In a simple clock generator example, I see the following code:
always #(cycle/2) clk ~= clk;
I've seen always #(*) before but not pound (#). I tried to find it in the documentation, but all I could find was some reference to "real-valued ports" with no further elaboration.
It's a delay operation. It essentially just reads
always begin
#(cycle/2) //wait for cycle/2 time
clk ~= clk;
end
You might sometimes see this used with raw values, like #5 or #10, which means to wait 5 or 10 units of your timescale.
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.
For Example, what is the difference between these two implementations?
with initial statment:
module with_initial();
reg clk,reset,enable,data;
initial begin
clk=0;
reset=0;
enable=0;
data=0;
end
always #(…)
..
end
endmodule
and here without using an initial statement:
module without_initial();
reg clk,reset,enable,data;
clk=0;
reset=0;
enable=0;
data=0;
always #(…)
..
end
endmodule
The code in without_initial is not legal syntax, and it should produce compile errors for you, as it does for the simulators on edaplayground.
Procedural assignments must be inside an initial or always block. Refer to IEEE Std 1800-2017, section 10.4 Procedural assignments.
Also, a reg can be declared with an initial value outside of an initial block:
module without_initial;
reg clk=0;
reg reset=0;
reg enable=0;
reg data=0;
endmodule
Refer to IEEE Std 1800-2017, section 6.8 Variable declarations.
Initial block is where your execution starts and they execute at
time 0. This is a non synthesiable procedural statement ( in the sense it cannot be modelled to hardware) . It is just used for initializing all the variable and drives them with specific values.
In veriog (HDL) it's however mandatory to have atleast one initial block . Maybe if you skip adding that block the error caused would be due to following reasons :
1) The compiler never knowing what to drive into the modelled gates used in your code .
2) The assignment to the variables never happen.
Not so damn sure about it. But this is what I feel happens behind the scenario.
I write simple code using ISE 14.7 to generate clock signal, but the output in iSim is always 1, until I change = with <= then the clock works, could you tell me why?
module nonblocking( clk );
output reg clk;
initial
#10 clk =0;
always # (clk)
#10 clk = ~ clk; // change to <= to work
endmodule
This code honestly doesn't make any sense.
Your current code states that based on clk, the value of clk should be ~clk.
When you use <= you are at least specifying a nonblocking assignment, which makes slightly more sense, but still doesn't.
If you want a clock generator for simulation, use the following within an initial block (giving a 10ns period):
forever begin
#5 clk <= ~clk;
end
If you want a clock generator for FPGA synthesis, either connect a physical oscillator to your device's clock input pin, or use a built-in oscillator (if available on your device). You can always use a DCM/PLL/CMT/MMCM/whatever your FPGA calls it to adjust the clock frequency.
I am new to Verilog, so this question might be quite dumb.
What I am trying: I have a component that has a clk, an 8 bit input and an 8 bit output. What it should do, is:
If the clock event is negative edge, it should set the output to 0
If the clock event is positive edge, it should set the output to whatever input is at this moment of the edge event. During the high phase of the clock, the output should NOT change, regardless changes on the input.
What I tried so far:
always #(negedge clk)
_ledOut <= 0;
always #(posedge clk)
_ledOut[RowSize-1:0] <= ledIn[RowSize-1:0];
This tells my, that it can't resolve multiple constant drivers for net _ledOut.
However, putting this together in an always #(negedge clk, posedge clk) tells me, it can't test for both conditions.
So I tried to make just one always #(clk) block and then used an if statement:
always #(clk) begin
if(clk == 0)
_ledOut <= 0;
else if(clk == 1)
_ledOut[RowSize-1:0] <= ledIn[RowSize-1:0];
end
But this didn't just switch on a clk event. During the high phase of the clock, it links _ledOut with ledIn, so that changes on ledIn do also have effect on _ledOut. What am I doing wrong here?
Best regards,
Michael
This tells my, that it can't resolve multiple constant drivers for net
_ledOut.
For synthesis you cannot assign reg types from multiple always blocks.
However, putting this together in an always #(negedge clk, posedge
clk) tells me, it can't test for both conditions.
This essentially describes a DDR register. While many FPGA devices have these they typically cannot be synthesized. Xilinx uses ODDR2 and IDDR2 primitives if you really need this functionality.
If the clock event is negative edge, it should set the output to 0 If
the clock event is positive edge, it should set the output to whatever
input is at this moment of the edge event. During the high phase of
the clock, the output should NOT change, regardless changes on the
input.
If this is all you need then you can use a D flip flop with an AND gate on the output. The flip-flop will sample ledIn on each rising edge of clk and the AND gate will mask the output whenever the clock is zero. This is not ideal as you generally do not want clocks to touch non-sequential logic but avoiding this would likely mean changing your requirements.
As toolic indicated, the code you posted will work but you should understand that code will synthesize to a multiplexer controlled by clk.
