Hello,

I am a newbie to VHDL programming and want to test my FPGA board with
a code which lights a LED every second. To do this I need a VHDL code
for 1 Hz signal generator. Unfortunately, I cannot find such in Web.
Also I am not experienced in VHDL programming and not sure how to
write such a code. If you have any similar code could you put it in
this thread, please or give me some idea how to write this. I also
wonder - is it necessary to use a clock for such signal generator?

Vagant schrieb:

> I am a newbie to VHDL programming and want to test my FPGA board with
> a code which lights a LED every second. To do this I need a VHDL code
> for 1 Hz signal generator. Unfortunately, I cannot find such in Web.

Take your clock (which is running at X MHz) and divide it by X.


process(reset_n,clk)
variable   cnt   : integer;
begin
if (reset_n='0') then
	clk_out<='0';
	cnt:=0;
elsif rising_edge(clk) then
	if (cnt=divider_half-1) then
		clk_out<=NOT(clk_out);
		cnt:=0;
	else
		cnt:=cnt+1;
	end if;
end if;
end process;

Note that divider_half is X/2.

The code is not tested - just written in the mail program.

Ralf

On Jan 8, 10:22=A0pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
> Vagant schrieb:
>
> > I am a newbie to VHDL programming and want to test my FPGA board with
> > a code which lights a LED every second. To do this I need a VHDL code
> > for 1 Hz signal generator. Unfortunately, I cannot find such in Web.
>
> Take your clock (which is running at X MHz) and divide it by X.
>
> process(reset_n,clk)
> variable =A0 cnt =A0 : integer;
> begin
> if (reset_n=3D'0') then
> =A0 =A0 =A0 =A0 clk_out<=3D'0';
> =A0 =A0 =A0 =A0 cnt:=3D0;
> elsif rising_edge(clk) then
> =A0 =A0 =A0 =A0 if (cnt=3Ddivider_half-1) then
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 clk_out<=3DNOT(clk_out);
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3D0;
> =A0 =A0 =A0 =A0 else
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3Dcnt+1;
> =A0 =A0 =A0 =A0 end if;
> end if;
> end process;
>
> Note that divider_half is X/2.
>
> The code is not tested - just written in the mail program.
>
> Ralf

Thank you. I just wonder where I could get a complete listing (which
includes 'entity' part)?

Vagant wrote:

> Thank you. I just wonder where I could get a complete listing (which
> includes 'entity' part)?

Do you have a text editor?
Didn't the board come with some examples?

Here's a related example including an entity:
http://home.comcast.net/~mike_treseler/count_enable.vhd

        -- Mike Treseler

On Jan 8, 10:22=A0pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
> Take your clock (which is running at X MHz) and divide it by X.
>
I guess it should be divided by X*10^6 to get 1 Hz, isn't it?

On Jan 8, 11:12=A0pm, Mike Treseler <mike_trese...@comcast.net> wrote:
> Vagant wrote:
> > Thank you. I just wonder where I could get a complete listing (which
> > includes 'entity' part)?
>
> Do you have a text editor?
> Didn't the board come with some examples?
>
> Here's a related example including an entity:http://home.comcast.net/~mike=
_treseler/count_enable.vhd
>
> =A0 =A0 =A0 =A0 -- Mike Treseler

No, it's not what I am asking about.

On Jan 8, 10:22=A0pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
> Vagant schrieb:
>
> > I am a newbie to VHDL programming and want to test my FPGA board with
> > a code which lights a LED every second. To do this I need a VHDL code
> > for 1 Hz signal generator. Unfortunately, I cannot find such in Web.
>
> Take your clock (which is running at X MHz) and divide it by X.
>
> process(reset_n,clk)
> variable =A0 cnt =A0 : integer;
> begin
> if (reset_n=3D'0') then
> =A0 =A0 =A0 =A0 clk_out<=3D'0';
> =A0 =A0 =A0 =A0 cnt:=3D0;
> elsif rising_edge(clk) then
> =A0 =A0 =A0 =A0 if (cnt=3Ddivider_half-1) then
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 clk_out<=3DNOT(clk_out);
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3D0;
> =A0 =A0 =A0 =A0 else
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3Dcnt+1;
> =A0 =A0 =A0 =A0 end if;
> end if;
> end process;
>
> Note that divider_half is X/2.
>
> The code is not tested - just written in the mail program.
>
> Ralf

However the syntax check gives an error message:

Parameter clk_out of mode out can not be associated with a formal
parameter of mode in.

for the line:
clk_out<=3DNOT(clk_out);

On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
> On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
> > Take your clock (which is running at X MHz) and divide it by X.
>
> I guess it should be divided by X*10^6 to get 1 Hz, isn't it?

