Solar Charging Night Light not working

Hi all,

I am trying to implement a very simple solar charged night garden light. Schematics is here:

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Basically a solar panel charges a battery. When the solar panel output is low (at night), a joule thief circuit lights a led.

The joule thief part works just fine, the problem is that the dark detector (2n3906 + 4.7K resistor) never triggers the joule thief and I cannot understand why.

Here are the pictures of the breadboard:

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If I bypass the dark detector connecting the positive wire from the battery to the center of the transformer (moving the yellow wire from row 16 to raw 14) the led lights up.

Thanks a lot for your time.

Reply to
Alberto
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It would seem to me that the 5K resistor is not going low enough when the solar panel is covered. Try covering or disconnecting the panel and putting the 5k resistor to battery neg ( the end that was connected to the diode). That should enrgise the led, if not test the transistor.

Reply to
RheillyPhoull

Hi, thanks for your reply. I tried without the panel connected and covering it with black tape, nothing happened. I did not try connecting the 5k resistor to the battery negative as I cannot do it now, but I will give it a try in the next week and I will update you.

Thank you very much!

Reply to
alberto.m.scattolo

When you say the blocking oscillator works are you saying the light works as long as the battery is charged, and the cell has full sunlight on it? So the light stays on all the time and doesn't turn off when there is light on the cell?

It would be better to think of it as a light detector that stops the oscillator while the cell has light on it.

(the "famous" joule thief, is ancient technology from the days of vacuum toobes)

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Reply to
default

Hi!

When I say "The joule thief part works just fine" I mean that, with a charg ed 1.2 AA battery, and bypassing the light detector, a 5mm blue led lights up very bright. So, the oscillator works, otherwise the led would not switc h on at all. "Bypassing the light detector" means that I connect the battery positive to the center of the transformer directly, without going through the first tr ansistor so that the transformer is always powered by the battery as far as it is charged. This works with or without the solar panel, with or without light on it.

I am interested in making the light detector work, to switch on the light o nly when meaningful and to maintain the battery charged.

Eventually, I would also try to replace the battery with a capacitor but I am not 100% sure this is a good idea. And I would like to know the frequenc y of oscillation, I don't know if there is a practical way to calculate it, I do not an oscilloscope.

Thanks link too!

Reply to
alberto.m.scattolo

OK now I think I get it.

It appears as if the solar panel is part of the bias circuit for the PNP transistor.

Solar panels normally supply power when the sun shines, but when the sun is absent they turn into resistors and allow power to flow backwards. Which biases the PNP transistor "on" allowing power to go to the oscillator circuit etc.. The light presumably turns ON and all is happy with the world....

got it?

If you were to use a panel that doesn't "leak" or has a built-in diode to prevent reverse flow, that circuit would not work.

So what are you using for a solar cell?

I'm assuming you are using the specified panel and your battery is two cells just like the schematic shows, the 1N914 diode has the right polarity, etc.. Right?

Try connecting a 5-10K resistor between the base of the PNP to ground, that should get it conducting and turn the light on, and that will provide a clue to why it isn't working. (solar cell dark during the experiment) It must have a resistor to protect the base from drawing too much current and killing the PNP transistor.

The idea is to provide a leakage path that may be lacking in your solar cell.

Reply to
default

Switching the main current to the Joule Theif is probably the wrong approach, switch the current to the base instead, that is only switch the current that flows through the transformer branch that goes to the base. this is a much smaller current, and so it takes less energy to run the switch. (you can use a smaller base current to the switch)

A second problem is that the joule thief may not start automatically when power is gradually applied. Some sort of positive feedback to ensure that the power snaps on should help there.

--
  Jasen.
Reply to
Jasen Betts

How do you plan to switch only the base of the oscillator transistor?

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Reply to
default

the transformer has two windings, connect the other one to the supply.

--
  Jasen.
Reply to
Jasen Betts

The windiings have the turn on bias a well as the inductive feedback and turn-off riding on them. The center-tap is integral to the operation..

There should be a way to do it: pass the AC feedback while isolating the DC bias, but it is a little more complicated than separating the center tap and it would take a few more components.

Reply to
default

it's just a node.

current into the base is close enough to monodirectional that it doesn't matter.

a capacitor across the added transistor could maybe help with charge scavenging a little but that 1K resistor is going to slow that down anyway.

--
  Jasen.
Reply to
Jasen Betts

So you have... nothing. Post a schematic.

Reply to
default

No, asshole, I have built it and it works. I am not wasting any further time responding to your clueless taunts with useful information.

--
  Jasen.
Reply to
Jasen Betts

ROFLMAO

sure you built it

Reply to
default

Nice to see you guys getting along :-)

Reply to
RheillyPhoull

Oh guys, come on. Unfortunately I still cannot try your suggestions as I am not at home, but please, let's try to be pragmatic. If any of you have ideas, schematics or whatever, post it. I have limited components at home and some are just scavengered but I would give a try to any good idea that would allow me to keep 4 bright led on all night long.

BTW, is replacing the battery with a capacitor an idea worth trying/meaningful?

Thank you all

Reply to
alberto.m.scattolo

Yet another demostration of your stupidity.

--
  Jasen.
Reply to
Jasen Betts

That would depend on how many hours of light you expect to get out of one charge. Some guy had posted a series of fountain lights he was running from a cap. Farad size caps aren't cheap as a rule.

I think Jasen has the glimmering of an idea, bias the oscillator itself to stop during daylight rather than have the voltage drop of the PNP transistor, to contend with when it's on. I suspect it may be trickier than he suggests though. The circuit works because when the oscillator transistor switches off abruptly, it causes a high voltage spike in the inductor. Diddling the bias may have other consequences in addition to shutting it down during daylight. But, I have my own projects to play with.

BTW there are Chinese IC's that do exactly what you want, using a single inductor. see:

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The YX8182 IC does the same thing without an inductor.

I built several circuits of that type to illuminate a buoy, so I could kayak before the sun came up. A flashing light attracts attention faster and uses less power.

Reply to
default

Ok, how would be the schematic?

Any specific schematic to suggest?

I would like a steady light effect (high frequency on/off is also fine as far as it seems constant at human eye).

Thanks

Reply to
alberto.m.scattolo

When it come to the Chinee, circuits are given for the chip type in the datasheet. (often written in Chinese, but the numbers and symbols can sometimes be gleaned and the schematic is readable)

BTW I did put together a stack of super-caps in a box to start my truck. 500 F 2.7v each. It is currently doing duty as a wifi router backup battery.

Some months ago one of the surplus outlets had some 50F 2.7 volt beauties on sale for $0.59 each. I see them priced at ~$3 each from US sources and as low as $0.50 each from China. I got several of the

50 F (~5/8" diameter ~1-1/2" long) and subbed two/series in place of a battery on my buoy flasher (runs 6 months on 3-AA cells) and in one day it could slurp up enough energy to run for 4 days.
Reply to
default

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