请问RCD嵌压电路的设计公式的疑问?

miklystone

Figure 3. MOSFET turn-off waveforms with clamping.

This implies that the power dissipated in the clamp at turn-off is

PCLAMP = (1/2) × VCLAMP × ICLAMP × ∆t × f

Where, f is the switching frequency.  

However,

∆t = (LLP × IP)/(VCLAMP - VOUT/N).

The power dissipated in the clamp is through the resistor. Therefore  

RCLAMP = [2 × VCLAMP × (VCLAMP - VOUT/N)]/( LLP × IP² × f)  (a)

It is important to minimize the ripple, Vripple, superimposed on VCLAMP to keep the MOSFET drain voltage close to
the clamp voltage. The minimum value for the capacitor, CCLAMP, is therefore CCLAMP = VCLAMP/ (Vripple × RCLAMP
× f

我想问公式（a）怎么理解，电阻上消耗的功率为什么那样计算的...我看的raidly也是这样写得，但是我没理解到，请大侠解释下...
参考资料：maxim AN848

china_ic

能不能具体到公式的某一个地方，“为什么这么写？”，你认为应该怎样写，这样大家会更容易理解/。

miklystone

回复 2# china_ic


    1/2 LP*(Ip^2)*f这个应该是在漏感上的功率吧，平衡时，这些功率应该在电阻上消耗掉。把表达式A移下就发现电阻上消耗的功率为
[Vclamp*（Vclamp-Vo/N）]/R，就是这个电阻上消耗的功率我不明白为什么是这样写得。
谢谢 china_ic!