I understand that DC is powering the lights. That is a given. Why are you using a wallwart when you have solar panels??????? There is no need for a solar panel at all then. And no need for a battery unless you wanted lights during a power failure and with those lights it should be no big deal if they were not lite during infrequent power failures.

If you need both wallwart and solar panels to make this all work then this is an awfully poor design.

Spinnaker: You are right in the fact that if I have a wallwart then I don't need the solar panels and batteries, but, the solar panels and battery are already in each individual light so I might as well use them. The solar lights are fine most of the time. However there are cloudy days where the solar panel doesn't charge up the battery enough. It is on those days that I would like to use the wallwart to get the lights on at night.

 

As the guy said "nice lights".
I think what I would do is buy a 5 volt wall wort for $5 that was pretty well regulated. The use a 317 voltage regulator to drop this voltage to 1.18 volts. Wire all you batteries in parallel and hook them up to the regulator. This will keep the batteries from going below 1.18 volts which should still run the lights and not let the batteries go "dead".
With the high voltage supply you run the risk of overcharging the batteries, though I think that risk is low since they come on every night.
Either way I would get a lower voltage power supply.

 

Those cheap garden lights are really not worth the plastic they are made of. IMHO, you are way over thinking this. It will cost pennies to run those lights just using a wallwart. If you really want to be efficient then you could get one of those mechanical timers to shut the wall wart off during the day.

If you are doing this for fun then why not design a proper system. Forget those panels on the lights. They are worthless. You can get one big panel or a few small ones pointed in a couple directions to maximize sunlight. You would need a controller. I designed a micro-controller operated controller but that was really overkill. You could do it with just a few analog components. Then you will need a battery charger.

 

Those cheap garden lights are really not worth the plastic they are made of. IMHO, you are way over thinking this. It will cost pennies to run those lights just using a wallwart. If you really want to be efficient then you could get one of those mechanical timers to shut the wall wart off during the day.

If you are doing this for fun then why not design a proper system. Forget those panels on the lights. They are worthless. You can get one big panel or a few small ones pointed in a couple directions to maximize sunlight. You would need a controller. I designed a micro-controller operated controller but that was really overkill. You could do it with just a few analog components. Then you will need a battery charger.

I like your idea of using a timer. I also like wayneh's idea of the photo light controller. I think I'll put in the timer, after the timer the photo light controller and then the wallwart. The timer would be set to turn on at 6PM and turn off at 3am. The photo controller would prevent the wallwart from turning on until after the LED is on. And by turning the timer off at 3am it would give enough time for the LED to drain at least some of the battery before the sun comes up.
As for your second idea of solar panels, controller, battery charger or battery: that is getting too complicated for me at this stage of learning. I am doing the aforementioned in order to learn about electronics and also 'improve' my existing solar lights.

 

As the guy said "nice lights".
I think what I would do is buy a 5 volt wall wort for $5 that was pretty well regulated. The use a 317 voltage regulator to drop this voltage to 1.18 volts. Wire all you batteries in parallel and hook them up to the regulator. This will keep the batteries from going below 1.18 volts which should still run the lights and not let the batteries go "dead".
With the high voltage supply you run the risk of overcharging the batteries, though I think that risk is low since they come on every night.
Either way I would get a lower voltage power supply.

View attachment 79793

ronv: I think I like your idea however, my knowledge is very limited at this time. I just learned about LEDs last month. I assume you meant a LM317 of which a drawing is attached. I have numbered the junction points and pins 1-3 so may be you can tell me what to hook up where (in layman's terms).
1) A and pin 3: + of wallwart?
2) C: - of wallwart?
3) Can I eliminate the 0.1 microF capacitor?
4) Can I eliminate the 1.0 microF capacitor and junction points D and E?
5) F: + of output?
6) Pin 2 is connected to heat sink. Do I electrically insulate the heat sink from everything?

That is it! Where do I go from here?

 

Yes, but I forgot the LM317 only goes down to 1.25 volts. But I think it is still ok, you can just remove R2 and tie pin 1 and the bottom side of R1 to ground.
The connections are as you describe. Ground and F go to each battery - F to positive side.. But lets add a 1 or 2.2 ohm resistor between F and the + side of each battery at each light. I would use both capacitors, they won't be large.
You can bolt the regulator directly to the heat sink, but don't let the heat sink touch anything else. If you use a 5 volt wall wort the heat sink can be small (maybe 2" of surface area) if you use the high voltage one it will need to be quite large.
Ground also goes to - of the wall wort.

