This is a basic question but it seems to lack of clear explanation to me.
In many of code examples,one style to write FSM output is
assign a = (current_state==DONE)?1:0;
I confuse this with definition of latches. Will this combinational logic infer latches as "a" holds its previous value if current_state != DONE? It seems no warnings from my compiler.
sometimes, i would have
always#(posedge clk)
begin
if(reset)
a<= 1'b0;
else
if(current_state == DONE)
a <=1'b1;
end
This is certainly a sequential logic(tho my output does not depend on my chains of past input) and a would keep its previous value until my control signal current_state == DONE.I would guess this logic will synthesize to a mux to the input of a flipflop.
so if in the 2nd case that I actually have a clocked FSM, i would have my mux with FSM states output as the select signal input.
Until now, can i say anything that is not combinational logic will not generate any latches?
However,when i have a structure like the following,
always#(posedge DCO or posedge reset or posedge enable)
begin
if(reset)
begin
end
else if(enable)
begin
end
else
begin
end
end
I get a warning in my FPGA that i have inferred a latch with control signal enable.
Why?
My enable changes based on another state machine for example,
assign enable = (pcurrent_state == START)?1:0;
Moreover, we have unintentional latches and intentional latches. but design rule basically says to avoid latches to avoid timing arches. Can someone give some examples of where intentional latches should be used in design rather than clock gating example?
Plus,
The output of all the storage elements (flip-flops) in the circuit at any given time, the binary data they contain, is called the state of the circuit. The state of a synchronous circuit only changes on clock pulses. At each cycle, the next state is determined by the current state and the value of the input signals when the clock pulse occurs.(from https://en.wikipedia.org/wiki/Sequential_logic)
This sounds like describing a mealy machine to me rather than a typical sequential logic. My simplest sequential logic does not need my output change determined by my current_state
Thank you for any help. I am doing this coding everyday and reading its definition but it seems that i am confusing myself without discuss with others.
To answer your question in parts:
The given assign statement will not infer a latch as a does not retain it's value. It will be 1 if current_state == DONE and otherwise be 0. So it's pure combinational logic.
The second block of code implements a flip-flop with synchronous reset and only loads itself with 1'b1 if current_state == DONE so there is retention in that code. This code shouldn't generate a latch due to the edge sensitivity on a single clock.
The last block would be difficult for any synthesis tool to handle due to the sensitivity on serveral signals, which is not common in hardware. Moreover, if say enable is asserted but not an edge when a positive edge of DC0 comes along, the code would have the body of the else if (enable) run, thus simulating some sort of latching behavior. Synthesis tools generally allow for a single clock and single reset to be specified in the sensitivity list of an always block to indicate a flip-flop with asynchronous reset. While Verilog certainly allows for more complicated sensitivity lists, their physical meaning gets complicated quickly, thus inferring latches. In most designs, you shouldnt ever need these complex sensitivity lists are you are then getting into asynchronous design for which most synthesis tools are not well suited on a behavioral level. FPGA tools especially are poor at asynchronous elements and even latches as the logic cells in the fpga to which the design must be mapped are designed specifically for synchronous designs using flip-flops; that's how fpgas are implemented.
Finally, in non-fpga designs, it is sometimes desirable to use latches if edge sensitivity isn't required, as latches are physically smaller and can allow a design to be smaller, faster and more power efficient in some cases. However, you need to have a firm grasp on what you are designing and understand potential trade offs and timing requirements when doing so. Here's an example of when a latch is a useful element: https://electronics.stackexchange.com/questions/255009/what-is-application-of-latch-in-vlsi-design
Related
When incompletely assigning a value I get a latch. But why did I get a latch in the example below? I think there is no need for the latch of F output because it is defined at all values of SEL.
Verilog code:
always # (ENB or D or A or B pr SEL)
if (ENB)
begin
Q=D;
if (SEL)
F=A;
else
F=B;
end
Inferred logic:
Although it is defined at all values of SEL, it is not defined for all values of ENB. If ENB = 0, your code says that both Q and F should hold the value from the previous cycle. This is also what is inferred in the image you are linking: only update Q and F if ENB = 1.
