For example a have to show 0 , 2 , 4 , 0 , 2 , 4 .. I use an output with 8 segments - less important .
output reg [7:0] data
always # (*) begin
case
1:data= 8'b00000011; //number 0
2:data= 8'b00100101; //number 2
3:data= 8'b10011001; //number 4
default:data=8'b00000011;
endcase
end
And the counter :
input clock,
input reset,
output [7:0] out
reg [31:0] counter;
always # (posedge clock) begin
if(reset==1) counter <= 0;
else counter <= counter + 1;
end
My question is can I increment the case value instead of counter ?
Like :
always # (posedge clock) begin
if(reset==1) case <= 1;
else case <= case + 1;
if(case==3) reset<=1;
end
If not then how can I do this ?
Your case statement is wrong. It should select based on the counter value:
always # (*) begin
case(counter)
1: data= 8'b00000011; //number 0
2: data= 8'b00100101; //number 2
3: data= 8'b10011001; //number 4
default:data=8'b00000011;
endcase
end
Your counter should be:
always # (posedge clock) begin
if(reset==1 || counter == 3) counter <= 1;
else counter <= counter + 1;
end
Note that case is a keyword. You can't use it in an expression.
Related
module count(clk,rst,cnt);
want to wrire the verilog code which counts upto 7 and then down to
0 and repeats forever as follows. 0,1,2,3,4,5,6,7,6,5,4,4,3,2,1
endmodule
Blockquote
Repeat state 4 for two times while counting down:
module count(
input clk,
input rst,
output cnt
);
reg [2:0] counter;
assign cnt = counter;
// Count up if flag is 1'b0, count down if flag is 1'b1
wire flag;
assign flag = (counter == 3'b1) ? 1'b1 : (counter == 3'b0) ? 1'b0;
// Repeat counter once if special_flag is 2'b01
reg [1:0] special_flag;
always #(posedge clk)
begin
if (flag == 1'b0)
begin
counter = counter + 1;
end
else if (counter == 3'b100 && special_flag == 2'b01)
begin
// Do not decrease counter, reset special_flag back to 2'b0
special_flag = 2'b0;
end else
begin
counter = counter - 1;
end
// Set special_flag to be 2'b01 when counter is 3'b1
if (counter == 3'b1)
begin
special_flag = 2'b01;
end
end
endmodule
module count(
input clk,
input rst,
output cnt
);
reg [2:0] counter;
assign cnt = counter;
// Count up if flag is 1'b0, count down if flag is 1'b1
wire flag;
assign flag = (counter == 3'b1) ? 1'b1 : (counter == 3'b0) ? 1'b0;
always #(posedge clk)
begin
if (flag == 1'b0)
begin
counter = counter + 1;
end
if (flag == 1'b1)
begin
counter = counter - 1;
end
end
endmodule
I'm new to Verilog and it is maybe a dumb question but what is the preferred codeflow in Verilog to solve this problem:
Simple counter, counting external clk (INP) up to a particular value. If the counter matches the value it rises an output wire (DRDY) for one clk period then lowers it to 0. There is an external input (SR) where I'd like to set the comparison value, so if SR = 0, then the counting is up to 500000, if SR = 1 then up to 1000000. I can do it with one value, but I'd like to expand the functionality of my module.
Thank you in advance.
My code so far with one value comparison:
module ec(INP, RST, SR, DRDY, DRDY2);
input INP, RST, SR;
output reg DRDY, DRDY2;
reg [23:0] Q;
always #(posedge INP or negedge RST)
begin
if(!RST)
begin
Q <= 24'd0;
DRDY <= 1'b0;
end
else if( Q == 24'd1000000)
begin
Q <= 24'd0;
DRDY <= 1'b1;
DRDY2 <=~DRDY2;
end
else
begin
Q <= Q + 1;
DRDY <= 1'b0;
end
end
endmodule
An easy way to handle 2 options would be an expand the if statement:
always #(posedge INP or negedge RST) begin
if(!RST) begin
Q <= 24'd0;
DRDY <= 1'b0;
end
else if(
( (SR ==1'b0) && (Q == 24'd1000000) ||
(SR ==1'b1) && (Q == 24'd500000)
) begin
//...
