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I am converting some solar LED garden lights to emit infrared light for a raspberry pi critter camera to see what kind of wildlife trifles through my garbage at night.

I just have one question.

These solar garden lights have a small 3v solar panel charging a AAA battery during the day.

How can a single AAA battery with just 1.2 volts power an LED?

When I take it out of the circuit board it fails to work.

There are no visible joule thief's or any other work arounds.

Ideas?

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    \$\begingroup\$ make a schematic of the circuit you have. That said, your probably confusing an inductor for a resistor. All these circuits work on a basic inductor dual transistor pair \$\endgroup\$
    – Passerby
    Commented Nov 30, 2014 at 9:25
  • \$\begingroup\$ Infrared LED needs less voltage than visible LED. For example, this one mouser.com/ds/2/311/… works with 1.25 V, and could be connected directly to the AAA without resistor, or preferably with a very small one. \$\endgroup\$
    – Roger C.
    Commented Nov 30, 2014 at 10:42

3 Answers 3

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There are two pieces that work together in order to light the LED.

The first is an IC such as the QX5251. This is a solar battery charger and LED driver in a TO-94 package. It rectifies the solar input as well as detects when it is dark in order to activate the LED.

The second is an inductor which the driver uses to boost the voltage from the battery high enough to light the LED. The driver uses the same mechanism as a boost regulator but without any of the filtering.

The result is an LED that is imperceptibly flickering. This flickering may interfere with your vision attempts, so do plenty of testing before deploying it in the wild.

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The 3V Solar Panel

A "3V Solar Panel" is the open-circuit voltage. In practice, the battery (1.2V) and the solar panel (3V, open circuit) will have to agree on where to operate. This is call the Q-point (Q = quiescent, latin for "being still" = equilibrium).

In the case of a weak panel and a comparative strong battery, the battery will win. The panel will be forced to operate at the battery voltage, driving current into the battery, charging it.

There will always be some form of directional element (usually a diode) between them to prevent the battery from "charging" the solar panel when the panel is insufficiently insolated (illuminated by the sun). This directional element will add a voltage drop further allowing the solar panel voltage to be closer to it's optimum power output point (typically about 60-80% of the open circuit voltage).

Here is data from a paper I wrote showing the location of the maximum power point for various levels of insolation on a small polycrystalline silicon solar panel.

enter image description here source: V. Raghunathan, A. Kansal, J. Hsu, J. Friedman, and M. Srivastava, "Design Considerations for Solar Energy Harvesting Wireless Embedded Systems"

The 1.2V LED

As you are, correctly, suggesting, it is very unlikely that the circuit drives the LED directly. You don't mention the LED's illumination color, but here's a broad overview of the various color chemistries:

enter image description here

As you can see, there isn't much output for any color-type at 1.2V. There is likely either a capacitive or inductive boost circuit in play that you simply haven't identified yet.

Further help

Post a schematic or, failing that, at a least several pictures of the device and we can try to help you identify the circuit you are dealing with.

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I think here here the solution

https://www.youtube.com/watch?v=VG-DKuSjo9k

it basically converts, with a yx801 or similar, voltage (2V) from the solar panel to higher voltage ( 3,3 ) -> DC current to AC current

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  • \$\begingroup\$ Can you add some info to this post so that it's still useful when the video gets deleted/removed/etc? \$\endgroup\$
    – Greg d'Eon
    Commented Mar 29, 2015 at 15:59

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