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I just got a outdoor solar powered LED lamp to illuminate our house number at night. It is supposed to charge up its battery during the day, and automatically turn the LED on at dusk. The problem I'm discovering is that the solar panel on this lamp doesn't get enough natural light to charge it so that it can power the LED even for a few hours at night. I like the light fixture well enough so I'm thinking of keeping it but giving it some help.

I have a door bell and opener circuit located under the lamp that runs off a transformer. I was thinking about giving the LED lamp some help from this circuit. Here are the specs for the LED lamp and the doorbell circuit:

LED lamp: SMD 5630 LED 0.5 W, 3.2V 1000mAh Li-ion battery

Doorbell transformer output: 16V AC 30VA (but I measure 18.25V AC at the output)

Would this be a complicated exercise? I have some soldering skills but I need help figuring out how to step down the voltage and amperage to feed the light from the doorbell. I'm assuming that with the proper step down I can wire in the power from the doorbell to the same place the little solar panel is attached to augment the same charging circuit in the light?

Any help would be greatly appreciated!

Back of PV and battery (blue and yellow wires from PV, blue and red from battery

Circuit board. The IC is labeled: 8182 4K154

During the day the battery charges from the PV, at dusk the LED switches on and is powered from Battery. Could I maintain the switching capability and just power the LED with appropriate 3.2V power supply? Will the PV be attempting to charge my power supply during the day?! I don't want to cut out the PV since it appears to be the dawn/dusk switch as well...

Back of PV and battery (blue and yellow wires from PV, blue and red from battery

Circuit board. The IC is labeled: 8182 4K154

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  • \$\begingroup\$ Do you want to power the LED only from the transformer? Can you show us the inside of the light fixture so that we have some idea of how it is wired and the circuitry? \$\endgroup\$ – Andrew Morton Oct 11 '16 at 17:31
  • \$\begingroup\$ Thanks for all of the great tips gents! I managed to pull the fixture apart today. I'm posting some photos... \$\endgroup\$ – Soheil Zahedi Oct 12 '16 at 17:58
  • \$\begingroup\$ I think that powering the LED from the transformer is probably the right thing to do. Cutting out the PV and battery probably makes sense. I could also use a different transformer since I've found the AC mains wires in the wall and could just wire up a different transformer that would suit this light. It would be nice to be able to keep the photovoltaic switch aspects of this lamp since it goes on nicely at dusk and isn't just on all the time. \$\endgroup\$ – Soheil Zahedi Oct 12 '16 at 18:17
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I picked up a bunch of doorbell transformers (just for my junkbox) at $1 each, a while back. About the same as you describe. I'd probably use a bridge rectifier followed by a ripple capacitor to get the basic DC part done in preparation for an LM2596 buck converter. (You can get these for under a dollar on ebay, if you don't want a fancy voltage and current display, and they can easily handle your LED requirements. Most include potentiometers to use in setting the voltage and current limit, too.) Once you get the voltage into the right vicinity (say 5 V), you can then drop the rest of the voltage with a current limiting resistor in the usual way.

The first thing to do is to test your transformer under load. Just to make sure it works, mostly. I use one of those big 25 W resistors -- \$100\:\Omega\$ if you have one around? Load it down and see what it does. Chances are, it's fine though. But I like to know what the transformer does under load just to be sure.

But that's what I'd probably do (partly because I have a small box of buck converter boards laying around, too.)

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  • \$\begingroup\$ Many of the DC-DC LM2596 buck converters have a constant current mode, so no resistor would be needed. Just set the current for 150mA. I would also replace the LED with a brighter one. \$\endgroup\$ – Misunderstood Feb 25 '17 at 17:43
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Bare minimum, a diode ac full wave rectifier, maybe a step down regulator and a lipo charger circuit. The solar and battery part of the lamp is likely just a solar cell, a diode and the battery pack. Dumb and depends on the low output of the solar cell, matching the charged voltage of the battery.

The LED half has a LDR circuit as part of it that turns on automatically, so no extra modification is really needed. At that point though, you might as well ditch the solar cell and battery and directly wire it to the AC wire. Or run a wire for low voltage outdoor lighting which is often 12V DC.

Highly depends on the actual wiring and circuitry of your lamp.

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  • Assuming 12h LED duration, current is ~156mA (0.5W/3.2V)
  • thus 1900 mAh is needed (156mA*12h)
  • you are getting a few hours with a 1000 mA LiPo battery charged only to 3.2V
  • thus not getting full capacity of up to 6h (minus series losses from 3.7 to 3.2 or ~14% (0.5/3.7)

    • If 16Vac Tfmr is unloaded, it is reading 14% higher due rated V at 86% efficiency.
    • the transformer can supply <2A @30VA @16Vac only to a linear load and not to a bridge rectified battery but a series R can make it more linear yet lossy with < 15% efficiency 3.5/24V in powering the LED.
    • thus another wire pair from a full bridge diode at transformer or single wire to bridge and series R at load must be wired to charge the battery using current load for expected voltage R drop.

    • Using 16Vdc avg to 3.2Vdc LEDs with up to 156mA

    • with no PV, R=82 Ω drawing 2W (12.5*0.156A)
    • however when LED is off, Vbat continues to charge up during day time with no overvoltage protection.

Thus you can leave LED on all the time, and choose a 5W resistor for cooler operation or buy a proper supply or add more complexity to regulate LiPo charger.

Unfortunately both your PV and storage battery are under-designed to meet the load requirements.

One simple solution is power the battery and LED from a 3.3V source such as tap into an ATX PSU with wire pair to series Sch. diode to LiPo, or buy one ~3.2Vdc >150mA wall supply to feed circuit.

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  • \$\begingroup\$ Regarding the battery charging, if the OP replaced the PV panel with a DC supply of the appropriate voltage fed from the existing doorbell transformer, wouldn't the fixture's battery charging circuit take care of that, and provide a few hours of mains failure backup? \$\endgroup\$ – Andrew Morton Oct 11 '16 at 18:01
  • \$\begingroup\$ One must know the effects of leakage current on doorbell in series. If present system provides only few hrs or 25% of Ah need, then what does 120mA do ? activate doorbell or just heat up coil with 1.5W of heat (75%*156mA* (16-3.2)V) \$\endgroup\$ – Sunnyskyguy EE75 Oct 11 '16 at 18:07
  • \$\begingroup\$ In the late 70's, I illuminated my doorbell switch with an LED and series diode +R but that had no effect on bell coil due to 30mA half wave \$\endgroup\$ – Sunnyskyguy EE75 Oct 11 '16 at 18:12
  • \$\begingroup\$ Isn't that back when LEDs were red, and they were tough enough to wire in series with a capacitor across the mains without a reverse voltage protection diode? I hid behind something the first time switching on my PSU after changing its neon to an LED + C, just in case. \$\endgroup\$ – Andrew Morton Oct 11 '16 at 18:26
  • \$\begingroup\$ yes GaAs LEDs had much lower leakage currents<<1uA and higher PIV could be achieved but not suggested, ... with aging mine lasted 5yrs, your C acted impedance divider and LED as a DC clamp with diode response, not good practice but ... \$\endgroup\$ – Sunnyskyguy EE75 Oct 11 '16 at 18:42

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