Not really. You're already using your clk to increment cnt. You just
need to divide it by X.

For instance, if you had a clock running at 50 MHz, cnt would
increment 50,000 times a second. Divide it by 50 and you're left with
a cnt that increments a thousand times a second - a 1 MHz clock.

On Jan 9, 8:23=A0am, neeraj2...@gmail.com wrote:
> On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
> > > Take your clock (which is running at X MHz) and divide it by X.
>
> > I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>
> Not really. You're already using your clk to increment cnt. You just
> need to divide it by X.
>
> For instance, if you had a clock running at 50 MHz, cnt would
> increment 50,000 times a second. Divide it by 50 and you're left with
> a cnt that increments a thousand times a second - a 1 MHz clock.

Well, but I need 1 Hz clock not 1 MHz.

On Jan 9, 11:23=A0am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
> On Jan 8, 10:22=A0pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
>
>
>
>
> > Vagant schrieb:
>
> > > I am a newbie to VHDL programming and want to test my FPGA board with
> > > a code which lights a LED every second. To do this I need a VHDL code
> > > for 1 Hz signal generator. Unfortunately, I cannot find such in Web.
>
> > Take your clock (which is running at X MHz) and divide it by X.
>
> > process(reset_n,clk)
> > variable =A0 cnt =A0 : integer;
> > begin
> > if (reset_n=3D'0') then
> > =A0 =A0 =A0 =A0 clk_out<=3D'0';
> > =A0 =A0 =A0 =A0 cnt:=3D0;
> > elsif rising_edge(clk) then
> > =A0 =A0 =A0 =A0 if (cnt=3Ddivider_half-1) then
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 clk_out<=3DNOT(clk_out);
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3D0;
> > =A0 =A0 =A0 =A0 else
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 cnt:=3Dcnt+1;
> > =A0 =A0 =A0 =A0 end if;
> > end if;
> > end process;
>
> > Note that divider_half is X/2.
>
> > The code is not tested - just written in the mail program.
>
> > Ralf
>
> However the syntax check gives an error message:
>
> Parameter clk_out of mode out can not be associated with a formal
> parameter of mode in.
>
> for the line:
> clk_out<=3DNOT(clk_out);- Hide quoted text -
>
> - Show quoted text -

You can not have a port of mode "out" being assigned to another
signal. That makes sense if you think in terms of hardware, doesn't
it? :)

I can give you two solutions:
1. Change the mode of clk_out to "inout."
2. A better solution would be to declare an internal signal and assign
it to clk_out after the process. For example,

signal clk_sig : std_logic;
process(reset_n,clk)
variable   cnt   : integer;
begin
if (reset_n=3D'0') then
        clk_sig<=3D'0';
        cnt:=3D0;
elsif rising_edge(clk) then
        if (cnt=3Ddivider_half-1) then
                clk_sig<=3DNOT(clk_sig);
                cnt:=3D0;
        else
                cnt:=3Dcnt+1;
        end if;
end if;
end process;

clk_out <=3D clk_sig;

On Jan 9, 11:25=A0am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
> On Jan 9, 8:23=A0am, neeraj2...@gmail.com wrote:
>
> > On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
> > > > Take your clock (which is running at X MHz) and divide it by X.
>
> > > I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>
> > Not really. You're already using your clk to increment cnt. You just
> > need to divide it by X.
>
> > For instance, if you had a clock running at 50 MHz, cnt would
> > increment 50,000 times a second. Divide it by 50 and you're left with
> > a cnt that increments a thousand times a second - a 1 MHz clock.
>
> Well, but I need 1 Hz clock not 1 MHz.

