So, I have these garden lights powered by solar cells. I also happen to have a power-LED over from a broken flashlight, I think its a Cree R2 or Q5 or similar. These lights charge a 1.5V 600mah Nimh battery during the day, and then light up these rectangular low-power leds.

Now I would want to experiment with simply driving a high-power LED at similar, or slightly lower power levels, meaning a very very low current. I have not measured the drive current for these, but I imagine its well below 50ma. The theory being that it will be brighter still because of the excellent efficiency of power LEDs in this power band.

I know power leds generally require current limiting or control, and that the battery voltage is not enough so I will need some sort of boost circuit if not already present.

What I am wondering is what is the most efficient way of doing this, without being overly complicated? Is it possible to do this with just a resistor, or am I still prone to current run-away at these low levels?

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    \$\begingroup\$ I think your 2nd to last paragraph needs revision. If (since) 1.5V is not enough to drive high-power white LED, you'll require a boost circuit. A buck converter lowers the voltage, a boost converter increases it. If you were using "buck circuit" to mean a generic DC-DC LED driver, you may want to clarify this. \$\endgroup\$ – Kevin Vermeer May 3 '11 at 14:25
  • \$\begingroup\$ @reemrevnivek: Out of curiosity, are there any good flyback-mode LED drivers that can operate independent of whether the supply is above or below the LED voltage? I have a project where the LED voltage is about 5 volts and the supply is about 4-15 (depending upon battery configuration and condition). At present, the LED driver shares the main supply switcher (which converts to 3.3 volts) so it always gets 3.3 volts, but it would be nicer if it could run directly off the raw supply. \$\endgroup\$ – supercat May 3 '11 at 15:49
  • \$\begingroup\$ @supercat - I've not used any personally, but I'm sure there are. Flyback is almost always a buck-boost configuration. See this National article, specifically the Flyback/Buck-Boost/Boost ICs heading, where they explain that a flyback buck-boost converter is essentially the same as a generic boost converter. Here's an example of a Linear Tech part for this purpose. But we're getting off topic. \$\endgroup\$ – Kevin Vermeer May 3 '11 at 16:13
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    \$\begingroup\$ @reemrevnivek: In some applications, a flyback and boost converter would be similar, but LED drivers I've seen use a low-side current sensing input which they would expect to measure current below the positive rail. The LT part looks interesting, though it requires more external components than some other drivers. As for topicality, the OP's goal seems to be to drive an LED using some type of switcher; having a switching topology which is agnostic to whether the input voltage is above or below the LED forward voltage would seem a useful option. \$\endgroup\$ – supercat May 3 '11 at 18:03

The amount of light LEDs put out has a very linear relationship with the current they are driven with, and the voltage to current graph is relatively constant as they are diodes with a characteristic junction voltage. Taken together, this means that the efficiency can rise about 25% if an LED is driven at 10% of rated brightness (in this LED, but others should be fairly similar).

In any case, regardless of the level you choose to drive the LED at, what you want is a "single cell LED driver". These are usually current-regulated, eliminating the need for any resistors which do nothing for your efficiency, and also have the ability to step-up the voltage, so you can drive a 3 V nominal LED with one or two 1.2 V NiMH cells.

Linear, Micrel, and Diodes Inc all make drivers for this purpose, and you can find many more in a search for "single cell LED driver".

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    \$\begingroup\$ Resistors do a lot for your efficiency - They degrade it by the ratio of the voltage across the resistor to the voltage across the resistor - LED combination! \$\endgroup\$ – Kevin Vermeer May 3 '11 at 14:27
  • \$\begingroup\$ I don't follow that 25% rise in efficiency. In your first statement you say luminance is proportional to current, thus to power, which is correct. Then efficiency is more or less constant. You can't raise it by 25%. \$\endgroup\$ – Federico Russo Aug 22 '13 at 8:31
  • \$\begingroup\$ @FedericoRusso, you're assuming the forward voltage remains fixed--as they aren't ideal diodes, there is a minor dependence with current. \$\endgroup\$ – Nick T Aug 22 '13 at 22:55
  • \$\begingroup\$ Agreed, a minor dependence, but not enough to get that 25% difference in efficiency. It's not like the voltage changing from 2 V to 2.5 V, which is 25%. \$\endgroup\$ – Federico Russo Aug 23 '13 at 16:02
  • \$\begingroup\$ In the datasheet I link it's 15%. Must have been a bit loose with my math a couple years ago. Oops. \$\endgroup\$ – Nick T Aug 23 '13 at 16:44

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