CODE UPDATED ON 01/11/2017 AFTER IMPROVISING!
THE ARTICLE WAS UPDATED AGAIN ON 04/03/2024!
There are many situations in which you may need to activate a process after a certain delay or at fixed time intervals.If you want to do simulation alone for your design then you can simply use "wait for" statement to call a delay routine.But this keyword is not synthesizable. So what will you do in such situations?
A simple delay routine,which is synthesizable, can be designed using the properties of a "MOD-n counter". A MOD-n counter can be used to generate a frequency of (f / n) using a frequency 'f'. The code for generating such a delay is given below:
VHDL Design:
library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.NUMERIC_STD.ALL; entity delay_gen is port (clk,reset : in std_logic; delay_cycles : in unsigned(31 downto 0); --delay to be generated. flag : out std_logic --this is a pulse to notify that time interval equal to delay is over. ); end delay_gen; architecture Behavioral of delay_gen is signal count : unsigned(31 downto 0) := (others => '0'); begin process(clk,reset) begin if(reset = '1') then flag <= '0'; count <= (others => '0'); elsif(rising_edge(clk)) then --when the necessary delay is achieved, reset count and set flag high if(count = delay_cycles-1) then count <= (others => '0'); flag <= '1'; else flag <= '0'; count <= count +1; --increment counter otherwise. end if; end if; end process; end Behavioral;
The module has 3 inputs including a clock and a reset signal. The delay_cycles input determines the amount of delay generated by the code.
We can write,
delay_cycles = [Delay in seconds] / [Clock period of clk].
Suppose you want a delay of 100 ns. Let's say your clk frequency is 100 MHz. That means your clock period is 10 ns. So the value of delay_cycles should be equal to (100 ns / 10 ns) = 10. When the counter counts from 0 to 9, it outputs a pulse on the flag output port.
Few more example values of delay_cycles are given below for different delay times. Lets assume that system clock frequency is 100 MHz.
The following testbench code demonstrates how this can be done in practice.
Testbench Code:
LIBRARY ieee; USE ieee.std_logic_1164.ALL; USE ieee.numeric_std.ALL; ENTITY tb IS END tb; ARCHITECTURE behavior OF tb IS -- Component Declaration for the Unit Under Test (UUT) COMPONENT delay_gen port (clk,reset : in std_logic; delay_cycles : in unsigned(31 downto 0); --delay to be generated. flag : out std_logic --this is a pulse to notify that time interval equal to delay is over. ); END COMPONENT; --Inputs signal clk : std_logic := '0'; signal reset : std_logic := '0'; signal delay_cycles : unsigned(31 downto 0) := (others => '0'); --Outputs signal flag : std_logic; -- Clock period definitions constant clk_period : time := 10 ns; BEGIN -- Instantiate the Unit Under Test (UUT) uut: delay_gen PORT MAP ( clk => clk, reset => reset, delay_cycles => delay_cycles, flag => flag ); -- Clock process definitions clk_process :process begin clk <= '1'; wait for clk_period/2; clk <= '0'; wait for clk_period/2; end process; -- Stimulus process stim_proc: process begin reset <= '1'; wait for clk_period*10; reset <= '0'; --by setting the input as "10", I want a delay of 10 clock cycles. --in real time,the delay = 10*period of clock. --so here in simulation, delay = 10*10 ns = 100 ns. delay_cycles <= to_unsigned(10,32); wait; end process; END;
Simulation Waveform:

This code can be used in many situations, for example:
- To run some parts of your design at a lesser clock frequency than your system clock.
- To create a delay between some of the processes.
- To create a 'wait for X seconds' function.
4 Comments
Thanks for the code,
ReplyDeleteCan I have the same in verilog?
@kartik : yeah. You can use the basic idea here to write a similar code in verilog.
ReplyDeletebut how do i use this in a desgin code? for example if i want to implement a flip flop how do i encode this mod-n counter in the code for flip flop...can u please give an example? I'm a novice.
ReplyDelete@nebula : any MOD-n counter can be designed using flip flops.You just have to reset all the flip flops when the output reaches the particular count.If it is 3 bit counter , and you want the max count to be "5" then the reset input of all FF's should be connected to R= count(0) xor (not count(1)) xor count(2).
ReplyDelete