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Counter from 0 to 11 then back to 0 (system verilog)

  • Thread starter hoheiho
  • Start date
  • #1
47
0
Hi all,
i am trying to do a counter using system verilog. The counter will count from 0 to 11 then back to 0 and start the counting again.

Input = rdy, if rdy = 0, start counting.
Output = count

I have done the simulation for my code and it is not working as i prefer. It will continues count until it's overflow but not back to 0 if the count = 12.
It seems like " else if (i==12)" doesnt work. How can i fix the problem? or i should not use else if here?

Thanks for the help
Ivan

Code:
always_ff @ (posedge clk,negedge rst)
  if(!rst)
      begin
      count<=0;
      i<=0;
      end
      else 
        begin
        if (!rdy)
          begin
          count<=count+1;
          i<=i+1;
          end
          else if (i==12)
            begin
              count<=0;
              i<=0;
            end
        else
          begin
          count<=count;
          i<=i;
          end
        end
        endmodule
 

Answers and Replies

  • #2
I dont know verilog, but the if ... else if ... control flow typically won't attempt to evaluate the second conditional "else if (i ==12)" in this case, if the first is true.

there are a million and one ways to implement a counter in a hardware descriptive language. i would probably structure VHDL as
Code:
if(rising_edge(clock) and not RST) then
      if(i<12) then
          i <= 1 + 1;
      else
          i <= 0;
      end if
else if ( RST ) then
         count <= 0;
end if
its been a long time, so my syntax is probably all wrong.

p.s, whats the point in the ready signal?
 
  • #3
47
0
I dont know verilog, but the if ... else if ... control flow typically won't attempt to evaluate the second conditional "else if (i ==12)" in this case, if the first is true.

there are a million and one ways to implement a counter in a hardware descriptive language. i would probably structure VHDL as
Code:
if(rising_edge(clock) and not RST) then
      if(i<12) then
          i <= 1 + 1;
      else
          i <= 0;
      end if
else if ( RST ) then
         count <= 0;
end if
its been a long time, so my syntax is probably all wrong.

p.s, whats the point in the ready signal?


Thanks for your reply
what does RST mean?
 
  • #4
Thanks for your reply
what does RST mean?
sorry, its reset. ive set it up to be asynchronous.
 
  • #5
47
0
I dont know verilog, but the if ... else if ... control flow typically won't attempt to evaluate the second conditional "else if (i ==12)" in this case, if the first is true.

there are a million and one ways to implement a counter in a hardware descriptive language. i would probably structure VHDL as
Code:
if(rising_edge(clock) and not RST) then
      if(i<12) then
          i <= 1 + 1;
      else
          i <= 0;
      end if
else if ( RST ) then
         count <= 0;
end if
its been a long time, so my syntax is probably all wrong.

p.s, whats the point in the ready signal?

ready is the input, if the ready=0 and last for 3 clock cycle, it will count from 0 to 2.
 
  • #6
47
0
sorry, its reset. ive set it up to be asynchronous.
sorry, 1 more question

can i use count to the counting?not i.
like
if count < 12
...
..
i have try that in my code but it doesnt work. is that becuase count is the output so i cannot use it for internal counting?
 
  • #7
sorry, 1 more question

can i use count to the counting?not i.
like
if count < 12
...
..
i have try that in my code but it doesnt work. is that becuase count is the output so i cannot use it for internal counting?
again, I have zero experience with verilog, but since it compiles down to the same thing as VHDL, i would imagine the structure of what you can do is similar.

in the code posted below, i and count are dealt with in exactly the same way. so unless you have declared them as different datatypes, theres no reason count would not work instead of i.

if you declare a variable as an output, then make some conditional statements based on its value, you must also be able to read it as an input. VHDL requires that this variable be declared as "inout" or "buffer" (or perhaps "signal" or "variable")

I assume that verilog has a similar restriction?
 

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