Whoops, my mistake. The line should have been:
Divide it by 50 and you're left with
a cnt that increments a thousand times a second - a 1 Hz clock.

Sorry.

On Jan 9, 11:30=A0am, neeraj2...@gmail.com wrote:
> On Jan 9, 11:25=A0am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
>
>
>
>
> > On Jan 9, 8:23=A0am, neeraj2...@gmail.com wrote:
>
> > > On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > > On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote=
:
>
> > > > > Take your clock (which is running at X MHz) and divide it by X.
>
> > > > I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>
> > > Not really. You're already using your clk to increment cnt. You just
> > > need to divide it by X.
>
> > > For instance, if you had a clock running at 50 MHz, cnt would
> > > increment 50,000 times a second. Divide it by 50 and you're left with
> > > a cnt that increments a thousand times a second - a 1 MHz clock.
>
> > Well, but I need 1 Hz clock not 1 MHz.
>
> Whoops, my mistake. The line should have been:
> Divide it by 50 and you're left with
> a cnt that increments a thousand times a second - a 1 Hz clock.
>
> Sorry.- Hide quoted text -
>
> - Show quoted text -

No, on second thought, you're right. For a 50 MHz clock, you would
count to 24999 before toggling clk_out.

On Jan 9, 8:32=A0am, neeraj2...@gmail.com wrote:
> On Jan 9, 11:30=A0am, neeraj2...@gmail.com wrote:
>
>
>
>
>
> > On Jan 9, 11:25=A0am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > On Jan 9, 8:23=A0am, neeraj2...@gmail.com wrote:
>
> > > > On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > > > On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wro=
te:
>
> > > > > > Take your clock (which is running at X MHz) and divide it by X.
>
> > > > > I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>
> > > > Not really. You're already using your clk to increment cnt. You just=

> > > > need to divide it by X.
>
> > > > For instance, if you had a clock running at 50 MHz, cnt would
> > > > increment 50,000 times a second. Divide it by 50 and you're left wit=
h
> > > > a cnt that increments a thousand times a second - a 1 MHz clock.
>
> > > Well, but I need 1 Hz clock not 1 MHz.
>
> > Whoops, my mistake. The line should have been:
> > Divide it by 50 and you're left with
> > a cnt that increments a thousand times a second - a 1 Hz clock.
>
> > Sorry.- Hide quoted text -
>
> > - Show quoted text -
>
> No, on second thought, you're right. For a 50 MHz clock, you would
> count to 24999 before toggling clk_out.- Hide quoted text -
>
> - Show quoted text -

Thank you. I have corrected the code and now it looks like this (is
this correct?):

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;

entity sig_gen is
   Port (clk : in STD_LOGIC;
     reset_n : in STD_LOGIC;
     clk_out : out STD_LOGIC);
end entity sig_gen;

architecture Behavioral of sig_gen is
signal clk_sig : std_logic;
begin
process(reset_n,clk)
    variable   cnt   : integer;
    begin
    if (reset_n=3D'0') then
        clk_sig<=3D'0';
        cnt:=3D0;
        elsif rising_edge(clk) then
        if (cnt=3D2499-1) then
                clk_sig<=3DNOT(clk_sig);
                cnt:=3D0;
               else
                cnt:=3Dcnt+1;
        end if;
     end if;
end process;

clk_out <=3D clk_sig;

end Behavioral;

