Hi team,

I have discovered this is a somewhat complex task, I thought it would be easy. I have a bunch of Solar LED string lights (multiple sets), and all the batteries are now dead after a year or so of use. Rather than replace the batteries, I thought why not run them off a dc wall wart as I have a plug nearby.

Using one set as my example, it has 30 bulbs in parallel, the current draw is 16mah which I believe I need to times by 30 equalling 0.48 mah. The battery is 1.2v 600mah, with an 8 pin unmarked chip doing voltage boosting, I am taking a somewhat educated guess the pulsing voltage is 3.0v but I can't say for sure.

I would really like to keep the features of the solar module - auto on/off and the different flashing modes. So my plan was to attach the DC mains supply to the battery terminals rather than to the led's.

I may be way off here, but what I have thought was to best replicate the battery was use a 5v, 1.2a wall wart, with a resistor in series and connect it to the battery terminal, I am aware the led's will take all the current that's on offer to them so to avoid burning anything out I have calculated like so:

30 x 0.016 = 0.48 mah
5v - 1.2v = 3.8v drop
3.8 / 0.48 = 8 ohm resistor
3.8 x 0.48 = 1.8 watt resistor
and the closest the shops sells: 8.2 Ohm 5 Watt Wire Wound Resistor

I am sure there is a better (maybe more energy efficient) way of doing this, so open to any suggestions. I Also need to stop the solar panel sending charging current back down the dc supply during the day, I thought either a diode or trial and error with unsoldering pins.

For reference the circuit board pics are attached.

 

I think you are mixing units here. The current requirement is in mA whereas mAh is the capacity of the battery, so 600mAh implies the battery can theoretically deliver 600mA for one hour, or 300mA for two hours and so on. You can certainly repace the battery with a 1.2V power supply - it'll probably be happy with a slightly higher voltage so you could test with a standard 1.5V AA battery.

How about using something like https://www.ebay.co.uk/itm/374198233781 to drop directly from AC mains to 1.5V with minumum losses. I'm not sure if the solar panel would disturb this but a diode in series will drop 0.6V and you are then sure to protect the supply.

You could certainly use resistors to drop the voltage but I'm not really following your maths - it'd be nice to know the actual current in mA drawn from the battery. If it's 200mA and you have a 5.0V supply and are dropping 0.6V across a diode you'd need to lose 3.2V across a resistor which calculates to 16 ohms. Maybe try 15 ohms which is standard and near enough. Power lost in the resistor is then current squared times resistance = 0.6W so maybe replace the resistor with two half watt 33 ohm resistors in parallel. You can tease te voltage up a bit by putting another resistor of higher value, e.g. 150R in parallel with the two already in parallel.

 

One problem with your approach is that the resistor only drops the voltage when the LEDs are on. When they are off, it is essentially supplying 5V to a circuit designed for 1.2V. What could go wrong?

The other problem is that the supply does not act like a battery. Who knows what the charging circuit will do when there is no battery?

Without knowing the exact circuit, I will not attempt to recommend a solution.

 

You could use a 1.5V supply made from an old phone charger, the chip boosts the 1.2V using an Inductor to drive the leds, the solar panel charges the battery via a diode and also tell the chip to switch off during the day as it feeds voltage to the sense pin, so as it gets darker the chip puts the leds on as the sense has no voltage.

 

Hi all, appreciate all the replies, really useful.!


So what I am thinking, should I just get a buck convertor, I'll connect a power supply with enough amp to drive all the lights, turn the voltage down to 1.2 and connect a diode in between the buck and the battery terminal to prevent reverse voltage. that way I can easily tweak the voltage if needed and the voltage will never go higher without load?

Is there any risk of the led's pulling too much current or can I assume the circuit will prevent that?

 

The chip does all the current monitoring, you'll find that all the leds are in parallel and the voltage is about 3V , just put the supply to the battery terminals, and remove the diode on the pcb .

 

I think it is highly unlikely that the lights are drawing a total of 480mA. These devices use a circuit called a 'joule thief', which is a very cunning way of increasing the 1.2V from, a rechargeable cell to around 3V needed to power LEDs.

You say in your original post that "...the current draw is 16mah [sic]..." but you then multiply this by 30. I suspect that the TOTAL current draw of the string of lights is around 16mA, which is why they are able to run all night on a single 1.2V cell. If they were really consuming 480mA, they would only run for about an hour on your 600mAh cell!

Joule thief circuits are quite temperamental. If you really want to do this, I'd suggest doing away with that circuitry altogether and simply supplying your strings of lights with 3V DC from a power supply. Add in a simple photo-switch to turn them on and off and leave it at that!

 

The chip does all the current monitoring, you'll find that all the leds are in parallel and the voltage is about 3V , just put the supply to the battery terminals, and remove the diode on the pcb .

Thank you very much this is what I will do. Just for my learning, what is the purpose of removing the diode that is on the circuit now?

 

I think it is highly unlikely that the lights are drawing a total of 480mA.

I think that each of the 30 solar garden lights draws 16mA from 1.25V. Then the 600mAh battery can power it for about 30 hours if charged from sunlight or for a few hours if charged under clouds.

 

These devices use a circuit called a 'joule thief',

I don't see any magnetics or other energy storage component. A DC/DC boost converter of *any* type must have one.

ak

 

A solar garden light stores a charge in one Ni-MH battery cell.
The IC in one is a QX5252.

 

I'm not talking about energy storage in a battery, I'm talking about energy storage in an inductor, such as L1 in your schematic. Without that (and a non-DC switching circuit), you cannot boost the battery terminal voltage to a higher voltage for the LED.

ak

 

Thank you very much this is what I will do. Just for my learning, what is the purpose of removing the diode that is on the circuit now?

The diode is to charge the battery from the solar panel.

 

The inductor in the solar garden light circuit is small (like a 1/4W resistor) and cannot store a large charge. It is fed low voltage high frequency current pulses which create a higher voltage each time the inductor current pulse turns off.

 

The inductor in the solar garden light circuit is small (like a 1/4W resistor) and cannot store a large charge. It is fed low voltage high frequency current pulses which create a higher voltage each time the inductor current pulse turns off.

I know how a boost converter works, and, again, not the point I was making.

ak

 

Look on the top side of the board. The blue-green component could be an inductor.

 

Look on the top side of the board. The blue-green component could be an inductor.

Yes, it is a 56uH inductor looking like the 1/4W resistor I mentioned.

 

The diode is to charge the battery from the solar panel.

So I removed D1, everything still worked however I still got 1.5v across the terminals with the cell removed and panel facing the sun (very overcast day mind you)

 

So I removed D1, everything still worked however I still got 1.5v across the terminals with the cell removed and panel facing the sun (very overcast day mind you)

Each segment of a little solar panel produces 0.5V in bright sunlight with its rated load. Less voltage and current on an overcast day.
D1 is needed since it prevents the battery being discharged by the solar panel at night.

 

Each segment of a little solar panel produces 0.5V in bright sunlight with its rated load. Less voltage and current on an overcast day.
D1 is needed since it prevents the battery being discharged by the solar panel at night.

Ok understood, is there a trace/component I can break on the board to stop the charge reaching the battery terminals (now that I will have a dc supply in there instead of the battery) - Without losing the auto on/off function of the solar panel