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Is there a right way of implementing a T flip flop in verilog wrt using reset signal?

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Is there a right way of implementing a T flip flop in verilog wrt using reset signal?

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3

$begingroup$

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

share|improve this question

asked 1 hour ago

YashaYasha

524

$endgroup$

add a comment | 

3

$begingroup$

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

share|improve this question

asked 1 hour ago

YashaYasha

524

$endgroup$

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

share|improve this question

asked 1 hour ago

YashaYasha

524

$endgroup$

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 

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

share|improve this question

asked 1 hour ago

YashaYasha

524

share|improve this question

asked 1 hour ago

YashaYasha

524

asked 1 hour ago

YashaYasha

524

asked 1 hour ago

YashaYasha

524

1 Answer
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4

$begingroup$

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.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

$endgroup$

add a comment | 

4

$begingroup$

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.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

$endgroup$

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.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

$endgroup$

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269

answered 1 hour ago

Dave Tweed♦Dave Tweed

125k10155269