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Just what the title says. An LED series should probably go dead after an LED goes OC. There are strings out there in the market where the series continues to function even though one, or more LED dies.

I'm guessing the series is not a real series but something more complex... How would I achieve something like this? I'm guessing perhaps the series provides multiple paths somehow..

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  • \$\begingroup\$ Don't diodes usually fail open, not shorted? \$\endgroup\$ – Michael Kohne Oct 26 '11 at 11:01
  • \$\begingroup\$ I've seen a TVS Zener fail short. Got hot enough to desolder itself. \$\endgroup\$ – Mike DeSimone Oct 27 '11 at 2:54
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LEDs that fail catastrophically usually fail open circuit but some occasionally fail short circuit.

Large groups of LEDs are often arranged as say X strings of Y LEDs in series in each string.
If no attempt is mde to do otherwise, open-circuit failure of a single LED will result in all Y LEDs in the string going out.

If X strings of Y LEDs are connected in parallel and the whole array is fed with constant current, if one string goes OC then the current wu=ill be shared by the remaining Y-1 strings and current will rise by a factor of Y/(Y-1) eg if there are 3 strings in parallel then o/c failure of one string will increase the current in the other two strings by 50% each. Whether this matters depends on the LEDs and how close they were being run to their rated value.
Modern White (and certain other) LEDS have a very low margin of Imaxmax and Imaxoperating. This may be 1.2:1 or 1.3:1 . Simple circuits can be provided which maintain the string currents if one string fails BUT the safest measure is one current source per series string. This requires as little as eg an LM317 and a sense resistor (see text and diagram below). LM317s can be had for about 50 cents /1 and 8 cents in 10,000 quantities at Digikey.

The excessively keen can wire a component across an LED such that if the LED goes OC the shunting circuit carries its current. eg for White LDS with a Vf operating of say 3.3V then connecting a 3V9 zener across each LED will allow O/C LEDS to be allowed for. As below. This would usually only be done dor upmarket systems. A transistor and resistor divider circuit can perform the same function.

enter image description here


In any serious product with LEDs providing illumination and not just acting as "indicators" the LEDs MUST be driven with constant current to achieve consistent brightness and good lifetimes. A constant current circuit can be very very simple and quite cheap.
In the circuit below
Iled = V/R = 1.25/R
or R = Vled/Iled.

For 3R9 as shown Iled = V/R = 1.25/3.9 = 320 mA.

One LED is shown but as many as desired can be placed in series and will share the same current.

The limiting value is that Vin must supply Vleds + 1.25V drop across Rcurrent_set + regulator dropout voltage (about 2.5V.)

So eg for 5 x 3.3V LEDS Vin needs to be >= Vl + Vrc + Vdo = 5 x 3.3 + 1.25 + 2.5 = 20.25 V minimum.

enter image description here

.

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Most strings of LEDs that I have seen are not a simple series.

Instead you have multiple series of 2* LEDs plus resistors in parallel:

enter image description here

If one LED goes open circuit then the other LED in the same line will go out. If one LED goes short-circuit then the other LED will light up brighter, and will die sooner - if not almost immediately depending on voltages and tolerances etc.

Example:

enter image description here

The exact order of resistor and diodes is irrelevant - in that one the resistor comes between the diodes instead of before them, but that makes no difference at all to it electrically.

*or more depending on voltages in use

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    \$\begingroup\$ Depending on voltage the chains are often longer, as longer chains improve efficiency. \$\endgroup\$ – starblue Oct 26 '11 at 10:23
  • \$\begingroup\$ Such series I've seen didn't exhibit the presence of a resistance at that point. Any other ways it could be done? \$\endgroup\$ – Everyone Oct 26 '11 at 10:46
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    \$\begingroup\$ If you have LEDs you either have a resistor (known as a "Current Limiting Resistor" or a constant current source. The resistor is considerably easier and cheaper than the constant current source. \$\endgroup\$ – Majenko Oct 26 '11 at 10:53
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    \$\begingroup\$ @Everyone - The constant-current source could be embedded in the driver. Many LED drivers have (adjustable, even!) drivers included. However, if you're using a standard shift register or microcontroller IO port, then the resistor is likely to be easier and cheaper, as Majenko noted. A reference to your specific LED strip would make this much more answerable! \$\endgroup\$ – Kevin Vermeer Oct 26 '11 at 11:41
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The best you can do is to connect each LED in parallel in other to make the same voltage to get across each LED.

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  • \$\begingroup\$ thank you for taking the time to try to help, but I think the question may not have been clear to you. Currently, the user believes they have series diodes in a chain but the chain is not failing when one diode fails. \$\endgroup\$ – Kortuk Oct 27 '11 at 11:31

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