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31 Threads found on edaboard.com: Clock Testbench Vhdl
You don't show your testbench. What's the relationship between readReq and the clock? Is it synchronous? If readReq is coincident with the clock edge, what do think will happen?
How about using a counter with a 10 Hz clock (1 ms period) or use whatever system clock you have, e.g. 100 MHz clock and create a counter that counts from 0-9,999,999-0 and increment a second counter when you reach 9,999,999 (or at 0,000,001 for a starting increment). Instead of giving the simulator lots of work to do
LIBRARY IEEE; USE ieee.std_logic_1164.all; --USE ieee.std_logic_arith.all; USE ieee.numeric_std.all; library ieee_proposed; use ieee_proposed.fixed_pkg.all; ENTITY dwt IS PORT ( clk : IN STD_LOGIC; vid_in: IN STD_LOGIC_VECTOR(7 DOWNTO 0); -- Pixels from main memory hor_sync: IN STD_LOGIC;
I am trying to write a code in vhdl with a given time period of clock in testbench .When i execute it i get proper output , but if i modify the code slightly i.e including more functions (more operations) then although am not changing anything in the test bench , but still i the clock in the test bench waveform appears to (...)
What exactly is the problem? have you got a testbench to simulate the code? One point : the enable signal should be inside the clock. Your current code could be implemented as a gated clock enable, which is not a good idea in FPGA.
if you use d1 and d2 as variables instead of signals it will instead of signals, it will be on the same clock cycle.
Sounds like a "My first testbench - part 2" problem, because every testbench for sequential logic involves clock generation, and incrementing an integer or real signal together with clock generation would be the next step.
Hi friends I have the Test Enable signal(TE) 111680 which should be write in vhdl, clkperiod : integer := 4; -- system clock period signal clk : std_logic := '0'; signal te : std_logic := '0'; constant ct : integer := clkperiod/2; clk is already written Can I k
ISE is a synthesis tool and ignores any simulation timing statements. To generate timing in hardware, you need an input clock and sequential statements referring to it.
I am trying to convert this vhdl testbench code to verilog testbench, kindly help me to convert this part of code slave_clkedge <= '1' when SLAVE_CPHA = SLAVE_CPOL else '0'; -- Define a 3-bit counter to count SCK edges and data into register so that parallel -- register is loaded. Use same clock (...)
Ok, comments on the code. 1. You should only have clk and reset in the sensitivity list. 2. You should ONLY have if statements for clock and reset in the synchronous process, not anything else. If you want asynchronous code then make another process that is asynchronous. 3. It is best not to and anything with the clock. Put it in a nested if insid
The input is transferred to ouput on the rising edge of clk. It is not shifted. The problem is either in the testbench or a lack of understanding synchronous logic. Depending on the design purpose, you might want to code a pure combinational process without a clock.
Hi, I saw your replies in the forum and felt encouraged to post this question. Thanks for providing your expertise. I have a vhdl test bench that is associated with both implementation and simulation (as I would like to run P&R simulation). I just want to define my clock with a certain time period. Say clock = '1', wait for (...)
1. Yes 2. Its a lot easier to do these checks inside your HDL testbench. For vhdl, the quickest and dirtiest way to stop a testbench is: assert (not end_of_sim) report "Simulation finished!" severity failure; Although the cleanest way is to stop all stimulus - ie. halt input processes and turn the clock off. 3. You cant (...)
I only have the clock and reset as inputs - - - Updated - - - I only have the clock and reset as inputs
What is the clock rate?
Hi all, one of my vhdl modules produces "X" values even with the slowest clock; there were no problems with its behavioral simulation though. Here's the code; basically it's an adder (it sums its single data input A2 to a constant FO1 after reshaping A2 into the variable A1) whose output is forced to 0 if the result is negative ent
I guess, the problem is missing knowledge of the vhdl textio package. Generally ASCII formatted data files are a straightforward way to read in stimulation data. You'll read e.g. one line for each clock cycle in your testbench, and decode one or multiple values and assign it to the stimulus vector. The testbench scans (...)
If you are using normal clock frequencies it will take a very long time to simulate one second. Use a shorter time for simulation and debugging. When you have it working you can build it for 1 second and test it in a real FPGA.
Basic answer: dont create a clock in logic. Use clock enables instead.