I'm presently trying to use VHDL to design a traffic light controller, which I'm programming on an Altera EPM240T100C5 with a custom expansion board for displaying the traffic lights. As the slowest clock setting on the board is still faster than I would like, I've needed to write a clock divider which I did as so:
LIBRARY ieee;
USE ieee.std_logic_1164.all;
entity clockdivider is
port
(
clkin : in std_logic;
dividedclk : out std_logic
);
end clockdivider;
architecture divider of clockdivider is
signal J : std_logic;
signal K : std_logic;
begin
J <= '1';
K <= '1';
process(clkin)
variable tempdividedclk : std_logic;
begin
if (rising_edge(clkin)) then
tempdividedclk := (NOT(tempdividedclk) AND J) OR (tempdividedclk AND (NOT(K)));
end if;
dividedclk <= '0';
dividedclk <= tempdividedclk;
end process;
END divider;
This runs fine on the board but in the simulator (ModelSim) the "dividedclk" output fails to ever initialise to anything. I was wondering if anyone had any idea why?
At the beginning of the simulation, "tempdividedclk" is initialized to "unitialized".
When a clock edge occurs, tempdividedclk will be assigned to (not(U) and 1) or (U and 0), which is "undefined". To simulate correctly, tempdividedclk must be initialized either by a reset or just at simulation level. It works find on silicon because the "U" state will be either a 1 or a 0.
Related
I'm an amateur when it comes to VHDL and hardware in general but I've been working on a project for school and I came across something that I can't understand.
I have a register (type D FF) that's processing a clock signal to store the input value and, in simulation, it works fine if I use a "Force clock" in it's clk input but if I try to "simulate" a clock by manually changing it with "Force Constant" from zero to one and so forth it doesn't "pick up" on the rising edge.
Is this normal behavior? I assumed it would still detect the rising edge when going from 0 to 1.
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
entity register_D is
generic (
WIDTH : POSITIVE := 1
);
Port ( CLK : in STD_LOGIC;
RST : in STD_LOGIC;
EN : in STD_LOGIC;
D : in STD_LOGIC_VECTOR(WIDTH-1 downto 0);
Q : out STD_LOGIC_VECTOR(WIDTH-1 downto 0));
end register_D;
architecture Behavioral of register_D is
begin
process (CLK, RST, EN)
begin
if (RST='1') then
Q <= (others=>'0');
elsif (rising_edge(CLK) and EN = '1') then
Q <= D;
end if;
end process;
end Behavioral;
Screenshot of me trying to trigger the FF by manually setting the clock (and not working):
Here you can see it working when I switch clk from "Force constant" to "Force clock":
I have the following code (a primitive "RS-232 signalling" transmitter)...
LIBRARY ieee;
USE ieee.std_logic_1164.all;
entity SerialTX is
port(
baud_clk : in std_logic;
data : in std_logic_vector(7 downto 0);
send : in std_logic;
serial_out : out std_logic := '0';
busy : out std_logic := '0'
);
end entity;
----------------------------------------
architecture behavioural of SerialTX is
constant IDLE_BITS : std_logic_vector(10 downto 0) := "00000000001";
signal shifter : std_logic_vector(10 downto 0) := IDLE_BITS;
signal shift : std_logic := '0';
signal internal_busy : std_logic := '0';
begin
-------- ALWAYS HAPPENING --------
serial_out <= shifter(0);
busy <= internal_busy;
internal_busy <= '1' when (shifter /= IDLE_BITS) else '0';
----------------------------------
shifting_handler:
process(baud_clk) is
begin
if rising_edge(baud_clk) then
if (send = '1') and (shifter = IDLE_BITS) then
shifter <= "11" & data & '0';
elsif (shifter /= IDLE_BITS) then
shifter <= '0' & shifter(10 downto 1); -- shifter >>= 1;
end if;
end if;
end process;
end architecture behavioural;
... it works well (in simulation) but has a limitation. The send signal (that causes a transmission to begin) has to be a '1' level for longer than at least one full cycle of the baud_clk in order for the transmitter to see it reliably.
I have been trying to find a way to convert this code so that it responds to the rising edge of the send signal instead of testing its level at the rising edge of baud_clk. I want to be able to respond to a send pulse less than 100ns in duration even when the baud_clk is running at a much slower rate (115200 hz for instance).
I've tried (naively) altering the process thus...
shifting_handler:
process(baud_clk) is
begin
if rising_edge(baud_clk) then
if (shifter /= IDLE_BITS) then
shifter <= '0' & shifter(10 downto 1); -- shifter >>= 1;
end if;
elsif rising_edge(send) and (shifter = IDLE_BITS) then
shifter <= "11" & data & '0';
end if;
end process;
Here I was hoping to change the logic to test for a rising edge on send when there isn't a rising edge on baud_clk.