end
else begin
//..
end
This can look quite messy in the code so could be separated out into a count target logic, if more options are to be supported then switch to a case statement instead of if.
reg [23:0] cnt_target ;
always #* begin
if (SR == 1'b1) begin
cnt_target = 24'd1000000 ;
else begin
cnt_target = 24'd500000 ;
end
end
always #(posedge INP or negedge RST) begin
if(!RST) begin
Q <= 24'd0;
DRDY <= 1'b0;
end
else if( Q == cnt_target) begin
//...
end
else begin
//..
end
NB: You might want to consider using Q >= cnt_target that way if SR changed on the fly you do not have to wait for Q to overflow. Plus >= for me tends to synthesis smaller than ==.
I'm having trouble implementing a controller block for an 8-bit multiplier. It works normally, but only if I turn the reset wire on, then off, such as in the following stimulus (which works fine):
`timescale 1ns / 100ps
module Controller_tb(
);
reg reset;
reg START;
reg clk;
reg LSB;
wire STOP;
wire ADD_cmd;
wire SHIFT_cmd;
wire LOAD_cmd;
Controller dut (.reset(reset),
.START(START),
.clk(clk),
.LSB(LSB),
.STOP(STOP),
.ADD_cmd(ADD_cmd),
.SHIFT_cmd(SHIFT_cmd),
.LOAD_cmd(LOAD_cmd)
);
always
begin
clk <= 0;
#25;
clk <= 1;
#25;
end
initial
begin
LSB <= 0;
START <= 0;
reset <= 1;
#55;
reset <= 0;
#10;
START <= 1;
#100;
START <= 0;
LSB <= 1;
#200;
#20;
#100;
end
initial
$monitor ("stop,shift_cmd,load_cmd, add_cmd: " , STOP,SHIFT_cmd,LOAD_cmd,ADD_cmd);
endmodule
Here's the simulation result for the working stimulus:
Now, when I set the reset to zero, without ever bringing it high, here's what happens:
Clearly, I'm using the reset wire to bring my Controller to the IDLE state. Here's the code for the controller block:
`timescale 1ns / 1ps
module Controller(
input reset,
input START,
output STOP,
input clk,
input LSB,
output ADD_cmd,
output SHIFT_cmd,
output LOAD_cmd
);
//Five states:
//IDLE : 000 , INIT: 001, TEST: 011, ADD: 010, SHIFT: 110
localparam [2:0] S_IDLE = 0;
localparam [2:0] S_INIT = 1;
localparam [2:0] S_TEST = 2;
localparam [2:0] S_ADD = 3;
localparam [2:0] S_SHIFT = 4;
reg [2:0] state,next_state;
reg [3:0] count;
// didn't assign the outputs to wire.. if not work, check this.
assign ADD_cmd = (state == S_ADD);
assign SHIFT_cmd = (state == S_SHIFT);
assign LOAD_cmd = (state == S_INIT);
assign STOP = (state == S_IDLE);
always #(*) begin
case(state)
S_INIT: begin
count = 3'b000;
end
S_SHIFT: begin
count = count + 1;
end
endcase
end
always #(*)
begin
next_state = state;
case (state)
S_IDLE: next_state = START ? S_INIT : S_IDLE;
S_INIT: next_state = S_TEST;
S_TEST: next_state = LSB ? S_ADD : S_SHIFT;
S_ADD: next_state = S_SHIFT;
S_SHIFT: next_state = (count == 8) ? S_IDLE : S_TEST;
endcase
end
always #(posedge clk)
begin
//state <= S_IDLE;
if(reset) state <= S_IDLE;
else state <= next_state;
end
reg [8*6-1:0] statename;
always #* begin
case( state )
S_IDLE: statename <= "IDLE";
S_INIT: statename <= "INIT";
S_TEST: statename <= "TEST";
S_ADD: statename <= "ADD";
S_SHIFT: statename <= "SHIFT";
default: statename <= "???";
endcase
end
endmodule
I don't know how to fix this. As you can see from the code above, there is a commented portion which is basically always initializing the state to IDLE. But even that doesn't work. Here's the simulation for the code above removing the comment from '//state <= S_IDLE;':
It's going into a different state than any listed above, and I have no idea why.
