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I building multiple parallel strings of 4 LEDs in series.

Since the four LEDs in series have a combined forward voltage of 12.56V at 20mA, and it's powered by 2 AA batteries, I have added a voltage booster.

Now my question is, since I can both choose the output voltage of the boost converter, and the resistor added to each string of 4 LEDs, what would be a reasonable value for both?

I could set the voltage to exactly 12.56V and not use a resistor, but that would probably lead to uneven current spread between the sets of 4 LEDs, as their forward voltage can differ slightly due to manufacturing differences.

On the other hand, I don't want to set the voltage too high, and wast a lot of power.

So what would be an acceptable voltage drop over a current limiting resistor?

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  • I think the The answer depends the on what you mean exactly by a "lot of power" (to work out the LED resistor value) and how bright (very bright in daylight, ok to be visible in dark only?) you want the LED to be.
    – RS2322016
    Commented Sep 19, 2016 at 20:07
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    @qwerty10: Brightness is dependent on current, which has been fixed at 20mA. The only power consideration (for this question) is power loss in the ballast resistor. Commented Sep 19, 2016 at 20:08
  • @Ignacio Vazquez-Abrams I didnt notice it said 20ma. But I do now. I guess it could be 19.5ma rounded up to 20ma.
    – RS2322016
    Commented Sep 19, 2016 at 20:21

3 Answers 3

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Since the relation between voltage and power is quadratic, the obvious answer is "as little as possible".

Allowing for 3% variance in VF gives us just under 13V, so that is a reasonable starting point. Solving for R gives 22ohms, a E24 value. This will result in a 8.8mW loss, which even a 0201 resistor can handle.

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  • Thank you for your explanation. I was already leaning towards 22 ohm as a somewhat wild guess. 0201 is still a bit too small for me, but I plan on ditching the 0805 for 0603 pretty soon. 0805 is quite big if your want to create small PCBs.
    – Gerben
    Commented Sep 19, 2016 at 20:32
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I could set the voltage to exactly 12.56V and not use a resistor, but that would probably lead to uneven current spread between the sets of 4 leds, as their forward voltage can differ slightly due to manufacturing differences.

While that is a completely true statement, by having multiple LEDs in a chain you effectively average out the forward voltages. The more LEDs you have in your chain the more "average" each chain gets.

It's still not good to rely on the forward voltage to set the current, since it also varies with temperature (as the LEDs heat up the current will rise, leading to more heat and therefore more current, until you get thermal runaway).

As Ignacio says, the less the better when it comes to the resistance, and has recommended 22Ω and 13V, which I would recommend too.

Another option is to use a low-side constant current sink on each chain. This will guarantee that the current is exactly what you want (within reason) with no thermal drift in either the LEDs or the resistor.

You can find some example circuits here: http://www.radio-electronics.com/info/circuits/transistor/active-constant-current-source.php

Or you can use constant-current sink LED driver chips, which also often gives you the advantage of being able to control each chain of LEDs individually - either on/off, or PWM.

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Using a boost converter and a resistor to limit current seems rather inefficient. The resistor is used to approximate a constant current source, but it would be far more efficient to use a switch mode supply to generate a constant current.

There are dedicated LED controllers which do this, although usually used with higher power LED e.g. CREE LED.

It may be possible to modify the feedback circuit in the boost converter to regulate the current, monitored with a small series resistor.

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    The issue is that there are multiple parallel strings, which would mean that a) ballast resistors would be required regardless, and b) failure of a single string will cascade to the other strings. Commented Sep 20, 2016 at 4:16

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