Let's take the example code below:
always #(posedge clock)
begin
if (reset == 1)
begin
something <= 0
end
end
Now let's say reset changes from 0 to 1 at the same time there's a posedge for the clock. Will something <= 0 at that point? Or will that happen the next time there's a posedge for the clock (assuming reset stays at 1)?
It depends on exactly how reset is driven.
If reset and something are both triggered off the same clock, then something will go to 0 one clock cycle after reset goes to 1. For example:
always #(posedge clock)
begin
if (somethingelse)
begin
reset <= 1;
end
end
If reset is synchronous and based on clock, The simulatore will defiantly see reset on the next clock and not the current. Physical design has clock-to-Q, therefor a rise in reset will not be observed in the same clock that caused it. You may see reset at the same time as clock in waveform. reset <= 1'b1; make the assignment happen near the end of the scheduler (after all code has executed).
To not have to worry about this when looking at a waveform, some logic designers like to put a delay on the assignment creating an artificial clock-to-Q delay (ex reset <= #1 1'b1; and something <=#1 0;). Synthesis tools will ignore the delay, but some will give warnings. That warning can be avoided by using a macro.
`ifdef SYNTHESIS
`define Q /* blank */
`else
`define Q #1
`endif
...
reset <= `Q 1'b1;
...
something <=`Q 1'b1;
...
If reset is asynchronous and being use with synchronous reset, setup time requirements need to be respected. In simulation if clock and reset rise at the same time, it is up to your verilog scheduler to decide if reset will be the new value or old value. Usually it will take the left-hand side value (old value), which means the reset will be missed on the current clock. Physical design uncertainly as well with a meta-stability risk.
The code you have written infers a flip-flop with synchronous reset. This means it is assumed that the "reset" signal is synchronised to the "clock" domain before being used in this way. If the "reset" signal is not synchronised then you should modify the code to infer a flip-flop with asynchronous reset as below:
always#(posedge clock or posedge reset)
begin
if (reset)
something <= 0
else
something <= something_else
end
Coming back to your question and assuming the code you have written is what you want, the outcome depends on how the reset is driven. If it is synchronous then the simulator will see it in the next clock edge. If it is asynchronous then the simulator can assume anything, it can vary from simulator to simulator. Please note that in simulator everything is a sequence of events and there is no such thing as happening at the same time.
In the physical world, what you have coded will result in a flip-flop with reset signal being one of the inputs to the combo driving the input of this flop. Now if the reset is synchronous, you are guaranteed that there will be no setup or hold violation at this flop. Whether the flop will 'see' the reset in this clock or the next depends on the various delays of the synthesised circuit (Usually this is the main reason that the reset is always held for few clock cycles to make sure all the flops in your design sees the reset). If reset is asynchronous then the flop will go into a metastable state. You will never want this in your design.
Hope this clarifies.
The short answer is that either of your two outcomes (immediately, or next cycle) could happen. This is a standard race condition, and simulators are free to handle this any way they want; some will give one answer, and others will give the other one.
For the long answer, look up any introductory text on how VHDL delta cycles work. Verilog doesn't specify 'delta cycles', but any Verilog simulator will work in exactly the same way, with some (irrelevant) changes in the overall scheduling algorithm. In this case, the scheduler finds that it has two events on the queue in a specific delta - reset rising, and clock rising. This is what "at the same time" means. It chooses one in an unspecified way (it might be earlier in the text source, or later, for example), works through all changes associated with that edge, and then goes back and works through all changes associated with the other edge.
Related
I'm trying to add an edge_triggered reset to my module, meaning that only if a transient from 0 to 1 detected in reset, the reset operation should be perform.
and further more, I don't want to use any initial block in my module (because I think initial blocks have issues in FPGAs).
finally, I decided to following something like this:
always # (posedge clock, posedge reset) ...
but the problem I encountered, is that I need to know which element of always block sensitivity list is activated.
also I think this method isn't a proper solution:
always # (posedge clock, posedge reset) begin
if(reset) begin
//doing reset operations
end
.
.
.
end
because if reset is 1 from previous clocks (not transient from 0 to 1) in next posedge clock the reset operation will be perform! but we didn't want this.
so, is there any way to find out which element of sensitivity list activated the always block?
or even any other solution for implementing edge_triggered reset?
There almost certainly isn't a primitive in your FPGA that will do exactly what you want. And even in simulation, you wouldn't be able to tell which of the edges in the sensitivity list triggered the update.
The code you posted should synthesize, but as you noted it will behave like a standard asynchronous reset and continue to reset as long as reset is high. In order to reset only on the reset rising edge, you'll need some clock domain crossing logic to detect the edge and output a single pulse on the clock domain.
For example:
// This module can assert reset_pulse asynchronously but must output a pulse
// for at least one 'clock' cycle and deassert synchronously.
async_edge_to_pulse e2p (
.in_edge (reset),
.out_clock (clock),
.out_pulse (reset_pulse)
);
always #(posedge clock or posedge reset_pulse) begin
if (reset_pulse) begin
// reset logic
end else begin
// operational logic
end
end
I have just started using VCS for compiling and simulating my TB written in verilog.
