VHDL code to light up LED???

N

network lines

Hello...i am new to VHDL, and learning how to
program using a Cypress CPLD.
Anybody got some simple code to turn on and turn off
an LED???
please email sample code to (e-mail address removed)

TIA
 
J

Jim Lewis

For active high:
LedOut <= '1' ;

For a simple on/off with a switch:
LedOut <= SwitchIn ;

Of course you probably should debounce the switch input
and register the LedOut output. For debounce,
do a search of the newsgroup. Assume that your
switch is bouncing if it has not been stable for
anywhere from 5 ms to 15 ms.

If you bought your board, there may be a test
code that your vendor provides that does this.

Cheers,
Jim

network said:
Hello...i am new to VHDL, and learning how to
program using a Cypress CPLD.
Anybody got some simple code to turn on and turn off
an LED???
please email sample code to (e-mail address removed)

TIA


--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jim Lewis
Director of Training mailto:[email protected]
SynthWorks Design Inc. http://www.SynthWorks.com
1-503-590-4787

Expert VHDL Training for Hardware Design and Verification
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
R

Ray Andraka

If the switch just results in turning of and off a LED used as an
indicator, there is no need to debounce it. Debounce becomes
important if the fact that the switch turns on and off many times
within a millisecond or so after it is actuated disturbs the
operation of your circuit. Something like a counter that counts
how many times the switch is operated would need debounce.
Simply turning on and off the led to follow the switch movement
does not.

Likewise, dipswitches used to set modes often do not require
debouncing. They can either be accepted as is, or in many cases
are read at one particular instant during some initialization, so
debounce is not needed. Only if the operation of the switch
causes an immediate action is the debounce needed.

Jim said:
For active high:
LedOut <= '1' ;

For a simple on/off with a switch:
LedOut <= SwitchIn ;

Of course you probably should debounce the switch input
and register the LedOut output. For debounce,
do a search of the newsgroup. Assume that your
switch is bouncing if it has not been stable for
anywhere from 5 ms to 15 ms.

If you bought your board, there may be a test
code that your vendor provides that does this.

Cheers,
Jim



--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jim Lewis
Director of Training mailto:[email protected]
SynthWorks Design Inc. http://www.SynthWorks.com
1-503-590-4787

Expert VHDL Training for Hardware Design and Verification
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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

Jim Lewis

My conservative suggestion to debounce is concern over
noise causing the lifetime of the LED to be shorter.
Perhaps this is not a concern?

I know in my house some light bulbs seem to burn out
faster than others and I have always wondered if it
was due to excessive noise from the switches (look
nice, but knowing the builder they are probably
low cost).

Cheers,
Jim


If the switch just results in turning of and off a LED used as an
indicator, there is no need to debounce it. Debounce becomes
important if the fact that the switch turns on and off many times
within a millisecond or so after it is actuated disturbs the
operation of your circuit. Something like a counter that counts
how many times the switch is operated would need debounce.
Simply turning on and off the led to follow the switch movement
does not.

Likewise, dipswitches used to set modes often do not require
debouncing. They can either be accepted as is, or in many cases
are read at one particular instant during some initialization, so
debounce is not needed. Only if the operation of the switch
causes an immediate action is the debounce needed.

Jim Lewis wrote:




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

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


--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jim Lewis
Director of Training mailto:[email protected]
SynthWorks Design Inc. http://www.SynthWorks.com
1-503-590-4787

Expert VHDL Training for Hardware Design and Verification
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
S

Symon

Hi Jim,
I don't think it matters mate. I know almost all the 7 segment LED displays
I've seen, e.g. on frequency counters, are multiplexed. Thus they're
flashing on and off 'faster than the human eye'! You can see this if you
snort like a pig while watching them! Honest! It's something to do with the
snort vibrating your head.
Cheers, Syms.
 
P

paris

Jim Lewis said:
My conservative suggestion to debounce is concern over
noise causing the lifetime of the LED to be shorter.
Perhaps this is not a concern?

I know in my house some light bulbs seem to burn out
faster than others and I have always wondered if it
was due to excessive noise from the switches (look
nice, but knowing the builder they are probably
low cost).

Cheers,
Jim

actually i think it's because of the light bulb being turned on when the
electric current sinus feeding it is not zero or near zero volts, which
causes stress on the bulb filament (the thin tunsgten wire that lights)
LEDs being solid state devices (and using a different light emitting
pheanomenon than bulbs), i dont think they have this problem
 
R

Ray Andraka

LEDs don't burn out due to turn on/off. In fact, a common way of
varying the brightness in LED displays is to pulse width modulate the
driving voltage.

Light bulbs burn out faster on turn on/off cycles because the wire
resistance is lower when it is cold, and when turned on it has a surge
of current much higher than the on current. LEDs are a different
mechanism for producing the light. I doubt that noise on the switch is
going to alter incadesant lamp life much, the on-off cycles are way to
fast to get the shock heating that happens when you turn them on.


Jim said:
My conservative suggestion to debounce is concern over
noise causing the lifetime of the LED to be shorter.
Perhaps this is not a concern?

I know in my house some light bulbs seem to burn out
faster than others and I have always wondered if it
was due to excessive noise from the switches (look
nice, but knowing the builder they are probably
low cost).
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Jim Lewis
Director of Training mailto:[email protected]
SynthWorks Design Inc. http://www.SynthWorks.com
1-503-590-4787

Expert VHDL Training for Hardware Design and Verification
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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

Symon

Indeed, but it's not the excess turn on current per se that blows the bulb,
it's the thermal stress on the filament which expands as it heats up. The
lower filament resistance at cold means it heats up faster than it would if
the resistance stayed constant. There are stories of light bulbs in fire
stations that are permanently on that last for many decades.
check out http://www.firehouse.com/news/2001/6/11_bulb.html . 100 years and
still going.
cheers, Syms.
 

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