So I'd like to know:
Why is it going into an unknown state? Why doesn't my uncommented code work?
What can I change for it to work as I intend?
Your problem is that without a reset or initial value, state and next_state will be X. Your case statement assigning to statename will take the default branch and decode to ???. Since your process that assigns next_state does not handle cases where state is X it will get stuck in this state forever.
Your attempt to fix this will not work:
state <= S_IDLE;
if(reset) state <= S_IDLE;
else state <= next_state;
When reset is low you are making two assignments to state, the first as S_IDLE and the second as next_state. This is not a race condition. The Verilog standard states that:
Nonblocking assignments shall be performed in the order the statements were executed.
Since no re-ordering of the event queue occurs for sequential statements within a process this translates to last assignment wins. Therefore your state <= S_IDLE; is effectively optimised away since regardless of the value of reset the assignment will be overridden.
There are two ways you could fix this so that you don't need a reset:
1. Use the default clause to make your state machine safe
always #(*)
begin
next_state = state;
case (state)
S_IDLE: next_state = START ? S_INIT : S_IDLE;
S_INIT: next_state = S_TEST;
S_TEST: next_state = LSB ? S_ADD : S_SHIFT;
S_ADD: next_state = S_SHIFT;
S_SHIFT: next_state = (count == 8) ? S_IDLE : S_TEST;
default: next_state = S_IDLE;
endcase
end
This will ensure that your state-machine is 'safe' and drops into S_IDLE if state is a non-encoded value (including X).
2. Initialise the variable
reg [2:0] state = S_IDLE;
For some synthesis targets (e.g. FPGAs) this will initialise the register to a specific value and can be used alongside or instead of a reset (see Altera Documentation on power-up values).
A couple of general points:
Depending on your synthesis tool it may be better to use an enumeration rather than explicitly defining values for your states. This allows the tool to optimise based on the overall design or use a global configuration for encodings (for example safe, one-hot).
Using a reset registers holding state is standard practice so you should carefully consider whether you really want to avoid using a reset.
The uncommented code is an example of poor coding practice because you are making 2 nonblocking assignments to state in the same timestep. Synthesis linting tools are likely to warn you of this situation.
Since using a reset is a common, good practice, I don't think you need to fix anything.
i am making a midi interface. UART works fine, it sends the 8 bit message along with a flag to a control unit. When the flag goes high, the unit will store the message in a register and make a clr_flag high in order to set the flag of UART low again. The problem is that i can not make this clr_flag one period long. I need it to be ONE period long, because this signal also controls a state machine that indicates what kind of message is being stored (note_on -> key_note -> velocity, for example).
My question here is, how can a signal (flag in this case) trigger a pulse just for one clk period? what i have now makes almost a pulse during a clock period, but i does it twice, because the flag has not become 0 yet. ive tried many ways and now i have this:
get_data:process(clk, flag)
begin
if reset = '1' then
midi <= (others => '0');
clr_flag <= '0';
control_flag <= '0';
elsif ((clk'event and clk='1') and flag = '1') then
midi <= data_in;
clr_flag <= '1';
control_flag <= '1';
elsif((clk'event and clk='0') and control_flag = '1') then
control_flag <= '0';
elsif((clk'event and clk='1') and control_flag = '0') then
clr_flag <= '0';
end if;
end process;
the problem with this double pulse or longer than one period pulse(before this, i had something that made clr_flag a two period clk pulse), is that the system will go though two states instead of one per flag.
so in short: when one signal goes high (independent of when it goes low), a pulse during one clock period should be generated.
thanks for your help.
The trick to making a single cycle pulse is realising that having made the pulse, you have to wait as long as the trigger input is high before getting back to the start. Essentially you are building a very simple state machine, but with only 2 states you can use a simple boolean to tell them apart.
