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Forum: FPGA, VHDL & Verilog Clock frequency reducer


Clock frequency reducer
von Eric J. (coderic)

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I am trying to divide my input clock signal by 10 - effectively setting 
my out HIGH 5 of ten clock pulses and LOW the other 5.

However, my counter idea does not work - the clk_ou2 signal doesn't 
update at all, and I do not know why.


1
library ieee;

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use ieee.std_logic_1164.all;

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use ieee.numeric_std.all;

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entity Taktteiler is

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port (

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    clk_in : in std_logic;

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    clk_out : out std_logic;

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    reset : in std_logic

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);

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end Taktteiler;

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architecture rtl of Taktteiler is

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signal counter : unsigned(3 downto 0);

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signal clk_out2 : std_logic := '0';

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begin

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process(clk_in)

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    begin

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        if(rising_edge(clk_in)) then

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            counter <= counter + 1;

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            if std_logic_vector(counter) = "101" then --reset the counter at 101

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                counter <= (others => '0');

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                clk_out2 <= clk_out2 XOR '1'; --toggle the out2 signal

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            end if;

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        end if;

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        if reset = '1' then --reset on HIGH reset input

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            counter <= (others => '0');

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            clk_out2 <= '0';

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        end if;

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end process;

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clk_out <= clk_out2;

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end architecture rtl;


  


  




  

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    Re: Clock frequency reducer
  


  

    

      von
      
        Achim S. (Guest)
      
      


      
    


    

      
    

    

    

  
  


  

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counter is a 4 bit vector. It can never be equal to "101", because that 
ist a 3 bit vector. So std_logic_vector(counter) = "101" will never be 
true...

Try to replace line 24 by
    if counter="0101" then

If it works, then check, if you really divide by 10 or by some other 
value..

  


  




  

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    Re: Clock frequency reducer
  


  

    

      von
      
        Duke Scarring (Guest)
      
      


      
    


    

      
    

    

    

      

      

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I prefer a simple solution with just a single shift register:
1
library ieee;

2
use ieee.std_logic_1164.all;

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entity clk_divider is

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    port

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    (

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        clk_in  : in  std_logic;

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        clk_out : out std_logic

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    );

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end entity clk_divider;

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architecture rtl of clk_divider is

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    signal clk_reg : std_logic_vector( 9 downto 0) := "0000011111";

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begin

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    process

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    begin

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        wait until rising_edge( clk_in);

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        clk_reg <= clk_reg( clk_reg'high-1 downto 0) & clk_reg( clk_reg'high);

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    end process;

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    clk_out <= clk_reg( clk_reg'high);

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end architecture rtl;


Duke

  


  




  

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    Re: Clock frequency reducer
  


  

    

      von
      
        Lothar M.
            (Company: Titel)
        (lkmiller)
        (Moderator)
      
      


      
    


    

      
    

    

    

  
  


  

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Eric J. wrote:
> signal counter : unsigned(3 downto 0);
Achim S. wrote:
> counter is a 4 bit vector. It can never be equal to "101"
To get rid of such kind of problems I would use an integer range 0 to 9 
for counting.
Then the compare for 0 and 5 is easily human readable:
1
:

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signal counter : unsigned(3 downto 0);

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:

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:

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   process(clk_in)

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   begin

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      if(rising_edge(clk_in)) then

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         if counter<9 then counter <= counter + 1;

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         else              counter <= 0;

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         end if;   

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         if counter = 0 then clk_out <= '0'; end if;

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         if counter = 5 then clk_out <= '1'; end if;

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   end process;

14
:


Duke Scarring wrote:
> I prefer a simple solution with just a single shift register
I prefer the very same without any process ;-)

1
library ieee;

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use ieee.std_logic_1164.all;

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entity clk_divider is

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    port ( clk_in  : in  std_logic;

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           clk_out : out std_logic);

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end entity clk_divider;

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architecture rtl of clk_divider is

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    signal clk_reg : std_logic_vector(9 downto 0) := "0000011111";

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begin

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    clk_reg <= clk_reg(8 downto 0) & clk_reg(9) when rising_edge(clk_in);

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    clk_out <= clk_reg(9);

14
end architecture rtl;


  


  



  

  
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