Re: Using a single resistor with several LEDs

When using an LED from a power source whose voltage is higher than the

>rating of the LED, a resistor is typically used.

Actually you still need the resistor even when the voltage is at the rating of the LED albeit a small one. It serves the purpose of current limitation in addition to dropping voltage.

If more than one LED is to be used (let's say three), >each one would have its own respective resistor, >and the schematic would look something like this ... > > > +----------+-----------+-----------+-------------- > | | | | > | | | | > | LED-1 LED-2 LED-3 >power | | | >source | | | > | resistor-1 resistor-2 resistor-3 > | | | | > | | | | > +----------+-----------+-----------+--------------

Right. It's the correct way to get uniform brightness among the set.

>But if all the LEDs were of the same voltage and amperage >(e.g., if they were all of the typical 3.6V 20mA), >isn't there a simpler way to do this by using only ONE resistor? >In other words, can't it be done like the following ? > > > +--------------+-----------+-----------+--- > | | | | > | | | | >power LED-1 LED-2 LED-3 >source | | | > | | | | > | | | | > +---resistor---+-----------+-----------+---

It serves the safety purpose, but then you get variability on the brightness because each LED will not draw exactly the same current. So some will be dimmer than others.

The second thing is that the total current through the resistor will be a factor of the current through each LED. So for your example the resistor above would need to allow 60ma of current to flow through to feed the LEDs downline. The problem is that if one LED blows out then that current will have to be pushed through the other working LEDs, which can cause the others to fail also.

And what would be the proper value of the resistor?

Single resistor value divided by the number of LEDs across it.

>(I'm guessing it would be one third of what it was in the first diagram >because the required voltage reduction would be the same but the >effective current draw of the 3-LED assembly would be three times what it >was before. >So if the power source were 6V and the LEDs were 3.6V and 20mA each, >then the required resistance would be 120 ohms in the first case >and 40 ohms in the second case. Correct?)

Right. But see the caveat above.

A better way to do it is to bump the voltage up and put the LEDs in parallel like this:

+V -> resistor -> LED1 -> LED2 -> LED3 -> GND

You treat the equation the same as a single LED except that you add the Vf voltages of the LEDs. So for your example the total voltage of the LEDs would be 10.8V. So the resistor you'd use with a 12V +V would be:

(12V - 10.8V) / .02A = 60 ohms.

But there are significant differences:

1) Each LED will draw the same current, so you'll get uniform brightness. 2) If a single LED (or the resistor) fails, then the whole string goes out with only the single failed component, instead of going into cascade failure due to increasing overload.

When doing a single resistor, that's the better way to handle it.

BTW, thanks for crossposting this message. There are many who don't understand the concept. But I think I'll limit my replies to basics and misc, which are appropriate.

BAJ

Reply to
Byron A Jeff
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parallel

That is NOT parallel, that is SERIES.

[snip]
Reply to
Watson A.Name - "Watt Sun, th

...

Wrong answer. There is no "proper" value for this resistor, because it's just plain wrong. It _can_ be done, if you want to spend a day or so matching LEDs, but any perturbation in LED current will gave a positive feedback effect, so the LEDs _will_ fail - it's only a matter of time.

Put a resistor in series with each, or with a series string.

Figuring out the resistor value is left as an exercise for the reader. :-)

Cheers! Rich

Reply to
Rich Grise

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