Morten is correct about the need to adopt one of the standard patterns for a clocked process; I have chosen a different one that works equally well.
get_data:process(clk, reset)
variable idle : boolean;
begin
if reset = '1' then
idle := true;
elsif rising_edge(clk) then
clr_flag <= '0'; -- default action
if idle then
if flag = '1' then
clr_flag <= '1'; -- overrides default FOR THIS CYCLE ONLY
idle <= false;
end if;
else
if flag = '0' then
idle := true;
end if;
end if;
end if;
end process;
There are several issues to address in order to make the design for a one cycle
pulse using flip flops (registers).
First, the use of flip flops in hardware through VHDL constructions typically
follows a structure like:
process (clk, reset) is
begin
-- Clock
if rising_edge(clk) then
-- ... Flip flops to update at rising edge
end if;
-- Reset
if reset = '1' then
-- Flip flops to update at reset, which need not be all
end if;
end process;
So the get_data process should be updated accordingly, thus:
Sensitivity list should contain only clock (clk) and reset
The nested structure with if on event should be as above
Only rising edge of clk should be used, thus no check on clk = '0'
Making a one cycle pulse on clr_flag when flag goes high can be made with a
synchronous '0' to '1' detector on flag, using a version of flag that is
delayed a single cycle, called flag_ff below, and then checking for (flag =
''1) and (flag_ff = '0').
The resulting code may then look like:
get_data : process (clk, reset) is
begin
-- Clock
if rising_edge(clk) then
flag_ff <= flag; -- One cycle delayed version
clr_flag <= '0'; -- Default value with no clear
if (flag = '1') and (flag_ff = '0') then -- Detected flag going from '0' to '1'
midi <= data_in;
clr_flag <= '1'; -- Override default value making clr_flag asserted signle cycle
end if;
end if;
-- Reset
if reset = '1' then
midi <= (others => '0');
clr_flag <= '0';
-- No need to reset flag_ff, since that is updated during reset anyway
end if;
end process;
Synchronisation and Edge Detection for FSM
The Rise, Edge and Fall outputs will strobe for one cycle when those events are detected. Inputs and outputs are synchronised for use with a Finite State Machine.
entity SynchroniserBit is
generic
(
REG_SIZE: natural := 3 -- Default number of bits in sync register.
);
port
(
clock: in std_logic;
reset: in std_logic;
async_in: in std_logic := '0';
sync_out: out std_logic := '0';
rise_out: out std_logic := '0';
fall_out: out std_logic := '0';
edge_out: out std_logic := '0'
);
end;
architecture V1 of SynchroniserBit is
constant MSB: natural := REG_SIZE - 1;
signal sync_reg: std_logic_vector(MSB downto 0) := (others => '0');
alias sync_in: std_logic is sync_reg(MSB);
signal rise, fall, edge, previous_sync_in: std_logic := '0';
begin
assert(REG_SIZE >= 2) report "REG_SIZE should be >= 2." severity error;
process (clock, reset)
begin
if reset then
sync_reg <= (others => '0');
previous_sync_in <= '0';
rise_out <= '0';
fall_out <= '0';
edge_out <= '0';
sync_out <= '0';
elsif rising_edge(clock) then
sync_reg <= sync_reg(MSB - 1 downto 0) & async_in;
previous_sync_in <= sync_in;
rise_out <= rise;
fall_out <= fall;
edge_out <= edge;
sync_out <= sync_in;
end if;
end process;
rise <= not previous_sync_in and sync_in;
fall <= previous_sync_in and not sync_in;
edge <= previous_sync_in xor sync_in;
end;
Below is a way of creating a signal (flag2) that lasts exactly one clock period from a signal (flag1) that lasts at least one clock period.
