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Forum: FPGA, VHDL & Verilog Phase detector in Xilinx


Author: Rock B. (rocko445)
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Hello , i am trying to run in Xilinx the following code of phase 
detector
as shown in the attached link and attached photo.

i am not sure why up,dn should be reg?
why ff_rst is a wire why its not a reg?
what is the role of vcxo,rstN?


http://www.dtic.mil/get-tr-doc/pdf?AD=ADA483891


module phase_detector(ref,vco,up,dn);
input ref;
input vco;
output up;
output dn;

wire ref;
wire vco;
reg up;
reg dn;


assign ff_rst = dn & up; // the AND gate
always @( posedge ref or posedge ff_rst or negedge rstN )
begin
 if (!rstN) up<=1'b0;
 else if (ff_rst) up <= 1'b0;
 else up <= 1'b1;
end
always @( posedge vcxo or posedge ff_rst or negedge rstN )
begin
 if (!rstN) dn<=1'b0;
 else if (ff_rst) dn <= 1'b0;
 else dn <= 1'b1;
end

endmodule

: Edited by User
Author: Duke Scarring (Guest)
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Rock B. wrote:
> what is the role of vcxo,rstN?
rstN ist a low active reset. I think this is only necessary for 
simulation or asic design.

For vcxo show here:
https://en.wikipedia.org/wiki/Phase-locked_loop#Bl...

Duke

Author: Rock B. (rocko445)
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i cant see vcox in the diargram on the article, i am trying to implement 
the code in Xilinx but its not working

module phase_detector(ref,vco,up,dn);
input ref;
input vco;
output up;
output dn;

wire ref;
wire vco;
reg up;
reg dn;


assign ff_rst = dn & up; // the AND gate
always @( posedge ref or posedge ff_rst or negedge rstN )
begin
 if (!rstN) up<=1'b0;
 else if (ff_rst) up <= 1'b0;
 else up <= 1'b1;
end
always @( posedge vcxo or posedge ff_rst or negedge rstN )
begin
 if (!rstN) dn<=1'b0;
 else if (ff_rst) dn <= 1'b0;
 else dn <= 1'b1;
end

endmodule

Author: Duke Scarring (Guest)
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Rock B. wrote:
> i am trying to implement
> the code in Xilinx but its not working
That is not a valid error description.

What did you try?
Will you just simulate the code?
Where is your testbench?

Duke

Author: Lothar M. (lkmiller) (Moderator)
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Rock B. wrote:
> i cant see vcox in the diargram on the article
Its absolute obviously a typo like vcxo = vco...

> i am trying to implement the code in Xilinx
Then forget about that rstN. You don't need it.

> but its not working
How did you find that out? What do you expect? And what do you get 
instead? Does your simulation look fine?

I would try it that way:
module phase_detector(ref,vco,up,dn);
input ref;
input vco;
output up;
output dn;

wire ref;
wire vco;
reg up = 1'b0;
reg dn = 1'b0;

assign ff_rst = dn & up; // the AND gate

always @(posedge ref or posedge ff_rst)
begin
 if (ff_rst) up <= 1'b0;
 else up <= 1'b1;
end
always @(posedge vcxo or posedge ff_rst)
begin
 if (ff_rst) dn <= 1'b0;
 else dn <= 1'b1;
end 

endmodule

Author: Rock B. (rocko445)
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Hello,i thought to simulate it without a test bench ,in XILINX we can 
force a signal and clock in the inputs.

ok , so for the variables ,why up,dn are reg type?
why ff_rst is a wire?
i cant see it the from the attached diagram


Thanks

Author: ossi-2 (Guest)
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why up,dn are reg type? why ff_rst is a wire?

Lookup the description of D-Flip-Flops. What type of signal is denoted 
by "Q", what type of signal is denoted by rst ? That should answer your 
question.(Remember a "Flip-Flop" closely resembles a "register")

Author: Rock B. (rocko445)
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from the manual i know that the inputs and outputs should be wires
that they are not storing data.

in this case i cant see what variable stores data and what variable does 
not store data ?

Thanks

Author: Lothar M. (lkmiller) (Moderator)
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Rock B. wrote:
> in this case i cant see what variable stores data and what variable does
> not store data ?
Only the two flipflops can store values.

Author: Rock B. (rocko445)
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Hello , i tried  to implement the reg everywhere i think it stores 
memory ,
i did defined dn
why ff_rst cannot be input?

Thanks

 Line 22: Port dn is not defined
 Line 23: Non-net port ff_rst cannot be of mode input
 Line 21: Module <phase_d> ignored due to previous errors.
module phase_d(ref,vco,ff_rst);
output up,dn;
input ref,vco,ff_rst;

reg dn,ff_rst;
reg up;
reg ref;
reg vco;
always @(posedge ref or posedge ff_rst)
begin
assign ff_rst=dn&up;
if (!ff_rst) up<=1'b0;
else up<=1'b1;
end

always @(posedge vco or posedge ff_rst)
begin
assign ff_rst=dn&up;
if (!ff_rst) dn<=1'b0;
else dn<=1'b1;
end

endmodule

Author: Andy (Guest)
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Rock B. wrote:
> i did defined dn

but you don't have it in the port list.

> why ff_rst cannot be input?

because it's also the output of the AND gate. Just make it a wire, no 
in- or output.

Look at the code in the picture of your first post to place the assign 
at the right place and only once.

Andy

Author: Rock B. (rocko445)
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the code doesnt give me any errors but i am curios regarding few issues.
why assigh have to be outside always?
why the input cannot be a reg?
why ff_rst has to be a wire for being an output of AND
dn and up are output too?


