I am generating a divided clock, something like this:
logic div_clk;
always_ff #(posedge clk or negedge rstb) begin
if(!rstb) div_clk <= 1'b0;
else div_clk <= !div_clk;
end
I then launch data on clk and capture on div_clk. Something like this:
always_ff #(posedge clk) begin
clk_data <= something;
end
always_ff #(posedge div_clk) begin
div_clk_data <= clk_data;
end
In my simulations I am getting a race condition since clk_data updates coincident with div_clk and the div_clk_data gets the wrong value.
In synthesis I define these two clocks to be synchronous by creating a generated clock:
create_clock -name CLK [get_ports clk]
create_generated_clock -name GEN_DIV_CLK -source [get_ports clk] -divide_by 2 [get_pins div_clk]
Is there something equivalent that I can put into my RTL, or something I can do to tell my simulator that div_clk is synchronous to clk and prevent the race condition from happening?
This is a case where NBA's should not be used. You should not have any NBA's in your clock tree (including gated clocks) if you want the clocks to remain synchronous.
Related
I have two 2-bit inputs and an output of 1-bit. So what I'm trying to do is code a next state value with taking AND of the two inputs, then use non-blocking <= to assign that value into register xy_r, which is now two D-flipflops. So I am seeking to get always the XOR-value of the xy_r previous clock edge as my output xor_out. I suppose that the way below is not the right one?
It worked in the simulation but then again in RTL Synthesis I didn't end up with an XOR gate but a third flip-flop so it seems like xor_out is treated as a register.
I suppose I can't use assign outside of the if-else-statements because in that case the output wouldn't follow the xy_r previous state but the present.
Can you please assist me how to solve this issue, if there is a somewhat simple way.
module 2ffsxor
(input logic clk,
input logic rst_n,
input logic [1:0] x_in,
input logic [1:0] y_in,
output logic xor_out
);
logic [1:0] xy;
logic [1:0] xy_r;
always # (posedge clk or negedge rst_n)
begin
if (rst_n == '0)
xy_r <= '0;
else
begin
xy = x_in & y_in;
xy_r <= xy;
xor_out = xy_r[0] ^ xy_r[1];
end
end
endmodule
And here's a schematic what it should be:
Schematic 2ffsxor
Synthesis treats this code as if you had wrote
begin
xy_r <= x_in & y_in;
xor_out <= xy_r[0] ^ xy_r[1];
end
That's two stages of registers. But since you assigned xor_out with a blocking assignment, you have a simulation race condition for any process that tries to read xor_out. The race would be from the result from the previous values of xy_r and the result before that!
If you use an continuous assign statement outside this always block, any process reading xor_out on the clock edge will see the result from the previous state of xy_r.
But why not write this as on register:
always # (posedge clk or negedge rst_n)
begin
if (rst_n == '0)
xor_out <= '0;
else
begin
xy = x_in & y_in;
xor_out <= xy[0] ^ xy[1];
end
end
I tried using expect with the following property
module tb;
logic a;
logic clk=0;
default clocking #(posedge clk); endclocking
always
#5ns clk = ~clk;
initial begin
$dumpfile("dump.vcd"); $dumpvars;
$display("START");
a = 0;
#100ns;
a = 1;
#100ns;
$finish;
end
initial begin
#10ns;
expect(#(posedge clk) 1 ##1 $changed(a) |-> 1) $display("SUCCESS"); else
$display("FAIL");
end
endmodule
Is the expect going to block until a change from 0 to 1 at 100ns ?
No, it will block until the second (posedge clk), regardless of the value of a, and will always pass.
The expect statement does not start evaluating the property until the first clk edge. The antecedent takes two cycle to either match or not match. Since the consequent is always true, the property passes on match. If there is no match, the property also passes, vacuously.
What is the difference between:
1
forever begin
**#posedge(clk) begin**
if(vif.sof == 1) begin
//some code here
end
end
end
2
forever begin
**#posedge(clk);**
if(vif.sof == 1) begin
//some code here
end
end
Does the begin..end that goes with #(posedge clk) make a difference ?
event_controls like # and # in procedural code are not statement by themselves; they are prefixes to the statements that follow. And a statement can be a simple statement, like an assignment, or a block like begin/end or fork/join. And a block is allowed wherever a single statement is allowed.
When you write #(posedge clk); it is really #(posedge clk) null_statement;
I should have given you enough information to answer your question, but here is another variation:
forever
#posedge(clk)
if(vif.sof == 1) begin
//some code here
end
Now there is a big difference if a semicolon follows #(posedge clk) or not.
Assume SV interface contains a concurrent assertion property.
Is it possible to enable this assertion only when an event is triggered? I tried writing property inside a task in interface file, but I ended up with error: near "property": syntax error, unexpected property.
Interface intf;
task e1();
-> e1;
endtask
wait(e1.triggered); // something like this, where property waits for trigger
property prop1;
#(posedge clk) a ##1 b;
endproperty
endinterface
Thank you.
I think you need to consider that properties are best written synchronously and are evaluated on every clock cycle. You property says that one cycle after every clock cycle where a occurs b occurs. I guess what you're after is something like:
if e1 occurs, then a should be true (on the next rising edge of
clk?) and then b should be true on the rising edge of clk after
that
So, one way of doing that would be to create an always block that generates a pulse one clock wide should e1 occur, eg:
always begin
#(e1);
#(posedge clk) e = 1'b1;
#(posedge clk) e = 1'b0;
end
property prop1;
#(posedge clk) e |-> a ##1 b;
endproperty
or some variation of that. I feel like I should be worried about races, however.
I've trouble designing assertions(SVA) for this scenario.
When a mux sel is asserted, the data_in is expected to be stable for 2 clocks ; clock prior to mux sel being asserted, and the current clock when mux sel is asserted.
Now the data_in is a wide vector/bus signal, whereby some bits of the bus are Z and X during functional mode (this is expected), while this bits may carry value during non-functional mode.
This then implies, the approach to design the SVA would be to compare bit by bit of data bus when mux sel is asserted.
This is my approach, but SVA fails and am not sure why.
generate
for( genvar i=1 ; i<BUS_WIDTH ; i++ ) begin
always # (posedge clk) begin
if(!$isunknown(data_in[i]) && reset) begin
data_in_temp_prev[i] <= data_in[i];
if (mux_sel==1 && reset==1 && i>0) begin
SVA_TEST: assert property (data_in[i] == data_in_temp_prev[i-1]) else `uvm_error("TRIAL_SVA",$sformatf("datain expected to be stable for 2 clks prior to mux sel"));
end //if
end //if isunknown
else begin
din0_temp_prev[i] <= 0;
end
end //always
end // for genvar
endgenerate
Any suggestions on how to approach designing this SVA ?
Thanks.
If I understood correctly your problem, you don't need all this structure for this kind of problem, look at my example below and tell me if helps or not.
property mux_compare;
disable iff(!reset)
#(posedge mux_sel)
!$isunknown(data_in)
|->
#(posedge clk)
$stable(data_in) [*2];
endproperty: mux_compare
My approach to SVA checkers is to use a standard structure to properties to avoid all kinds of problems. the structure is to always use clocked properties with "disable iff" and always use an implication operator, where the left hand side is the trigger and the right hand side is what we want to verify.
Here's an example from the LRM:
property abc(a, b, c);
disable iff (c) #(posedge clk) a |=> b;
endproperty
similarly to the answer above, I would do this:
property mux_compare;
disable iff(!reset) #(posedge clk)
mux_sel===1 |->
!$isunknown(data_in[i]) || $stable(data_in) && data_in[i] == $past(data_in[i],2);
endproperty: mux_compare