I am stumped by how it treats the 0->1 or 1->0 transitions. I believe that for evaluation purposes a 0->1 transition is infact seen as a 0 and a 1->0 is seen as a 1. I would expect this behavior in my sims for my design to behave correctly.
But for the life of me I cant figure out why its treating the generated input stimulus' 0->1 transition as 1 and 1->0 as 0
Below is the portion of code relevant for this. Its a simple flop with an enable signal - "en" which is the input stimulus that gets triggered randomly from an event controlled by the same clock, clk. clk is used everywhere in TB and design
always #(posedge clk or negedge reset_) begin
if (!reset_) begin
q <= {5{1'b0}};
end else begin
if (!en) begin
q <= d;
end
end
end
Now what happens is that when the event gets triggered, i.e a 0->1 transition happens on en, then I want the flop to see it still as a 0 and flop d. And on the falling edge of en which coincides with another posedge of clk I need to ensure flopping doesnt happen (it should be seen as a 1).
The simulations are doing opposite of what I am expecting unless i slightly delay the en triggering which I dont think is the ideal solution.
Is it wrong to expect such a behavior? If no, then how can I go about fixing this? Any insights would be greatly appreciated.
Please let me know if you need more details.TIA
Raising Greg's comment to an answer so this question can be marked as answered:
There will not be a race condition if you assign clk with a blocking statement (=) and assign en with a non-blocking statement (<=). This is is the proper coding style. Then the only thing you need to remember is the sampled value of en and d is the left side if the clock edge. – Greg Oct 26 '17 at 14:44
This code is supposed to increment a counter (outputting to LEDs) when one button is pushed and decrement it when the other one is pushed. It works OK with decrementing but on incrementing it changes LEDs to random configuration instead.
module LED_COUNTER( CLK_50M, LED, Positive, Negative );
input wire CLK_50M;
input wire Positive;
input wire Negative;
output reg[7:0] LED;
always#( posedge Positive or posedge Negative )
begin
if(Positive == 1)
LED <= LED + 1;
else
LED <= LED - 1;
end
endmodule
I am using this board: http://www.ebay.com/itm/111621868286. The pin assignment is:
The connections:
After swapping pin assignments for buttons the behavior stays the same.
As others have already pointed out, you should be clocking with the CLK_50M and you should de-bounce your inputs (some FPGAs do it for you automatically, check your manual).
The reason you see partial functionality is from the way the synthesizer interprets the RTL. If the sensitivity list is edge triggered and that signal is referenced in the body of the always block, then the synthesizer will think it is an asynchronous level sensitive signal. This is intend for asynchronous reset & set (sometimes named clear & preset). ASIC design typically use flops with asynchronous reset in most the design. FPGAs tend to have a limited number of flops with asynchronous set or rest, so check your manual and use sparingly.
With your code, Negative is the clock and Positive is treated as an active high asynchronous input.
Modify the code to a functional behavioral equivalent (in simulation) seen below, then Positive will be the clock and Negative will be the active high asynchronous input.
always#( posedge Positive or posedge Negative )
begin
if(Negative == 1)
LED <= LED - 1;
else
LED <= LED + 1;
end
There are several resource for learning Verilog available online (use your favorite search engine), and I have some resources posted on my profile though more aimed at SystemVerilog.
Here is some pseudo code to point you in the correct direction for your project:
always #(posedge CLK_50M)
begin
past_Positive <= Positive;
// ...
case({/* ... , */ past_Positive,Positive})
4'b0001 : LED <= LED + 1;
4'b0100 : LED <= LED - 1;
// ...
endcase
end
First, update the design to a synchronous design where state only changes at rising edge of the CLK_50M, like
always#( posedge CLK_50M)
begin
...
end
Then add de-bounce logic logic for the two switch inputs; see contact bounce. This can be done with a small module you write yourself; that is a good exercise.
Output from the contact de-bounce logic can then be used for change detection, to make a single cycle indication each time a contact is pressed, and this indication can then be used to update the counter.
You have no debounce circuitry or logic. A mechanical switch will physically bounce a lot, so your 50MHz clock is going to see many, many transitions on the input signal, leading to erratic behavior.
I forgot to mention that you're not even using that 50MHz clock in a synchronous design. Rather you're asynchronously looking for transitions.
You need a low-pass filter somewhere. Either implemented with analog components on the input signal, or as a counter in hardware.
I'm writing some Verilog code to be programmed on an Altera Cyclone II FPGA board, and I have an always block which should be activated on the press of a key switch:
reg START;
...
...
always # (negedge key[3]) begin
if (START != 1) START = 1;
end
I'm writing a program for a finite state machine and this key press is supposed to indicate that the user would like to begin using the program and it should move from its initial state to the next state. Since the initialization of registers is not synthesizable, I can't assume that START begins at 0.
The problem is that once I program the board and turn it on, this always block has already run once before I press the key assigned to key[3]. I've done checks for the value of START at program execution and it is already at 1. I can't figure out why this would be happening, as the key is at its negative edge only upon key press. I've used always blocks with the same condition in previous situations and it worked fine, so I assume this has something to do with the initialization of START?
