I believe that SystemVerilog is a much higher level of abstraction in coding. Is it possible to interface a SystemVerilog module with a verilog module? Are they any aspects that should be kept in mind when trying to integrate them?
Verilog and SystemVerilog are the same language - that is, anything you know about Verilog exists in SystemVerilog. From a synthesis point of view, you will sill be connecting bit of signals with other bits of signals. Its just that with SystemVerilog, you will have more advanced ways of declaring those signals, and many more operators to manipulate those signals.
Without knowing any SystemVerilog, I suggest that you learn it by itself before trying to integrate older Verilog modules with SystemVerilog modules. It will be difficult to explain what to look out for.
One thing that does carry over from Verilog to SystemVerilog is the concept of nets(wires) and variables(regs). Make sure you have a clear understanding of that, plus the new semantics SystemVerilog adds. I have a small article on it. Verilog only allowed wires to pass through ports and did not enforce directions. SV allows variables to pass through ports (meaning variables on both sides of the port connection) but strongly enforces directionality.
Yes, it is possible to interface system verilog module to verilog module.
Before that you must have understanding regarding signals (variable) which are used in verilog module.
You have to create interface from which your verilog signals are connected to system verilog module.
So, data transfer between verilog and system verilog module is possible.
Here I provide verilog module and system verilog module. Main part of code is interface from which verilog and system verilog module are connected.
verilog module code :
module dff(qn,d,clk,reset);
output qn;
input d,clk,reset;
reg qn;
always#(posedge clk,negedge reset)
begin
if (!reset)
begin
qn=1'bx;
end
else if (d==0)
begin
qn=0;
end
else if (d==1)
begin
qn=1;
end
end
endmodule
System verilog module code :
interface melay_intf(input bit clk);
logic o,clk,rst,i;
clocking c1#(posedge clk);
input o;
output i,rst;
endclocking
endinterface
module top;
bit clk;
always
#1 clk = ~clk;
melay_intf i1(clk);
dff d1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i));
melay_tes(i1.tes);
endmodule
program melay_tes(melay_intf i1);
initial
#100 $finish;
initial
begin
i1.rst <= 0;
#4 i1.rst <= 1;
#4 i1.rst <= 0;
i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
repeat(10)
begin
i1.i = 1;
#2 i1.i = $urandom_range(0,1);
end
end
initial
$monitor("output = %d clk = %d rst = %d i = %d",i1.o,i1.clk,i1.rst,i1.i);
initial
begin
$dumpfile("mem.vcd");
$dumpvars();
end
endprogram
Here important part is interface and in it I used clocking block for synchronization purpose.
Here clocking c1#(posedge clk); so all signals which are mention inside the clocking block which are i,o,rst.All this signal change its value at every posedge of clk signal.
Here dff d1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i));
Important thing that you find in top module I made connection between verilog signals and system verilog signals.
You can find verilog module name is "dff".
I took the instance of dff verilog module and made the connection.
Here i1.o,i1.clk,i1.rst,i1.i is system verilog signals which are connected to o,clk,rst,i signals of verilog module of with dot convention.
System verilog (SV) is mainly used for design verification, so the thing is that the DUT (Device Under Test) is written in verilog mainly because verilog can be synthesized mostly. SV is then used to write verification environment for that DUT.
Interfaces are needed in between to connect the SV with the DUT which is verilog. Separate file is written is SV to specify what are the different connection between the two files. This file say INTERFACE file is included in all the separate blocks which need those connections.
You can refer the IEEE standard https://standards.ieee.org/getieee/1800/download/1800-2012.pdf for further information on SV.
Related
Note this question is not for when I am simulating. I have found numerous resources as to how to use readmemh which does not solve my problem. What I am trying to do is load the RAM for a processor that I designed with the program data. I believe that the FPGA is not getting any of the program data, just the HDL description of the processor.
I tried using Verilog's readmemh function, which I now realize is only useful for simulation. I have also tried using the /* synthesis ram_init_file = "file.mif" */; instruction (?). Either way, there was no difference in how the device worked. I simulated all these cases in ModelSim so I know the design works. I just am stumped as to how I can preload the data.
