setup vs. clock-to-output time vs. hold time

Hello,

I am new to VHDL and FPGA design.

My question:

I want to make a synchronous design.

What is the condition for a synchronous design to work
with regard to the setup and clock-to-output and hold time of
the FFs?

Does the setup time have to be smaller than the clock-to-output
time?
What about the hold time in respect to setup and clock-to-output?


Thanks to you.

Rgds

Bochumfrau@gmx.de

Set up time is the amount of time before the clock edge that the
input signal needs to be stable to guarantee it is accepted properly
on the clock edge. Hold time is the amount of time after the clock
edge that same input signal has to be held before changing it to make
sure it is sensed properly at the clock edge. The clock to output
time is the worst case time it takes for a signal to propagate out
after the clock edge.

Now that you know that, it is fairly easy to see that the clock to
output time plus signal propagation time on the route to the other
flip-flop plus the set up time must be less than the period of the
clock. Tco + Tpd + Tsu < Tclk

To satisfy hold time, the clock to output time plus the propagation
delay must be greater than the hold time. Chip makers try to make
the hold times negative because it is hard to guarantee minimum
propagation times.


"" wrote:

> Hello,
>
> I am new to VHDL and FPGA design.
>
> My question:
>
> I want to make a synchronous design.
>
> What is the condition for a synchronous design to work
> with regard to the setup and clock-to-output and hold time of
> the FFs?
>
> Does the setup time have to be smaller than the clock-to-output
> time?
> What about the hold time in respect to setup and clock-to-output?
>
> Thanks to you.
>
> Rgds

--
--Ray Andraka, P.E.
President, the Andraka Consulting Group, Inc.
401/884-7930 Fax 401/884-7950
email
http://www.andraka.com

"They that give up essential liberty to obtain a little
temporary safety deserve neither liberty nor safety."
-Benjamin Franklin, 1759

Ray Andraka

<> wrote in message
news: om...
> Hello,
>
> I am new to VHDL and FPGA design.
>
> My question:
>
> I want to make a synchronous design.
>
> What is the condition for a synchronous design to work
> with regard to the setup and clock-to-output and hold time of
> the FFs?
>
> Does the setup time have to be smaller than the clock-to-output
> time?
> What about the hold time in respect to setup and clock-to-output?
>
>
> Thanks to you.
>
> Rgds
I'm not sure what you mean my "clock-to-output" time. Latch clock input
pin to latch output pin?

Anyway, for a synchronous design to work the worst-case latch-to-latch
path delays must be less than (clock period - latch setup requirement).
The latch-to-latch path delays include the delay through a latch (clock
input to latch output) and all the delays through the combinational
logic feeding the next latch, including the delays through the gates
themselves and the delays through the wires connecting them.

Also, the best-case latch-to-latch delay must be greater than the latch
hold time requirement.

In addition, the primary input to latch setup and hold times must be met
as well as the required arrival times at the primary outputs.

Skew in the clock tree also needs to be considered.

Charles Bailey

Bochumfrau wrote:
> I want to make a synchronous design.

Excellent choice.

> What is the condition for a synchronous design to work
> with regard to the setup and clock-to-output and hold time of
> the FFs?

1. All inputs are synchronized to the system clock.
2. There is only one system clock. Any other inputs
that "feel" like clocks are synchronized and become
regular inputs or synchronous clock enables.
3. The VHDL code uses the synchronous template for all processes.
4. Constrain the place+route reg to reg fmax to sys_clk
5. Verify that fmax has been met by the static timer.

That's it. The computer will worry about the setup/hold requirements
to meet fmax. You don't need any other constraints.

-- Mike Treseler

Mike Treseler

> To satisfy hold time, the clock to output time plus the propagation
> delay must be greater than the hold time. Chip makers try to make
> the hold times negative because it is hard to guarantee minimum
> propagation times.

Negative hold time? What does that mean?

Ken

Ken

Ken a écrit:
>>To satisfy hold time, the clock to output time plus the propagation
>>delay must be greater than the hold time. Chip makers try to make
>>the hold times negative because it is hard to guarantee minimum
>>propagation times.
>
> Negative hold time? What does that mean?

The value on the input can change *before* the clock edge and still be
carried to the output.

--
____ _ __ ___
| _ \_)/ _|/ _ \ Adresse de retour invalide: retirez le -
| | | | | (_| |_| | Invalid return address: remove the -
|_| |_|_|\__|\___/

Nicolas Matringe

it means that the window in which the input signal has to be held stable
ends before the arrival of the clock edge. This happens if the
propagation delay on the data path is guaranteed to be longer than the
propagation of the clock to the same flip-flop. It is far easier to
guarantee relative delays on a monolithic chip than it is to guarantee
minimum delays at the board level where you have different chips. For
that reason, it is desirable for the chip manufacturers to make the hold
times zero or negative by careful use of delays. On the other hand,
adding delays to get a zero or negative hold time degrades maximum
performance, so if the component is one that is used in an application
where they know you will have some minimum clock to out delay (or think
you will anyway), it is common to have a positive hold time. An example
of that is memories.

Ken wrote:

>
> Negative hold time? What does that mean?
>
> Ken

--
--Ray Andraka, P.E.
President, the Andraka Consulting Group, Inc.
401/884-7930 Fax 401/884-7950
email
http://www.andraka.com

"They that give up essential liberty to obtain a little
temporary safety deserve neither liberty nor safety."
-Benjamin Franklin, 1759

Ray Andraka

Thank you very much for your answers : )

Bochumfrau@gmx.de