Re: Anybody Seen a Simple LED "Fail-Over" Circuit ?

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?

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.

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.

  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.

  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.

[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.

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".

  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.