I'm having trouble finding direct answers to how FPGA-level LUTs are accessed/implemented in simple modules (I'll provide an example below). Referring to the top answer on this post, https://electronics.stackexchange.com/questions/163961/creating-a-verilog-code-for-4-bit-multiplier-using-lookup-table, the type of LUT I'm trying to understand is the first one listed (FPGA-level).
For example, let's say I had the following module:
module RightRotation
{
input logic clk
input logic [2:0] din
output logic [2:0] dout
};
always#(clk) begin
dout[0] <= din[2];
dout[1] <= din[0];
dout[2] <= din[1];
end
endmodule
If I were to implement this module with a FPGA-level LUT with 3 inputs and 1 output, how many lookups would it take (for instance, is a value looked up every time a value is assigned with a nonblocking statement in the always#)? How would this answer change if I had 4, or 5, or 6 inputs?
After fixing syntax errors, Vivado produces these results for utilization and implementation.
N inputs would use N flops in your design.
Vendor datasheets can help understand look up tables, CLBs, slices etc.
Example here:
https://www.xilinx.com/support/documentation/user_guides/ug474_7Series_CLB.pdf
Related
module RingCounter(
input logic Clock,
input logic Reset,
output logic [3:0] Count
);
always_ff #(posedge Clock, posedge Reset)
begin
if (Reset)
Count <= 4’d1;
else
Count <= {Count[2:0], Count[3]};
end
endmodule
I have the working code above for a 4-bit ring counter in SystemVerilog, but I am unsure how one line of it works as it wasn't clearly explained in the lecture.
Count <= {Count[2:0], Count[3]};
Any help on explaining exactly what this line does would be much appreciated.
The curly braces, {}, are the concatenation operator. They concatenate multiple bits into a bus.
On the left hand side of the nonblocking assignment (<=), you have Count, which is a short-hand way of writing the 4-bit bus: Count[3:0].
On the right hand side of the assignment, you have the 3-bit signal Count[2:0] concatenated with the 1-bit signal Count[3].
Another way to write the RHS is as 4 separate bits in the following order:
{Count[2], Count[1], Count[0], Count[3]}
Another way to write the LHS is as 4 separate bits in the following order:
{Count[3], Count[2], Count[1], Count[0]}
Therefore, the assignment sets the new Count[3] to the old Count[2], etc.
Refer to IEEE Std 1800-2017, section 11.4.12 Concatenation operators.
My project is to use a CPDL, which I am programming in verilog to commutate a BLDC motor. Part of that process is to read in hall sensors A,B,C.
I want to count the amount of positive edges on A,B or C. I have a 3 bit variable [2:0] hallIn to store these inputs. The code below works in a modelsim simulation but not on an actual chip. How come? What is the proper way to do this? The error message I get is: 73:12:73:55|Can't mix posedge/negedge use with plain signal references
always # (posedge hallIn[2] or hallIn[1] or hallIn[0])
Synthesize code is a subset of what Verilog will let you do. Synthesizable code requires a coding structure that can be mapped to logic cells. Non-synthesizable code is intended for behavior modeling and test benches.
Flip-flops should normally be synchronized by a common clock source; not data. With a clock, you could detect the posedge data by comparing the inputs current value against it previous value.
Example:
assign posedge_detected = |(hallIn & ~prev_hallIn);
always #(posedge clock) begin
if (reset) begin
prev_hallIn <= 3'b000;
count <= 16'h0000;
end
else begin
prev_hallIn <= hallIn;
count <= count + posedge_detected;
end
end
end
I am trying to change a c++ code into verilog HDL.
I want to write a module that changes one of its inputs. (some how like call by reference in c++)
as I know there is no way to write a call by reference module in verilog (I can't use systemverilog)
Here is a code that I wrote and it works. are there any better ways to do this?!
my problme is that the register I want to be call by reference is a big array. this way duplicates the registers and has a lot of cost.
module testbench();
reg a;
wire b;
reg clk;
initial begin
a = 0;
clk = 0;
#10
clk = 1;
end
test test_instance(
.a(a),
.clk(clk),
.aOut(b)
);
always#(*)begin
a = b;
end
endmodule
module test(
input a,
input clk,
output reg aOut
);
always #(posedge clk) begin
if (a == 0)begin
a = 1;
aOut = a;
end
end
endmodule
Verilog is not a software programming language; it is a hardware description language. The inputs to a module are pieces of metal (wires, tracks, pins); the outputs from a module are pieces of metal. If you want a port that is both an input and an output you can use an inout. However, inout ports are best avoided; it is usually much better to use separate inputs and outputs.
A Verilog module is not a software function. Nothing is copied to the inputs; nothing is copied from the outputs. A Verilog module is a lump of hardware: it has inputs (pieces of metal carrying information in) and outputs (pieces of metal carrying information out).
Your are right to say that you can use either pass-by-copy or pass-by-reference in SystemVerilog. If you wish to pass a large data structure into a function or into/out of a task, then passing by reference may save simulation time.
By reference means by address, so to translate this to hdl directly you would either need to provide a way for the module to get on that bus and perform transactions based on that address.