 

I like your idea of using a timer. I also like wayneh's idea of the photo light controller. I think I'll put in the timer, after the timer the photo light controller and then the wallwart. The timer would be set to turn on at 6PM and turn off at 3am. The photo controller would prevent the wallwart from turning on until after the LED is on. And by turning the timer off at 3am it would give enough time for the LED to drain at least some of the battery before the sun comes up.
As for your second idea of solar panels, controller, battery charger or battery: that is getting too complicated for me at this stage of learning. I am doing the aforementioned in order to learn about electronics and also 'improve' my existing solar lights.

Yes there is no reason you can't do both timer and sensor. The purpose of the timer is that wallwarts are always consuming some power even when they are not connected to anything. You will want to do your math to see if cost of the timer exceeds the cost running the wallwart for X time to see what your pay off is. Of course if you want to be environmentally friendly then that cost does not matter.

I would just forget those panels on the lights. They are just next to worthless. Get everything working that way and then you can go with a more robust solar panel only lighting system.

 

Yes, but I forgot the LM317 only goes down to 1.25 volts. But I think it is still ok, you can just remove R2 and tie pin 1 and the bottom side of R1 to ground.
The connections are as you describe. Ground and F go to each battery - F to positive side.. But lets add a 1 or 2.2 ohm resistor between F and the + side of each battery at each light. I would use both capacitors, they won't be large.
You can bolt the regulator directly to the heat sink, but don't let the heat sink touch anything else. If you use a 5 volt wall wort the heat sink can be small (maybe 2" of surface area) if you use the high voltage one it will need to be quite large.
Ground also goes to - of the wall wort.

Ronv:
"Ground and F go to each battery - F to positive side". That "-" or is it a dash? Does F go to positive of battery and ground and D and G go to negative of battery?
R1 is going to be 240 ohms?
To confirm: The negative of the wallwart Vin is directly connected to Pin 1 and negative of battery(s)(Vout)?
I'll have more questions later but I have to call it quits for today. Thanks again.

 

Maybe a picture.
The diode is probably not needed, but we don't know anything about your power supply, so better safe than sorry.

 

Maybe a picture.
The diode is probably not needed, but we don't know anything about your power supply, so better safe than sorry.

Presently I am experimenting with the LM317 and two solar lights. My layout is the same as the one in #29 above except I omitted D1, tran.1, R2 and R3. I added a potentiometer between adj and ground and added a diode at Vout because it seems that the 250 ohm and/or LM317 is ‘pulling’ 3 ma from the battery when the wallwart (WW: 5VDC, 1.5A) is off. My intention is to turn on the WW only at night when the lights go out and I am still outside and turn off the WW when I go inside.

1 It looks like that the amps going to the solar light is dependent on the difference in voltage between Vout and the battery volts.

2 If the lights are off and Vout is charging the batteries the ma drops as the batteries charge. At very low battery (0.85 volts when the lights go out) the WW is turned on and the draw is 50 ma and drops fairly quickly to 10 ma (the lights are still out)

3 If I have Vout at 1.25 volts (no load, after the diode) then when the lights go out and I turn on the WW there is insufficient voltage differential and insufficient amps are supplied to turn on the lights.

4 I cranked up the Vout to 1.70 volts (no load) and then the two lights will turn back on and I am getting close to 100 ma to both lights (each light draws 55ma from the battery when on or 55 ma from Vout if I pull the batteries) I assume at this stage all the ma are for the lights and almost none is going to recharging the batteries.

5 When Vout is 1.70 volts (no load) then Vout drops to about 1.15 volts when hooked to the lights.

6 On the rechargeable battery (1.2V, Ni-Mh AA, 900mAh) it says “standard charge of 90 ma for 15 hours”). I think I read some place that it is not good for the battery to be ‘trickle’ charged.

7 I think Vout (under load) would drop as I put my 10 solar lights in parallel. I am guessing. Would this be correct?

8 If I am correct on #7 then I think I would have to crank up Vout when hooking up all the lights.

9 If I do this and have a ‘failure’ in some of my lights I am concerned that then the Vout under load would be too high for the remaining lights and possibly blow my circuit boards. Any suggestions on this?

10 A side question: What happens if I change the value of the 250 ohm resistor?

11) What is "tran.1" and what does it do?

Am I thinking right? Any comments suggestions?