The answer is going to be tool specific because the initial blocks are, in general, not synthezisable. If you can do it, it is just because the tool has a specific template that is matched with your initial block. Initializing a memory is one of these special cases, where the 'initial' block is discarded from your logic but it is used to generate the initialization data passed along the bitstream.
From the Intel Quartus documentation we can see that there are slightly differences on the actual implementation of the two kinds of memories, dedicated RAM and MLABs, however the general idea is to use an initial block:
module ram_with_init(
output reg [7:0] q,
input [7:0] d,
input [4:0] write_address, read_address,
input we, clk
);
reg [7:0] mem [0:31];
integer i;
// Init the memory with these values
initial begin
for (i = 0; i < 32; i = i + 1)
mem[i] = i[7:0];
end
always # (posedge clk) begin
if (we)
mem[write_address] <= d;
q <= mem[read_address];
end
endmodule
Or, for the quartus pro, you can use readmemh, readmemb:
reg [7:0] ram[0:15];
initial
begin
$readmemb("ram.txt", ram);
end
I suggest you look at the documentation linked as the most updated reference.
This is my main code. The function is switching the light on or off when I toggle.
Maybe the t filp-flop can help?
module demo(
input switch1,switch2,
output reg light=1'b0
);
always#(switch1 or switch2)
begin
light <= ~light;
end
endmodule
And the testbench
`timescale 1ns/100ps
module demo_test;
reg switch1,switch2;
wire light;
demo d1(.switch1(switch1),
.switch2(switch2),
.light(light));
initial
begin: in_set_blk
#0 switch1 =1'b0;
switch2 =1'b0;
$display ($time, "light=%d \n",light);
#10 switch1 =1'b1;
switch2 =1'b0;
$display ($time, "light=%d \n",light);
#10 switch1 =1'b0;
switch2 =1'b0;
$display ($time, "light=%d \n",light);
#10 switch1 =1'b0;
switch2 =1'b1;
$display ($time, "light=%d \n",light);
#10 switch1 =1'b0;
switch2 =1'b0;
$display ($time, "light=%d \n",light);
#10 $stop;
#10 $finish;
end
endmodule
This is the modelsim waveform diagram; I think it works well:
This is the diagram in RTL viewer; that is the strange part:
Diagram in RTL viewer is so strange; the input even didn't connect to the inverter.
Any error in my main Verilog code?
Yes, the problem is that your demo module Verilog code is not synthesizable.
Not all legal Verilog code is synthesizable. Synthesis tools recognize specific coding patterns, but your code does not match any of those patterns. There should be documentation with your FPGA tool set that shows what syntax is supported for synthesis.
This is why your RTL viewer tool gives you a strange diagram.
Here is how the simulation works. At time=0, you set light and the 2 switch inputs to 0. At time=10, you change switch1 to 1. This triggers the always block, executing the light <= ~light; assignment. This sets light=1.
Every time either of the inputs change, the always block is triggered, causing light to toggle.
If that is how you want the design to work, you should consider using edge detectors. However, that requires flip-flops and a clock.
Your code represents an implied latch, which is usually not synthesizable. This is a common mistake when you try to use always blocks with sensititivity lists (verilog 95). Your code will simulate correctly, but synthseis requres that all inputs of the always block are listed in the sensitivity list and all elements from the sensitivity list must be inputs to the block (flops excluded).
always#(switch1 or switch2)
begin
light <= ~light;
end
In the code above the result is not a function of the switch1,2 listed in the sensitivity list and sensitivity list does not include light.
In your case a flop is needed. light <= ~light represents a zero-delay loop if used in combinatorial logic or in a latch. It can only be resolved with a flop.
Flop needs a clock. Clock must be generated in test bench and flagged in synthesis.
The code can look like the following:
always#(posedge clk) begin
if (switch1 || switch2)
light <= ~light;
end
Flops are recognized by synthesis tools due to specific programming pattern and do not require listing of all inputs.
As a general suggestion: do not use sensitivity lists for combinatorial logic or latches (flops excluded). Modern verilog (v2k) allows you to use always #* construt instead.
How can you drive internal signals of a DUT verilog code from testbench?