Or better, if you need this as an input take each of the items in the struct and make individual inputs from them. If it is pass by reference because it is an output or is also an output, then you create individual outputs for each of the items in the struct. The module then distinguishes between the input version of that sub item and output version of that sub item.
my.thing.x = my.thing.x + 1;
becomes something like
my_thing_x_output = my_thing_x_input + 1;
I'm new to Verilog, ISE, FPGAs. I'm trying to implement a simple design into an FPGA, but the entire design is being optimized away. It is basically an 2D array with some arbitrary values. Here is the code:
module top(
output reg out
);
integer i;
integer j;
reg [5:0] array [0:99][0:31];
initial begin
for(i=0;i<100;i=i+1) begin
for(j=0;j<32;j=j+1) begin
array[i][j] = j;
out = array[i][j];
end
end
end
endmodule
It passes XST Synthesis fine, but it fails MAP in the Implementation process. Two Errors are given:
ERROR:Map:116 - The design is empty. No processing will be done.
ERROR:Map:52 - Problem encountered processing RPMs.
The entire code is being optimized away in XST. Why? What am I doing wrong?
The reason your design is being synthesized away is because you have not described any logic in your module.
The only block in your design is an initial block which is typically not used in synthesis except in limited cases; the construct mainly used for testbenches in simulation (running the Verilog through ModelSim or another simluator).
What you want is to use always blocks or assign statements to describe logic for XST to synthesize into a netlist for the FPGA to emulate. As the module you provided has neither of these constructs, no netlist can be generated, thus nothing synthesized!
In your case, it is not entirely clear what logic you want to describe as the result of your module will always have out equal to 31. If you want out to cycle through the values 0 to 31, you'll need to add some sequential logic to implement that. Search around the net for some tutorials on digital design so you have the fundamentals down (combinational logic, gates, registers, etc). Then, think about what you want the design to do and map it to those components. Then, write the Verilog that describes that design.
EDIT IN LIGHT OF COMMENTS:
The reason you are get no LUT/FF usage on the report is because the FPGA doesn't need to use any resources (or none of those resources) to implement your module. As out is tied to constant 31, it will always have the value of 1, so the FPGA only needs to tie out to Vdd (NOTE that out is not 31 because it is only a 1-bit reg). The other array values are never used nor accesses, so the FPGA synthesized them away (ie, not output needs to know the value of array[0][1] as out is a constant and no other ports exist in the design). In order to preserve the array, you need only use it to drive some output somehow. Heres a basic example to show you:
module top( input [6:0] i_in, // Used to index the array like i
input [4:0] j_in, // Used to index the array like j
output reg [5:0] out // Note, out is now big enough to store all the bits in array
);
integer i;
integer j;
reg [5:0] array[0:99][0:31];
always #(*) begin
// Set up the array, not necessarily optimal, but it works
for (i = 0; i < 100; i = i + 1) begin
for (j = 0; j < 32; j = j + 1) begin
array[i][j] = j;
end
end
// Assign the output to value in the array at position i_in, j_in
out = array[i_in][j_in];
end
endmodule
If you connect the inputs i_in and j_in to switches or something and out to 6 LEDs, you should be able to index the array with the switches and get the output on the LEDs to confirm your design.
I am trying to write an "inverter" function for a 2's compliment adder. My instructor wants me to use if/else statements in order to implement it. The module is supposed to take an 8 bit number and flip the bits (so zero to ones/ones to zeros). I wrote this module:
module inverter(b, bnot);
input [7:0] b;
output [7:0]bnot;
if (b[0] == 0) begin
assign bnot[0] = 1;
end else begin
assign bnot[0] = 0;
end
//repeat for bits 1-7
When I try and compile and compile it using this command I got the following errors:
vcs +v2k inverter.v
Error-[V2005S] Verilog 2005 IEEE 1364-2005 syntax used.
inverter.v, 16
Please compile with -sverilog or -v2005 to support this construct: generate
blocks without generate/endgenerate keywords.
So I added the -v2005 argument and then I get this error:
vcs +v2k -v2005 inverter.v
Elaboration time unknown or bad value encountered for generate if-statement
condition expression.
Please make sure it is elaboration time constant.
Someone mind explaining to me what I am doing wrong? Very new to all of this, and very confused :). Thanks!
assign statements like this declare combinatorial hardware which drive the assigned wire. Since you have put if/else around it it looks like you are generating hardware on the fly as required, which you can not do. Generate statements are away of paramertising code with variable instance based on constant parameters which is why in this situation you get that quite confusing error.
Two solutions:
Use a ternary operator to select the value.
assign bnot[0] = b[0] ? 1'b0 : 1'b1;
Which is the same as assign bnot[0] = ~b[0].
Or use a combinatorial always block, output must be declared as reg.
module inverter(
input [7:0] b,
output reg [7:0] bnot
);
always #* begin
if (b[0] == 0) begin
bnot[0] = 1;
end else begin
bnot[0] = 0;
end
end
Note in the above example the output is declared as reg not wire, we wrap code with an always #* and we do not use assign keyword.
Verliog reg vs wire is a simulator optimisation and you just need to use the correct one, further answers which elaborate on this are Verilog Input Output types, SystemVerilog datatypes.