Making a frequency reducer

von Eric J. (coderic)

2022-03-19 15:13

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Screenshot_2022-03-19_151237.png
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I am working with an online learning tool for VHDL, trying to make a 
frequency reducer using generics.
The basic idea is: Set 'DIVIDE_BY' to a value, then get 'clk_en_out' = 1 
on every 'DIVIDE_BY'th rising edge of 'clk_in' for one pulse, before 
setting it back to 0.

E.g.: DIVIDE_BY = 8: wait for 7 rising edges, on the 8th, set clk_en_out 
= 1, on the 9th, set it back to 0, repeat.

If 'reset' = 1, reset the cycle and start at 0.

Unfortunately, I do not know what values the tool tests, so I cannot 
check why it doesn't work/on what pulse it wants me which test instance 
to be 1.
I do know though that 'DIVIDE_BY' is less than 16, so an overflow of my 
'counter' signal should not be the problem.

In the attached screenshot, the 'valid' line stops being 1 on rising 
edge 4, yet when I twisted the code so as to have 'clk_en_a' (test 
instance one; the second one is 'clk_en_b') be 1 there, it didn't 
change. I assumed that trying to get 'b' to be 1 there would be 
pointless, as that would make 'a' be 1 much earlier, which in theory 
should set the valid line to 0 much earlier.

Can someone help me figure out why it does not work?

library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity ClockEnableGenerator is
generic (
    DIVIDE_BY : integer
    clk_in : in std_logic;
    clk_en_out : out std_logic;
    reset : in std_logic
end ClockEnableGenerator;
architecture rtl of ClockEnableGenerator is
signal counter : unsigned(3 downto 0) := (others => '0');
signal clk_out2 : std_logic := '0';
process(clk_in)
    begin
        if(rising_edge(clk_in)) then
            counter <= counter + 1;
            if (to_integer(counter) = DIVIDE_BY) then --reset the counter at 'DIVIDE_BY'
                counter <= (others => '0');
                clk_out2 <= '1'; --set the out2 signal
            else
                clk_out2 <= '0';
            end if;
        end if;
        if reset = '1' then --reset on HIGH reset input
            counter <= (others => '0');
            clk_out2 <= '0';
        end if;
end process;
clk_en_out <= clk_out2; --update out to out2
end architecture rtl;

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Re: Making a frequency reducer

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

2022-03-19 22:29

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It looks fairly fine, all in all.

One problem is the well known one-by-one error: you are counting one 
step to far. With DIVIDE_BY-1 set to 8 you will see:

0,1,2,3,4,5,6,7,8

Try it this way:

if (to_integer(counter) = DIVIDE_BY-1) then...

An additional hint: why don't you simply use an integer as counter?

Eric J. wrote:
> In the attached screenshot
What's the testbench code for that waveform?

2022-03-19 22:29: Edited by Moderator

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Re: Making a frequency reducer

von Eric J. (coderic)

2022-03-20 07:24

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Hi, thanks for the reply.
I tried the -1 in the if... didn't work either, unfortunately.

Uh, good question with the integer. Most of the code was copied from my 
solution for a previous task, that's why it is a binary vector.

For the screenshot, I do not know what the testbench code is... that's 
the problem with the online tool.

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Re: Making a frequency reducer

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

2022-03-20 13:33

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Eric J. wrote:
> I do not know what the testbench code is...
The testbench code is useless for counters more than 7..8 clocks, 
because then the external reset pops up.

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Re: Making a frequency reducer

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

2022-03-21 08:35

Attached files:

cegenerator_unsigned.vhd (807 Bytes) | highlighted code
cegenerator_integer.vhd (714 Bytes) | highlighted code
tb_cegenerator.vhd (2.14 KB) | highlighted code

WF_cegenerator.PNG
9.7 KB

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I had a chance to put your code on a testbench. What I found: you have a 
weird async reset structure:

process(clk_in)
    begin
        if rising_edge(clk_in) then
        end if;
        if reset = '1' then
        end if;
end process;

That means: your sensitivity list is incomplete, reset is missing. Due 
to that you see a strange behaviour of your design: although reset pop 
up at a absolutely random time, the counter value is reset with the next 
change of the clk_in and therefore at the falling(!!) edge of this 
clk_in.