 

The intent of the circuit was to never let the batteries go "dead" which is hard on them. So the simple answer is to not unplug the WW. That way the battery voltage won't go below 1.25 volts even on multiple cloudy days. But if you do unplug it, yes the 240 ohm will discharge the batteries at 5 ma. total - no matter how many lights are hooked up. If you only want to use it when the lights go out the way you have it will be ok because there is a voltage drop across the diode of about .6 volts (depending on the diode), and yes it will go up slightly as you add more lights, but not much.
There is some risk that if you forget to unplug it the way you have it that the batteries may overcharge - hard to say. I don't think there is a risk of burning up your boards the way you have it as long as you don't set the voltage above 1.7 or 1.8 volts.
So to sum it up.
1-Best way- the original - leave it plugged in. 1.25 volts won't overcharge the batteries. You waste 5 or 10 ma a day.
2-Leave it as original. This will discharge the batteries with 5 ma. Not a big problem since in essence you have 10 batteries and solar panels in parallel. (You only waste the 5 ma once it is not cumulative)
3- The way you have it now, just unplug it when your done. Set the voltage as low as you can to light all the lights.
The Tran .1 is just part of the simulation.

 

It's bad form to design a scenario where the LM317 will see a higher voltage on its output than on its input. This can damage the IC and the datasheets always recommend a shunt diode from output to input around the IC, to protect it. It's probably not much of a concern at the low voltage of this project. Just sayin'.

 

It's bad form to design a scenario where the LM317 will see a higher voltage on its output than on its input. This can damage the IC and the datasheets always recommend a shunt diode from output to input around the IC, to protect it. It's probably not much of a concern at the low voltage of this project. Just sayin'.

WayneH:

Am I missing something? The WW is output 5vdc(input of LM317) and LM317 is outputing 1.70vdc. I assume with the diode after the LM317 output, there should not be much concern over the solar panel 'backfeeding' excessive voltage into the LM317 especially with the battery 'absorbing' some of the voltage.

 

Sorry, I wasn't clear. Yes, if there is a diode after the LM317, then it is protected from the battery driving current backwards through it. Using a diode after the regulator is a bit unusual, so my comment was more directed to the case where it is not present. Note that instead of a single diode feeding all the lights, I still advocate a diode at each unit, so that they are all independent of each other.

 

Funny how one little detail can complicate things.

wayneh, I think the protection diode is to prevent failure if the input is shorted to ground and the output cap is suddenly discharged thru the regulator, not to prevent voltage from back feeding, but your right if you look at the circuit for the 317 if the output goes higher than the input the output transistor just turns off. But if the voltage were above say 5 volts the emitter base junction might break down. I don't think we have that problem since the voltage is low and there is a diode on the input.
But the problem we do have if you let the batteries go dead before plugging in the "charger" is that with 10 dead batteries the current will be very high.
In the initial design I added the resistors to limit the current and since the batteries were never dead the charge current is lower. Those resistors will be to small if the batteries are dead and there will be no limit with just diodes.
Spinnaker may have had the best idea. Take out the batteries and plug it in at night.
But maybe you can take one more current reading with a dead battery and you current setup.

 

Funny how one little detail can complicate things.

wayneh, I think the protection diode is to prevent failure if the input is shorted to ground and the output cap is suddenly discharged thru the regulator, not to prevent voltage from back feeding, but your right if you look at the circuit for the 317 if the output goes higher than the input the output transistor just turns off. But if the voltage were above say 5 volts the emitter base junction might break down. I don't think we have that problem since the voltage is low and there is a diode on the input.
But the problem we do have if you let the batteries go dead before plugging in the "charger" is that with 10 dead batteries the current will be very high.
In the initial design I added the resistors to limit the current and since the batteries were never dead the charge current is lower. Those resistors will be to small if the batteries are dead and there will be no limit with just diodes.
Spinnaker may have had the best idea. Take out the batteries and plug it in at night.
But maybe you can take one more current reading with a dead battery and you current setup.

In my testing the light goes off when the battery goes to about 0.85 volts. After the light goes off the battery, within a few seconds, goes back up to 1 or 1.1 volts. Under the worse case scenario each light was pulling 60 ma if I turned it on again right away. That is 55 ma for the LED circuit board and a few extra ma for starting to recharge the battery. The ma for recharging the battery is minimal because the differential voltage is fairly small. It was only when I didn't have a diode after Vout did the battery drain significantly overnight from the 4 ma draw.
Reading up on 'batteryuniversity.com' it does not recommend to trickle charge Ni-Mh batteries because it causes memory. Supposedly best rate is .2 C (180 ma) which may be achieved in bright sunlight or trickle charging at .05 C (45 ma). Because of the low trickle charging in ronv option 1 I won't be going that way. I think option 2 is similar so I will go with option 3 and try to remember to turn off the WW when done. I'll look at installing a timer later on.

