What Powers Solar Outdoor Path Lights?

[RestlessMind]

First of all, I'll say that I do not have much experience in electronics, but I am relatively competent.

So, to the point; have any of you seen those outdoor path lights that have a solar cell on the top of them? They charge during the day and light up with LEDs at night. Does anybody know what kind of power source is being charged and depleted with each cycle? Capacitors, or some kind of battery?

Thanks very much!

 

[mgb_phys]

Usually just a pair of cheap AA nicad or nimh batteries - they don't work very well

 

[RestlessMind]

Would would work well? I'm thinking about a project using the same system, and I want it to work.

 

[mgb_phys]

Depends how much you want to spend.
The cheap garden lights don't have very complex circuitry, the solar panel puts out 3V an is connected directly to 2x1.2V batteries in series. The charging is limited simply by the number of hours you get power, which probably only partially fills the battery.

The re is then a very simple circuit (just a zener diode ?) to detect when the voltage from the cells is more than from the panel = night = connect the LED.

There isn't much energy generated from a few cm diameter cheap panel which is why you can only light a dim led for a few hours. You could add more panels and a bigger battery along with a smart charger.

There are solar powered marine bouys that do the same thing 'properly'

 

[RestlessMind]

I see. What do you mean by "zener diode"? A photodiode?

What is a "smart charger"?

 

[mgb_phys]

A zener diode is just a way of comparing a voltage to a fixed level.

It uses the solar panel as a light sensor, when the panel voltage is higher than the battery the battery charges and it must be day, When the voltage from the panel falls below the output of the battery it must be night.

To extend the life of the battery you should control the charging rate - the microchip that does this is an intelligent or 'smart' charger.

 

[RestlessMind]

Oh I see.

The method you just said is much better than the ones I was originally thinking of. But how does a zener diode "know" the fixed level?

 

[gnurf]

But how does a zener diode "know" the fixed level?

It "knows" the fixed level because it has read its own http://en.wikipedia.org/wiki/Zener_diode" where the zener breakdown voltage is explained. More importantly, you know what the breakdown voltage is, because you chose a zener diode with a particular breakdown voltage for your application. Good luck.

 

[RestlessMind]

Ah. Thanks, I think.

 

[vk6kro]

I found this page which has circuit diagram of a solar light on it.

http://www.instructables.com/id/STW6I05FJOH98XJ/

This one uses a single AA cell and an oscillator that steps up the voltage to run a white LED (which has to have about 3 volts on it).

Changeover from charging the battery to operating the light is achieved with the single 1 M bias resistor which turns the PNP transistor on if the solar panel voltage drops below the battery voltage.
The two transistors form an oscillator if the 1 M bias resistor has little or no voltage on it at the solar panel end.

I include this just for the schematic, but if you wanted to reproduce it you can just follow the article.

The real star of this circuit is the driver transistor, FJN965, from Fairchild Semi. Capable of handling up to 5-amps of current in a TO-92 case, it will start at .9v and run the light until the battery drops to under 0.3-volts. You can get them from http://fairchildsemi.com

The FJN965 is Fairchild's rebranded version of the 2SD965. You can substitute the 2SC2500 here as well.

 

[RestlessMind]

That looks very helpful, thank you!

Here's a question I am going to throw out there; if I wanted to have 2, or 3 LEDs instead of 1, how would I have to modify the circuit? Simply greater batteries/solar cells, or would it get more complicated?

 

[vk6kro]

In that article, there are questions at the bottom.

One of them asks the same question and the author suggests a solution. Have a read of it.

 

[RestlessMind]

Alright!

 

[RestlessMind]

Here's something I've been wondering about choosing solar cells. The schematic for a path light here: http://www.anybodyburns.com/pathlight/schematics.htm require a solar array with ~4.4 volts and ~250mA. However, the solar panels I've been looking at on Solarbotics such as this one:

http://www.solarbotics.com/products/scc2433b-mse/

Have ~4 volts but only 18mA. How would I ever reach the requirements of the circuit with that? Am I missing something, or are these solar cells simply not suitable?

 

[vk6kro]

The LEDs are powered by the batteries, so the solar panel only affects the charging time of the batteries.

However 18 mA is not very good. A 600 mAh battery would take 33 hours to charge fully with one of those.
They could be used if you wanted to put a LED inside your mailbox and only switch it on when you opened the door of the mailbox. The batteries would eventually charge fully and only get used when you switched them on.

But, used as a path light, the battery would be flattened quite quickly. NiCd batteries can be damaged by being partially charged and then flattened like this.

 

[RestlessMind]

I see. Is amperage proportional to voltage with solar cells? Is it possible to find one, or create an array of several wired in series, that are ~4.4 volts but at the same time ~250mA?

By the way, are there any other more reliable power sources for this sort of thing besides NiCad? They seem to be very fussy about their charging and discharging, and I'd like what I'm working on to be a bit more flexible in regards to that.

 

[mgb_phys]

I see. Is amperage proportional to voltage with solar cells? Is it possible to find one, or create an array of several wired in series, that are ~4.4 volts but at the same time ~250mA?

Wired in series increases the voltage, wired in parallel increases the current.
So if you have 4.4V cells that are 18mA each you would need 250/18 = 15 of them in parralel to give 250mA.
That's the problem with these garden lights, to recharge a 2500mA battery in a day would need a large and expensive panel to give 250mA.