I know that this is not a valid approach to the problem (the synthesizer moans of course) but I was hoping that someone could explain in simple terms why this cannot be done. What would happen if it was possible to use two edge detectors in a process? There is a concept here I cannot grasp and I always seem to end up writing the code in the same way and producing this problem. I'm fighting hard against years of ingrained software programming habits, which doesn't help much!
It sounds like send is asynchronous with respect to baud_clk. You therefore need to perform some form of clock domain crossing (CDC) in order to correctly implement your design, otherwise you will have a design that cannot pass timing and has the potential to not function correctly. CDC is a standard term that you should be able to find more information about in other questions, and elsewhere.
As you have found, you cannot have a design realised in real hardware if it has a process sensitive to edges on two different signals. There's no one 'right' way to do what you want, but here is one example that uses a simple 'toggle' CDC. This is very simple, but note that the design could miss sending a byte if one send request arrives before a previous byte has been transmitted. There will also be some delay introduced between assertion of the send signal, and the transmission starting. It's not clear if these issues matter in your system.
Create another process sensitive to send:
-- The initial state doesn't matter, but we want the design to work in simulation
signal send_toggle : std_logic := '0';
process(send)
begin
if (rising_edge(send)) then
send_toggle <= not send_toggle;
end if;
end process;
Now another process to synchronize this to the baud_clk domain. Use two cascaded registers to produce a design that is largely immune to any metastability (this is another standard term that you can look up) that can result from sampling a signal generated from a different clock domain:
signal send_toggle_r1 : std_logic;
signal send_toggle_r2 : std_logic;
process(baud_clk)
begin
if (rising_edge(baud_clk)) then
send_toggle_r1 <= send_toggle;
send_toggle_r2 <= send_toggle_r1;
end if;
end process;
The above is a very standard circuit block that you can use in many single-bit CDC scenarios.
Your transmit process can then register the send_toggle_r2 signal in order to look for a transition, in order to determine whether it should start sending. This signal is in the correct clock domain:
signal send_toggle_r3 : std_logic;
process(baud_clk) is
begin
if rising_edge(baud_clk) then
send_toggle_r3 <= send_toggle_r2;
if ((send_toggle_r3 /= send_toggle_r2) and (shifter = IDLE_BITS)) then
shifter <= "11" & data & '0';
elsif (shifter /= IDLE_BITS) then
shifter <= '0' & shifter(10 downto 1); -- shifter >>= 1;
end if;
end if;
end process;
Lastly, you will need to implement timing constraints to tell your tool chain not to worry about timing of the send_toggle_r1 register.
You might spot that if you are targeting hardware where the initial states of registers are random, you might get an erroneous byte transmission after the first few baud_clk cycles. To prevent this, you might choose to hold your baud_clk process in reset for some clock cycles after start up, but as I don't know if this is relevant for you, I won't detail this part.
This whole answer addresses your question directly, but my personal approach would be to use whatever higher-rate clock is generating your send signal to drive the entire design. The serial transmission would then in fact use the higher rate clock, with shifting enabled by a CDC > edge detector chain driven from the baud_clk. The bit timing would not be absolutely perfect, but this should not matter for a standard 'UART' scenario.
In the past I asked a question about resets, and how to divide a high clock frequency down to a series of lower clock square wave frequencies, where each output is a harmonic of one another e.g. the first output is 10 Hz, second is 20 Hz etc.
I received several really helpful answers recommending what appears to be the convention of using a clock enable pin to create lower frequencies.
An alternative since occurred to me; using a n bit number that is constantly incremented, and taking the last x bits of the number as the clock ouputs, where x is the number of outputs.
It works in synthesis for me - but I'm curious to know - as I've never seen it mentioned anywhere online or on SO, am I missing something that means its actually a terrible idea and I'm simply creating problems for later?
I'm aware that the limitations on this are that I can only produce frequencies that are the input frequency divided by a power of 2, and so most of the time it will only approximate the desired output frequency (but will still be of the right order). Is this limitation the only reason it isn't recommended?
Thanks very much!
