ECE 551 Digital System Design & Synthesis Lecture 09 Synthesis of Common Verilog Constructs

Overview  Don’t Cares and z’s in synthesis  Unintentional latch synthesis  Synthesis & Flip-Flops  Gating the clock… and alternatives  Loops in synthesis 2

Synthesis Of x And z  Only allowable uses of x is as “don’t care”, since x cannot actually exist in hardware  in casex  in casez  in defaults of conditionals such as if, case, ?  Only allowable uses of z:  As a “don’t care”  Constructs implying a 3-state output 3

Don’t Cares  x assigned as a default in a conditional, case, or casex.  x, ?, or z within case item expression in casex  Does not actually output “don’t cares”!  Values for which input comparison to be ignored  Lets synthesizer choose between 0 or 1  Choose whichever helps meet design goals  z, ? within case item expression in casez  Same conditions as listed above except does not apply to x 4

Don’t Cares Example [1]  Give synthesizer flexibility in optimization 5 case (state) 3’b000: out = 1’b0; 3’b001:out = 1’b1; 3’b010:out = 1’b1; 3’b011: out = 1’b0; 3’b100: out = 1’b1; // below states not used 3’b101:out = 1’b0; 3’b110:out = 1’b0; 3’b111: out = 1’b0; endcase case (state) 3’b000: out = 1’b0; 3’b001:out = 1’b1; 3’b010:out = 1’b1; 3’b011: out = 1’b0; 3’b100: out = 1’b1; default: out = 1’bx; endcase VS

Don’t Cares Example [2] 6  Synthesis results for JUST output calculation: VS Area = Area = 15.24

Don’t Cares Example [3]  casex can reduce number of conditions specified and may aid synthesis 7 casex (state) 3’b000: out = 1’b1; 3’b001:out = 1’b1; 3’b010:out = 1’b1; 3’b011: out = 1’b1; 3’b100: out = 1’b0; 3’b101:out = 1’b0; 3’b110:out = 1’b1; 3’b111: out = 1’b1; endcase casex (state) 3’b0??: out = 1’b1; 3’b10?:out = 1’b0; 3’b11?:out = 1’b1; endcase VS

Synthesis Of z  Example 1: (Simple Tri-State Bus) assign bus_line = (en) ? data_out : 32’bz;  Example 2 (Bidirectional Bus) // bus_line receives from data when en is 0 assign bus_line = (!en) data : 32’bz; // bus_line sends to data when enable is 1 assign data = (en) ? bus_line : 32’bz; 8

Combinational Logic Hazards  Avoid structural feedback in continuous assignments, combinational always assign z = a | y; assign y = b | z;// feedback on y and z signals!  Avoid incomplete sensitivity lists in combinational always always (*) // helps to avoid latches  For conditional assignments, either:  Set default values before statement  Make sure LHS has value in every branch/condition  For warning, set hdlin_check_no_latch true before compiling 9

Synthesis Example [1] 10 module latch2(input a, b, c, d, output reg out); b, c, d) begin if (a) out = c | d; else if (b) out = c & d; end endmodule Area = 44.02

Synthesis Example [2] 11 module no_latch2(input a, b, c, d, output reg out); b, c, d) begin if (a) out = c | d; else if (b) out = c & d; else out = 1’b0; end endmodule Area = 16.08

Synthesis Example [3] 12 module no_latch2_dontcare(input a, b, c, d, output reg out); b, c, d) begin if (a) out = c | d; else if (b) out = c & d; else out = 1’bx; end endmodule When can this be done? Area = 12.99

Registered Combinational Logic  Combinational logic followed by a register clk) y <= (a & b) | (c & d); clk) case (select) 3’d0: y <= a; 3’d1: y <= b; 3’d2: y <= c; 3’d3: y <= d; default y <= 8’bx; endcase  The above behaviors synthesize to combinational logic followed by D flip-flops.  Combinational inputs (e.g., a, b, c, and d) should not be included in the sensitivity list 13

Flip-Flop Synthesis 14  Each variable assigned in an edge-controlled always block implies a flip-flop.

