qs

LED1 and LED2 acts like a 3.6v zener, so you can replace the 2 with a Zener (anode pointing DOWN) to limit the charging voltage.

The circuit shown in part 1 is there to charge the battery, so that alone would be good enough for your use. Replace LED1 and LED2 with a 3.6v Zener (anode pointing DOWN) to limit the charging voltage to the proper one.

The most direct way is to start by calculating the power (V * I) of the LEDs, then match that to the capacity of the batteries and, finally to the solar batteries.For example, if you're using a white 1A LED, then its power requirement is 3.3W. - To run it for 3 hours, you'll need 10WHr. - So the batteries will need to be at least that capacity, while the capacity of solar cells will be 3/4 of that.Obviously, you will need to double if not triple the capacity of the batteries and solar power for rainy days.

Unless it's a misprint or a 240v transformer working on 110v, I can't explain it.If you are sure your readings are right, you can short out the 100-ohm (or so) resister over every third LED on the strip, it will now run at 10.5v, closer to the output of the xformer, and give you 85+% of full brightness

Most general purpose transistors (BC3x7, 2N2222 etc) are available through sales channels as well as eBay. Then you would be sure of getting the correct performing components.Yes, larger inductors do build up a greater charge, but it also takes time to do so. Lessening the inductor may yield a lower charge, but it does it at a higher frequency, so the resulting light may be greater.

NTE components tend to be much lower performers than their "equivalents". For example, if you're using the NTE-298, it only has less than half the gain of the "real" thing! To offset that you can try (1) reducing the 100k to 75k of 82k, or (2) reducing the coil to 28 or 30 turns. Change ONE element and test.

Hi OldGuy, Very briefly, the circuit starts when the PNP begins conducting, which turns on the NPN, and this starts charging the coil. At the same time, current is drawn through the 150pF capacitor, which further increases the drive to the PNP and, subsequently, the NPN. When the capacitor is charged, the drive is cut off, and the magnetic field built up in the coil collapses and cause a voltage to appear across the ends. This voltage is placed in series with the NiCd battery to turn on the LED. Simultaneously, a small portion of this voltage is diverted through the diode (D2) to charge the 1uF cap. This provides about 3.5v to drive the PNP, which improves the brightness and efficiency. The cycle then restarts. As far as your circuit, double check the specs for the transistor you're usi…

Hi OldGuy, Very briefly, the circuit starts when the PNP begins conducting, which turns on the NPN, and this starts charging the coil. At the same time, current is drawn through the 150pF capacitor, which further increases the drive to the PNP and, subsequently, the NPN. When the capacitor is charged, the drive is cut off, and the magnetic field built up in the coil collapses and cause a voltage to appear across the ends. This voltage is placed in series with the NiCd battery to turn on the LED. Simultaneously, a small portion of this voltage is diverted through the diode (D2) to charge the 1uF cap. This provides about 3.5v to drive the PNP, which improves the brightness and efficiency. The cycle then restarts. As far as your circuit, double check the specs for the transistor you're using - the datasheet I have rates the C3202 for 500mA, not 2A. And the gain at that level may be a tad low. This may be offset by reducing the 1M-ohm to 720K, which will likely help with the startup reliability. If the output is low, you can try to increase the drive to the NPN (by going down to 680k), but remember you are already pushing the transistor to its operating limits. Fwiw, the FJN965 is also available as the 2SD965 on Ebay  The 2SC2500 also works here. Another thing I'd watch for is that the pinouts for the 2SC series is different from the BC series, which is different from the 2N series! Good luck! qs

No -- the resister is there to help the transistors work, and doesn't have a direct function in the LED's brightness. TO use a 1W LED, the transistors will also need to be changed to handle the current to 'charge' the coil.Also, most AA batteries will not support the 1000 mA draw to power the light.

Any diode that can handle 150mA or more is suitable here. I've used the 1N4148 and they're still running after 5 years. If you can afford Schottky diodes they are even more efficient.

Here is the image of the LED board shown in Step 5 with a bit more description. The resistors are hidden by Layer 2 but the green lines show where they connect.

Good point - if the wiring is exposed, a 4.7k 1/2w resistor across the capacitor will reduce chances of electrical shock, but since the circuit was intended to be hidden away in an enclosure, I did not include that.

Hello there!It's always questionable whether a battery is being charged on overcast days. In the northern latitudes, even on sunny days, it is necessary to point the solar cell in a southerly direction, angled approximately 35-degrees to get the maximum exposure to sunlight. A good rule of thumb is that typical solar cell will produce about 50mA to charge the NiCad under full sunlight. So that translates to about the same operating time, since the JT here uses about 50mA while operating. So basically, you'll need 10 hours of sunlight to run the light through the night. Less sunlight? Less time.That's the reason I developed alternate circuits, notably the 'reverse' Joule Thief and the Blinking Joule Thief. Both are attempts to wring a bit more operating time out of whatever sunlight we see…

Hello there!It's always questionable whether a battery is being charged on overcast days. In the northern latitudes, even on sunny days, it is necessary to point the solar cell in a southerly direction, angled approximately 35-degrees to get the maximum exposure to sunlight. A good rule of thumb is that typical solar cell will produce about 50mA to charge the NiCad under full sunlight. So that translates to about the same operating time, since the JT here uses about 50mA while operating. So basically, you'll need 10 hours of sunlight to run the light through the night. Less sunlight? Less time.That's the reason I developed alternate circuits, notably the 'reverse' Joule Thief and the Blinking Joule Thief. Both are attempts to wring a bit more operating time out of whatever sunlight we see each day.I'm a fan of the blinking JT circuit since it can multiply run times by a factor of over 3, meaning I get 3 days of "reserve" for each day of sunlight.Let me know if this answers your q!qs