Search Engine


Add Question

161 Threads found on Negedge
Hi, Normally, you wouldn't add so much of logic to the reset condition in RTL and you wouldn't use the sequential clocking block always @ (posedge or negedge) for signals other than clocks and resets. If you want to sample the address pins , assign the incoming address to a register on the rising edge of clock and then in an always_comb block yo
I have no lint errors but few lint warnings in my design example :- assign enable=clk & b; always @(negedge clk) begin . . . . . end But the lint throws a warning that the clk is used for different purposes. so can I avoid this warning ? if no,what are the consequences that have to be faced in real time circuit?
if a clock is only posedge, why would the sdf capture it's negedge behavior?
As a concept, it is odd to intentionally create an async reset that is actually a sync reset. The Verilog always construct describes an asynchronous reset. You might say that it's odd in some way that an asynchronous input is described by a negedge event. The synthesis template for a edge triggered register with asynchronous
For example, it can not be used for Key press rise/fall edge detect, but why it can be used for clock or reset? Thank you!
What I want to do is to detect the raise of signal1 then at the fall of signal2 I want to do some logic but it must be in this order, something like this. always@(posedge signal1) begin @(negedge signal2) begin \\ do some logic end end how to synthesize this with Verilog? Thanks in advance. [url=obrazki.elek
My Comments: 1. Why have you used the negedge of the clock? it is more usual to use the posedge. It would mean this block would not be usuable by the majority of designs. 2. I would rather the addr_out is registered. This means other users dont need to worry about the combinatorial path delay in to their logic. 3. Have you tried synthesising this
Hi, Please let me know , what will be the optimized verilog code of below:- reg control_pin_ton_delay_cntr; always @ (posedge clk_i or negedge nrst_i) begin if (!nrst_i) control_pin_ton_delay_cntr <= 33'd0; else if (enable_i) begin if (control_on_i) control_pin_ton_delay_c
Hi All, How to stitch a couple of negedge flops in the design where most of the flops are posedge? How to balance chains where only few flops are negedge but most flops are posedge? Is it possible to combine posedge and negedge flops in the same chain? Thank you!
If ur clock is 10 ns and if the launch is at 0ns then the capture will be at 10 ns. If you are seeing capture at 5 ns, then it must be capturing data on the negedge of clock. Otherwise you must have constrained it accordingly. Go back to the RTL and check this particular timing path.
You didn't follow the Verilog template for synthesizable sequential code always @ (posedge clk or negedge rst) begin if(!rst) begin clk_out <= 1'bz; count <= 8'b0;; end else begin count <= count + 1; // all additional clock edge sensitive actions must go here[
Hi , the below code uses dff for mod3 counter design . rst generated (using expression1) does not work,where as rst generated (using expression 2) works . what is the reason ? module dff(output reg q,output qn,input d,clk,reset); always @(posedge clk or negedge reset) begin if(!reset) q<=1'b0; else q<=d; end assign qn=~
always @ (posedge a or posedge b or negedge c...) models edge sensitive sequential, not combinational logic.
Is it OK to use both positive and negative edge flip flops on the same clock in the same design from perspective of place & route tools? Is it going to make timing analysis impossible? Note that it won't have combinational logic depending on the outputs of both a posedge and a negedge sensitive flop at the same time.[/QUO
None of the three complies to the templates for synthesizable Verilog (e.g. IEEE Std 1364.1) - the event list of an always block modelling edge sensitive logic must contain only posedge and negedge events - an event can't act on both edges
initial begin // has race condition. reset = 1’b0; #20 reset = 1’b1; #40 reset = 1’b0; end reset = 1’b0 will cause race condition only in case the design is modeled to use asynchronous active low reset as follows. always @ (posedge clk or negedge reset) if (!reset) q <= 1'b0; else q <= d; Because the tool does not guara
