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NNTP-Posting-Date: Thu, 28 Nov 2024 04:47:37 +0000
Subject: Re: Anybody Seen a Simple LED "Fail-Over" Circuit ?
Newsgroups: comp.os.linux.misc
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From: "186282@ud0s4.net" <186283@ud0s4.net>
Organization: wokiesux
Date: Wed, 27 Nov 2024 23:47:36 -0500
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On 11/27/24 1:47 AM, Rich wrote:
> 186282@ud0s4.net <186283@ud0s4.net> wrote:
>> On 11/26/24 8:34 AM, Rich wrote:
>>> 186282@ud0s4.net <186283@ud0s4.net> wrote:
>>>> Critical Redundancy - One LED fails, another takes over ?
>>>>
>>>> Consider traffic lights, warning lights, similar.
>>>>
>>>> It's not as simple as dividing the drive current in half because LED
>>>> brightness is not strictly linear to the current.
>>>
>>> LED's are, at a low level, 'current' responsive lights.  Driving them
>>> with a current source is the best way to drive them.
>>
>>    Agreed ... but it's extremely COMMON to voltage or PWM them.
> 
> And because all the other Lemmings are walking off the cliff, that
> means you should too?


   Depends on the cost/benefit ratio  :-)

   If you can get the Desired Effect with two parts, or 20,
   which do you choose ? That last one-percent of 'perfection'
   may NOT be worth it.

   It has been said of good automobiles that the Germans
   achieve 'perfection' through deep complexity whereas
   the Japanese achieve it through simplicity - reducing
   things to their most basic elements.

   It's why RAM chips don't cost $10,000 .....

> You asked for 'robust' (albeit in combination with other factors).
> Constant current drive will provide the most robust (from an LED
> failure standpoint, but adding constant current drive brings in
> 'robustness' for the driver).
> 
>>> The simplest (if you can assume the upstream power supply will be
>>> functional [1]) is to drive each in parallel with their own current source
>>> (fixed current driver). I.e.:
>>>
>>>             PSU
>>>              |
>>>      +-------+-------+
>>>      |               |
>>>    driver          driver
>>>      |               |
>>>     LED             LED
>>>      |               |
>>>      +-------+-------+
>>>              
>>>             Gnd
>>>
>>>
>>> Then if one led (or its driver) fails, the other continues to operate,
>>> because it does not depend upon the first one.
>>
>>    Yep - except for a couple of things. First off, brightness
>>    goes down by 50% when an LED dies. In outdoor apps you may
>>    not even be able to see it clearly. Second, you're burning
>>    both LEDs - meaning LED-2 may also be near fail-time.
> 
> No different than if you resistor limited or pwm'ed both.  If both are
> lit at the same time, and one fails (and its failure does not take out
> the other) you get 50% reduction in brightness.

   My worry is that 50% less - in bright daylight - may
   equate to "invisible". Not every bit of tech functions
   in some darkened lab corner .....

> Now, if you meant #2 was an idle spare waiting for #1 to fail before
> turning on, well, then in that case, assuming the 'detection and
> failover' circuit operated properly, no drop in brightness, and no
> operating age on #2.  Which you want is dependent on /what/ you really
> want, and your initial post is ambigious enough that either of "run both,
> but keep one going if the other fails" or "hold an idle spare off, turn
> it on if main goes out" can fit the description.
> 
>>> Most LED's that fail do so because they are being driven hard [2]
>>> (right at the limits that they are rated for, if not well beyond
>>> sometimes).  If you derate your drive by a fair amount you'll find
>>> they do, in fact, appear to last nearly forever.  But then you will
>>> need more LED's for an equivalent amount of lumens of light output.
>>
>>
>>    Derating is most wise.  Even the recommended power levels are often
>>    'optimistic'.
> 
> Yes, even the value in the datasheet (assuming a part for which you can
> get a datasheet, and that includes 'recommended operating values' is
> often optimistic.  Esp. for white LED's used for general illumination.


   Yep, decidedly noticed THAT.


>>    Of course if you derate then you have to use bigger/more LEDs.
> 
> It is very hard to have your cake and eat it too.  You can have few
> parts (i.e.  Shenzen like cheapo designs) but you very likely won't be
> terribly obust against failure.  Or you can add more parts for more
> robustness and longer lifespan, but then you won't have fewer parts.


   But I want my cake !!!


>>    Also, LEDs can Just Die for no immediately obvious reason - bad
>>    manufacturing or maybe a nearby lightning hit.  MTBF is a "mean"
>>    after all.
> 
> True, and if the device takes a lightning hit (or nearby one) it is
> likely going to fail.  But you'll find if you derate a fair amount (and
> provide proper adequate cooling) that once you shake out the infant
> mortality portion of the bathtub curve, that the ones that make it past
> infantcy will run a very long time afterward.

   I've seen electronics ruined by a NEARBY lightning
   hit, one that went overhead, never touched the
   building or power grid. Pure EMP.

   Such effects CAN sometimes be moderated by using nothing
   but a zener ... the amps are ultra-small, it's the volts
   that kill the semiconductors.

>>> [1] redundant PSU's are a different matter
>>>
>>> [2] And they are being driven hard because the Shenzen engineer
>>> optimized for lowest BOM cost posible without regard to lifespan of
>>> the device.
>>
>>    The simplest thing I can think of starts off with just
>>    the current-limiting resistor and the LED. If the LED
>>    is working properly the voltage at its + terminal will
>>    be rather low,
> 
> More correctly, it will be whatever the LED's forward voltage drop
> value is, which is different depending what "color" LED is being used.

   Yep.

> But "low" is relative, and depends upon supply voltage.  Some LED's
> have forward voltage drops of 2.5v or 3v.  On a 3.3v supply, 2.5v and
> 3v are not at all "low".

   Well, gotta do SOME calx :-)

>>    the LED is sucking-up most of the power.  If you bias an FET and
>>    attach it to said + terminal then so long as the voltage there is
>>    low the FET won't turn on.  If the LED dies then the high voltage
>>    will turn on the FET - which is attached to LED-2.  COULD latch the
>>    FET so it fer-sure goes to 100%
> 
> Yeah, you could setup a suitably biased FET to turn on if the voltage
> across the LED goes too high -- indicating an open in the LED.  That
> won't catch an LED that fails short however.  So you'd only catch half
> the possible failure modes.  Provided you know your LED's always fail
> open, it would work.  But do you know they always fail open?
> 
>>    HAVE seen an LED or two fail mostly as a dead short ...  but almost
>>    never.
> 
> Diodes failing short do occur (think bridge rectifier diodes that do
> sometimes fail short).  Granted, they are different "chip dopings" than
> LED's, but an LED is just a 'special diode'.  There's no guarantee a
> given LED will always fail open.  It may be only 1% that fail short,
> but if you are talking sufficient numbers of units even 1% becomes a
> rather large physical count.

   IMHO here, you design for the "most likely" cases and
   trust to luck. Not "ideal", but simple and inexpensive.
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