp multiplied by Ipeak. Given that Ipeak approximately falls to zero linearly it should be clear. \$\endgroup\$
    – Andy aka
    Mar 15, 2021 at 17:34
  • \$\begingroup\$ @Andyaka Thanks, but I want to know how to derive this equation, because I used the 1/2*L*i^2 can't get the answer. and this document said this is from inductor energy. \$\endgroup\$
    – Jitter456
    Mar 15, 2021 at 17:54
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    \$\begingroup\$ Precisely and half Vclamp multiplied by Ipeak is the average power dissipated by the snubber per switching cycle. Have you considered the waveforms at all? \$\endgroup\$
    – Andy aka
    Mar 15, 2021 at 18:57
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    \$\begingroup\$ This brings fond memories of my early days with MOT : ) You can have a look at my more recent seminar The Dark Side of Flyback Converter where this is explained in details. \$\endgroup\$ Mar 15, 2021 at 20:56
  • \$\begingroup\$ @Andyaka I know the waveform, but I don't know why need to times 1/2 \$\endgroup\$
    – Jitter456
    Mar 16, 2021 at 1:43

2 Answers 2

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Look at the formula: -

enter image description here

It's energy multiplied by frequency and that energy is the energy dissipated by the clamp circuit per switching cycle. In other words Power = energy x frequency.

And, that energy is \$\frac{1}{2} \cdot V_{CLAMP}\cdot I_P\cdot \Delta_t\$

We also know that power x time = energy hence,

Power = \$\frac{1}{2} \cdot V_{CLAMP}\cdot I_P\$

And this is the average power dissipated by the RC clamp in a switching cycle. It's basically half of "V x I" (power) where \$V_{CLAMP}\$ is constant (as said in the linked document) and the current waveform will be a linearly ramping down current starting at a peak of the initial current dcue to the stored energy in the inductor.

When you multiply those two terms you get an average power of \$\frac{1}{2} \cdot V_{CLAMP}\cdot I_P\$.

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  • \$\begingroup\$ I don't understand why need to times 1/2, the power P= I*V, why have the 1/2 in this equation. \$\endgroup\$
    – Jitter456
    Mar 16, 2021 at 1:43
  • \$\begingroup\$ @Jitter456 because Vclamp is constant and I starts at Ip and falls roughly linearly to zero like a sawtooth. \$\endgroup\$
    – Andy aka
    Mar 16, 2021 at 10:18
  • \$\begingroup\$ Thanks, it is more clear rught now \$\endgroup\$
    – Jitter456
    Mar 16, 2021 at 22:51
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That formula comes from a balance of Energy at each cycle, in flyback topologies in discontinuous conduction mode.

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  • \$\begingroup\$ Thanks, Sir. but can you use the math equation to explain to me? I am not clear. \$\endgroup\$
    – Jitter456
    Mar 15, 2021 at 17:55

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