If you want Q to be a latch and F not, you can do this:
always # (ENB or D or A or B or SEL)
begin
if (ENB)
Q=D;
if (SEL)
F=A;
else
F=B;
end
Edit: additional information
As pointed out in the comments, I only showed how you could realize combinational logic and a latch, without modifying your code too much. There are, however, some things which could be done better. So, a non-TL;DR version:
Although it is possible to put combinational logic and latches in one procedural block, it is better to split them into two blocks. You are designing two different kinds of hardware, so it is also better to seperate them in Verilog.
Use nonblocking assignments instead of blocking assignments when modeling latches. Clifford E. Cummings wrote an excellent paper on the difference between blocking and nonblocking assignments and why it is important to know the difference. I am also going to use this paper as source here: Nonblocking Assignments in Verilog Synthesis, Coding Styles That Kill!
First, it is important to understand what a race condition in Verilog is (Cummings):
A Verilog race condition occurs when two or more statements that are scheduled to execute in the same simulation time-step, would give different results when the order of statement execution is changed, as permitted by the IEEE Verilog Standard.
Simply put: always blocks may be executed in an arbitrary order, which could cause race conditions and thus unexpected behaviour.
To understand how to prevent this, it is important to understand the difference between blocking and nonblocking assignments. When you use a blocking assignment (=), the evaluation of the right-hand side (in your code A, B, and D) and assignment of the left-hand side (in your code Q and F) is done without interruption from any other Verilog statement (i.e., "it happens immediately"). When using a nonblocking assignment (<=), however, the left-hand side is only updated at the end of a timestep.
As you can imagine, the latter assignment type helps to prevent race conditions, because you know for sure at what moment the left-hand side of your assignment will be updated.
After an analysis of the matter, Cummings concludes, i.a., the following:
Guideline #1: When modeling sequential logic, use nonblocking assignments.
Guideline #2: When modeling latches, use nonblocking assignments.
Guideline #3: When modeling combinational logic with an always block, use blocking assignments.
A last point which I want to highlight from the aforementioned paper is the "why". Except from the fact that you are sure the right hardware is infered, it also helps when correlating pre-synthesis simulations with the the behaviour of your actual hardware:
But why? In general, the answer is simulation related. Ignoring the above guidelines [about using blocking or nonblocking assignments on page 2 of the paper] can still infer the correct synthesized logic, but the pre-synthesis simulation might not match the behavior of the synthesized circuit.
This last point is not possible if you want to strictly adhere to Verilog2001, but if you are free to choose your Verilog version, try to use always_combfor combinational logic and always_latch for latches. Both keywords automatically infer the sensitivity list, and it is easier for tools to find out if you actually coded up the logic you intended to design.
Quoting from the SystemVerilog LRM:
The always_latch construct is identical to the always_comb construct except that software tools should perform additional checks and warn if the behavior in an always_latch construct does not represent latched logic, whereas in an always_comb construct, tools should check and warn if the behavior does not represent combinational logic.
With these tips, your logic would look like this:
always_latch
begin
if (ENB)
Q <= D;
end
always_comb
begin
if (SEL)
F = A;
else
F = B;
end
Let's say there's a code that runs like this
reg [4:0] data;
always # (posedge clk, posedge clr)
begin
if(clr)
data <= 0;
else
data <= data +1;
end
How would this look like in circuit level? My guess is roughly
but then that wouldn't help if Clk goes from 0 to 1 while Clr is 1......
Also, is it good practice to have multiple elements in the sensitivity list? From what I see, there's som overhead going on here..
Verilog excerpt will infer DFF (D Flip-Flop) with async reset. This happens due to the fact that reset signal is a part of sensitivity list.
NOTE1: as per LRM for Verilog, adding the reset to the sensitivity list is what makes the reset asynchronous.
NOTE2: each Verilog procedural block should model only one type of flip-flop. In other words, a designer should not mix resetable (sync or async) flip-flops with follower flip-flops (flops with no resets) in the same procedural block.
Your diagram is incorrect, 'clr' signal will be connected to extra input of the DFF called as CLEAR (it is basically an async reset). I suggest to start with some sort of Verilog tutorial, this is very basic thing and it is well explained in materials that are generally available. To grasp on concept of reset in HDL code I recommend the following material:
http://www.sunburst-design.com/papers/CummingsSNUG2003Boston_Resets.pdf
The schematic is not accurate. 4 D-FF will be implemented for each bit of data declared.
