Small DIY PV system to power LED's: DC-DC to limit current as well?

[casper.bang]

I'm creating a small PV system in my greenhouse, with 10W panel + charger + small 12V/4Ah AGM. It will only power some LED's at night, which will be high-brightness (12000 MCD) 2.2V/80mA LED's I will install into 8-10 modified garden lamps and run a small/cheap wire between. Some of the lamps may be as much as 100m away from the battery and I'd like to use as thin a cable as I can get away with, so I'm thinking of outputting the full 12V onto the "grid" and use separate DC-DC converters in each lamp (~14 times less power loss in wire at 12V@0.2A vs. 3V@0.8A). Also, I obviously want to avoid using resistors to burn of the precious stored energy as heat.

Using a DC-DC converter, will current limiting be necessary for the LED? I realize there could be a temperature issue. If so, can DC-DC converts to coupled to limit current as well as voltage?

Also, what is the simplest mechanism (except for 1A fuse) to guard against short-circuiting? Wiring up an LM317 as a current limiter doesn't seem very efficient.

[TerminalJack505]

Your post got me thinking about a way to buck a voltage down to power an LED and eliminate the current-limiting resistor.  This can be done when boosting--that's what the Joule Thief circuit does.  I've never seen it done bucking, though.

I don't know if it will work but the attached circuit might do what you want.  In theory, it should buck or boost the voltage to the LED's V_F. 

The closest I've seen to something similar is the TRANSISTOR and LED TESTER - 3 circuit here.  You'll notice that the right-hand side of that circuit is similar to mine, with the exception that mine has the LED's anode tied to ground. 

One potential problem with the circuit is that the LED's V_R would have to withstand 12V.  A diode in series with the LED might be necessary.  You'd also have to wind your own transformers if you use this circuit.

I'll test this circuit and let you know what I find out.  Like a lot of my ideas it may be "too clever by half", if you know what I mean.

[daveatol]

If you put an individual SMPS on each each lamp, yes you should regulate the current. You'd use a buck converter and feed a voltage from a current-sense resistor (in series with the LED) into the feedback pin of the SMPS IC. You may wish to amplify the voltage across the sense resistor so that it doesn't have to drop as much voltage (and waste as much power).

Another option is to use a single boost, buck-boost or flyback SMPS at the battery to supply all lights. Again, the SMPS regulates the current (with an overvoltage shutdown), but all lights are connecting in-line with a single loop of wire. Because the current through the loop is regulated, any LED connected inline gets the appropriate drive current and the output voltage of the SMPS adjusts accordingly (it also automatically accounts for wire resistance).

[Kevin.D]

If you want them all on at the same time ,why don't you just wire them is series , string's of 6.  ?

[TerminalJack505]

I played around a bit with the idea I had.  I changed the circuit around so that it uses an NPN transistor. 

After some trial and error I came up with the attached circuit.  It bucks the +12V down to power a ~3.2V white LED.  (The circuit should work with any color of LED.) The circuit can boost too but I doubt that boosting with this circuit is as efficient as using a normal Joule Thief circuit.

The cap and diode attached to the base might not be necessary if you use a 1:1 turn ratio on the transformer.  I added the cap because without it the transistor wasn't shutting off hard enough.  Once the cap was added the diode was necessary since the transistor's base was shooting way below ground.  To me the extra two components are worth the additional space on the transformer that you can use for primary windings.  The schematic says the turn ratio is 3:1, primary to secondary, but I didn't count turns so that's just an approximation.

The primary side of the transformer (where the LED is attached) was measured as ~5mH by a junk Chinese meter I have.  I don't know if that's anywhere close to the case but that's the only meter I have that can measure inductance.  I first used a different transformer measured as ~250uH and it didn't work so well.  The transistor got really hot using it.

Edit: Added pic of breadboard so you can get an idea about the size of the transformer.

[TerminalJack505]

I played around with my Buck/Boost Joule Thief circuit some more and made a few improvements.  I changed the diode that's in series with the LED to a Schottky diode (1N5818) and that improved efficiency. 

Changing the capacitor size adjusts the LED brightness--at the expense of using more current, obviously.  The circuit drew an average of 35mA when the capacitor was sized at 10nF.  It drew just 13mA when sized at 100nF and--incredibly--the LED was just as bright as a similar LED powered from a constant 15mA.  Which seems to suggest the circuit can be made very efficient.

[casper.bang]

Interesting solution TerminalJack505. However, what happens if the supply goes up to 13.5V or down to 10.5V?

[TerminalJack505]

I didn't test it beyond 12V.  The power supply I was using maxed-out at 12V. 

I tested it from 3V to 12V.  It was dimmer at the lower voltages than the higher voltages but I doubt there would be much of a noticeable difference between 10.5V and 13.5V.  The circuit will work, though, that shouldn't be a problem.

I might try a Darlington transistor when I get a chance.  This might help keep the brightness more uniform.  The lower voltages may not be able to drive the transistor hard enough.

[TerminalJack505]

It turns out a Darlington isn't such a good idea.  There's too much switching loss in a Darlington.  The circuit oscillates between 90kHz and 200kHz depending on the size of the cap so a Darlington is too sluggish.  The BC337 switches on and off in about 20ns.  The Darlington was somewhere in the neighborhood of 300ns.  That, combined with the higher V_ce(sat) made the Darlington get really hot.  The BC337 doesn't even break a sweat.