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On one of my early morning walks, I spotted a broken solar powered path light, which had been tossed out into the weeds near a railroad track. I picked up the thing and put it in my pocket. I thought some of the parts may come in handy. When I got home and I took a closer look at the thing, I noticed something strange. As shown below, at first glance, the 1.75 inch by 1.75 inch photovoltaic panel looked like many of the other devices I’ve seen on solar powered path lights. It looks like four overlapping solar cells with their shinny metallic collector strips.

However, when I looked at the thing a bit closer, I saw that instead of filling in the 3 square inch available area with solar cells, this unit cheated by using four skinny 0.13 inch x 1.6 inch cells as shown below. This yielded an active area of only 0.83 square inches. This corresponded to a fill factor of only 28%.

I measured the open circuit voltage at 2.3v and measured the short circuit current at 37ma in direct sunlight. I started tracing out the circuit used in this path light assembly. It appears that they used a 1N5819 schottlky diode between the solar panel and a single 1.2v NiMH rechargeable AA cell. The 1.2 volts from the single cell is too low to power any LED directly, so they used a transistorized (no ICs) DC to DC converter, to boost the voltage high enough to drive a single, most likely yellow, LED. During sunlight charging, the four solar cells act as a constant current source, feeding a maximum 37ma of current to the single battery cell. Other solar powered path lights use a higher solar panel voltage, which can charge two AA cells. The advantage of two cells is that the LED driver circuit is nothing but a current limiting resistor connected to a simple circuit, which turns off the LED when there is sunlight. This is typically done by monitoring the voltage from the solar cells. The thing I picked up seemed to be a cheap knock-off. But, would it work?

Let’s make a few assumptions. Let’s say that this thing was placed in a sunny area and got a full 6 hours of sunlight during the day. With a current of 37ma, the current pumped into the rechargeable battery would be 37 x 6 or 222 milliamp-hours into a 1.2v battery cell. That is 1.2 volts times 0.22 amps for 0.27 watt-hours of energy. Now, let’s assume that the yellow LED had a forward voltage drop of 2 volts at 20ma and was driven by this battery. We will not even factor in the losses of the DC to DC converter or the battery charging losses. Two volts times 0.02 amps is 0.04 watts. If we divide 0.27 watt-hours by 0.04 watts, we get 6.75 hours. That means that the energy during those 6 hours of sunlight would be enough to keep the path light lit for about 6 hours at night. At least here in Colorado, that would not be long enough to make it through the night. If the sun went down at 5:00 in the afternoon, the LED path light would turn on but would be off again by 11:00 PM. The battery would be dead. Maybe they drive the LED with less current to extend the operating time. But, overall, I think the design stinks.

Now, suppose that they used larger full size cells. A well designed photovoltaic panel should have a fill factor of about 90%. If we were only getting about 37 milliamps of charge current from the skinny cells, then full size cells would generate about 120 milliamps. By using full size solar cells, this would mean the path light would operate for a good 20 hours. That would be more than enough to make it through the night with some spare energy, to compensate for overcast skies.

Overall, I would say that this light I found was a bad design. The manufacturer was trying to save money by using only one rechargeable cell, and a very small solar cell area. Sure, the light lights but not for very long.

The next time you are shopping for path lights, look at the solar panel and only buy units that have full size solar cells.

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David Johnson, PE
www.discovercircuits.com