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I have a combinational circuit and I would like to find its critical path in design compiler. Essentially, I want to find out by how much the combinational logic will reduce the maximum clock frequency of the larger sequential design. For this purpose, I have added registers along the input of the combinational circuit (a simple multiplier in this case) which are clocked on the rising edge of a clock as advised in How to find the critical path delay of a big combinational block. I then run create_clock clk -period 5 -name clk and report_qor in DC, but I'm getting a Critical Path Length of 0.00 ns. This looks odd. If I move the multiplier directly to the test module, I get a more reasonable-looking Critical Path Length of 4.88 ns however.

module my_multiplier(
    output reg [31:0] out,
    input      [15:0] in1, in2,
    input             enable
);

always @(*) begin
   if (enable) begin
         out = in1 * in2;
   end
end

endmodule

I've created a separate module to instantiate the multiplier circuit and also clock the inputs to the multiplier:

module Test_multiplier_Tcrit(
    output  [31:0] out,
    input   [15:0] in1, in2,
    input          clk, enable
);

reg  [15:0] in1_reg, in2_reg;

my_multiplier my_multiplier(.out(out), .in1(in1_reg), .in2(in2_reg), .enable(enable)); 

always @(posedge clk) begin
   in1_reg <= in1; 
   in2_reg <= in2;
end

endmodule
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Try putting a register on the output as well. Generally the timing analysis is done register-to-register, so without an output register it may not be able to give you a good answer.

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  • \$\begingroup\$ I've already tried the register to output but get the same result. I used: always @* out <= out1; after redeclaring the output port out as a register. \$\endgroup\$ – Mowgli Nov 14 '14 at 10:39
  • \$\begingroup\$ Move the assignment to the output port out into a sequential always block to get a register always@(posedge clk) out <= out1; \$\endgroup\$ – andrsmllr Nov 14 '14 at 22:24
  • \$\begingroup\$ @damage Doesn't work. Have already tried that. \$\endgroup\$ – Mowgli Nov 14 '14 at 23:03

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