VHDL-problem with symmetrical frequency divider by 3

W

Winfried Salomon

Guest
Hello,

I am a beginner in VHDL and use Xilinx ISE 6.103i. I tried to program a
symmetrical frequency divider by 3 in VHDL, which exists as hardware
tested schematic solution with 3 D-FFs. But the VHDL version doesn't
work at all, only the behavioral simulation with ModelSim works right,
but the post fit simulation and the hardware not at all. The chip is an
CPLD XC9572-15PC84, an Webcase at Xilinx had not yet success. I attached
the code in div3_vhdl.vhd, which is very short and easy to understand.
Why does it not work? The generated circuit is not a synchronous clocked
state machine, there are no memory elements at all like D-FFS and it has
no input, so it must be totally wrong.

To generalize the question, there must be triggered on both rising and
falling edge of clock in one process. Is this possible in VHDL? I
suppose not! But I cannot find in books that this is illegal. Can it be
a compiler dependent problem? I use the XST, perhaps another can solve
the problem? Where is the mistake? Many thanks in advance,

Winfried
 
You are trying to infer a FF that is sensetive to both the rising and
falling edge - such a FF is (mostly) not synthesizeable.
the Modelsim can simulate it however..

In order to design such a divider you need to use two sets of FFs each
sensetive to a different edge (rising / falling) and then
combine their outputs (using combinatorical logic - probably an "and"
gate) to get your divided clock.

Regards, Moti.
 
Hi Moti,

<moti@terasync.net> schrieb im Newsbeitrag
news:1106066690.821488.24970@c13g2000cwb.googlegroups.com...
You are trying to infer a FF that is sensetive to both the rising and
falling edge - such a FF is (mostly) not synthesizeable.
the Modelsim can simulate it however..
Click to expand...
it seems to be so, but there is no error message, although the circuit
synthezising process failes.

In order to design such a divider you need to use two sets of FFs each
sensetive to a different edge (rising / falling) and then
combine their outputs (using combinatorical logic - probably an "and"
gate) to get your divided clock.

Regards, Moti.
Click to expand...
Yes, I made this on paper by myself without computer, it needs 3 D-FFS, 2
inverters and 1 and2, 1 clock is inverted, a very simple looking circuit and
it works fine. So I must think, in VHDL is no symmetrical odd frequency
divider possible, because in 1 process not on both edges can be triggered.

regards, Winfried
 
Winfried Salomon wrote:

But ..... it's not the solution of the problem. Of course you can write
a schematic circuit in VHDL an then it must work, only you must find the
circuit yourself. I hoped that VHDL or the implementation tool can find
this circuit, but it cannot.
Click to expand...
You have to distinguish between VHDL as behavioral language, and RTL
subset of VHDL that can be synthesized. To create good HW implementations
VHDL must be written in a certain way that different tools recognize.
The tool might create something out of behavioral code, but the
end result can be horrible in terms of speed and area (or might not work
in real world). Also there are quite big differences between tools.
Tools like Synplify can process more complex structures compared to
synopsys DC.

VHDL or Verilog are not a magical bullet. State machines can be coded
in such a cryptic way that the downstream tools won't see it as a state
machine. And after that the implementation can be much slover because
SM optimisation tehqniques could not be used. But it is usually just
as easy to code the SMs using the normal way of 2 or 3 processes
(on clocked process and 1-2 combinatiorial ones).

Clock dividers are quite special blocks where the actual gate implementation is
important. All chips I have seen have used manual instantiation of
library cells in clock structures. Also those cells usually are manually
placed quite often. For example to get accurate 50:50 clock manually placed
buffers might be needed because 1->0, 0->1 transitioning can have different
propagation times in gates (that is usually layout work tough).

Since Xilinx application note also have only a schematic and no
VHDL-code I'm almost shure, that this problem is not solvable. You
cannot trigger on both edges of a signal in one process, although you
can simulate it behavioral. I will pursue the suggestion with splitting
Click to expand...
Of course some tool might even support that. But triggering at both edges
is quite rare in normal synchronous logic. It makes also STA etc. more
difficult. It's easier to use the common style in coding.


--Kim
 
Hello Kim,

Kim Enkovaara wrote:

[....]

Since Xilinx application note also have only a schematic and no
VHDL-code I'm almost shure, that this problem is not solvable. You
cannot trigger on both edges of a signal in one process, although you

can simulate it behavioral. I will pursue the suggestion with splitting

Of course some tool might even support that. But triggering at both edges
is quite rare in normal synchronous logic. It makes also STA etc. more
difficult. It's easier to use the common style in coding.