David
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.ALL;
library UNISIM;
use UNISIM.VComponents.all;
use IEEE.math_real.all;
ENTITY CLK_DIVIDER IS
GENERIC(INPUT_FREQ : INTEGER; --Can only divide the input frequency by a power a of 2
OUT1_FREQ : INTEGER
);
PORT(SYSCLK : IN STD_LOGIC;
RESET_N : IN STD_LOGIC;
OUT1 : OUT STD_LOGIC; --Actual divider is 2^(ceiling[log2(input/freq)])
OUT2 : OUT STD_LOGIC); --Actual output is input over value above
END CLK_DIVIDER;
architecture Behavioral of Clk_Divider is
constant divider : integer := INPUT_FREQ / OUT1_FREQ;
constant counter_bits : integer := integer(ceil(log2(real(divider))));
signal counter : unsigned(counter_bits - 1 downto 0) := (others => '0');
begin
proc : process(SYSCLK)
begin
if rising_edge(SYSCLK) then
counter <= counter + 1;
if RESET_N = '0' then
counter <= (others => '0');
end if;
end if;
end process;
OUT1 <= counter(counter'length - 1);
OUT2 <= not counter(counter'length - 2);
end Behavioral;
Functionally the two outputs OUT1 and OUT2 can be used as clocks, but that method of making clocks does not scale and is likely to cause problems in the implementation, so it is a bad habit. However, it is of course important to understand why this is so.
The reason it does not scale, is that every signal used as clock in a FPGA is to be distributed through a special clock net, where the latency and skew is well-defined, so all flip-flops and memories on each clock are updated synchronously. The number of such clock nets is very limited, usually in the range of 10 to 40 in a FPGA device, and some restrictions on use and location makes it typically even more critical to plan the use of clock nets. So it is typically required to reserve clock nets for only real asynchronous clocks, where there is no alternative than to use a clock net.
The reason it is likely to cause problems, is that clocks created based on bits in a counter have no guaranteed timing relation. So if it is required to moved data between these clock domains, it requires additional constrains for synchronization, in order to be sure that the Clock Domain Crossing (CDC) is handled correctly. This is done through constrains for synthesis and/or Static Timing Analysis (STA), and is usually a little tricky to get right, so using a design methodology that simplifies STA is habit that saves design time.
So in designs where it is possible to use a common clock, and then generate synchronous clock enable signals, this should be the preferred approach. For the specific design above, a clock enable can be generated simply by detecting the '0' to '1' transition of the relevant counter bit, and then assert the clock enable in the single cycle where the transition is detected. Then a single clock net can be used, together with 2 clock enables like CE1 and CE2, and no special STA constrains are required.
Morten already pointed out the theory in his answer.
With the aid of two examples, I will demonstrate the problems you encounter when using a generated clock instead of clock enables.
Clock Distribution
At first, one must take care that a clock arrives at (almost) the same time at all destination flip-flops. Otherwise, even a simple shift register with 2 stages like this one would fail:
process(clk_gen)
begin
if rising_edge(clk_gen) then
tmp <= d;
q <= tmp;
end if;
end if;
The intended behavior of this example is that q gets the value of d after two rising edges of the generated clock clock_gen.
If the generated clock is not buffered by a global clock buffer, then the delay will be different for each destination flip-flop because it will be routed via the general-purpose routing.
Thus, the behavior of the shift register can be described as follows with some explicit delays:
library ieee;
use ieee.std_logic_1164.all;
entity shift_reg is
port (
clk_gen : in std_logic;
d : in std_logic;
q : out std_logic);
end shift_reg;
architecture rtl of shift_reg is
signal ff_0_q : std_logic := '0'; -- output of flip-flop 0
signal ff_1_q : std_logic := '0'; -- output of flip-flop 1
signal ff_0_c : std_logic; -- clock input of flip-flop 0
signal ff_1_c : std_logic; -- clock input of flip-flop 1
begin -- rtl
-- different clock delay per flip-flop if general-purpose routing is used
ff_0_c <= transport clk_gen after 500 ps;
ff_1_c <= transport clk_gen after 1000 ps;
-- two closely packed registers with clock-to-output delay of 100 ps
ff_0_q <= d after 100 ps when rising_edge(ff_0_c);
ff_1_q <= ff_0_q after 100 ps when rising_edge(ff_1_c);
q <= ff_1_q;
end rtl;
The following test bench just feeds in a '1' at input d, so that, q should be '0' after 1 clock edge an '1' after two clock edges.
library ieee;
use ieee.std_logic_1164.all;
entity shift_reg_tb is
end shift_reg_tb;
architecture sim of shift_reg_tb is
signal clk_gen : std_logic;
signal d : std_logic;
signal q : std_logic;
begin -- sim
DUT: entity work.shift_reg port map (clk_gen => clk_gen, d => d, q => q);
WaveGen_Proc: process
begin
-- Note: registers inside DUT are initialized to zero
d <= '1'; -- shift in '1'
clk_gen <= '0';
wait for 2 ns;
clk_gen <= '1'; -- just one rising edge
wait for 2 ns;
assert q = '0' report "Wrong output" severity error;
wait;
end process WaveGen_Proc;
end sim;
But, the simulation waveform shows that q already gets '1' after the first clock edge (at 3.1 ns) which is not the intended behavior.
