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I am using D flip flops in my clock divider circuit. I have started with one FF and moving up with the number of divisions I want to have in my clock. This is how I want my D ffs to work.

enter image description here

Now I have my Verilog code for one FF.

   module dff (clk, reset, d, q, qb); // D flip flop one level
   input      clk;
   input      reset;
   input      d;
    output     q;
    output     qb;
    reg        q;
    assign qb = ~q;

    always @(posedge clk or posedge reset)
    begin
      if (reset) begin
      // Asynchronous reset when reset goes high
      q <= 1'b0;
    end else begin
      // Assign D to Q on positive clock edge
      q <= d;
    end
   end
  endmodule

and the test bench

    module testbench;
    reg clk_t;
    reg reset_t;
    reg d_t;
    wire q_t;
    wire qb_t;  
   // Instantiate design under test
   dff dff(.clk(clk_t), .reset(reset_t), .d(d_t), .q(q_t), .qb(qb_t));          
  initial begin
  // Dump waves
  $dumpfile("dump.vcd");
  $dumpvars(1);

  $display("Reset flop.");
  clk_t = 0;
  reset_t = 1;
  d_t = 1'bx;
  display;

  $display("Release reset.");
  d_t = 1;
  reset_t = 0;
  display;

  $display("Toggle clk.");
  clk_t = 1;
  display;

 // Dump waves
 $dumpfile("dump1.vcd");
 $dumpvars(1);

 $display("Reset flop.");
 clk_t = 0;
 reset_t = 1;
 d_t = 1'bx;
 display;

 $display("Release reset.");
 d_t = 1;
 reset_t = 0;
 display;

 $display("Toggle clk.");
 clk_t = 1;
 display;  
 end

 task display;
 #5 $display("d_t:%0h, q_t:%0h, qb_t:%0h", d_t, q_t, qb_t);    
 endtask

endmodule

and the simulation enter image description here

When I add the second flip flop with this module:

     module halfclk(clk, reset, d, q, qb); // D flip flop two levels: 1/2 clock
     input clk;
     input reset;
     input d;
     output q;
     output qb;
     wire w1, w2;   

     dff dff1(clk, reset, d, q, qb);
     dff dff2(w1, reset, d1, w2, qb1);

    endmodule    

and changing the module line in test bench to:

dff halfclk(.clk(clk_t), .reset(reset_t), .d(d_t), .q(q_t), .qb(qb_t));

which I get this:

enter image description here

which is basically the same signal with no change. This is what I want to see: enter image description here

What am I missing here?

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  • \$\begingroup\$ Why not implement it with a counter? \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 4 '15 at 0:17
  • \$\begingroup\$ I like to know why I am not getting the right result and then move on to other options. But that is something that I will consider while on it. \$\endgroup\$ – user10291355 Feb 4 '15 at 0:33
  • \$\begingroup\$ I don't see any connection between "dff2" and "dff1" (except the RESET input) ? \$\endgroup\$ – TEMLIB Feb 4 '15 at 2:02
  • \$\begingroup\$ As @TEMLIB said, I too sense a problem in module instantiation of second dff. As per your ckt, shouldnt q be assigned the clock for second FF ? \$\endgroup\$ – Plutonium smuggler Feb 4 '15 at 2:45
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At a glance to get your circuit, the wiring seems wrong, in halfclk you had:

dff dff1(clk, reset, d, q, qb);
dff dff2(w1, reset, d1, w2, qb1);

Shouldn't it be (renaming a little bit for clarity):

input clk, reset;
output q1, q2;
wire  qb1, qb2;

dff dff1(clk,  reset, qb1,  q1, qb1);
dff dff2(q1,   reset, qb2,  q2, qb2);

However a warning: If you use the q output to the clock, a real circuit will add up a significant delay. That's bad for a number of reasons. In some circumstances it's ok though, if you don't care about the phase relationship. If you do care, then look into designing a synchronous counter. Also, for a real circuit, buffer the output from the counter, otherwise if you just pass out q or qb it might see a big load and your counter will (at best) slow down even more.

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