EN2911X: Reconfigurable Computing Lecture 06: Verilog (3)
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1 EN2911X: Lecture 06: Verilog (3) Prof. Sherief Reda Division of Engineering, Brown University Fall 09
2 Level sensitive latch (D-Latch) The Verilog implementation of a D-latch is an always block that makes a nonblocking assignment ( < = ) of d to q when the g input is nonzero. When g input is zero, then the always block does not make any assignment to q, causing the synthesis tool to infer a latch on the q output as the q output must retain its last known d value when g was nonzero Nonblocking assignments ( <= ) as opposed to blocking assignments ( = ) should be used in always blocks that are used to synthesize sequential logic; [from Thornton & Reese]
3 Edge-triggered storage element (D-FF) symbol is used to specify an event control. Statements can be executed on changes in signal value or at a positive (posedge) or negative (negedge) transition of the signal. In general, edge-triggered storage elements are preferred to levelsensitive storage elements because of simpler timing requirements The 1-bit edge-triggered storage elements provided by FPGA vendors are DFFs because of their simplicity and speed. [Thornton & Reese]
4 DFF chains [Thornton & Reese] Each nonblocking assignment synthesizes to a single DFF whose input happens to be the output of another nonblocking assignment. The ordering of these nonblocking assignments within an always block does not matter
5 DFF with asynchronous/synchronous inputs [Thornton & Reese]
6 Blocking vs. non-blocking statements [Thornton & Reese] Zero-delay blocking assignments are so named because the assignment of the right-hand side (RHS) to the left- hand side (LHS) is completed without any intervening Verilog code allowed to execute, i.e., the assignment blocks the execution of the other Verilog code. For nonblocking assignments within an always block, all RHS expressions are evaluated, and are only assigned to the LHS targets after the always block completes.
7 Avoid combinational loops [Thornton & Reese]
8 Guidelines to avoid frustration Use blocking assignments ( = ) in always blocks that are meant to represent combinational logic. Use nonblocking assignments ( <= ) in always blocks that are meant to represent sequential logic. Do not mix blocking and nonblocking assignments in the same always block. If an always block contains a significant amount of combinational logic that requires intermediate wires (and thus, intermediate assignments), then place this logic in a separate always block. If an always block for combinational logic contains complicated logic pathways due to if-else branching or other logic constructs, then assign every output a default value at the beginning of the block. This ensures that all outputs are assigned a value regardless of the path taken through the logic, avoiding inferred latches on outputs. Do not make assignments to the same output from multiple always blocks [Thornton & Reese]
9 Loops in Verilog integer count; integer y=1; integer x=2; initial for (count = 0; count < 128; count = count + 1) begin x <= x + y; y <= x; end initial count = 0; while (count < 128) begin.. count = count +1; end initial count = 0; repeat(128) begin.. count = count +1; end Must contain a number or a signal value; only evaluated once at the beginning
10 D2 example: A 1 second blinking light module sec (input CLOCK_50, output reg [8:0] LEDG); integer count=0; initial LEDG[0]=1'b0; CLOCK_50) begin count=count+1; if(count == 50_000_000) begin count=0; if(ledg[0]) LEDG[0]=1'b0; else LEDG[0]=1'b1; end end endmodule
11 Tasks Tasks are declared with the keywords task and endtask. Tasks can have input, inout, and output arguments to pass values (different than in modules). Tasks can have local variables but cannot have wires. Tasks can only contain behavioral statements. Tasks do not contain always or initial statements but are called from always blocks, initial blocks, or other tasks and functions. Can operate directly on reg variables defined in the module module always begin BOP (AB_AND, AB_OR, AB_XOR, A, B); BOP (CD_AND, CD_OR, CD_XOR, C, D); end task BOP; output [15:0] ab_and, ab_or, ab_xor; input [15:0] a, b; begin ab_and = a & b; ab_or = a b; ab_xor = a ^ b; end endtask endmodule
12 Example: clock display on DE2 Last lecture we had a simple example of a 1 second blinking LED Let s generalize it to a clock display minutes seconds HEX3 HEX2 HEX1 HEX0
![Task to display digits task digit2sev(input integer digit, output [6:0] disp); begin if (digit == 0) disp = 7'b1000000; else if (digit == 1) disp = 7'b1111001; else if (digit == 2) disp = 7'b0100100;](/docs-images/91/105731273/images/13-2.jpg "else if (digit == 3) disp = 7'b0110000; else if (digit == 4) disp = 7'b0011001; else if (digit == 5) disp = 7'b0010010; else if (digit == 6) disp = 7'b0000011; else if (digit == 7) disp = 7'b1111000;")
13 Task to display digits task digit2sev(input integer digit, output [6:0] disp); begin if (digit == 0) disp = 7'b ; else if (digit == 1) disp = 7'b ; else if (digit == 2) disp = 7'b ; else if (digit == 3) disp = 7'b ; else if (digit == 4) disp = 7'b ; else if (digit == 5) disp = 7'b ; else if (digit == 6) disp = 7'b ; else if (digit == 7) disp = 7'b ; else if (digit == 8) disp = 7'b ; else if (digit == 9) disp = 7'b ; end endtask
14 One way module clock(clock_50, HEX0, HEX1, HEX2, HEX3); output reg [6:0] HEX0, HEX1, HEX2, HEX3; input CLOCK_50; integer count=0; reg [3:0] d1=0, d2=0, d3=0, d4=0; CLOCK_50) begin count=count+1; if (count == 50_000_000) begin count=0; d1 = d1+1; if(d1 == 10) begin d1 = 0; d2 = d2+1; if(d2 == 6) begin d2 = 0; d3 = d3 + 1; if(d3 == 10) begin d3 = 0; d4 = d4 + 1; if(d4 == 6) d4 = 0; end end end digit2sev(d1, HEX0); digit2sev(d2, HEX1); digit2sev(d3, HEX2); digit2sev(d4, HEX3); end end task digit2sev; endtask endmodule
15 Resource utilization is 152 LEs
16 Second code module clock(clock_50, HEX0, HEX1, HEX2, HEX3); input CLOCK_50; output reg [6:0] HEX0, HEX1, HEX2, HEX3; integer count=0; reg [15:0] ticks=16'd0; reg [5:0] seconds=6'd0, minutes=6'd0; initial display_time(seconds, minutes); CLOCK_50) begin count = count+1; if (count == 50_000_000) begin count=0; ticks = ticks + 1; seconds = ticks % 60; minutes = ticks / 60; display_time (seconds, minutes); end end
17 task display_time(input [5:0] s, input [5:0] m); begin digit2sev(s%10, HEX0); digit2sev(s/10, HEX1); digit2sev(m%10, HEX2); digit2sev(m/10, HEX3); end HEX3 HEX2 HEX1 HEX0 endtask task digit2sev(input integer digit, output [6:0] disp); begin if (digit == 0) disp = 7'b ; else if (digit == 1) disp = 7'b ; else if (digit == 2) disp = 7'b ; else if(digit == 3) disp = 7'b ; else if(digit == 4) disp = 7'b ; else if(digit == 5) disp = 7'b ; else if(digit == 6) disp = 7'b ; else if(digit == 7) disp = 7'b ; else if(digit == 8) disp = 7'b ; else if(digit == 9) disp = 7'b ; end endtask endmodule
18 Resource utilization for 2nd code Circuit consumes 611 LEs (2% of the chip logic resources). You have to be careful! Changing ticks, seconds and minutes to integer increases area to become 2500 LEs (8% of the utilization)
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