Is there a right way of implementing a T flip flop in verilog wrt using reset signal? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Verilog inital value for flip flopModelling Circuit from FSM using VerilogT - Flip Flop Using D Flip Flop (Verilog)T-flip flop in VerilogHow is the Truth Table of Positive edge triggered D Flip-Flop constructed?Explanation of Edge Triggered D type flip flop triggered at positive edge of the clock pulse cycle (from Morris Mano Book)?How do I redirect/regenerate an input clock to an output pin in my FPGA design (Verilog)Verilog circuit not synchronousImplementing circuit with d-flipflop in verilogError with reference to scalar wire 'reset' is not a legal reg or variable lvalue
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Is there a right way of implementing a T flip flop in verilog wrt using reset signal?
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Verilog inital value for flip flopModelling Circuit from FSM using VerilogT - Flip Flop Using D Flip Flop (Verilog)T-flip flop in VerilogHow is the Truth Table of Positive edge triggered D Flip-Flop constructed?Explanation of Edge Triggered D type flip flop triggered at positive edge of the clock pulse cycle (from Morris Mano Book)?How do I redirect/regenerate an input clock to an output pin in my FPGA design (Verilog)Verilog circuit not synchronousImplementing circuit with d-flipflop in verilogError with reference to scalar wire 'reset' is not a legal reg or variable lvalue
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$begingroup$
I made a t flip flop using structural modeling in verilog.
module Tflip(input T,input clk,output Q,output Qbar) ;
wire S,R;
and(R,Qbar,T,clk);
and(S,Q,T,clk);
SRLatch srt(S,R,Q,Qbar);
endmodule
module SRLatch(input S,input R, output Q, output Qbar);
nand(Q,R,Qbar);
nand(Qbar,S,Q);
endmodule
And then I tried to make T flip-flop in behavioural modelling. It took me ages to find a neat way to initialise the flip-flop.The below code works great. Is it always required to use a reset of some kind ? Would that make the structural code above wrong?
module T_FF(input T,input rst,input clk,output reg Q);
always@(posedge clk)
begin
if (rst == 1)
Q <= 0;
else if (T)
Q <= !Q;
end
endmodule
I tried to implement a version without a reset signal, but it would only work if the test bench started off a certain way.
Every diagram of T flip-flop i looked for did not show a reset signal, so it took me a lot of time to even understand a reset input was required. It feels like structural modelling a foolproof way of creating things.
verilog flipflop
$endgroup$
add a comment |
$begingroup$
I made a t flip flop using structural modeling in verilog.
module Tflip(input T,input clk,output Q,output Qbar) ;
wire S,R;
and(R,Qbar,T,clk);
and(S,Q,T,clk);
SRLatch srt(S,R,Q,Qbar);
endmodule
module SRLatch(input S,input R, output Q, output Qbar);
nand(Q,R,Qbar);
nand(Qbar,S,Q);
endmodule
And then I tried to make T flip-flop in behavioural modelling. It took me ages to find a neat way to initialise the flip-flop.The below code works great. Is it always required to use a reset of some kind ? Would that make the structural code above wrong?
module T_FF(input T,input rst,input clk,output reg Q);
always@(posedge clk)
begin
if (rst == 1)
Q <= 0;
else if (T)
Q <= !Q;
end
endmodule
I tried to implement a version without a reset signal, but it would only work if the test bench started off a certain way.
Every diagram of T flip-flop i looked for did not show a reset signal, so it took me a lot of time to even understand a reset input was required. It feels like structural modelling a foolproof way of creating things.
verilog flipflop
$endgroup$
add a comment |
$begingroup$
I made a t flip flop using structural modeling in verilog.
module Tflip(input T,input clk,output Q,output Qbar) ;
wire S,R;
and(R,Qbar,T,clk);
and(S,Q,T,clk);
SRLatch srt(S,R,Q,Qbar);
endmodule
module SRLatch(input S,input R, output Q, output Qbar);
nand(Q,R,Qbar);
nand(Qbar,S,Q);
endmodule
And then I tried to make T flip-flop in behavioural modelling. It took me ages to find a neat way to initialise the flip-flop.The below code works great. Is it always required to use a reset of some kind ? Would that make the structural code above wrong?
module T_FF(input T,input rst,input clk,output reg Q);
always@(posedge clk)
begin
if (rst == 1)
Q <= 0;
else if (T)
Q <= !Q;
end
endmodule
I tried to implement a version without a reset signal, but it would only work if the test bench started off a certain way.
Every diagram of T flip-flop i looked for did not show a reset signal, so it took me a lot of time to even understand a reset input was required. It feels like structural modelling a foolproof way of creating things.
verilog flipflop
$endgroup$
I made a t flip flop using structural modeling in verilog.
module Tflip(input T,input clk,output Q,output Qbar) ;
wire S,R;
and(R,Qbar,T,clk);
and(S,Q,T,clk);
SRLatch srt(S,R,Q,Qbar);
endmodule
module SRLatch(input S,input R, output Q, output Qbar);
nand(Q,R,Qbar);
nand(Qbar,S,Q);
endmodule
And then I tried to make T flip-flop in behavioural modelling. It took me ages to find a neat way to initialise the flip-flop.The below code works great. Is it always required to use a reset of some kind ? Would that make the structural code above wrong?
module T_FF(input T,input rst,input clk,output reg Q);
always@(posedge clk)
begin
if (rst == 1)
Q <= 0;
else if (T)
Q <= !Q;
end
endmodule
I tried to implement a version without a reset signal, but it would only work if the test bench started off a certain way.
Every diagram of T flip-flop i looked for did not show a reset signal, so it took me a lot of time to even understand a reset input was required. It feels like structural modelling a foolproof way of creating things.
verilog flipflop
verilog flipflop
asked 1 hour ago
YashaYasha
424
424
add a comment |
add a comment |
1 Answer
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$begingroup$
The problem with a T FF is that its state always depends on its previous state. So if you have no way of initializing the state, the simulator will always propagate "unknown" on every clock edge.
Of course, in real life, the TFF will always definitely be in one state or the other, and you often don't care exactly which one it is at any given moment, which is why you see no initialization in typical applications.
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem with a T FF is that its state always depends on its previous state. So if you have no way of initializing the state, the simulator will always propagate "unknown" on every clock edge.
Of course, in real life, the TFF will always definitely be in one state or the other, and you often don't care exactly which one it is at any given moment, which is why you see no initialization in typical applications.
$endgroup$
add a comment |
$begingroup$
The problem with a T FF is that its state always depends on its previous state. So if you have no way of initializing the state, the simulator will always propagate "unknown" on every clock edge.
Of course, in real life, the TFF will always definitely be in one state or the other, and you often don't care exactly which one it is at any given moment, which is why you see no initialization in typical applications.
$endgroup$
add a comment |
$begingroup$
The problem with a T FF is that its state always depends on its previous state. So if you have no way of initializing the state, the simulator will always propagate "unknown" on every clock edge.
Of course, in real life, the TFF will always definitely be in one state or the other, and you often don't care exactly which one it is at any given moment, which is why you see no initialization in typical applications.
$endgroup$
The problem with a T FF is that its state always depends on its previous state. So if you have no way of initializing the state, the simulator will always propagate "unknown" on every clock edge.
Of course, in real life, the TFF will always definitely be in one state or the other, and you often don't care exactly which one it is at any given moment, which is why you see no initialization in typical applications.
answered 1 hour ago
Dave Tweed♦Dave Tweed
125k10155269
125k10155269
add a comment |
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