Wayneh diode idea: For starters, when I set this up this summer, I'll go without the diodes and see how things work. If I run into issues I'll install them and crank up the Vout with the potentiometer to offset the approximately extra .7 volts from the additional diode.

1) The maximum current thru the LM317 (rated for 1.5 A) is expected to be 600 ma (10 times 60 ma). On the LM317 there is a small heat sink. Do I need to attach a larger one to it? The maximum WW on time would be 4 hours.
2) When the lights turn off at 0.85 volts, any guess at what percentage of the 900 mAh is remaining in the battery?
3) For my continuing learning: What does the 250 ohm resistor do? What happens if I increase it or decrease it?
4) Supposedly the LM317 is sensitive to static. If I short (with alligator clips) the 3 terminals, would I be safe to solder with an ordinary electric soldering iron?

 

In continuation of my threads “LED: current, voltage questions” and “Solar Light Circuit Board”, let’s see if I learned something:

I have 10 solar lights and would like to supply extra power to supplement the charging from the solar panel so that the lights stay on longer.

1) My power supply (WallWart) : 120VAC, 19 watt, output:12VDC, 800ma, 16.5VDC no load

2) Each light is 7 feet apart for a total run of 70 feet.

3) I would like to supply maximum 50ma to each light at a max of 1.3v. Since the power supply may be running at night I want to ensure that the inductor doesn’t get too high a voltage and in turn blow the LED. 10 lights at 50ma= 500ma which a safe load for the power supply.

4) I want to run 18 or 20 gauge stranded wire from the power supply to the middle light (go 35feet each way) and hook up the lights in parallel. Any significant concern about voltage drop over 35 feet?

5) Since the batteries may be full at some time I will assume a voltage output of 16.5 v and battery voltage of 1.3 volts. 16.5 – 1.3= 15.2v, 15.2/50ma=304 ohms, 15.2 X .05= 0.76 watt. Use 300 ohm 1 watt resistor installed on the positive end of the battery of each light (easiest place to solder)?

6) Do I require a diode at the power supply to prevent ‘feedback/draining battery’ when the power supply is disconnected from the 120AC?

Any other suggestions/comments?

You are missing the purpose of using solar panel, which is to save energy.

If you use 12V power supply and drop it to 1.3V using a resistor, you are only using 1.3/12=10.8% of the energy from the power supply, and wasting about 90% of the energy. this is not even considering the energy wasted from converting AC to 12V DC.

another way to look at it this: every hour your setup is running, you could have used the same amount of energy to run the led lights for 10 hours if you have the correct power supply for LED.

you may as well forget about the little energy you saved from solar panels.

 

You are missing the purpose of using solar panel, which is to save energy.

If you use 12V power supply and drop it to 1.3V using a resistor, you are only using 1.3/12=10.8% of the energy from the power supply, and wasting about 90% of the energy. this is not even considering the energy wasted from converting AC to 12V DC.

another way to look at it this: every hour your setup is running, you could have used the same amount of energy to run the led lights for 10 hours if you have the correct power supply for LED.

you may as well forget about the little energy you saved from solar panels.

jamesd168:
Appreciate your point of view. I wasn't thinking about efficiencies at this time (but I'll keep it in mind) as I am just learning and am concentrating on the other technical stuff. In the mean time, to get you up to speed: My new WW is outputing 5 vdc. My intention is to turn on the WW when I intend to stay outside after the lights turn off (from lack of juice from the battery). This will probably occur maybe, say 50 times a year. The rest of the time the solar panels will be required and the WW off.

 

jamesd168:
Appreciate your point of view. I wasn't thinking about efficiencies at this time (but I'll keep it in mind) as I am just learning and am concentrating on the other technical stuff. In the mean time, to get you up to speed: My new WW is outputing 5 vdc. My intention is to turn on the WW when I intend to stay outside after the lights turn off (from lack of juice from the battery). This will probably occur maybe, say 50 times a year. The rest of the time the solar panels will be required and the WW off.

if this is the case, maybe you should add a motion sensor to turn on the power supply when you are outside.

 

if this is the case, maybe you should add a motion sensor to turn on the power supply when you are outside.

Lot of options: timers, motion sensors, light sensors, low voltage sensors, etc. Presently I am trying to use the KISS principle. The more complicated things get the more likely that something 'breaks' down and nothing works when you need it.