I am currently in the process of designing a counter in SystemVerilog and I am unsure of how to design the D-flip flip module.
A flip-flop is inferred by synthesis tool for all signals assigned in an always_ff process.
So if you are designing an 8b counter, you would write:
logic [7:0] cnt;
always_ff # (posedge clk or negedge rst_n) begin : proc_cnt
if(!rst_n) cnt <= 0;
else cnt <= cnt + 1;
end
Related
I'm trying to understanding how the following code works, but struggling to put it together in my head. Could someone give me a more intuitive (visual) explanation of how this pipelined multiplier stage works?
// This is one stage of an 8 stage (9 depending on how you look at it)
// pipelined multiplier that multiplies 2 64-bit integers and returns
// the low 64 bits of the result. This is not an ideal multiplier but
// is sufficient to allow a faster clock period than straight *
module mult_stage(
input clock, reset, start,
input [63:0] product_in, mplier_in, mcand_in,
output logic done,
output logic [63:0] product_out, mplier_out, mcand_out
);
logic [63:0] prod_in_reg, partial_prod_reg;
logic [63:0] partial_product, next_mplier, next_mcand;
assign product_out = prod_in_reg + partial_prod_reg;
assign partial_product = mplier_in[7:0] * mcand_in;
assign next_mplier = {8'b0,mplier_in[63:8]};
assign next_mcand = {mcand_in[55:0],8'b0};
//synopsys sync_set_reset "reset"
always_ff #(posedge clock) begin
prod_in_reg <= #1 product_in;
partial_prod_reg <= #1 partial_product;
mplier_out <= #1 next_mplier;
mcand_out <= #1 next_mcand;
end
// synopsys sync_set_reset "reset"
always_ff #(posedge clock) begin
if(reset)
done <= #1 1'b0;
else
done <= #1 start;
end
endmodule
I am trying to assign value on a specific bit of a 2D array(code[i][k]). This is a net type. But the value not being assigned.reg [3:0] code[0:3] gets unknown logic value 'X'.
Here is the code snippet
for(k=0;k<len;k++) begin
if (tc[k] == 1'b0) begin
code[i][k]= 1'b0;//----> value is not assigning as expected
end else begin
code[i][k]= 1'b1;// ---> value is not assigning as expected
end
end
codeLen[i] = len;
This for loop belongs to always block.Here, code and codeLen is output type.
output [3:0] code[0:3];
output [3:0] codeLen[0:3];
reg [3:0] code[0:3];
reg [3:0] codeLen[0:3];
codeLen[i] is assigned correctly but not the code[i][k]. I was trying to assign k-th bit of i-th byte.
Details
I have created a module which takes 6 inputs and returns two 2-dimensional arrays as output.
Here is the module:
`timescale 1ns / 1ps
module generate_code(CLK,nRST,nodes,nodeCount,characters,charCount,code,codeLen);
input CLK;
input nRST;
input integer nodeCount;//Total nodes in huffman tree
input integer charCount;//Total unique characters
input [6:0] characters[0:3];
input [23:0] nodes[0:6]; // total characters
output [3:0] code[0:3]; //[2:0] max code length <= total characters
output [3:0] codeLen[0:3];
reg [3:0] code[0:3];
reg [3:0] codeLen[0:3];
reg[3:0] tc;//temprary code reg. Holds a single bit in each byte
integer len=0;//code length
reg [23:0] tNode;
function void FindRoot;
reg [23:0] aNode;//local
integer i;
begin
for (i=0; i<nodeCount;i++) begin // For all nodes
aNode= nodes[i]; // aNode is current node
if (tNode[23:16] == aNode[14:7]) begin
tc[len]= tNode[15];//15th bit of nodes is codebit
len++;
//aNode is parent of tNode. Is it root?
