Xilinx ISE 6.3 confusion with CPLD logic results

Feb 21, 2006 2 Replies

My VHDL project has out grown a XC95108 CPLD, so I'll be using a XC95144 instead. After running the ISE synthesizer and fitter, all of the XC95144's Function Block Inputs are used. Using exhaustive fit mode, 92% of the function block inputs are used. This still doesn't leave much room for additional features. I then told ISE to use a XC95144XL, instead. Only 64% of the function block inputs are used, and



the other resources look good, too. Even though the XL is a 3.3 volt chip, it's 5 volt tolerant, so it should work. The odd thing is that the XL version uses _8 more_ flip flops than the standard version, and the timing report shows that the XL is faster than the std part, even though I selected 10 ns speed grade for both parts. I haven't simulated both chips yet. There's lots of information in the Fitter Report, but I don't know what I should be looking for. I'm a bit overwhelmed. So far, the Xilinx docs haven't helped. TIA,



-Dave Pollum



The difference in flip-flop count could be due to the synthesiser using another kind of implementation for some feature for your older CPLD. For example, there are a couple of ways to encode a FSM, some being smaller and others being a bit bigger but faster.

Secondly, the 10 ns speed grade only says something about the pad-to-pad delay. If I recall correctly, this is the cumulative delay of a typical signal path. Since the -XL is a whole different chip (3,3V instead of 5V logic) the component delays making up the total delay are bound to be different.

Hope that helps.

Wouter

Replace "spamtrap" with first name for email address.

Thanks! Your comments do make me feel a bit better, but I'm still curious about the different results. I guess I'll re-read the XC9500 and XC9500XL docs.

-Dave Pollum

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