Add some ASCII art explaining mux decomposition

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Eddie Hung 2019-07-10 12:20:04 -07:00
parent e573d024a2
commit 521971e32e
1 changed files with 21 additions and 0 deletions

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@ -214,6 +214,27 @@ module \$__XILINX_SHIFTX (A, B, Y);
assign Ax = {A[1], A};
\$__XILINX_MUXF78 fpga_hard_mux (.I0(Ax[0]), .I1(Ax[2]), .I2(Ax[1]), .I3(Ax[3]), .S0(B[1]), .S1(B[0]), .O(Y));
end
// Note that the following decompositions are 'backwards' in that
// the LSBs are placed on the hard resources, and the soft resources
// are used for MSBs.
// This has the effect of more effectively utilising the hard mux;
// take for example a 5:1 multiplexer, currently this would map as:
//
// A[0] \___ __ A[0] \__ __
// A[4] / \| \ whereas the more A[1] / \| \
// A[1] _____| | obvious mapping A[2] \___| |
// A[2] _____| |-- of MSBs to hard A[3] / | |__
// A[3]______| | resources would A[4] ____| |
// |__/ lead to: 1'bx ____| |
// || |__/
// || ||
// B[1:0] B[1:2]
//
// Expectation would be that the 'forward' mapping (right) is more
// area efficient (consider a 9:1 multiplexer using 2x4:1 multiplexers
// on its I0 and I1 inputs, and A[8] and 1'bx on its I2 and I3 inputs)
// but that the 'backwards' mapping (left) is more delay efficient
// since smaller LUTs are faster than wider ones.
else if (A_WIDTH <= 8) begin
wire [8-1:0] Ax = {{{8-A_WIDTH}{1'bx}}, A};
wire T0 = B[2] ? Ax[4] : Ax[0];