I can not figure this vhdl logic out, help.

L

logia

Guest
Hi there,

I have been read a piece of vhdl code and could not understand
following part.
My question is that regLoadEn = '1' only when both stLoadEn and
qfnFingerEn
are '1'. And all these statements are inside a PROCESS, which means
they are
executed in sequence. The second if statement will turn regLoadDone =
'1' when
regLoadEn = '1' and stLoadEn = '0'. But is this possible? Not, right?
Because
regLoadEn = '1' then stLoadEn must be '1'; if regLoadEn = '0' the
second if
statement will never be true. Therefore per following statements
regLoadDone will
never be '1'.

Correct me if I am wrong. Thanks in advance!

FingerInitClock :
PROCESS ( crSysResetAX, sysDOFnDomClk )
...
...
regLoadEn <= stLoadEn AND qfnFingerEn;

IF (qfnFingerEn = '1') THEN


-----------------------------------------------------------------------
-- Falling Edge Detect

-----------------------------------------------------------------------

IF (regLoadEn = '1' AND stLoadEn = '0') THEN

regLoadDone <= '1';

ELSIF (stSymbolEn = '1' AND reg1stSymOnw = '0') THEN

regLoadDone <= '0';

END IF;

ELSE

regLoadDone <= '0';

END IF;
....
....
END PROCESS
FingerInitClock;
 
You left out an important part of the code, but the sensitivity list
implies it's a synchronous process, and assuming the right conditions,
regLoadDone will be '1' for one clock cycle after stLoadEn goes low.

And all these statements are inside a PROCESS, which means they are
executed in sequence.
Click to expand...
You might want to review how a process works, as that's not the case...
(i.e. the
"regLoadEn <= stLoadEn AND qfnFingerEn;"
statement can be moved after
"IF (qfnFingerEn = '1') THEN ... END IF;"
and it would still work the same.
 
logia wrote:
Hi there,

I have been read a piece of vhdl code and could not understand
following part.
My question is that regLoadEn = '1' only when both stLoadEn and
qfnFingerEn
are '1'. And all these statements are inside a PROCESS, which means
they are
executed in sequence. The second if statement will turn regLoadDone =
'1' when
regLoadEn = '1' and stLoadEn = '0'. But is this possible? Not, right?
Because
regLoadEn = '1' then stLoadEn must be '1'; if regLoadEn = '0' the
second if
statement will never be true. Therefore per following statements
regLoadDone will
never be '1'.

Correct me if I am wrong. Thanks in advance!

FingerInitClock :
PROCESS ( crSysResetAX, sysDOFnDomClk )
...
...
regLoadEn <= stLoadEn AND qfnFingerEn;

IF (qfnFingerEn = '1') THEN


-----------------------------------------------------------------------
-- Falling Edge Detect

-----------------------------------------------------------------------

IF (regLoadEn = '1' AND stLoadEn = '0') THEN

regLoadDone <= '1';

ELSIF (stSymbolEn = '1' AND reg1stSymOnw = '0') THEN

regLoadDone <= '0';

END IF;

ELSE

regLoadDone <= '0';

END IF;
...
...
END PROCESS
FingerInitClock;
Click to expand...
I'm guessing from the name that "sysDOFnDomClk" is the clock for the
process. However, I don't see that it's used in the process. There
are also several signals that should be in the process sensitivity
list, such as reg1stSymOnw. If regLoadDone is supposed to be a
register, the code is not correct for a synchronous process.
-Dave Pollum
 
Ok, I missed the ELSIF. I looked at online vhdl tutorial. It says
statements are executed in order
in process body. my understanding is that regLoadEn becomes '0' right
away when DomClk = '1'
and stLoadEn is '0'. Could you please explain a little bit more about
what is really going on here?

PROCESS

ELSIF (sysDOFnDomClk'EVENT AND sysDOFnDomClk = '1') THEN

regLoadEn <= stLoadEn AND qfnFingerEn;

IF (qfnFingerEn = '1') THEN


-----------------------------------------------------------------------
-- Falling Edge Detect

-----------------------------------------------------------------------

IF (regLoadEn = '1' AND stLoadEn = '0') THEN

regLoadDone <= '1';

ELSIF (stSymbolEn = '1' AND reg1stSymOnw = '0') THEN

regLoadDone <= '0';

END IF;

ELSE

regLoadDone <= '0';

END IF;
END IF;

END PROCESS
jens wrote:

You left out an important part of the code, but the sensitivity list
implies it's a synchronous process, and assuming the right conditions,
regLoadDone will be '1' for one clock cycle after stLoadEn goes low.



And all these statements are inside a PROCESS, which means they are
executed in sequence.



You might want to review how a process works, as that's not the case...
(i.e. the
"regLoadEn <= stLoadEn AND qfnFingerEn;"
statement can be moved after
"IF (qfnFingerEn = '1') THEN ... END IF;"
and it would still work the same.
Click to expand...
 
The sensitive list is just like that. And the code is working. I did not
post all the
process code here, it's too long. All the other signals including
regLoadEn are
decleared in the architecture.

Dave Pollum wrote:

I'm guessing from the name that "sysDOFnDomClk" is the clock for the
process. However, I don't see that it's used in the process. There
are also several signals that should be in the process sensitivity
list, such as reg1stSymOnw. If regLoadDone is supposed to be a
register, the code is not correct for a synchronous process.
-Dave Pollum
Click to expand...
 
My question is that regLoadEn = '1' only when both
stLoadEn and qfnFingerEn are '1'. And all these statements are inside
a PROCESS, which means they are executed in sequence. The second if
statement will turn regLoadDone = '1' when regLoadEn = '1' and
stLoadEn = '0'. But is this possible? Not, right? Because regLoadEn =
'1' then stLoadEn must be '1'; if regLoadEn = '0' the second if
statement will never be true. Therefore per following statements
regLoadDone will never be '1'.

Correct me if I am wrong. Thanks in advance!
Click to expand...
Hi logia.
Signal values are only updated at the end of a process.
(Reader-Driver Model of VHDL)

This means: While the assignment places a '0' into the driver of
regLoadEn a former '1' in the reader of that signal can still be
evaluated in a following if-statement.

The really confusing part is that the regLoadEn assignment is placed
before the clocked part of the process. Due to the sensitivity list,
this line will not be simulated unless clock or reset changes, but
synthesis doesn't care about the sensitivity list. So.. will this line
be synthesized into a register which is not explicitly written in the
code or will it become a simple and gate? in the later case it could be
written outside the process as a concurrent assignment. This again would
affect the simulation.
That's the real problem of this code.

have a nice simulation
Eilert
 
The really confusing part is that the regLoadEn assignment is placed
before the clocked part of the process. ...
Click to expand...
In post #3, a more complete version is shown and the regLoadEn
assignment is inside the clocked part, so there is no confusion and
simulation should match synthesis.

I just noticed my previous post was incomplete... yes, in a clocked
process statements are executed in order, but that is only for
determining what the next value of a signal is. All signals are then
simultaneously updated at the specified clock edge. So even though it
may appear as if one signal assignment is immediately using the value
of another signal assignment, that's not the case. From a hardware
perspective, picture the output of one flip-flop going through some
logic and then feeding into another flip-flop (that's what the process
is doing- creating one flip-flop for every signal assignment). Note
that variables are different, and the results would be much different
if the process used variables.

A good way to get more familiar with those concepts is to simulate (and
make changes and see what they do).
 
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