您无法真正避免延迟，因为DCM需要时间来锁定输入信号。
如果您只想将输入时钟除以2.5，那么您可以使用结构触发器代替
一个DCM。
这种方法的缺点是产生的20 MHz信号不会
相位与输入时钟对齐。
顺便说一句，你的应用程序需要在重置后如此快地运行时钟？
- Gabor
- Gabor
以上来自于谷歌翻译
以下为原文
You can't really avoid the delay because the DCM needs time to lock to the input signal.
If all you want is to divide the input clock by 2.5, then you could use fabric flip-flops instead
of a DCM. The down side to this method is that the resulting 20 MHz signal will not be
phase aligned with the input clock.
By the way, what is your application that needs the clock running so soon after reset?
-- Gabor
-- Gabor
您无法真正避免延迟，因为DCM需要时间来锁定输入信号。
如果您只想将输入时钟除以2.5，那么您可以使用结构触发器代替
一个DCM。
这种方法的缺点是产生的20 MHz信号不会
相位与输入时钟对齐。
顺便说一句，你的应用程序需要在重置后如此快地运行时钟？
- Gabor
- Gabor
以上来自于谷歌翻译
以下为原文
You can't really avoid the delay because the DCM needs time to lock to the input signal.
If all you want is to divide the input clock by 2.5, then you could use fabric flip-flops instead
of a DCM. The down side to this method is that the resulting 20 MHz signal will not be
phase aligned with the input clock.
By the way, what is your application that needs the clock running so soon after reset?
-- Gabor
-- Gabor
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