Thanks
[verilog]
module phase_d(ref,vco,dn,up);
output up,dn;
input ref,vco;

wire ff_rst;
reg up,dn;


assign ff_rst=dn&up;
always @(posedge ref or posedge ff_rst)
begin

if (!ff_rst) up<=1'b0;
else up<=1'b1;
end

always @(posedge vco or posedge ff_rst)
begin
if (!ff_rst) dn<=1'b0;
else dn<=1'b1;
end

endmodule
[/verilog]

: Edited by User
Author: Andy (Guest)
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Rock B. wrote:
> why assigh have to be outside always?

assign connects wires to inputs and outputs or defines combinatorial 
logic. Such signals are always active in opposite to registers which 
change their state only on an active clock edge and hold the state until 
the next edge.

> why the input cannot be a reg?

An input is just a wire with a defined direction, if you want to 
register the state of the input you need an additional register with a 
clock.

> why ff_rst has to be a wire for being an output of AND
> dn and up are output too?

ff_rts is an internal signal that connects something. The terms input 
and output are only used at the ports of a module to define the 
direction of the port. An output port can also directly be defined as a 
register:

output reg up,dn;

or as in your code, first as an output and then as a register.

Author: Rock B. (rocko445)
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Hello in order to see the effect of the phase detection i have defined 
ref and vco as shown in the attached photo.
ref as 60ps period and 50 duty cycle
vco as 50ps period and 50 duty cycle

as it shows in the attached simulation photo there is a phase shift 
between the ref and vco but the up and dn aren't changing.what went 
wrong?

Thanks
module phase_d(ref,vco,dn,up);
output up,dn;
input ref,vco;

wire ff_rst;
reg up,dn;


assign ff_rst=dn&up;
always @(posedge ref or posedge ff_rst)
begin

if (!ff_rst) up<=1'b0;
else up<=1'b1;
end

always @(posedge vco or posedge ff_rst)
begin
if (!ff_rst) dn<=1'b0;
else dn<=1'b1;
end

endmodule

: Edited by User
Author: Lothar M. (lkmiller) (Moderator)
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Rock B. wrote:
> if (!ff_rst) up<=1'b0;
The simulation does exactly what you have written down. If you want 
something else then look at my code or use that grey thing between your 
ears and answer the question:
Why the heck a NOT before ff_rst?

Author: Rock B. (rocko445)
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Hello , i think there is another problem  because as you can see in the 
attached photo the ff_rst is zero all the way although the ref and vco 
do change.it seems as if the ASIGN command ran only once,why is that?

Thanks
module phase_d(ref,vco,dn,up);
output up,dn;
input ref,vco;

wire ff_rst;
reg up=1'b0;
reg dn=1'b0;


assign ff_rst=dn&up;
always @(posedge ref or posedge ff_rst)
begin

if (ff_rst) up<=1'b0;
else up<=1'b1;
end

always @(posedge vco or posedge ff_rst)
begin
if (ff_rst) dn<=1'b0;
else dn<=1'b1;
end

endmodule

Author: Lothar M. (lkmiller) (Moderator)
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Rock B. wrote:
> the ff_rst is zero all the way
I don't think so, because the up-flipflop is reset at each rising edge 
of vco. The thing is: the ff_rst is high only for one timestep within 
the same simulation cycle.

Or with other words: as far as i can see the simulation does exactly 
what it should. The up flipflop goes high at the rising edge of ref, and 
it is reset when vco rises also.
The only problem is here, that the frequency of one input is always 
higher than the other. Change the both input frequencies and look whats 
happening.

: Edited by Moderator
Author: Rock B. (rocko445)
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Hello, why the 'dn' is always zero?
There are places where VCO rises and because ff_rest is always zero then 
dn should go to '1' every time VCO rise
it meet the criteria of the always@ part bellow.
Thanks
always @(posedge vco or posedge ff_rst)
begin
if (ff_rst) dn<=1'b0;
else dn<=1'b1;
end

: Edited by User
Author: Lothar M. (lkmiller) (Moderator)
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Rock B. wrote:
> Hello, why the 'dn' is always zero?
Did you swap the two input signals? What happened?

> There are places where VCO rises and because ff_rest is always zero then
> dn should go to '1'
Can't see such occurrences.
And when the two input signals are constant and stable, then why 
should the output change?

: Edited by Moderator
Author: Rock B. (rocko445)
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Hello ,Yes when i switched the signals up=0 all the way as you can see 
in the attached photo.

so for the first always@ ff_rst=0 all the time thus making up=0 the 
whole time as we can see in the result.

but for the second always@  ff_rst is not always zero because dn do get 
to be 1 some times

because we have only one ff_rst  then we should have or both 'dn' and 
'up' zero or both of them toggeling by the VCO and ref clocks?
always @(posedge ref or posedge ff_rst)
begin

if (ff_rst) up<=1'b0;
else up<=1'b1;
end 
always @(posedge vco or posedge ff_rst)
begin
if (ff_rst) dn<=1'b0;
else dn<=1'b1;
end

: Edited by User
Author: Lothar M. (lkmiller) (Moderator)
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Ok, let's resume: the code is working as expected. And let me propose, 
that the synthesized hardware will look as the hardware in the picture. 
So the only problem is, that you can't recognize and accept that.

With other words: of course ff_rst goes high! But you can't see it 
directly, because at the very same moment both flipflops go low and so 
at the very same time ff_rst goes low again. This happens in 0 (in 
words: zero) ps, because you are doing a behavioural simulation which 
has absolutely no (in numbers 0) delay!

Think about this: the only way to reset the flipflops is by ff_rst going 
high. At the very same time when dn or up goes from 1 to 0 the signal 
ff_rst must have been 1.

Think about that. It isn't that difficult...

When nothing works for you then try a post route simulation.

: Edited by Moderator
Author: Rock B. (rocko445)
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Thank you very much

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