if(aNode[23:16]==8'b0000_0000) begin//or frequency==nodeCount or node_id = 8'b1111_1111
return;
end else begin
tNode=aNode;
FindRoot();
end
end
end
end
endfunction
always#(posedge CLK or negedge nRST)
begin
if(!nRST) begin
// init
end
else begin
// Do code generation
integer i,j,k;
for(i= 0;i < charCount;i++) begin // For all character we are going to find codeword
for(j=0; j<nodeCount; j++) begin
tNode= nodes[j];//current node
if (characters[i] == tNode[6:0]) begin
// Got the character. tNode is a leaf nodes. Lets back track to root.
break;
end
end
len=0;
FindRoot();
for(k=0;k<len;k++) begin
if (tc[k] == 1'b0) begin
code[i][k]= 1'b0;
end else begin
code[i][k]= 1'b1;
end
end
//code[i]=2;
codeLen[i]= len;
end
end
end
endmodule
When I am assigning values to code[][], it is expected that following loop is executed. Though not all the bits of code[][] will be set. During debugging, when I come to assignment, I found that value is not being assigned (code[i][k] =1 or 0). Its getting unknown logic value X.
for(k=0;k<len;k++) begin
if (tc[k] == 1'b0) begin
code[i][k]= 1'b0;
end else begin
code[i][k]= 1'b1;
end
end
Testbench:
`timescale 1ns / 1ps
module generate_code_test;
// Inputs
reg CLK;
reg nRST;
integer nodeCount=7;//Total nodes in huffman tree
integer charCount=4;//Total unique characters
reg [6:0] characters[0:3];
reg [23:0] nodes[0:6]; // total characters
// Outputs
wire [3:0] code[0:3]; //[2:0] max code length <= total characters
wire [3:0] codeLen[0:3];
generate_code uut (
.CLK(CLK),
.nRST(nRST),
.nodes(nodes),
.nodeCount(nodeCount),
.characters(characters),
.charCount(charCount),
.code(code),
.codeLen(codeLen)
);
initial begin
// Initialize Inputs
CLK = 0;
nRST = 0;
nodeCount= 7;
charCount= 4;
characters[0]= 7'b110_0001;
characters[1]= 7'b110_0010;
characters[2]= 7'b110_0011;
characters[3]= 7'b110_0100;
nodes[0] = 24'b0000_0011_0_0000_0001_110_0001;
nodes[1] = 24'b0000_0011_1_0000_0010_110_0011;
nodes[2] = 24'b0000_0101_1_0000_0011_111_1111;
nodes[3] = 24'b0000_0101_0_0000_0100_110_0010;
nodes[4] = 24'b1111_1111_1_0000_0101_111_1111;
nodes[5] = 24'b1111_1111_0_0000_0110_110_0100;
nodes[6] = 24'b0000_0000_0_1111_1111_111_1111;
// Wait 10 ns for global reset to finish
#10;
nRST = 1;
end
parameter DELAY = 1;
always
#DELAY CLK = ~CLK;
endmodule
The code has been compiled in ModelSim 2016
I just started learning verilog. So I would really appreciate your help to show my mistakes.
Regards.
I got a fix for my problem. Not all the bits of code[][] has been set. This leads to unknown logic value in code[][] even after setting the bit. It gets solved after initializing all the bits of code[][] in always block.
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 ==.
Good coding convention says that we should use blocking assignments in a combinational block, and non-blocking assignments in a sequential block. I want to use the ++ operator in a combinatorial block, but I don't know if it is blocking. So is this code:
input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
count_ones = '0;
for(int i=0; i<4; i++) begin
if(some_bus[i])
count_ones++;
end
end
equivalent to this:
input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
count_ones = '0;
for(int i=0; i<4; i++) begin
if(some_bus[i])
count_ones = count_ones + 1;
end
end
or this:
input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
count_ones = '0;
for(int i=0; i<4; i++) begin
if(some_bus[i])
count_ones <= count_ones + 1;
end
end
I did look in the 1800-2012 standard but could not figure it out. An answer that points me to the appropriate section in the standard would be appreciated.
According to section 11.4.2 of IEEE Std 1800-2012, it is blocking.
SystemVerilog includes the C increment and decrement assignment operators ++i , --i , i++ , and i-- . These do not need parentheses when used in expressions. These increment and decrement assignment operators behave as blocking assignments.
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.