On Jan 9, 3:43 pm, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
> On Jan 9, 8:32 am, neeraj2...@gmail.com wrote:
>
>
>
> > On Jan 9, 11:30 am, neeraj2...@gmail.com wrote:
>
> > > On Jan 9, 11:25 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > > On Jan 9, 8:23 am, neeraj2...@gmail.com wrote:
>
> > > > > On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> > > > > > On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>
> > > > > > > Take your clock (which is running at X MHz) and divide it by X.
>
> > > > > > I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>
> > > > > Not really. You're already using your clk to increment cnt. You just
> > > > > need to divide it by X.
>
> > > > > For instance, if you had a clock running at 50 MHz, cnt would
> > > > > increment 50,000 times a second. Divide it by 50 and you're left with
> > > > > a cnt that increments a thousand times a second - a 1 MHz clock.
>
> > > > Well, but I need 1 Hz clock not 1 MHz.
>
> > > Whoops, my mistake. The line should have been:
> > > Divide it by 50 and you're left with
> > > a cnt that increments a thousand times a second - a 1 Hz clock.
>
> > > Sorry.- Hide quoted text -
>
> > > - Show quoted text -
>
> > No, on second thought, you're right. For a 50 MHz clock, you would
> > count to 24999 before toggling clk_out.- Hide quoted text -
>
> > - Show quoted text -
>
> Thank you. I have corrected the code and now it looks like this (is
> this correct?):
>
> library IEEE;
> use IEEE.STD_LOGIC_1164.ALL;
> use IEEE.STD_LOGIC_ARITH.ALL;
> use IEEE.STD_LOGIC_UNSIGNED.ALL;
>
> entity sig_gen is
>    Port (clk : in STD_LOGIC;
>      reset_n : in STD_LOGIC;
>      clk_out : out STD_LOGIC);
> end entity sig_gen;
>
> architecture Behavioral of sig_gen is
> signal clk_sig : std_logic;
> begin
> process(reset_n,clk)
>     variable   cnt   : integer;
>     begin
>     if (reset_n='0') then
>         clk_sig<='0';
>         cnt:=0;
>         elsif rising_edge(clk) then
>         if (cnt=2499-1) then
>                 clk_sig<=NOT(clk_sig);
>                 cnt:=0;
>                else
>                 cnt:=cnt+1;
>         end if;
>      end if;
> end process;
>
> clk_out <= clk_sig;
>
> end Behavioral;


Have you checked that your board provides you with a 50 MHz clock? I
quoted 50 just as an example. :)
If yes, then you should count from 0 to 24999 (i.e., 25000 50 MHz
clocks or half a second) and toggle clk_sig. Then you reset the count
to 0 and start again. At the end of the next 25000 50 MHz clocks (or
another half a second) you toggle clk_sig again and so on.

So, your code should be:
>         if (cnt=25000-1) then

Of course, (25000 - 1) is 24999. So you could also write:
>         if (cnt=24999) then

I hope the calculation is clear. To give another example, for a 10 MHz
clock, you count from 0 to 4999 (5000 10 MHz clocks).

While coding, you could also declare the count limit as a constant.

constant HALF_FREQ : INTEGER := 25000;

So,
>         if (cnt=HALF_FREQ-1) then

That way, if you decided to change the clock later on, you would only
have to change the constant at one place. This could be helpful if you
write a really large piece of code.

Vagant a �crit :
> On Jan 8, 11:12 pm, Mike Treseler <mike_trese...@comcast.net> wrote:
>> Vagant wrote:
>>> Thank you. I just wonder where I could get a complete listing (which
>>> includes 'entity' part)?
>> Do you have a text editor?
>> Didn't the board come with some examples?
>>
>> Here's a related example including an entity:http://home.comcast.net/~mike_treseler/count_enable.vhd
>>
>>         -- Mike Treseler
> 
> No, it's not what I am asking about.

The response of Mike, if I understood it rightly, means that a so simple 
question can be answered by yourself if you take the time to read the notice of 
your board, or if any of a VHDL book for beginner.

Do not expect the forum's guys to do your project at your place : it will not 
learn you anything.

That I read below let me out of mind...

50Mhz is 50 000 000 period per second. So you have to divide it by 50 000 000 go 
get 1Hz, or to count two times from 0 to 24 999 999.
I hope the error was to see if someone follows ?
:-)

Pascal

On Jan 9, 4:14=A0pm, Pascal Peyremorte
<p.peyremorte.remove_t...@free.fr> wrote:
>
> The response of Mike, if I understood it rightly, means that a so simple
> question can be answered by yourself if you take the time to read the noti=
ce of
> your board, or if any of a VHDL book for beginner.
>
> Do not expect the forum's guys to do your project at your place : it will =
not
> learn you anything.
>
>
> Pascal

Well, a reply with a long code which was not asked would not be
helpful either and if one does not like question then just reply
nothing.

