25 Responses

1. @Pirate88179 hey Bill watch my other video about recharging disposable batteries /watch?v=UDwMxISYojM :)

2. @MattBlytheTheOne

   Yes I have over 40 videos, most of which deal with the joule thief circuit experiments. By “dead” I mean down to the level that most electronics devices will not run from them. (cameras, phones, flashlights, etc) I collect “dead” batteries from my neighbors and then can use the joule thief to run them all the way down to .2 volts. Most batteries are considered dead at about 1.1 volts. These would normally be tossed away.

   Bill

3. @Pirate88179 OK by dead you mean no longer able to hold a charge? Then how can it light LED’s unless the Joule Thief is a free energy receiver? Did you make a video?

4. @canaanav

   This is not correct. I can light 400 leds from a single AA battery that was “dead” for many, many hours using a JT circuit.

   Try one and see for yourself.

   Bill

5. LOL! schoolboy science = win

6. Oi ! this aint no circus trick it is schoolboy science… if you don’t get it …. JUST DANCE ;)

7. MAn oh man.
   Y do such clowns play such circus crap musick in the bkground.
   Cut the “beats” and get on with it !!

   Oi !.

8. I know of an emergency lighting system that runs from as low as 0.95v at 145mA driving an 8w flourescent light with efficiency as high as 90% these run for hours!

9. Convoluted. If you wanted to get 115VAC off a 1.5V batt just run a 60Hz sin generator that converts 1.5VDC to AC (1VAC @ 67% effective) then through a 115:1 transformer. Just remeber, as voltage goes up, current goes down in a system. A 500A/hour AA battery wouls only last a minute at best, especially with such a high discharge rate.

10. You’re proposal needs a bit of work. Please read up on capacitance of an energy storage cell; farads and amp/hours. The simplest designs are most often, the most efficient.

11. from Wikipedia: “Today[update]’s flash units are often electronic xenon flash lamps. An electronic flash contains a tube filled with xenon gas, where electricity of high voltage is discharged to generate an electrical arc that emits a short flash of light.”

12. not really, eight capacitors would take much longer to charge and they would still discharge relatively quickly through the flash bulb draining the battery faster also. 8 caps in parallel would store more power but still only reach the same voltage, in series they would take an eon to charge and only reach their optimal voltage rating. Flash bulbs req. a high voltage “spike” to ionise the gas in the flash tube.

13. so an 8 capaictor circut would work

    the capacitor is 15k in micro farads

14. hi, it is not about step up transformers it the discharge time of the capacitor. As the cap discharges it pulls current through the trigger coil of the flash momentarily inducing approx. 5kV, ionising the gas in the flash tube and firing the flash. You would need a continuous high voltage source to run the flash continously.

15. hey can you get enough step up trans formers to keep a flash bulb on for 1 minute?

16. that would probably do it, but there are easier ways to run a lighting system. CFL bulbs require a lot of power. You can light a flourescent neon for less voltage and a fraction of the current. There are already circuits designed to run on a AA battery! I was thinking more in terms of inverting the output from the capacitor into and AC voltage…

17. Here is how you do it, You need 3 camera units, increase the size of the cap, large UF rating,,,then a circuit to switch between the camera units we will call A B C, in other words the circuit starts on A then before it goes out switch to B while B is running down A is charging the when B goes down to a set point switch to C during all this the 2 unused sections are charging, this should give you constant light,should not be hard use RC time constant and switching transistors

18. Such powerful technology hidden in a disposable camera!

19. the camera flash circuit is dangerous because when you discharge the capacitor it pulls current through the flash trigger coil which in turn ramps the output voltage momentarily to a whopping 5kV at around 55W of power! All from a harmless little AAA battery!

20. NO! This does NOT protect you from being the path of least resistance. This reduces the risk of conducting the current from arm to arm ONLY. You would be better off wearing a thick pair of rubber gloves and using properly insulated test equipment, and making sure the HV source was switched OFF and DISCHARGED before even going near it.

21. the best way to protect yourself from HV is NOT to play around with it !!

22. HV is dangerous if the current passes through your body. If a voltage source is high enough voltage it can jump to a lower potential by way of an arc (effectively conducting through air), if this “lower potential” is you it can be dangerous! Unprotected arcs can also damage the eyes (arc welding) and start fires! One hand in your pocket does not protect you from electric shocks, it does however reduce the risk of the current passing directly through your heart.

23. well you said it yourself, high voltage is potentially (pun intended) dangerous especially if it is carrying high current. There are special instances but generalising, if it is HV don’t touch it or even go near it without switching the power OFF first.

24. Having a hand in your pocket is causing the path of least resistance to be unavailable (arm to arm path)..is that correct?

25. But tell me if I have this right… High voltage is dangerous because it arc easely. This by itself is not dangerous. What is dangerous is when enough current is allow to pass through this arc especially when the frequency is too low. When playing aroung high voltage, always keep one hand in your pocket :) .. this from what I understand is to prevent electricity from passing through your body.