I have twenty LEDs that I want to light up. I planned out a few possible circuits, which I describe below along with a few questions. Would any of them not work, and why? Anything else I should be taking into consideration but haven't? The LED has the following specs:

Vmin: 2.8V
Vmax: 3.5V
If: 20 mA

From what I understand, the Vmin is essentially the same as Vd in this diagram, so the current will be 0 until it reaches that voltage. Is this correct?

If the LEDs are connected in series, that means that I'll need 20*2.8V=56V in total to get current passing through every LED. I calculated the approximate amount of power necessary (60V*0.02A=1.2W) and compared it to the amount that can be provided by a typical AA battery, which I've been told usually provide around 800mA of current (1.5V*0.8A=1.2W). Seeing as they're the same, I'm pretty sure it'll be possible to step up from a single AA to what's needed to light the LEDs. So far, I know of the SEPIC, which I'm reading up on right now. Is there a better way of accomplishing this? Possibly without needing inductors?

Another solution I've considered is connecting them in parallel, which I recall reading somewhere that it's a bad idea unless there's a resistor in series with each of the LEDs. Two AA batteries in series would be enough in this case, providing a 3.0V potential across each branch. Since the LED takes 2.8V, the resistor would take up the remaining 0.2V, and to get a 20mA current, I'll need a (0.2V/0.02A=10ohms) 10 Ohm resistor.

I could also plug it into an adapter which converts the AC electricity from the outlet to 15V (the current rating on the adapter means the maximum it can give and not the amount it'll always give, right?). In this case, I could separate the LEDs into four groups of five (which is convenient given their layout), each LED in a group of five is connected in parallel to each other, and the four groups are connected in series. That would give a voltage of (15V/4) 3.75V across each group of five, and since the LEDs in the group are in parallel, they'll each get 3.75V too. The current needed would then be (0.02A*5) 100mA. The voltage across the resistor in series with the LEDs is (3.75V-2.8V) 0.95V. The resistance needed would then be (0.95V/0.02A) about 50 ohms.


4 Answers 4


The most efficient method to do this would be with a boost converter, preferably one that has current control. Hundreds of LED drivers which are essentially modified power ICs are available nowadays, including ones that operate from single alkaline cells down to 0.9 V e.g. the Micrel MIC2282 (for more look up "single cell led driver") which should be able to drive up to 33 V strings (~8 LEDs), though asking it do that from a single cell may be pushing it

In a previous question, I did actually recommend a power IC, the Linear LT1618 (not outwardly marketed as an LED driver, but an application schematic showed it as one)

Whatever you choose, powering (20) 20 mA LEDs from a single alkaline cell is probably not feasible just because the voltage will dip so low with that load it will undervolt most controllers. If you could use two, you'd probably be in business.


56 V is inconveniently high, and in reality, your 20 * 2.8 V will vary as the LEDs heat up. You could do it with a switching voltage converter, but it's not the easiest way.

Your plan of putting many LEDs in parallel is good as long as you include the resistors you describe. As the battery discharges, the voltage will sag, and the LEDs will dim.

I'd start with one LED and get the resistor right so you get the brightness you want. Your 10 ohm calculation is a good start. Then add LED/resistor pairs in parallel. If the battery voltage sags, add more batteries in parallel.

I don't quite understand your last plan-- maybe provide a diagram? There is likely a parallel combination of series LED strings that will work. You'll definitely need a resistor in each string.


There are specialized ICs on the market that act like constant-current sinks, ensuring uniform LED brightness even with varying power supplies.

I would suggest going with the 15V adapter and driver ICs for the best performance with minimal headache.

Your maths seem to make sense at first glance. Also, a boost converter is most likely a better choice than a SEPIC in terms of simplicity. I don't think charge-pump (inductorless) voltage boosting would work well in this sort of application.


I haven't tried this myself, but I read an application note saying it is practicable to connect strings of LEDs in parallel and omit all those load-sharing resistors providing there are AT LEAST 4 LEDs in each string. It said that, when connected like this, LEDs tend to naturally share the current. (Of course, we are talking about LEDs of the same type.) So I understand them to be saying that if you had, say, 6 parallel strings, and each string comprised 4 LEDs in series, then if you pump 180mA through the lot, it will pretty well divide itself with ~30mA through each string. Easy enough to test this, especially if you include a small test resistor in series with each string and use it for current-sensing. If the currents are almost identical, you can remove the resistors! Careful with CURRENT sources though, if you really push the boundaries and any string goes open-circuit, then all of its current will divide as extra current through the other strings, maybe in turn burning them out! The more strings you have, the lesser the risk. (I'd only do this for cheap LEDs, not the $40 room-lighting LEDs!)

As every hobbyist knows, specialized ICs and modules are notoriously hard to get hold of in small quantities. But a boost module I know to be available by mail order is the BoostPuk 4019 from LUXdrive. It won't work down to single AA cell voltages, but because of this thread's general subject title, hobbyists with a similar need are likely to be reading this. Check out the LUXdrive 4019 and other handy driver modules here: http://www.ledsupply.com/led-drivers.php

  • \$\begingroup\$ Just a warning: if a string fails open-circuit, as you wrote, you're "lucky". But if one of the LEDs fails short-circuit (and it happens!) all LEDs of his string will become a bunch of SED and friodes... :-) \$\endgroup\$
    – Axeman
    Commented Aug 14, 2012 at 7:40

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