It should be noted that synthesizability of initial depends on a target you are coding for.
For example, if you code for simulator, initial works just nice. If your target is FPGA, the tool (quartus in your case) has full control over the schematics and over the initial state of every trigger inside it: actually, uploading the firmware to FPGA sets every trigger to a known state, and quartus is able to parse initials to derive each trigger's state.
In contrary, if your target is a bare silicon, every trigger is just a bunch of transistors and its state is totally indefinite upon power-on, so the only way to control it's power-on state is to apply some type of reset, like this:
always #(posedge clk, negedge rst_n)
if( !rst_n )
START <= 1'b0; // no start condition upon reset
else if( some_condition )
START <= 1'b1;
Another point on your code is that when parsing inputs from switches you should do first resynchronization to your design's clock, like this:
reg start_r, start_rr;
always #(posedge clk)
begin
start_r <= START;
start_rr <= start_r;
end
// now use start_rr instead of START
Resynchronization is a key element to avoid metastable states in your synchronous design.
The second point is that you should consider debouncing of any input from mechanic switches, unless your design is tolerant to multiple tripping of switches.
Returning to the original problem Ryan McClure asks for. It could be seen, that in the original code without initial START is undefined at startup, and can trip only to '1' state. Therefore, synthesizer just assumes START is constant, always having it '1'.
You should use an "initial" block to set the startup value of your signals. The value of START and key[3] has to be set.
initial begin
START = 1'b0;
key[3] = 1'b1;
end
You said
Since the initialization of registers is not synthesizable, I can't
assume that START begins at 0.
But this is not true! You can set a default value to any signal in your design with the method above. This value is included in the bitstream of your firmware and the signal startup with this very value.
cheers
I have a CPLD with a digital input representing a reset button. When the reset button is pressed, the signal goes high. What I need to do is have a register whose value tells if the button has ever been pressed. Basically a latch. When the button goes high, a latch register goes high and stays high forever.
I thought this would be straightforward, but I got a bunch of warnings when I tried to code it up. A little Googling showed "Don't make latches in HDL! Bad practice!", but I don't really see the alternative here.
Here's my attempt. clk_10m is a fast free-running clock, pwr_off_req is the button input.
reg pwr_off_req_latched = 0;
always # (clk_10m or pwr_off_req) begin
if (pwr_off_req == 1'b1)
pwr_off_req_latched <= 1'b1;
else
pwr_off_req_latched <= pwr_off_req_latched;
// I tried this to make sure it's always set to something
end
Can you assume that the pulse length of the button press is much longer than the clock frequency of your device? If it's a physical button I think that is a very safe assumption. In that case I think this would work perfectly fine:
always #(clk_10m)
pwr_off_req_latched <= power_off_req_latched | power_off_req;
Latches are not bad in HDL they just require some consideration, implied latches from forgetting to specify else clauses in combinatorial sections are bad because you do not end up with the hardware you expect, and can create timing problems.
If you are applying a reset you might need to specify a 'pragma' so that the synthesis tool correctly identifies it.
Also latches should use = not <=, when they are enabled they are combinatorial (open) and will not break feedback loops.
This is typical way to create a latch with an asynchronous reset:
//synopsys async_set_reset "rst_an"
always #* begin
if (~rst_an) begin
// Reset
x = 1'b0;
end
else if (latch_open) begin
//next datavalue
x = y ;
end
end
In your case you might want something like :
//synopsys async_set_reset "rst_an"
always #* begin
if (~rst_an) begin
pwr_off_req_latched = 1'b0;
end
else if ( pwr_off_req ) begin
pwr_off_req_latched = 1'b1 ;
end
end
Latches can create problems for timing analysis tools. They also don't map to certain (FPGA) architectures directly, so are much harder for the place-and-route tools. Hence the warnings.
However, what you are asking for is not a latch as I understand the digital logic sense - merely a flipflop which doesn't ever get reset.
So, it can be simplified to a simple d-type flipflop with the D input tied to 1 and the clk input connected to your pwr_off_req signal:
reg pwr_off_req_latched = 0;
always # (posedge pwr_off_req) begin
pwr_off_req_latched <= 1'b1;
end
You'll have no noise rejection on that at all - any positive going edge will latch the flipflop to 1.
If I were doing this, I would run the input into a double-flip-flop synchroniser and then count a few clock pulses of the synchronised signal to make sure it's not noise before setting the latched signal. Unless you are expecting real events shorter than a few clock pulses that'd the way to do it.
Aside:
A "latch" in the digital logic world usually means either
a circuit whose output holds whichever of the two inputs was last high (a Set/Reset or SR latch)
a circuit whose output holds the input value while a control signal is inactive, but follows the input when the control signal is low - a transparent latch
This is in comparison to a flipflop, whose output holds some aspect related to the input(s) when the control signal changes (usually) from low to high, and ignores the inputs except for a tiny time window around that rising edge. These are D-type, T-type and JK-type flipflops, depending on how the output behaves relative to the input.