Tuesday, February 26, 2008

Joule Thief

Here's a circuit I put together Sunday night. It's real simple and is designed to drain the last remains of power out of dead batteries.

It's impressive because you can run a 3V LED off of a battery giving something more along the lines of 1V. (I'm measuring 0.91V right now)


  • The cell is a dead battery, in my case a AAA.
  • The pair of inductors are oppositely wound 20 turn magnet wire on a toroid I pulled from a power supply.
  • The transistor is a 2N2222. The documentation I read said to use a BC547, but they're both NPN AF small signal transistors, so since I have 2N2222s, those are what I use.
  • The LED is a 3V high intensity red that I picked up at a local bulk electronics store last year for $10 for 40.

I've been trying to get some readings from the circuit, but my volt-ohm meter can only give me so much. I really need to figure out who to be friends with to get access to an oscilloscope.
  • The input voltage has run from 0.4V - 1.0V and runs fairly bright.
  • The output voltage measures the same as the input voltage, probably due to the duty-cycle. The volt meter measures the average voltage of a length of time, but this circuit probably runs a higher voltage a fraction of the time (50% duty cycle would be 1/2 the time). This does indicate that the output voltage and duty cycle are inversely proportional, which makes sense. (Shouldn't be able to make power out of nothing!)
  • The frequency measured on the output voltage was anywhere from 300 to 80,000 hertz, which pretty much means it doesn't mean anything.
So now that I've failed to make any meaningful measurements, other than that V(output)*DutyCycle(output) = V(input) * 100%, I'm going to venture a guess as to what is going on.

DISCLAIMER: I'm a freshmen level Mechanical Engineer. I've spent some time studying electrical circuits, but not enough to really make any kind of analysis of this. Everything from here on is total guesswork.

So I think the battery charges the inductor from the positive terminal, generating a magnetic field around the windings. Once the charge travels through the left winding and the 1k resistor, it will open the transistor and ground the right winding. This will cause the right field to collapse, which will in turn pull the left field down with it, creating the higher voltage (V(out) = V(right) + V(left)). As the collapsing right field pulls down the left field, this closes the transistor and shunts this energy into the LED instead, so it lights. As the right winding drains, the battery charges the left winding again and the cycle repeats.

1 comment:

  1. Hi!

    I've encountered the same problem here. Yes, it's because the frequency is too high for the tester to "see" the actual output voltage. try adding a diode (1N5818 or 1N4001)& a capacitor after the diode between the collector & the LED plus a 2k resistor after the led to the ground. You should be getting abt 18~24v at the positive terminal of the diode. Careful or use ordinary LED first-white LED are expensive.

    Jojo DP

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