Adding reset (i.e. clr) to sensitivity list will make the ckt async (Verilog LRM).
The D-FF will have an additional clear pin, there will be NO bubble to this pin as your reset (i.e. clr) is active high.
I have written the below code for a simple multiplication of 2 n-bit numbers(here n=16). It is getting simulated with desired output waveform but the problem is, it is not getting synthesized in vivado 17.2 even though I have written a static 'for loop'(ie., loop iteration is constant). I am getting below mentioned error.
[Synth 8-3380] loop condition does not converge after 2000 iterations
Note: I have written
for(i=i;i<n;i=i+1)
instead of
for(i=0;i<n;i=i+1)
because the latter one was executing once again after i reached n. So that is not a mistake. Kindly someone help. Thank you for your time
//unsigned integer multiplier
module multiplication(product,multiplier,multiplicand,clk,rset);
parameter n = 16;
output reg [(n<<1)-1:0]product;
input [n-1:0]multiplier, multiplicand;
input clk,rset;
reg [n:0]i='d0;
always #( posedge clk or posedge rset)
begin
if (rset) product <= 'd0;
else
begin
for(i=i;i<n;i=i+1)
begin
product =(multiplier[i] == 1'b1)? product + ( multiplicand << i ): product;
$display("product =%d,i=%d",product,i);
end
end
end
endmodule
First of all, it is not a good practice to use for, while kind of loops if you really want to implement your design on FPGA (Vivado is optimized to be used to implement your design on FPGA). Even if you can successfully synthesize your design, you might face with timing problems or unexpected bugs.
I think you can find your answer here.
Edit: I just wanted to inform you that generally controlling timing is very important in HW design, especially when you want to integrate your design with other system, loops can be nightmare for that.
Go back to using your original for (i=0 loop.
Your error is that you assume i=0 because of reg [n:0]i='d0; That is only true the very first time. Thus only once, at the start of the simulation.
because the latter one was executing once again after i reached n.
Yes, the loop will repeat again and again for every clock cycle. That is what #( posedge clk ...) does.
More errors:
You are using blocking assignment in the clock section, use non-blocking:
product <=(multiplier[i] == 1'b1)? product + ( multiplicand << i ): product;
Your product is only correct the first time after a reset (when it starts at zero). The second clock cycle after a reset you do the multiplication again but start with the previous value of product.
Your i is a bit big you use 17 bits to count to 16. Also global loop variables have pitfalls. I suggest you use the system Verilog syntax of: `for (int i=0; ....)
The error message is not about the iterations of your loop, but about the iterations of the internal algorithm of your synthesis program. It tells you that Vivado failed to create a circuit that could actually be implemented on your FPGA of choice at your clock speed of choice and which actually does what you're asking for.
Let me elaborate, after I mention two items of general import: don't use blocking assignments (=) inside always #(posedge clk) blocks. They almost never do what you want. Only non-blocking assignments (<=) should be clocked. Secondly, the correct way to synthesize a for loop is with generate, even though Vivado seems to accept the simple for.
You are building a fairly large block of combinatorial logic here. Remember that when you synthesize combinatorial logic you are asking for a circuit that can evaluate the expression that you've written within a single clock cycle. The actual expression taken into consideration is the one after the for loop has been unrolled. I.e., you are (conditionally) adding a 16bit number to a 32bit number 16 times, each times shifted one bit further to the left. Each of these additions has a carry bit. So each of these additions will actually have to look at all of the upper 16bits of the result of the previous addition, and each will depend on all but the lowest bit of the previous addition. An adder for one bit + carry needs O(5) gates. Each addition is conditional, which adds at least one more gate for each bit. So you are asking for at minimum of 16*16*6 = 1300 interdependent gates that all have to stabilize within a single clock cycle. Vivado is telling you that it cannot fulfill these requirements.
One way of mitigating this problem will be to synthesize for a lower clock frequency, where the gates have more time to stabilize, and you can thus build longer logic chains. Another option would be to pipeline the operation, say by only evaluating what corresponds to four iterations of your loop within a single clock cycle and then building the result over several clock cycles. This will add some bookkeeping logic to your code, but it is inevitable if one wants to evaluate complex expressions with finite resources at high clock frequencies. It would also introduce you to synchronous logic, which you will have to learn anyway if you want to do anything non-trivial with an FPGA. Note that this kind of pipelining wouldn't affect throughput significantly, because your FPGA would then be doing several multiplications in parallel.