By the way, are there any other more reliable power sources for this sort of thing besides NiCad?

NiMH are a bit better and about as cheap. Lead acid are easy to charge but are large for the capacity
You also need to take into account the temperature when they are used outside.

 

[vk6kro]

Solar cells also come in various sizes and the current capability of these is roughly proportional to the area of the cells. Big cells give more current than small cells.

Each cell delivers only about 0.6 volts regardless of its area, though, so you have to stack them in series to get useful voltages from them, even if they are large cells.

 

[RestlessMind]

Each cell delivers only about 0.6 volts regardless of its area, though, so you have to stack them in series to get useful voltages from them, even if they are large cells.

Then why does http://www.solarbotics.com/products/scc3733/" solar cell advertise to give 6.7 volts?

Wired in series increases the voltage, wired in parallel increases the current.
So if you have 4.4V cells that are 18mA each you would need 250/18 = 15 of them in parralel to give 250mA.
That's the problem with these garden lights, to recharge a 2500mA battery in a day would need a large and expensive panel to give 250mA.

That's good to know!

NiMH are a bit better and about as cheap. Lead acid are easy to charge but are large for the capacity
You also need to take into account the temperature when they are used outside.

In what was are NiMH better? I am looking for a battery that is reliable; that is, it lasts a long time, and is not picky and how much it is charged before it is used.

Temperature will be of no concern for this project.

 

[vk6kro]

Each cell delivers only about 0.6 volts regardless of its area, though, so you have to stack them in series to get useful voltages from them, even if they are large cells.

Then why does this solar cell advertise to give 6.7 volts?


That would be because the panel has 11 cells wired in series in it. (11 times 0.6 is 6.6 volts)

NiCd and NiMH cells both lose charge with time, even if they are not being used.
There is a new generation of NiMH cells (Eneloop is one kind) that are much better at holding a charge for longer.

But these batteries are all very fussy about charging and you can easily destroy them by overcharging. Not that this is a problem with the small solar cells.

They all also do not behave well if they are fully discharged. They should not be discharged to less than 1 volt (from 1.2 volts) or else they tend to grow conductive whiskers internally and these short out the battery so you can't recharge it.

I don't know of anything better though.

 

[RestlessMind]

That would be because the panel has 11 cells wired in series in it. (11 times 0.6 is 6.6 volts)

Oh!

They all also do not behave well if they are fully discharged. They should not be discharged to less than 1 volt (from 1.2 volts) or else they tend to grow conductive whiskers internally and these short out the battery so you can't recharge it.

Oh, I see. How long (rough estimate assuming they are starting off fully charged) do you think it would take 4 NiCads to discharge to the point of damage when running 3 LEDs? More than 12 hours, or less?

 

[vk6kro]

If the NiCd batteries are in series and the white LEDs are each in series with a resistor, the LEDs will actually protect the NiCds.

This is because they need about 3.5 volts to turn on, so the NiCds will not continue to pour current into the LEDs once the total NiCd voltage drops below 3.5 volts.

That would mean each NiCd would have a voltage of 0.875 volts, if they split the voltage equally. This is below the 1 volt which is ideal, but avoids the total discharge situation which is most damaging.

 

[RestlessMind]

Ah, that is a good idea. Sorry for such a rudimentary question, but would the resistor be inserted on the positive or negative end of the LED-wired-in-series array (for lack of a better name)?

Oh, and at the 0.875 volts cutoff point, how long do you think it would take for the batteries to be damaged to the point of no longer functioning, if there were about 1 charge/discharge per day with that mechanism implemented?

 

[vk6kro]

Ah, that is a good idea. Sorry for such a rudimentary question, but would the resistor be inserted on the positive or negative end of the LED-wired-in-series array (for lack of a better name)?

Oh, and at the 0.875 volts cutoff point, how long do you think it would take for the batteries to be damaged to the point of no longer functioning, if there were about 1 charge/discharge per day with that mechanism implemented?

It wouldn't matter which end of the LED / resistor string was which. It seems traditional to put it in the + lead, but it is only there to limit the current, so it doesn't matter where it is.

I doubt if the batteries would be damaged by being discharged to 0.875 volts. I have a battery tester that discharges batteries to 0.6 volts and they seem to accept that OK.

Incidentally, a battery with a terminal voltage of 0.875 volts has very little charge left in it. NiCds and NiMH batteries keep a fairly constant 1.2 volts during discharge and then rapidly drop voltage as they reach the end of their stored charge.
So, 0.875 volts is just one point on a steep discharge voltage decline.

 

[RestlessMind]

Oh, excellent! I'm going to ask another newbie question (but I promise I don't ask questions twice ); what kind of resistor should be used (for example, a 2.2k)?

 

[vk6kro]

Assume the input voltage to each LED and resistor is 4.8 volts (4 NiCd cells at 1.2 volts each)
The current in each LED is 20 mA (this is a guess, ask when you buy it.)
The LEDs are white LEDs, and so they need 3.5 volts on them.

The voltage drop across the resistor is 4.8 minus 3.5 Volts or 1.3 volts.
So the resistance is 1.3 volts / 0.02 amps (ie 20 mA) or 65 ohms.
You would buy a 68 ohm resistor.

You would need to know how much power the resistor needed to dissipate.
Power = 1.3 times 0.02 = 26 mW.

 

[RestlessMind]

Oh, I see how that works. Thank you!