That's because FF 1 already sees the new value from FF 0 when the clock arrives there.
This problem can be solved by distributing the generated clock via a clock tree which has a low skew.
To access one of the clock trees of the FPGA, one must use a global clock buffer, e.g., BUFG on Xilinx FPGAs.
Data Handover
The second problem is the handover of multi-bit signals between two clock domains.
Let's assume we have 2 registers with 2 bits each. Register 0 is clocked by the original clock and register 1 is clocked by the generated clock.
The generated clock is already distributed by clock tree.
Register 1 just samples the output from register 0.
But now, the different wire delays for both register bits in between play an important role. These have been modeled explicitly in the following design:
library ieee;
use ieee.std_logic_1164.all;
library unisim;
use unisim.vcomponents.all;
entity handover is
port (
clk_orig : in std_logic; -- original clock
d : in std_logic_vector(1 downto 0); -- data input
q : out std_logic_vector(1 downto 0)); -- data output
end handover;
architecture rtl of handover is
signal div_q : std_logic := '0'; -- output of clock divider
signal bufg_o : std_logic := '0'; -- output of clock buffer
signal clk_gen : std_logic; -- generated clock
signal reg_0_q : std_logic_vector(1 downto 0) := "00"; -- output of register 0
signal reg_1_d : std_logic_vector(1 downto 0); -- data input of register 1
signal reg_1_q : std_logic_vector(1 downto 0) := "00"; -- output of register 1
begin -- rtl
-- Generate a clock by dividing the original clock by 2.
-- The 100 ps delay is the clock-to-output time of the flip-flop.
div_q <= not div_q after 100 ps when rising_edge(clk_orig);
-- Add global clock-buffer as well as mimic some delay.
-- Clock arrives at (almost) same time on all destination flip-flops.
clk_gen_bufg : BUFG port map (I => div_q, O => bufg_o);
clk_gen <= transport bufg_o after 1000 ps;
-- Sample data input with original clock
reg_0_q <= d after 100 ps when rising_edge(clk_orig);
-- Different wire delays between register 0 and register 1 for each bit
reg_1_d(0) <= transport reg_0_q(0) after 500 ps;
reg_1_d(1) <= transport reg_0_q(1) after 1500 ps;
-- All flip-flops of register 1 are clocked at the same time due to clock buffer.
reg_1_q <= reg_1_d after 100 ps when rising_edge(clk_gen);
q <= reg_1_q;
end rtl;
Now, just feed in the new data value "11" via register 0 with this testbench:
library ieee;
use ieee.std_logic_1164.all;
entity handover_tb is
end handover_tb;
architecture sim of handover_tb is
signal clk_orig : std_logic := '0';
signal d : std_logic_vector(1 downto 0);
signal q : std_logic_vector(1 downto 0);
begin -- sim
DUT: entity work.handover port map (clk_orig => clk_orig, d => d, q => q);
WaveGen_Proc: process
begin
-- Note: registers inside DUT are initialized to zero
d <= "11";
clk_orig <= '0';
for i in 0 to 7 loop -- 4 clock periods
wait for 2 ns;
clk_orig <= not clk_orig;
end loop; -- i
wait;
end process WaveGen_Proc;
end sim;
As can be seen in the following simulation output, the output of register 1 toggles to an intermediate value of "01" at 3.1 ns first because the input of register 1 (reg_1_d) is still changing when the rising edge of the generated clock occurs.
The intermediate value was not intended and can lead to undesired behavior. The correct value is seen not until another rising edge of the generated clock.
To solve this issue, one can use:
special codes, where only one bit flips at a time, e.g., gray code, or
cross-clock FIFOs, or
handshaking with the help of single control bits.
I am busy trying to code a ping pong type game into my FPGA Board (Altera Cyclone II model) and there are two clocks, 50MHz and 27MHz. A clock is required for the game to work. I want to use the 50MHz clock but need to slow it down to 1Hz (the "ball" should only move one LED for every oscillation). I am a very new user to VHDL and FPGA and have done some research online but none have really helped me regarding the slowing down of the clock in VHDL.
My thinking is to use two loops, one for low and one for high and using 25000000 clock cycles for high and 25000000 clock cycles for low which will end up generating a 1Hz clock when using the 50MHz clock. I have no idea how to go about implementing it in VHDL though even though I know the non-code logic for it.