Consider this following example:
module dut(input bit clk);
logic [7:0] data;
endmodule : dut
module top;
bit clk;
dut dut1(.*);
assign dut.data = '0; // this doesn't work.
endmodule
Cross module references do work. The catch, though, is that any signal in the DUT will already be driven. You need to override that driver. Force and release are the usual way of doing this but you can also just use a stronger driver strength.
The default drive strength is "Strong" so the only thing stronger is "supply".
For your example:
assign (supply0, supply1) data = '0;
Strictly speaking, the supply1 is unnecessary as you are only driving zero. However, it eliminates the surprise you might get if you ever need to change your code to drive '1'.
I am trying to write something like this:
always#(posedge bus_start)
begin
#(posedge scl) buffer[7] = sda;
#(posedge scl) buffer[6] = sda;
#(posedge scl) buffer[5] = sda;
#(posedge scl) buffer[4] = sda;
#(posedge scl) buffer[3] = sda;
#(posedge scl) buffer[2] = sda;
#(posedge scl) buffer[1] = sda;
#(posedge scl) buffer[0] = sda;
end
But it is not synthesizable according to the error I got.
I may use FSM to solve this, but that could make it complicated, is there any way to make it synthesizable?
You don't give enough information.
Answer from #Krabby127 is also not synthesisable.
I think you have a lot of more work to do. You should to learn the difference between Verilog and C,first. RTL code is to describe the hardware in FPGA.So the concept of synthesisable is very important for you.
And you have to take care of debounce of I2C signal.Take care of tristate signal.
After all I suggest you to get some I2C project from opencores.org to study.
Forgive my poor English.
It looks like you're trying to capture data from an I2C interface.
Here is a simple solution for capturing 8 bits of data from sda and storing that into a register called buffer.
// starts on rising edge of scl or bus_start
always#(posedge bus_start or posedge scl)
begin
buffer[0] <= sda; // loads sda into the lowest bit of buffer
buffer[7:1] <= buffer[6:0]; // shifts the lower 6 bits one to
// the left, acting as a shift register
end
More information about I2C in verilog can be found here
In addition, a basic I2C slave can be found here (you must register and login to view the source code).
I2C slaves mostly use a sampling clock other than SCL. I don't mean that it's impossible otherwise, but the common practice is to use a faster clock for sampling SCL and SDA signals.
I definitely recommend to start the design with the below ports at minimum. Since SDA is a tristate signal, I also recommend not to deal with it at this stage. You can assume that it's done out of your module and can continue with separate input and output ports.
input clk; // system clock
input rst; // system reset
input scl; // i2c clock
input sda_in; // i2c input data
output sda_out; // i2c output data
To be honest, I2C is not a good choice to learn an HDL (Verilog/SV/VHDL) and digital design. It will bring some complex issues.
I am trying to implement a microcontroller on an FPGA, and I need to give it a ROM for its program. If I use $readmemb, will that be correctly synthesized to a ROM? If not, what is the standard way to do this?
It depends on the synthesis tool whether or not $readmemb is synthesizable.
Altera's Recommended HDL Coding Styles guide includes example 10-31 (page 10-38), which demonstrates a ROM inferred from $readmemb (reproduced below):
module dual_port_rom (
input [(addr_width-1):0] addr_a, addr_b,
input clk,
output reg [(data_width-1):0] q_a, q_b
);
parameter data_width = 8;
parameter addr_width = 8;
reg [data_width-1:0] rom[2**addr_width-1:0];
initial // Read the memory contents in the file
// dual_port_rom_init.txt.
begin
$readmemb("dual_port_rom_init.txt", rom);
end
always # (posedge clk)
begin
q_a <= rom[addr_a];
q_b <= rom[addr_b];
end
endmodule
Similarly, Xilinx's XST User Guide states that:
The $readmemb and $readmemh system
tasks can be used to initialize block
memories. For more information, see:
Initializing RAM From an External File
Coding Examples
Use $readmemb for
binary and $readmemh for hexadecimal
representation. To avoid the possible
difference between XST and simulator
behavior, Xilinx® recommends that you
use index parameters in these system
tasks. See the following coding
example.
$readmemb("rams_20c.data",ram, 0, 7);