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Hack Solar Lights with Dual LEDs to Increase Run Time

This effective but counter-intuitive hack hinges on replacing the LED in your standard solar-powered garden light with two LEDs in order to extend the battery life.

When two LEDs are wired in a circuit they draw less total power but the output of the two slightly dimmed LEDs is virtually indistinguishable a single one. Tests at the project blog BigClive yielded a run time extension of roughly 25%. Hit up the link below to see just how easy it is to crack open a cheap solar light and upgrade the LEDs.

Hack Your Solar Garden Lights [via Hack A Day]

Jason Fitzpatrick is warranty-voiding DIYer and all around geek. When he's not documenting mods and hacks he's doing his best to make sure a generation of college students graduate knowing they should put their pants on one leg at a time and go on to greatness, just like Bruce Dickinson. You can follow him on if you'd like.

  • Published 05/25/12

Comments (4)

  1. Alex

    So…twice the voltage, more than half the current?

    P=IV

    If V becomes 2V and I becomes (1/2)*I that would be the same…I think I’m missing something..

  2. LadyFitzgerald

    Yes, You are. The voltage didn’t change. Putting two LEDs in series increases the resistance, thus lowering the current passing through them. Less current means less battery drain.

    While the amount of light put out by the two LEDs is brighter than I would have expected but I wouldn’t go so far as to say they are virtually indisinguishable. The difference in the areas illuminated in the photos is quite noticeable. However, if the end user feels that even the reduced illumination will meet the user’s needs, then this would be a very worthwhile hack.

    I have no need for those kind of lights so I don’t plan on experimenting (I have other things on my plate begging for attention) but I wonder if the PVs on the lights would be able to charge a larger NiMH (larger capacity, not higher voltage). An LSD NiMH probably wouldn’t be needed but Sanyo Eneloops are noted for their long life (up to 1500 discharge/recharge cycles now).

  3. Anonymous

    @Alex

    You seem to forget that the voltage doesn’t change. V (or E) has not changed, only the “load” (R) has. It might take a bit more current (I) to push “through” the load but according to the author it doesn’t. This might have something to do with the forward voltage bias that, last time I looked, is only 0.7 volts in order for any LED to start emitting light.

    The thing that makes it fishy is that according to Kirchoff”s voltage law (that the sum of all voltage drops in a complete circuit equals the total source voltage) that we keep thinking about LED’s as static resistors. And to some degree, they are. However, these “resistors” really do change in value the more or less of them you add in series. That’s what I think you (and I) keep forgetting.

  4. Adam

    First some simple relevant facts:
    - LED’s are essentially constant voltage devices, which makes them a little tricky to power efficiently. When emmitting light, they will have an essentially constant voltage which is dependent on the type of LED but is typically in the range 1.7V to 4.6V. Only normal (non light emitting) diodes have a 0.7V drop. In a simple battery circuit, you need to use a resister to limit the current, which means the energy being consumed in the resister is lost. If the LED has a 2V drop and you are using a resister and 2×1.5V batteries in series, 33% of the battery energy will be lost to the resister.

    - Alex is on the right track in the sense that by series’ing up the LED’s, the voltage doubles. So on the face of it, if the current isn’t halved, there will be more power consumed and hence greater battery drain.

    - LadyFitzgerald is also on the right track although seemingly saying something completely different (voltage doesn’t change).

    The reason is this is not a simple ohms law problem due to the fact that the power source is a switch mode power supply. A switch mode supply has a couple of good characteristics which is useful here. Firstly, it can vary its voltage output and secondly it’s efficient. Instead of burning power up in variable resisters (e.g. POTs) or voltage control IC’s (which are liked controlled resisters), the voltage is controlled by changing the frequency and using an inductor as an energy storage and recovery device to boost the voltage.

    Every power supply has a source impedance and you get maximum efficiency by matching the load impedance to the source impedance. My guess is the two LED’s in series must have a better matching “impedance” to the supply, although impedance is a more complicated concept when switch mode supplies and non-linear devices like diodes are in circuit.

    By the way, the author is measuring the current on the battery side of the switch-mode supply. The battery will be constant voltage, so reduced current means reduced power consumption and increased battery life, as stated by LadyFitzgerald.

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