7
\$\begingroup\$

I was measuring voltages in my circuit, and noticed something strange. I have an array of 8 LEDs. Each of them is in series with a 220 Ohms resistor. The circuit has a 5v source.

schematic

simulate this circuit – Schematic created using CircuitLab

The LEDs are all the same model. They all have ~1.90v across them. All but one, which has ~2.1v. At the same time, that LED has an inferior brightness. I can interchange this LED with another anywhere in the circuit, and it still behaves the same.

Is it normal / usual ? Does it mean I burnt it somehow or made a mistake in my circuit ? What's with the inferior brightness when it drops a higher voltage ?

\$\endgroup\$
  • \$\begingroup\$ When you change the lower brightness LED with one of the higher brightness LEDs, how does it behave then? Try to find out which part is 'different' by exchanging positions for several components, ultimately identifying where the difference is. It may be in your wiring too. \$\endgroup\$ – jippie May 1 '13 at 6:46
  • \$\begingroup\$ I can swap LEDs, it's always the same that has the lowest brightness. I also tried replacing it with a new one; in that case the new LED drops 1.9v just like any other. I'm pretty sure it's really the LED that causes this, but maybe there's something I can't see. \$\endgroup\$ – Antoine_935 May 1 '13 at 9:20
4
\$\begingroup\$

If there is a higher voltage across that LED, than it is drawing less current because the voltage across the 220 ohm resistor is less (5- 2.1 or 2.9 volts compared to 5 - 1.9 or 3.1 volts across the other 220 ohm resistors. Thus it is drawing 2.9/220 or 13.2 ma while the others are drawing 3.1/220 or 14.1 ma. Generally, LED brightness varies in step with the current through it. That being said, the difference between 14.1 and 13.2 is not much so that the brightness difference shouldn't be that much. Do you have the specs on the LED? It may be perfectly normal for that particular LED to have that variation of voltage drop. I don't believe you damaged the LED nor have any circuit errors.

\$\endgroup\$
  • \$\begingroup\$ I don't have the exact specifications of the LEDs sadly, but these are normal looking 5mm, orange color. But I guess not much can be inferred from these vague specs... \$\endgroup\$ – Antoine_935 Apr 30 '13 at 22:12
3
\$\begingroup\$

Variations in manufacturing can lead to parts that act slightly different. Component matching cost more, and even then, a given batch is not tested 100%. A statistically significant portion of the batch is selected, pass/failed based on target specs, and then labeled according to actual performance.

This happens with pretty much all types of devices. Some, like resistors, are more obvious. They come with certain tolerances, the average resistor being within ± 10% of it's given value (so your 220Ω resistors might be between 198Ω and 244Ω), with more expensive ones coming in ± 5/1/0.1%. Leds are just as likely. You will find that led drivers (Like the TI TLC5940) have options just for this situation, called Dot Correction. From App Note: TLC5940 dot correction compensates for variations in LED brightness

Dot correction is a method for managing pixel brightness by adjusting the current supplied through each individual LED in the array. The dot correction feature enables the processor to control full current to a panel of LEDs while the LED driver scales the current to each LED and creates uniform brightness. ... To implement dot correction, manufacturers measure the brightness of individual LEDs through photo capture. The dimmest LED in the system is designated as the “base” LED to which every other pixel is matched.

That different led could be a outlier in a single batch, come from a different batch, come from a different factory or come from a completely different manufacturer. Part matching needs to be done if a specific brightness is necessary. Binning is one method (Making sure that the Manufacturer gives you only leds that are paired) but it is not always the best in high led count. Dot Correction is the next method. And simply eyeballing it is the next. That App Note has some good information on this.

\$\endgroup\$

protected by W5VO May 1 '13 at 3:02

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.