Gated Clocks  Use only if necessary (e.g., for low-power)  Clock gating is for experts! module gated_dff(clk, en, data, q); inputclk, en, data; outputreg q; wiremy_clk = clock & en; my_clk) q <= data; endmodule  When clock_en is zero flip-flop won’t update.  What problems might this cause? 15

Alternative to Gated Clocks module enable_dff(clk, en, data, q); input clk, en, data; output reg q; clock) if (en) q <= data; endmodule  Flip-flop output can only change when en is asserted  How might this be implemented? 16

Clock Generators  Useful when you need to gate large chunks of your circuit.  Adding individual enable signals adds up module clock_gen(pre_clk, clk_en, post_clk); inputpre_clk, clk_en; outputpost_clk; assign post_clk = clk_en & pre_clk; endmodule  Key: If you do this, you cannot use pre_clk on any flip-flops! Exclusively use post_clk.  This is still mostly recommended only for experts. 17

Loops  A loop is static (data-independent) if the number of iterations is fixed at compile-time  Loop Types  Static without internal timing control  Combinational logic  Static with internal timing control  Sequential logic  Non-static without internal timing control  Not synthesizable  Non-static with internal timing control  Sometimes synthesizable, Sequential logic 18

Static Loops w/o Internal Timing  Combinational logic results from “loop unrolling”  Example begin andval[0] = 1; for (i = 0; i < 4; i = i + 1) andval[i + 1] = andval[i] & a[i]; end  What would this look like?  How can we register the outputs? 19

Static Loops with Internal Timing  If a static loop contains an internal edge-sensitive event control expression, then activity distributed over multiple cycles of the clock always begin for (i = 0; i < 4; i = i + clk) sum <= sum + i; end  What does this loop do?  This might synthesize, but why take your chances? 20

Non-Static Loops w/o Internal Timing  Number of iterations is variable  Not known at compile time  Can be simulated, but not synthesized!  Essentially an iterative combinational circuit of data dependent size! n) begin andval[0] = 1; for (i = 0; i < n; i = i +1) andval[i + 1] = andval[i] & a[i]; end  What if n is a parameter? 21

Non-Static Loops with Internal Timing  Number of iterations determined by  Variable modified within the loop  Variable that can’t be determined at compile time  Due to internal timing control—  Distributed over multiple cycles  Number of cycles determined by variable above  Variable must still be bounded always begin continue = 1’b1; for (; continue; ) clk) sum = sum + in; if (sum > 8’d42) continue = 1’b0; end 22 Timing controls inside the always block make design guidelines fuzzy… Confusing, and best to avoid.

Unnecessary Calculations  Expressions that are fixed in a for loop are replicated due to “loop unrolling.”  Solution: Move fixed (unchanging) expressions outside of all loops. for (x = 0; x < 5; x = x + 1) begin for (y = 0; y < 9; y = y + 1) begin index = x*9 + y; value = (a + b)*c; mem[index] = value; end  Which expressions should be moved? 23

Unnecessary Calculations  Expressions that are fixed in a for loop are replicated due to “loop unrolling.”  Solution: Move fixed (unchanging) expressions outside of all loops. value = (a + b)*c; for (x = 0; x < 5; x = x + 1) begin index = x*9; for (y = 0; y < 9; y = y + 1) begin mem[index+y] = value; end Not so different from Loop optimization in software. 24

FSM Replacement for Non-Static Loops  Not all loop structures supported by vendors  Can always implement a loop with internal timing using an FSM  Can make a “while” loop easily by rewriting as FSM  Often use counters along with the FSM  All (good) synthesizers support FSMs!  Synopsys supports for-loops with a static number of iterations  Don’t bother with “fancy” loops outside of testbenches 25

Review Questions  Draw the Karnaugh-map for each statement, assuming each variable is one bit. if (c) z = a | b; else z = 0; if (c) z = a | b; else z = 1`x; Give a simplified Boolean expression for each statement  What would you expect to get from the following expressions? if (c) z = a & b; 26

Review Questions  Rewrite the following code, so that it takes two clock cycles and can use only a single multiplier and adder to compute z. Draw potential hardware for each piece of code. reg [15:0] z; reg [7:0] a, b, c, d clock) z = a*b + c*d; 27

Review Questions  What hardware would you expect from the following code? module what(input clk, input [7:0] b, output [7:0] c); reg [7:0] a[0:3]; integer i; clk) begin a[0] <= b; for (i = 1; i < 4; i=i+1) a[i] <= a[i-1] + b; end assign c = a[3]; endmodule 28