HI. I'm using like following that dcm dcm_inst ( .CLKIN_IN ( fclk), .USER_RST_IN( 1'b0 ), .CLKFX_OUT( ), .CLKIN_IBUFG_OUT(clk_out_0 ), .LOCKED_OUT( lock ) ); ... assign reset_n = lock; ... always @(posedge clk_out_0, negedge reset_n) be
Dear all, I've been working with Modelsim SE6.6 for years and all my simulation designs work fine. But today when I immigrate to Modelsim SE 10.1c, they don't. For example, a process like this: always @(posedge clk or negedge reset_n) if(!reset_n) message_type = 0; else if(cnt_reg == Packet_length) message_type = `head;
well you have wrote 5 dividers by 2, so first flop divide by 2 second flop divide by 4 third flop divide by 8 fourth flop divide by 16 fifth flop divide by 32 If you want to divide by 10, it is more easy to made a counter on clock, at reset start to 0, and when it reachs 9, set back to 0... always(posedge clk, negedge nreset) if
Hi all, How we can use the parameter as bit width in verilog for a CONSTANT value representation. For example : parameter VAL = 11; always @ ( posedge CLK or negedge RESETn ) begin if ( !RESETn ) begin count <= {VAL{1'b0}}; end else begin if ( count == 11'h5FF[/C
Hello all, I wnat to know how to achieve a special case always block with delayed non blocking assignments. Consider the following code `timescale 1s/1ps always@(posedge clk) a<=1'b1; always@(negedge clk) a<=#100n 1'b0; In this code, a<=#100n 1'b0 will be executed and the change is scheduled to happen after 100n. but If th
Hi , I am working on scan insertion. I have some querries on the same. My understanding is lockup latch is not needed when we have negedge flop followed by pos edge flop. But when we have pos edge flop followed by neg edge flop, can I use lockup latch between pos edge -> lockup latch -> neg edge ? Will this work ? or lockup latches are used on
Hi Vinod, First open file using $fopen task in an initial block. Then you can use $fdisplay task. Let's assume your number changes on rising edge of the clock, then I would read its value on falling edge always @(negedge clk) $fdisplay(fid1,"%d", theNumber); Before ending simulation you'll need to close the file ID. Start doing, debugging error
Assume you need an edge detecting circuit. 95313 assign out = Q1 & (!Q2); always @(posedge clk or negedge rst) if (!rst) begin Q1 <= 1'b0; Q2 <= 1'b0; end else begin Q1 <= In; Q2 <= Q1; end
Hi, I am using irun for compilation and elaboration of my design and the testbench. I have a very simple interface file '' which contains, interface usbsys (input wire clk); covergroup write_limit @ (negedge clk); WriteLimit : coverpoint top_tb.read_write { bins write= {0}; bins read = {1}; } endgroup wr
module AC ( input data_in, input load, clk, output reg data_out ); always @( posedge clk or negedge clk ) if( load ) data_out <= data_in; endmodule module ALU_REG ( input data_in,input clk, output reg data_out ); always @( posedge clk or negedge clk ) data_out <= data_in; endmodule module psw_reg ( input [2
The difference between posedge x and posedge y is unequivocally declared by the if (x) statement. The signal that appears only in a posedge event is the clock. Problems may arise in constructs that con't correspond to the expected template structure. By the way, ambiguity of posedge/negedge is a pure Verilog problem. VHDL has e
The correct syntax is always @(posedge clk or posedge rst) if(rst) q<=0; else q<=d for active low reset always @(posedge clk or negedge rst) if(!rst) q<=0; else q<=d; Please consider that the required Verilog syntax for asynchronous register control hasn't to do with edges, it's actually describing level
Hi, 1) Could you please explain the reason why are we using negedge of flop to get a TDO value in the Boundary scan operation ? Why don't we get a value at posedge of clock? 2) Why Bypass Register is a 1-bit register? Without bypass register, why cant we bypass the operation? Thanks, Sathish
That's also how I'd do it (given the gun to the head). A set of FFs clocked on posedge, other set on negedge, and then xor the outputs. On an fpga with local clock inversion you would only need one clock net. As ads_ee pointed out, that does require some constraints on placement to make sure this thing meets timing. But by RLOCing the FFs (ie RL
Mmmh? Not to be pedantic, but a DFFR clocked on negedge can occur in fpga designs as well, no? Or maybe I am missing something... But given that one generally uses posedge clocked FFs in fpga designs chances are that the OP is indeed doing something asic-ey.