You may also be able to rewrite the expression to handle the carry bits and interdependencies in a smarter way that allows Vivado to find its way through the expression (there probably is such a way, I assume it synthesizes if you simply write the multiplication operator?).
Finally, many FPGAs come with dedicated multiplier units because multiplication is a common operation, but implementing it in logic gates wastes a lot of resources. As you found out.
This code fails to compile and gives the error as in the title at the "if(overflow)" line.
always #(posedge clk or negedge overflow) begin
if(overflow)
count_posedge = count_posedge + 1;
else
count_posedge = 0;
end
I've somewhere on the internet that I must change it like this:
always #(posedge clk or negedge overflow) begin
if(~overflow)
count_posedge = 0;
else
count_posedge = count_posedge + 1;
end
...and it works perfectly.
From my understanding, the 2 code should behave the same. What's the problem with the first one?
This is more likely an issue with your synthesizer then your simulator. The likely problem is that the first code does not match any of it's templates for a synchronous flip-flop with asynchronous reset.
The common coding practice is to assign your reset logic before any other logic. This coding practice has been around for decades. I assume the rational behind this particular coding practice stems from:
Reset logic is critical from many designs; especially as designs get larger and more complex. It is put at the top for it importance and the fact that it usually fewer lines of code then the synchronous logic.
Early synthesizers were very limited and could only synthesize specific code structures.
The coding style has become a precedence. No one is going to change it unless you can convince and prove something else is superior or has specific advantages.
In your case, your synthesizer is doing a lint check and has determined your code is not following the conventional coding practices. The creator of your synthesizer has decided to only support the common coding structure and there is little incentive to change.
FYI: you should be assigning synchronous logic with non-blocking assignments (<=). Blocking assignments (=) should be used for combinatonal logic. If you do not follow proper coding practices you increase the risk of introducing race-conditions, RTL vs gate mismatch, and other bugs.
I am reading the IEEE Standard Verilog Hardware Description Language (specifically IEEE Std 1364-2001) which unambiguously defines and discusses simulatable Verilog. Unfortunately, the document does not touch upon the notion of synthesis.
I haven't been able to find a similar reference for synthesisable Verilog. All I find is vague rules, or unnecessarily restrictive ones.
Where can I learn the formal language of synthesisable Verilog?
IEEE 1364.1 is an adjunct to the 1364 Verilog standard titled Verilog Register Transfer Level Synthesis, which attempts to define a common synthesizable subset. However, as Jerry points out, different tools support different constructs, and to determine tool-specific behavior, you need to consult the tool documentation.
There isn't a formal (BNF-style) syntax definition for synthesizable Verilog. Whether code is synthesizable depends on usage as well as syntax. For example, the behavior described by an always construct with incomplete sensitivity, like always #(a) o = a || b, isn't synthesizable. (Most tools will synthesize that code as if the sensitivity list were complete, resulting in a possible simulation/synthesis mismatch.)
Circuit constructs like latches and multiply-driven nets can be synthesized from a Verilog description, but are disallowed or discouraged under most design rules. There are also synthesizable constructs that are unsupported or inadvisable given the choice of target library. For example, describing a RAM that's larger than the maximum supported by a chosen FPGA technology, or describing tri-state drivers when they aren't present in the target library.
The general constructs to stick to for synthesizable Verilog are:
Combinational logic modeled with continuous assignments (assign statements)
Combinational logic modeled with always blocks, which should use blocking assignments, and either have a complete sensitivity list or use always #*
Sequential logic (flip-flops) modeled with always blocks, which should use non-blocking assignments, and have either posedge clock alone (for sync reset or non-reset flops) or posedge clock and posedge or negedge reset (for async reset flops) in the sensitivity list.
The safest coding style for sequential logic is to code only reset logic in the sequential always block:
always #(posedge clk or posedge reset)
if (reset)
q <= reset_value;
else
q <= next_value;
However, if you're careful, you can code additional combinational logic in the sequential block. A common case where it may make sense to do this is a mux in front of a flop:
always #(posedge clk)
if (!sel)
q <= sel0_value;
else if (sel)
q <= sel1_value;
else
q <= 'bx;