>Well, a reply with a long code which was not asked would not be
>helpful either and if one does not like question then just reply
>nothing.

As I understand it, you have a PhD in nuclear magnetic resonance
(Manchester 2004), so FPGAs and the associated HDLs are not beyond
your abilities.

But, as I said earlier, you need to learn some basic electronics and
logic before you can handle these more recent developments, because
most of those working with HDLs have that experience of older
technologies and what you read will almost always assume that.

Did you think it was going to be easy?

Mike

>Well, a reply with a long code which was not asked would not be
>helpful either and if one does not like question then just reply
>nothing.

off-topic :  Esli Vy esche v Manchestere, u nas vecherinka russkaya v
gorode Bolton v subotu (12.01.2008) ot 18:00 na pabe "Balmoral" (v
tsentre).  Stoit vsego 60 funtov, tak, esli pridut N chelovek, prosyat
c kazhdogo 60/N funtov.  Mike

On Jan 9, 4:56=A0pm, MikeShepherd...@btinternet.com wrote:
> >Well, a reply with a long code which was not asked would not be
> >helpful either and if one does not like question then just reply
> >nothing.
>
> off-topic : =A0Esli Vy esche v Manchestere, u nas vecherinka russkaya v
> gorode Bolton v subotu (12.01.2008) ot 18:00 na pabe "Balmoral" (v
> tsentre). =A0Stoit vsego 60 funtov, tak, esli pridut N chelovek, prosyat
> c kazhdogo 60/N funtov. =A0Mike

Spasibo :), no ia ne v manchestere uje s 2004

On Jan 9, 4:38=A0pm, MikeShepherd...@btinternet.com wrote:
> >Well, a reply with a long code which was not asked would not be
> >helpful either and if one does not like question then just reply
> >nothing.
>
> As I understand it, you have a PhD in nuclear magnetic resonance
> (Manchester 2004), so FPGAs and the associated HDLs are not beyond
> your abilities.
>
> But, as I said earlier, you need to learn some basic electronics and
> logic before you can handle these more recent developments, because
> most of those working with HDLs have that experience of older
> technologies and what you read will almost always assume that.
>
> Did you think it was going to be easy?
>
> Mike

It goes allright with me and I am familiar with electronics. I got
Diploma in radiophysics and electronics (undegraduate study) but FPGA
and VHDL are something really new. So sometimes I get stuck and then i
ask questions here.

Hurrah! It works! I have used 50MHz clock as input and my 1Hz output
signal was connected to a LED. So it flashes every second now! Thanks
to all for encouragement, criticism, real help and suggestions!

P.S. Final version of the VHDL code used is:

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;


entity sig_gen is
   Port (clk : in STD_LOGIC;
     reset_n : in STD_LOGIC;
     clk_out : out STD_LOGIC);
end entity sig_gen;


architecture Behavioral of sig_gen is
signal clk_sig : std_logic;
begin
process(reset_n,clk)
    variable   cnt   : integer;
    begin
    if (reset_n='0') then
        clk_sig<='0';
        cnt:=0;
        elsif rising_edge(clk) then
        if (cnt=24999999) then
                clk_sig<=NOT(clk_sig);
                cnt:=0;
               else
                cnt:=cnt+1;
        end if;
     end if;
end process;


clk_out <= clk_sig;


end Behavioral;

Vagant a �crit :
> Hurrah! It works! I have used 50MHz clock as input and my 1Hz output
> signal was connected to a LED. So it flashes every second now! Thanks
> to all for encouragement, criticism, real help and suggestions!
> 
> P.S. Final version of the VHDL code used is:
> 
> library IEEE;
> use IEEE.STD_LOGIC_1164.ALL;
> use IEEE.STD_LOGIC_ARITH.ALL;
> use IEEE.STD_LOGIC_UNSIGNED.ALL;


Drop these last two non-standard libraries and use ieee.numeric_std instead.
The code you posted doesn't even need it anyway.