Ok, here is my working solution now. Maybe it's not the best verilog code you have seen out there. ;) This is, however, my first thing I do with it, as a project at my university. So as long as it does what I want it to do, this is a great success to me! ;)
Here is the code I used now, thanks to Adam12:
parameter RowSize = 8;
input clk;
input [RowSize-1:0] ledIn;
output [RowSize-1:0] ledOut;
reg[RowSize-1:0] _ledOut;
assign ledOut = _ledOut & {RowSize{clk}};
always #(posedge clk) begin
_ledOut[RowSize-1:0] <= ledIn[RowSize-1:0];
end
Consider the following stimulus:
module tb;
parameter RowSize = 8;
reg clk;
reg [7:0] ledIn, _ledOut;
always #(clk) begin
if(clk == 0)
_ledOut <= 0;
else if(clk == 1)
_ledOut[RowSize-1:0] <= ledIn[RowSize-1:0];
end
initial begin
$monitor($time, " clk=%b ledIn=%h _ledOut=%h", clk, ledIn, _ledOut);
ledIn = 0;
#22 ledIn = 8'h55;
#20 $finish;
end
always begin
#5 clk <= 0;
#5 clk <= 1;
end
endmodule
It produces this output:
0 clk=x ledIn=00 _ledOut=xx
5 clk=0 ledIn=00 _ledOut=00
10 clk=1 ledIn=00 _ledOut=00
15 clk=0 ledIn=00 _ledOut=00
20 clk=1 ledIn=00 _ledOut=00
22 clk=1 ledIn=55 _ledOut=00
25 clk=0 ledIn=55 _ledOut=00
30 clk=1 ledIn=55 _ledOut=55
35 clk=0 ledIn=55 _ledOut=00
40 clk=1 ledIn=55 _ledOut=55
Notice at time 22, when ledIn changes, the _ledOut output does not change. _ledOut only changes at the next posedge of clk at time 30. Therefore, the always #(clk) solution is doing what you want: the output only changes at the clock edge, as you specified.
This is a pretty unusual question, and it makes me advise you need to give more information about what you are actually trying to achieve, since it may well impact the timing performance and clock constraints if this is targeting an FPGA. Synthesis has been mentioned, but what will you be feeding the clock-gated output into? If it's a pin-pad, then you should read the DDR pad buffers in the device specifications and infer the specific primitive to be able to drive a DDR signal.
If you are keeping this signal within the chip then this is a very bizarre request. If I needed to generate that waveform, I would probably use a PLL to generate a phase-locked clock at twice the base frequency and put the gated data into that domain, with a toggle to apply the mast, so that the tooling will be able to properly analyse the clock crossings and the resulting data path is still effectively transitioning on a single edge.
The answers above to infer a register with a combinatorial multiplexer forced on the output is interesting, but whatever you feed this into will have to deal with awkward setup/hold conditions, and if on-chip, would only be sampling one edge anyway, so this is kind of redundant.
Which code is better in writing a RAM?
assigning data_out inside always block:
module memory(
output reg [7:0] data_out,
input [7:0] address,
input [7:0] data_in,
input write_enable,
input clk
);
reg [7:0] memory [0:255];
always #(posedge clk) begin
if (write_enable) begin
memory[address] <= data_in;
end
data_out <= memory[address];
end
endmodule
assigning data_out using assign statement:
module memory(
output [7:0] data_out,
input [7:0] address,
input [7:0] data_in,
input write_enable,
input clk
);
reg [7:0] memory [0:255];
always #(posedge clk) begin
if (write_enable) begin
memory[address] <= data_in;
end
end
assign data_out = memory[address];
endmodule
Any recommendations?
It depends on your requirements.
This registers your memory output. If you are synthesizing this to gates, you will have 8 more flip-flops than in case 2. That means you use a little more area. It also means your output will have less propagation delay relative to the clock than case 2. Furthermore, the output data will not be available until the next clock cycle.
Your output data will be available within the same clock cycle as it was written, albeit with longer propagation delay relative to the clock.
You need to decide which to use based on your requirements.
A third option is to use a generated RAM, which is a hard macro. This should have area, power and possibly timing advantages over both case 1 and 2.
to add to toolic's answer - if you use the asynchronous read method (case 2), it won't map to a RAM block in an FPGA, as the RAM blocks in all the major architectures I'm aware of have a synchronous read.
Both forms are valid, depending on the type of pipelining you want. I always recommend following the Xilinx RAM coding guidelines -- it's a good way to ensure that the code synthesizes into proper FGPA constructs.
For example, your example 1 would synthesize into into Xilinx BRAM (i.e., dedicated Block Ram), since it is synchronous read, and your example 2 would synthesize into Xilinx Distributed Ram (since it is asynchronous read).
See the coding guidelines in Xilinx document UG901 (Vivado Design Suite User Guide), in the RAM HDL Coding Techniques section. It also has a good description of the difference between synchronous read and asynchronous read for RAMs.
In the second program, there would be compilation error as we can not 'Assign' a value to 'Reg'.
It will give an error saying: 'Register is illegal in left-hand side of continuous assignment'