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Altera_Forum's avatar
Altera_Forum
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18 years ago

Using CPLD/FPGA as a Timer

Hi.

I am interested to implement a timer that can count up to hours, in my FPGA/CPLD. The main purpose of the timer is to issue a pulse after a certain fixed duration, eg. every 2 hours. Normally I would use a counter to do this, if the period is not that long. But for the counter that can count up to hours, it needs too many LEs, since I am using a 5-MHz input clock.

Is the any better way to do this, without using too many LEs? I would like to keep the 5-MHz clock if possible. Any idea is much appreciated. Thanks!

Regards,

CW

4 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    18 years ago

    how about pre-scaling at the output? say with few ripple-counter ICs and a D-flipflop to square the waveform duty cycle to 50%

    sure, a bit of a solder job but it may be easier than tinkering with the 5Mhz especially if you are using pre-fab board with all-SMD parts...

    ~B
  • Altera_Forum's avatar
    Altera_Forum
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    18 years ago

    Thanks Frank.

    I guess without using a much slower clock, there is no way I can greatly reduce the amount of LEs used.

    Regards,

    CW
  • Altera_Forum's avatar
    Altera_Forum
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    18 years ago

    Hello,

    I just was aware, that the said MAX_COUNT overflows the integer range, thus my example is incorrect. Changing the count value to unsigned with full length, the LE consumption increases to 54. Sorry for the confusion.

    Frank 

    -- Modified Quartus II VHDL Template
-- Binary Counter
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity timer is
generic
(
	MAX_COUNT : unsigned(35 downto 0) := x"861C467FF"
);
port
(
	clk	  : in std_logic;
	reset	  : in std_logic;
	enable	  : in std_logic;
	q	  : out std_logic
);
end entity;
architecture rtl of timer is
begin
process (clk)
	variable   cnt	: unsigned(35 downto 0);
begin
	if (rising_edge(clk)) then
		if reset = '1' then
			-- Reset the counter
			cnt := (others => '0');
			q <= '0';
			elsif enable = '1' then
			-- Increment the counter if counting is enabled
			if cnt < MAX_COUNT then			   
				cnt := cnt + 1;
				q <= '0';
			else
				cnt := (others => '0');
				q <= '1';
			end if;
		end if;
	end if;
end process;
end rtl;

    P.S.: Rounding MAX_COUNT to x"860000000", the LE consumption decreases to 41
  • Altera_Forum's avatar
    Altera_Forum
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    18 years ago

    Hello,

    basically, the simple fact that the system needs (2*3600*5e6) states can't be bypassed. ceil(log2((2*3600*5e6)-1)) = 36 gives the number of bits for a binary counter, which is always the minimal state representation, I think.

    Using a modified vhdl binary counter template with a MAX II CPLD, I got an overall LE consumption of 46. The 10 additional LE's are a tribute to the particular MAX_COUNT, a 2**N value would allow a simple ripple-carry counter.

    Best regards and Happy New Year!

    Frank 

    -- Modified Quartus II VHDL Template
-- Binary Counter
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity timer is
generic
(
	MIN_COUNT : natural := 0;
	MAX_COUNT : natural := (5e6*3600*2)-1
);
port
(
	clk	  : in std_logic;
	reset	  : in std_logic;
	enable	  : in std_logic;
	q	  : out std_logic
);
end entity;
architecture rtl of timer is
begin
process (clk)
variable   cnt	: integer range MIN_COUNT to MAX_COUNT;
begin
	if (rising_edge(clk)) then
		if reset = '1' then
		-- Reset the counter
			cnt := MIN_COUNT;
			q <= '0';
		elsif enable = '1' then
		-- Increment the counter if counting is enabled
			if cnt < MAX_COUNT then			   
				cnt := cnt + 1;
				q <= '0';
			else
				cnt:=MIN_COUNT;
				q <= '1';
			end if;
		end if;
	end if;
end process;
end rtl;