Nicolas

> 50Mhz is 50 000 000 period per second. So you have to divide it by 50 000 000 go
> get 1Hz, or to count two times from 0 to 24 999 999.
> I hope the error was to see if someone follows ?
> :-)
>
> Pascal


Correct. My mistake, sorry. :(

neeraj2608@gmail.com schrieb:
> On Jan 9, 11:30 am, neeraj2...@gmail.com wrote:
>> On Jan 9, 11:25 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>>
>>
>>
>>
>>
>>> On Jan 9, 8:23 am, neeraj2...@gmail.com wrote:
>>>> On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>>>>> On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
>>>>>> Take your clock (which is running at X MHz) and divide it by X.
>>>>> I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
>>>> Not really. You're already using your clk to increment cnt. You just
>>>> need to divide it by X.
>>>> For instance, if you had a clock running at 50 MHz, cnt would
>>>> increment 50,000 times a second. Divide it by 50 and you're left with
>>>> a cnt that increments a thousand times a second - a 1 MHz clock.
>>> Well, but I need 1 Hz clock not 1 MHz.
>> Whoops, my mistake. The line should have been:
>> Divide it by 50 and you're left with
>> a cnt that increments a thousand times a second - a 1 Hz clock.
>>
>> Sorry.- Hide quoted text -
>>
>> - Show quoted text -
> 
> No, on second thought, you're right. For a 50 MHz clock, you would
> count to 24999 before toggling clk_out.

Well, not quitre right yet. 50MHz is 50 MILLION Hz, so for 1 Hz output 
clock you're a factor of a thousand off track...

HTH,
Florian

On Jan 10, 10:42=A0am, Florian Teply <onlinef...@usenet.cnntp.org>
wrote:
> neeraj2...@gmail.com schrieb:
>
>
>
>
>
> > On Jan 9, 11:30 am, neeraj2...@gmail.com wrote:
> >> On Jan 9, 11:25 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
>
> >>> On Jan 9, 8:23 am, neeraj2...@gmail.com wrote:
> >>>> On Jan 9, 2:57 am, Vagant <vladimir.v.koroste...@rambler.ru> wrote:
> >>>>> On Jan 8, 10:22 pm, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote=
:
> >>>>>> Take your clock (which is running at X MHz) and divide it by X.
> >>>>> I guess it should be divided by X*10^6 to get 1 Hz, isn't it?
> >>>> Not really. You're already using your clk to increment cnt. You just
> >>>> need to divide it by X.
> >>>> For instance, if you had a clock running at 50 MHz, cnt would
> >>>> increment 50,000 times a second. Divide it by 50 and you're left with=

> >>>> a cnt that increments a thousand times a second - a 1 MHz clock.
> >>> Well, but I need 1 Hz clock not 1 MHz.
> >> Whoops, my mistake. The line should have been:
> >> Divide it by 50 and you're left with
> >> a cnt that increments a thousand times a second - a 1 Hz clock.
>
> >> Sorry.- Hide quoted text -
>
> >> - Show quoted text -
>
> > No, on second thought, you're right. For a 50 MHz clock, you would
> > count to 24999 before toggling clk_out.
>
> Well, not quitre right yet. 50MHz is 50 MILLION Hz, so for 1 Hz output
> clock you're a factor of a thousand off track...
>
> HTH,
> Florian- Hide quoted text -
>
> - Show quoted text -

I used 24999999 (which is 50 millions divided by 2 minus 1) and it
really produces 1 Hz signal. :)

On Jan 10, 1:53 pm, Vagant <vladimir.v.koroste...@rambler.ru> wrote:

>
> I used 24999999 (which is 50 millions divided by 2 minus 1) and it
> really produces 1 Hz signal. :)

Actually, the difference between 25e6 and 25e6-1 is well under 1 part
per million. You're unlikely to be generating 1 Hz with either
divisor; I'll go out on a limb and guess you're not using a calibrated
(atomic, GPS) 50 MHz oscillator and are probably lucky to be within 10
ppm to begin with. Using N-1 is clearly the theoretically correct
solution (for a piece of code in an appropriate style), as well as
being necessary system-wide in a lot of other ways, but don't expect
your output to be any better in a practical sense, nor to be correct
in an absolute sense :-(

 - Kenn