Did changing "OTPUT" to "OUTPUT" resolve the warning about ser_OTPUT being assigned but not used? To avoid typos like this, I recommend using "`default_nettype none" at the start of every Verilog file you create. Your use of "negedge enable" as a clock is bad design practice. In this sort of design, everything should be synchronous to a single clo
Hi All, Let's say we have two flops connected back-to-back. One of the flops works on posedge, another on the negedge of clock. For the correct STA, should a multi-cycle path be defined between these flops? Are there any HOLD requirements for the second flop? Could you imaging a scenario where it has no Setup violations, but do ha
Hi~ i have question about opencores i2c slave simulation results. Q1) how can i use "sda_dly" for fix 0 hold margin? Q2) how can i fix timing violation in i2c simulation result? Q3) why happened timing violation Warning! Timing violation $setup( posedge scl:742686 NS, negedge sda &&& scl:745866 NS, 4700.00 : 4700 NS );
Hey, mr_vasanth raises a good point. This bit probably indicates a posedge clocked vs negedge clocked issue. Source Clock: sip_fmc104_0/fmc10x_inst/clk_ab falling at 2.500ns Destination Clock: sip_fmc104_0/fmc10x_inst/clk_ab rising at 5.000ns As for "which skew is correct, the positive or the negative bits?
Generate enable on negedge flop or use icg (latch plus AND gate)
Hi, I dont know whether u are in need of a digital circuit or a HDL. I have attached the HDL code. You can even translate this to a circuit. always@(posedge clk or negedge rst) begin if(~rst) a_f <= 1'b0; else a_f <= input_a; end assign zero_to_one_transition = !a_f && input_a; always@(posedge clk or negedge rst) begin if(~rst) b_f
I don't understand though why "reset" is under "negedge" in the sensitivity list... The aynchronous reset of a DFF is level sensetive and not edge I see it, the code should be always @ ( posedge clk or reset) and not always @ ( posedge clk or negedge reset) Would you agree ?[
always@(posedge clk 0r negedge rst) begin if(~rst) a_f <= 1'b0; else a_f <= input_a; end assign zero_to_one_transition = !a_f && input_a; always@(posedge clk 0r negedge rst) begin if(~rst) output_f <= 1'b0; else if(zero_to_one_transition) output_f <= 1'b1; else output_f <= 1
what are the problems on working on both posedge and negedge of a clock (by using an invertor on clock)... please explain or is this method fine?
Hi All, There is a multi if statement in licensed IP, which is like : reg array ; always @ (posedge clk or negedge rstn) begin if(~rstn) begin reset process end else begin if(condition A) array <= variable1; if(condition B) array <= variable2; .... if(condition X) each elem
This is frequency divide by 2 question without 50% duty cycle. One way to do is module divby2( input clk, input reset, output div_clk); reg div_clk_q; wire div_clk_d; always @ ( posedge clk, negedge reset) begin: clk process if(!reset) begin div_clk_q <= 1'b0;
Can anyone please explain system verilog event region? Actually I am asserting a signal at clock's negedge but in simulation's sequence time view due to system verilog event regions it shows some delay in setting the signal to 1. When I am writing a property to check that signal should be high at negedge of clock, it checks at clock's next nege
I am trying to write a system verilog property. Actually I am using these properties to write functional coverage. I have signal which goes high on clock negedge. And when I write property for it as following: A : cover property(@(negedge clock) $rose(signal)); Assertion checks the value at next clock negedge, but I want to check it (...)
I get different behavior depending on how I define a clock divider so I wrote some test code. I'm using modelsim-altera 10.0c. Here is the code. always @(posedge clk or negedge reset_n) if (~reset_n) begin aclk_div <= 0; // 2 bit counter bclk_div <= 0; // 2 bit counter end els
It totally depends on the synthesis algorithm. Each tool does its calculation differently. If needs to be done by hand, a simple example would be a posedge vs a negedge. If a design consists of posedge and negedge, it will use the cts inverters else most probably a cts buffer will be used.
i wrote this code but i'm gettin the error :" this signal is connected to multiple drivers " here's my code: module executer(ins,clk ); input ins; input clk; reg re,we,ie; reg i,ar1,ar2,aw,op; reg w,a,b; wire r1,r2,out; wire c,v,n,z; always @(negedge clk) if(ins==1) begin ar1=ins
input in reg seq; reg pass; always @ (posedge clk or negedge rstn) if (~rstn) seq <= 'b0; else begin seq <= {seq,in}; if (seq == 11001100) pass <= 1'b1; end // Add logic to deassert pass Please post a better solution or point out if there are any corrections to this
i have selected the scan style as -clocked scan in DFT compiler, instead of mux based design my flip-flop module is as below, module FF(clk1,clk2,si,data,q) wire clk; clk=clk1|clk2; always@(negedge clk) begin if(clk1)q=si; if(clk2)q=data; end endmodule. The above code fails in the compile -scan with the error "No SCAN equivalent Found"
because in verilog any signal default is X, intial begin A =0 end will bring A jump from X to 0, that's mean a negedge tigger