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Old 11-22-2005, 09:46 PM
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Default Re: Disabling Xilinx clock enable usage...

"Antti Lukats" <[email protected]> wrote in message
news:[email protected]
> "johnp" <[email protected]> schrieb im Newsbeitrag
> news:[email protected]
>> I'm working on a high speed design in a Xilinx V2Pro and I'm running
>> into a timing
>> problem. Instead of packing logic into LUTs, XST wants to use the
>> Enable
>> signal in the CLB. To use the Enable, it needs to use an extra LUT to
>> create
>> the Enable signal, so I get routing delays and an extra CLB delay.
>> Here's some sample code:
>> req [3:0] sig4;
>> wire [3:0] sig3;
>> always @(posedge clk)
>> if (sig1 & ~sig2)
>> sig4 <= sig3;
>> Xilinx could fit this into 4 CLBs total by simply using the 4 LUTs and
>> the 4 flip-flops.
>> Each LUT would handle one bit of sig4.
>> Instead, XST uses a LUT to create (sig1 & ~sig2), then feeds that
>> output to the
>> Enable pins on 4 flip-flops. I now get the delay through the LUT and
>> routing delays
>> to my flip-flops.
>> Any way to tell XST to not use the Enable signal and force it to use
>> the LUTs for
>> this section of logic?
>> Thanks!
>> John Providenza

> Hi John,
> in your example XST does exactly what it should do given your code.
> if you want the synthesis to avoid using clock enable then you should
> rewrite your code
> antti

I would respectfully disagree.

A decent synthesizer should *not* produce an extra level of logic with an
actual increase in area unless - and it's hard to see this as the case - the
extra fanout for a heavily loaded signal causes timing problems elsewhere in
the design.

In a properly constrained design, a decent synthesizer should *not* produce
logic that violates the timing constraints if there's an available solution
that meets the timing. Unfortunately we have to spend much of our time
tuning things manually to get the "obvious" to happen.

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