您
可以使用计时组来提高性能,让工具知道您的实际情况
从您的25 MHz时钟使能启用的任何触发器开始,在100 MHz时有4个时钟周期
在同一组中的任何其他触发器。
如果不这样做,那么所有25 MHz路径都是
由于它们的时钟输入,因此需要单个100 MHz时钟周期进行评估
现在与100 MHz相关联。
您可以使用时钟使能信号来定义25 MHz时序组。
只是确定
产生使能信号的触发器本身不会无意中落在其中
时间组。
如果在循环中使用信号,则会发生这种情况。
它通常发生在
你只是使用像ck_ena 以下为原文
Basically the tools are not smart enough to infer the timing for multicycle paths. You
can increase the performance using timing groups to let the tools know that you actually
have 4 clock cycles at 100 MHz to go from any flip-flop enabled by your 25 MHz clock enable
to any other flip-flop in that same group. If you don't, then all of your 25 MHz paths are
evaluated as requiring a single cycle of the 100 MHz clock, since their clock input
is tied to 100 MHz now.
You can use the clock enable signal to define your 25 MHz timing group. Just be sure
that the flip-flop that generates the enable signal itself doesn't inadvertently land in that
timing group. This can happen if you use the signal in a loop. It usually happens when
you're only dividing the clock by two using an assignment like ck_ena <= !clk_ena.
-- Gabor