I don't program in VHDL~ here is what I usually do for the same propose in Verilog:
always #(posedge clk or negedge rst) begin
if(~rst) flgD <= 1'b0;
else flgD <= flg;
end
assign trg = (flg^flgD)&flgD;
I am new to verilog and this is the sample code, which I used for triggering. Hope this serves your purpose. You can try same logic in VHDL.
module main(clk,busy,rd);
input clk,busy; // busy input condition
output rd; // trigger signal
reg rd,en;
always #(posedge clk)
begin
if(busy == 1)
begin
rd <= 0;
en <= 0;
end
else
begin
if (en == 0 )
begin
rd <= 1;
en <= 1;
end
else
rd <= 0;
end
end
endmodule
The below verilog code shall hold the value for the signals for one clock cycle exactly.
module PulseGen #(
parameter integer BUS_WIDTH = 32
)
(
input [BUS_WIDTH-1:0] i,
input clk,
output [BUS_WIDTH-1:0] o
);
reg [BUS_WIDTH-1:0] id_1 = 0 ;
reg [BUS_WIDTH-1:0] id_2 = 0 ;
always #(posedge clk)begin
id_1 <= i;
id_2 <= id_1;
end
assign o = (id_1 & ~id_2);
The way to achieve this is to create a debounce circuit. If you need a D flip-flop to change from 0 to 1, only for the first clock, just add an AND gate before its input like the image below:
So here you can see a D flip-flop and its debounce circuit.
P.S. Circuit created using this.
--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
--input of minimum 1 clock pulse will give output of wanted length.
--load number 5 to PL input and you will get a 5 clock pulse no matter how long input is.
--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
library ieee ;
use ieee.std_logic_1164.all ;
use ieee.std_logic_unsigned.all ;
entity fifth is
port (clk , resetN : in std_logic;
pdata : in integer range 0 to 5; --parallel data in. to choose how many clock the out pulse would be.
din : in std_logic;
dout : out std_logic
) ;
end fifth ;
architecture arc_fifth of fifth is
signal count : integer range 0 to 5;
signal pl : std_logic; --trigger detect output.
signal sample1 : std_logic;
signal sample2 : std_logic;
--trigger sync proccess.
begin
process(clk , resetN)
begin
if resetN = '0' then
sample1<='0';
sample2<='0';
elsif rising_edge(clk) then
sample1<=din;
sample2<=sample1;
end if;
end process;
pl <= sample1 and (not sample2); --trigger detect output. activate the counter.
--counter proccess.
process ( clk , resetN )
begin
if resetN = '0' then
count <= 0 ;
elsif rising_edge(clk) then
if pl='1' then
count<=pdata;
else
if count=0 then
count<=count;
else
count<=count-1;
end if;
end if;
end if ;
end process ;
dout<='1' when count>0 else '0';--output - get the wanted lenght pulse no matter how long is input
end arc_fifth ;
I'm new to VHDL and confused with this design
when Acknwledgement= '1' and clk='1' then
count should be count+1;
and when Acknwledgement= '0' my total counted value of clocks should be assigned to the 'output' and after that resetting count='0' and output='0'.
can anyone help with this.
Thanks in advance.
EDIT:
Code from comment pasted in:
library IEEE;
use IEEE.STD_LOGIC_1164.all;
entity acknw is
port (acknw : in std_logic;
clk : in std_logic;
output : out integer range 0 to 15);
end acknw;
architecture Behavioral of acknw is
begin
process(clk, acknw) variable c : integer range 0 to 15;
begin
if(clk'event and clk = '1') then
if(acknw = '1') then
c := c+1;
output <= c;
else
c := 0;
output <= c;
end if;
end if;
end process;
end Behavioral;
from your comment it sounds like you want an asynchronous acknw, try something like this:
library IEEE;
use IEEE.STD_LOGIC_1164.all;
entity acknw is
port (acknw : in std_logic;
clk : in std_logic;
output : out integer range 0 to 15);
end acknw;
architecture Behavioral of acknw is
begin
process(clk, acknw)
begin
if (acknw = '0') then
output <= 0;
elsif rising_edge(clk) then
-- rollover
if (output /= 15) then
output <= output + 1;
else
output <= 0;
end if;
end if;
end process;
end Behavioral;