Two solutions:
1. Add the reset to the sensitivity list

process(clk_in, reset)
    begin
        if rising_edge(clk_in) then
        end if;
        if reset = '1' then
        end if;
end process;

2. Change the structure to a synchronous reset:

process(clk_in)
    begin
        if rising_edge(clk_in) then
           if reset = '1' then
           end if;
        end if;
end process;

I prefer the unmentioned third one, so I get a synchronous design with 
no hazzle about a sensitivity list at all:

process begin
   wait until rising_edge(clk_in);
   if reset = '1' then
   end if;
end process;


See my solutions attached. I wrote a simple testbench generatin an 
100MHz clock and a reset signal every 331ns. Both of them behave the 
same and like I expect them to.

2022-03-21 08:35: Edited by Moderator

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Re: Making a frequency reducer

von Eric J. (coderic)

2022-03-21 19:13

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Thank you for the detailed explanation. Made my code prettier with the 
reset.
It didn't fix the problem, though; after some trial and error, I landed 
at:

library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity ClockEnableGenerator is
generic (
    DIVIDE_BY : integer := 1
    clk_in : in std_logic;
    clk_en_out : out std_logic;
    reset : in std_logic
end ClockEnableGenerator;
architecture rtl of ClockEnableGenerator is
signal counter : integer range -1 to 15 := 0;
process(clk_in, reset)
    begin
        if(rising_edge(clk_in)) then
            counter <= counter + 1;
            if (counter = DIVIDE_BY-2) then --reset the counter at 'DIVIDE_BY'
                counter <= -1;
                clk_en_out <= '1'; --set the out2 signal
            else
                clk_en_out <= '0';
            end if;
        end if;
        if reset = '1' then --reset on HIGH reset input
            counter <= 0;
            clk_en_out <= '0';
        end if;
end process;
end architecture rtl;


Note that counter is set to -1 once the DIVIDE_BYth rising edge is 
reached - because the interval clk_out_en is 1 apparently should not be 
counted. By giving the counter one more rising edge to count, the 
problem is fixed.

Thank you for the help!

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Re: Making a frequency reducer

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

2022-03-22 10:03

Attached files:

WF_cegenerator_divideby1.PNG
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Eric J. wrote:
> the problem is fixed.
What "problem"? As far as I see the problem is your understanding of 
what should happen.

> because the interval clk_out_en is 1 apparently should not be counted.
Look on any µC or any other computing device with a counter inside all 
over the world: each of them continues counting when reporting an 
overflow or a match or something else.

> because the interval clk_out_en is 1 apparently should not be counted.
Based on what "appearance" do you come to this conclusion? What should 
happen, when DIVIDE_BY is 1? If you have a clock of 10 MHz and DIVIDE it 
BY  1, then the clockenable must be '1' all the time. Because dividing 
by 1 means "run with the clocks frequency". And exactly thats what my 
code does (see the screenshot)...


Running your code against my testbench leads to

# RUNTIME: Fatal Error: RUNTIME_0067 design.vhd (24): Value 16 out of range (-1 to 15).
# KERNEL: Time: 165 ns,  Iteration: 0,  Instance: /tb_cegenerator/uut,  Process: line__20.

And its obviously why: you initialize counter with 0. Then you set 
DIVIDE_BY to 1. Then you assign a clock and do this comparison "if 
(counter = DIVIDE_BY-2) then..." which is the same as "if (counter = -1) 
then...". And as counter value is "0 and increasing" that compare 
against -1 is always wrong and the counter counts on and on until it 
reaches the end of its range generating an error message.

Check it out yourself: https://www.edaplayground.com/x/DZuL

2022-03-22 10:08: Edited by Moderator

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