I checked a previous similar question: Clock divider simulation but the person who posted the question did not include comments and I can't really follow what goes on in that code. Any help is highly appreciated!
Below is the code I have but I am not sure if it will work or not.
library ieee;
use ieee.std_logic_1164.all;
use ieee.std_logic_unsigned.all;
entity pingpong is
port(
clk : in std_logic;
gameclk : out std_logic
);
end pingpong;
architecture behave of pingpong is
begin
process(clk)
variable a : integer := 0;
signal b : std_logic := '0';
begin
if rising_edge(clk) then
a := a+1;
if a = 25000000 and b = '0' then
b := '1'; a := 0;
elsif a = 25000000 and b = '1' then
b := '0'; a := 0;
end if;
end if;
gameclk <= b;
end process;
end behave;
I want to take samples of digital data coming externaly to FPGA spartan 3.
I want to take 1000 samples/sec initially. How to select a clock frequency in vhdl coding?
Thanks.
Do not use a counter to generate a lower frequency clock signal.
Multiple clock frequencies in an FPGA cause a variety of design problems, some of which come under the heading of "advanced topics" and, while they can (if necessary) all be dealt with and solved, learning how to use a single fast clock is both simpler and generally better practice (synchronous design).
Instead, use whatever fast clock your FPGA board provides, and generate lower frequency timing signals from it, and - crucially - use them as clock enables, not clock signals.
DLLs, DCMs, PLLs and other clock managers do have their uses, but generating 1 kHz clock signals is generally not a good use, even if their limitations permit it. This application is just crying out for a clock enable...
Also, don't mess around with magic numbers, let the VHDL compiler do the work! I have put the timing requirements in a package, so you can share them with the testbench and anything else that needs to use them.
package timing is
-- Change the first two constants to match your system requirements...
constant Clock_Freq : real := 40.0E6;
constant Sample_Rate : real := 1000.0;
-- These are calculated from the above, so stay correct when you make changes
constant Divide : natural := natural(Clock_Freq / Sample_Rate);
-- sometimes you also need a period, e.g. in a testbench.
constant clock_period : time := 1 sec / Clock_Freq;
end package timing;
And we can write the sampler as follows:
(I have split the clock enable out into a separate process to clarify the use of clock enables, but the two processes could be easily rolled into one for some further simplification; the "sample" signal would then be unnecessary)
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.all;
use work.timing.all;
entity sampler is
Port (
Clock : in std_logic;
Reset : in std_logic;
ADC_In : in signed(7 downto 0);
-- signed for audio, or unsigned, depending on your app
Sampled : out signed(7 downto 0);
);
end sampler;
architecture Behavioral of Sampler is
signal Sample : std_logic;
begin
Gen_Sample : process (Clock,Reset)
variable Count : natural;
begin
if reset = '1' then
Sample <= '0';
Count := 0;
elsif rising_edge(Clock) then
Sample <= '0';
Count := Count + 1;
if Count = Divide then
Sample <= '1';
Count := 0;
end if;
end if;
end process;
Sample_Data : process (Clock)
begin
if rising_edge(Clock) then
if Sample = '1' then
Sampled <= ADC_In;
end if;
end if;
end process;
end Behavioral;
The base clock must be based on an external clock, and can't be generated just through internal resources in a Spartan-3 FPGA. If required, you can use the Spartan-3 FPGA Digital Clock Manager (DCM) resources to scale the external clock. Synthesized VHDL code in itself can't generate a clock.
Once you have some base clock at a higher frequency, for example 100 MHz, you can easily divide this down to generate an indication at 1 kHz for sampling of the external input.
It depends on what clock frequency you have available. If you have a 20MHz clock source, you need to divided it by 20000 in order to get 1KHz, you can do it in VHDL or use a DCM to do this.
This is from an example on how to create a 1kHz clock from a 20MHz input:
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
entity clk20Hz is
Port (
clk_in : in STD_LOGIC;
reset : in STD_LOGIC;
clk_out: out STD_LOGIC
);
end clk200Hz;
architecture Behavioral of clk20Hz is
signal temporal: STD_LOGIC;
signal counter : integer range 0 to 10000 := 0;
begin
frequency_divider: process (reset, clk_in) begin
if (reset = '1') then
temporal <= '0';
counter <= 0;
elsif rising_edge(clk_in) then
if (counter = 10000) then
temporal <= NOT(temporal);
counter <= 0;
else
counter <= counter + 1;
end if;
end if;
end process;
clk_out <= temporal;
end Behavioral;