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

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.
   Minimum parts count. For arg here, assume like a 12v or 24v
   source with way more amp capacity than you need. Minimum setup
   is the LED and a current-limiting resistor.

Searches really don't bring up much here.
>
Yea, there are more complex solutions ...  but what can be done with
the fewest, simplest, most robust parts ?

 fewest, simplest, most robust -- you get to pick two....

   Wow - TWO !  :-)

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.

But this is far from 'fewest' parts, as you need one driver per led.
While some driver chips can be had for pennies each in 1K quantities,
that still adds to the BOM cost in the end.
 

Not strictly Linux, but we DO sometimes wanna drive external
displays.  Usenet electronics groups ...  dismal at this point.
>
LEDs are great, but never "forever".  They DO fail - but for some
safety apps you can't just HAVE things go black.

 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'. Of course if you derate then you
   have to use bigger/more LEDs. Also, LEDs can Just Die for
   no immediately obvious reason - bad manufacturing or
   maybe a nearby lightning hit. MTBF is a "mean" after all.

[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, 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%
   HAVE seen an LED or two fail mostly as a dead short ...
   but almost never.