--Kim
Click to expand...
what is STA? Shure, this divider by 3 is a training example for
designing logic and learning VHDL. But we already needed a clock divider
by 3 in System Generator, which generates it automatically, though I
cannot recognize the structure of it.

So thank you and the others for replying, I'll procceed with the
simplified workaround, a counter which counts edges, rising or falling
edges. If I fail, I will return back with this problem.

Regards, Winfried
 
Hi Winfried,
Just for the fun I made such a divider - Ii belive that it should be
synthesizeable..
it uses 6 FFs and I'm sure that a reduced version is possible to create
but that's the first thing I thought about.
I hope it will help.

p.s - I needed the reset so I added it..



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

entity div3_vhdl is
Port ( clk : in std_logic;
reset : in std_logic;
q_out : out std_logic);
end div3_vhdl;

architecture Behavioral of div3_vhdl is

signal rise,fall: std_logic;

begin

process(clk,reset)
variable r_cnt : integer range 0 to 2;
begin
if reset = '1' then
r_cnt := 0;
rise <= '1';
elsif clk'event and clk ='1' then
if r_cnt = 2 then
r_cnt := 0;
rise <= not rise;
else
r_cnt := r_cnt + 1;
end if;
end if;
end process;

process(clk)
variable f_cnt : integer range 0 to 2;
begin
if reset = '1' then
f_cnt := 2;
fall <= '1';
elsif clk'event and clk ='0' then
if f_cnt = 2 then
f_cnt := 0;
fall <= not fall;
else
f_cnt := f_cnt + 1;
end if;
end if;
end process;

q_out <= fall xnor rise;

end Behavioral;
 
Hi Moti,

<moti@terasync.net> schrieb im Newsbeitrag
news:1106068400.390735.176080@z14g2000cwz.googlegroups.com...
Hi Winfried,
Just for the fun I made such a divider - Ii belive that it should be
synthesizeable..
it uses 6 FFs and I'm sure that a reduced version is possible to create
but that's the first thing I thought about.
I hope it will help.

p.s - I needed the reset so I added it..
Click to expand...
thanks, I tested your code and it is implementable. Now it seems to be
almost shure, that in VHDL not on both edges can be triggered. But
separating this into 2 processes makes it complicated, you must synchronize
them. In your example you try to synchronize it with reset, i will think
about self synchronizing.

The synthezised circuit is much more complicated than the 'paper' version
and I fear, that VHDL cannot find an optimized solution, because the
programmer must set a precondition, which IMHO is a contradiction to the
sense of VHDL. Since I looked in books and asked several people, also
experts, it is strange that no one knows this drawback.

In another (beginners) problem I had short time ago, I wanted to build an
up-down counter with two independent clocks in 1 process. Why this is not
possible is easier to understand, because no synchronous network is possible
to do that. But with the divider by 3 it is possible and therefore not to
understand.

Regards, Winfried
 
"Winfried Salomon" <wsalomon@t-online.de> wrote in message
news:csmnut$51e$01$1@news.t-online.com...
Hi Moti,

moti@terasync.net> schrieb im Newsbeitrag
news:1106066690.821488.24970@c13g2000cwb.googlegroups.com...
You are trying to infer a FF that is sensetive to both the rising and
falling edge - such a FF is (mostly) not synthesizeable.
the Modelsim can simulate it however..


it seems to be so, but there is no error message, although the circuit
synthezising process failes.

In order to design such a divider you need to use two sets of FFs each
sensetive to a different edge (rising / falling) and then
combine their outputs (using combinatorical logic - probably an "and"
gate) to get your divided clock.

Regards, Moti.


Yes, I made this on paper by myself without computer, it needs 3 D-FFS, 2
inverters and 1 and2, 1 clock is inverted, a very simple looking circuit
and
it works fine. So I must think, in VHDL is no symmetrical odd frequency
divider possible, because in 1 process not on both edges can be triggered.

regards, Winfried
Click to expand...
It isn't a VHDL issue; it's a tools and target device issue. As was
previously mentioned, it can simulate fine, but not be synthesizable. If
your target architecture supports flip-flops that clock on each edge, and if
your synthesis tools support that architecture, then it would be
synthesizable. There are many other (conceptually more complicated)
structures/building blocks that VHDL can simulate; many times, however, you
must instantiate the desired component (such as Xilinx's DCMs, certain DDR
(hah! both edges of the clock) registers which are available in Virtex
family devices.) In the past, you would have had to instantiate adders and
multiplier, but today many tools will synthesize the behavioral code and
infer the appropriate underlying architecture(s.) Maybe in another few
revisions, some will support inferred DDR, but don't hold your breath; it's
a specialized function, and not difficult to just instantiate. However,
that complicates the question of portability.

