首页 > 微波/射频 > RFIC设计学习交流 > 请教RCD钳压电路的设计公式的一个疑问?

请教RCD钳压电路的设计公式的一个疑问?

录入:edatop.com    阅读:
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,我理解是漏感电压的能量会在电阻上消耗掉,但是为什么电阻上消耗的功率那样计算的...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

申明:网友回复良莠不齐,仅供参考。如需专业解答,请学习本站推出的微波射频专业培训课程

上一篇:关于电荷泵的效率
下一篇:如何求出UnCox

射频和天线工程师培训课程详情>>

  网站地图