Jason
 
Hi Jason,

"jtw" <wrightjt @hotmail.invalid> schrieb im Newsbeitrag
news:ytGHd.12731$wZ2.9729@newssvr13.news.prodigy.com...
It isn't a VHDL issue; it's a tools and target device issue. As was
previously mentioned, it can simulate fine, but not be synthesizable. If
your target architecture supports flip-flops that clock on each edge, and
if
your synthesis tools support that architecture, then it would be
synthesizable. There are many other (conceptually more complicated)
Click to expand...
I just changed the chip to Virtex XCV400HQ240-4 anf the result is even
worse. The RTL-viewer shows only nonsense and as I understand, this is a pre
optimized circuit, which will be further optimized and fitted into hardware.
So IMHO the XST is not able to synthesize a logical circuit to divide
symmetrical by 3. In this case only 1 clock input must be inverted and the
XST does't find the solution. If VHDL has not that restriction, then the XST
must have it.

A hardware restriction is not to see as in GALs. In the GALs I used all
clocks are connected, so you cannot invert one, but in FPGAs this is not so.

structures/building blocks that VHDL can simulate; many times, however,
you
must instantiate the desired component (such as Xilinx's DCMs, certain DDR
(hah! both edges of the clock) registers which are available in Virtex
Click to expand...
I discussed with a collegue about DCM in Virtex, but this is not the same as
the the divider I want. I'm not yet familiar with DCM, but I suppose it is
not a digital circuit and has a lock in behavior like PLL.

family devices.) In the past, you would have had to instantiate adders
and
multiplier, but today many tools will synthesize the behavioral code and
infer the appropriate underlying architecture(s.) Maybe in another few
revisions, some will support inferred DDR, but don't hold your breath;
it's
a specialized function, and not difficult to just instantiate. However,
that complicates the question of portability.
Click to expand...
I don't know how to multiply in VHDL, there are also cores in core
generator. But I'm not shure if my problem is the same as to implement a
complex function. The XST does't recognize the formulation of the problem
and totally failes in solution. This is not a function block but a way to
synthesize. It were interesting if other sythesize tools then XST can solve
this problem, but I have none.

regards, Winfried
 
On Fri, 21 Jan 2005 01:37:27 +0100, "Winfried Salomon"
<wsalomon@t-online.de> wrote:

Hi Jason,

"jtw" <wrightjt @hotmail.invalid> schrieb im Newsbeitrag
news:ytGHd.12731$wZ2.9729@newssvr13.news.prodigy.com...


It isn't a VHDL issue; it's a tools and target device issue. As was
previously mentioned, it can simulate fine, but not be synthesizable. If
your target architecture supports flip-flops that clock on each edge, and
if
your synthesis tools support that architecture, then it would be
synthesizable. There are many other (conceptually more complicated)

I just changed the chip to Virtex XCV400HQ240-4 anf the result is even
worse. The RTL-viewer shows only nonsense and as I understand, this is a pre
optimized circuit, which will be further optimized and fitted into hardware.
So IMHO the XST is not able to synthesize a logical circuit to divide
symmetrical by 3. In this case only 1 clock input must be inverted and the
XST does't find the solution. If VHDL has not that restriction, then the XST
must have it.
Click to expand...
The restriction isn't in VHDL. It's not in XST either.

XST is quite capable of synthesising divide by 3 counter with
symetrical output, given appropriate source code.

You should be trying to end up with the design presented in this
Xilinx app note:
http://www.xilinx.com/xcell/xl33/xl33_30.pdf

Regards,
Allan
 
Hi Allan,

Allan Herriman wrote:

The restriction isn't in VHDL. It's not in XST either.

XST is quite capable of synthesising divide by 3 counter with
symetrical output, given appropriate source code.



ok, but the source code is in my first mail, you can simulate but not
Click to expand...
implement it.

You should be trying to end up with the design presented in this
Xilinx app note:
http://www.xilinx.com/xcell/xl33/xl33_30.pdf

Regards,
Allan


Thanks for your tip, I just tested it and it works. It's another
Click to expand...
approach with 2 D-FFs and 1 RS-FF, the expenditure is almost the same as
with my approach with 3 D-FFs, but my solution is somewhat faster ;-).

But ..... it's not the solution of the problem. Of course you can write
a schematic circuit in VHDL an then it must work, only you must find the
circuit yourself. I hoped that VHDL or the implementation tool can find
this circuit, but it cannot.

Since Xilinx application note also have only a schematic and no
VHDL-code I'm almost shure, that this problem is not solvable. You
cannot trigger on both edges of a signal in one process, although you
can simulate it behavioral. I will pursue the suggestion with splitting
up in 2 communicating processes triggered on rising and falling edges,
this could be a workaround.

Regards, Winfried
 
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