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A Joule thief is a simple step-up converter with which you can drive a 3.6 V LED with a 1.5 V battery even if it is spending the last drops of its life. It merely consists of a transistor, a transformer and a resistor.

The circuit is very forgiving especially concerning the number of coils of the transformer. But I was wondering if there is a way to precisely calculate the voltage produced. It would be nice to use an old 1.5 V battery to power a microcontroller. I read people reporting that Zeners would interfere with the oscillating behaviour of the circuit.

Does anyone have deeper knowledge of these kind of circuits? And would it be possible to create a stable 5 volts to power a microcontroller?

The discussion here is the closest I could find and the request is very similar to mine.

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  • \$\begingroup\$ Short answer is yes, it is possible. I've powered an Arduino Mega from an 'empty' penlight cell and a step up converter. But I had the things lying around on my desk and was just fiddling with them, don't know how to spec them. \$\endgroup\$
    – jippie
    Commented Dec 21, 2012 at 22:23
  • \$\begingroup\$ did you use a voltage regulator or did you attach the power 'before' the onboard regulator of the arduino? \$\endgroup\$
    – nansen
    Commented Dec 21, 2012 at 22:41
  • \$\begingroup\$ just attached it to the 6-20V input. Didn't want to kill the AVR on it, which is spec'd like max. 5V1 and the max. boost converter output was 7V. \$\endgroup\$
    – jippie
    Commented Dec 21, 2012 at 22:49
  • \$\begingroup\$ Please ignore my answer, I have voted it for delete. It just another fix. electronics.stackexchange.com/questions/52007/… and I'm too waiting to see the answer how to calculate the real voltage output of joule theft. I think it has more things to do with dynamic resistance of tank circuit and many maths. Hope somebody soon answer this question. \$\endgroup\$
    – Standard Sandun
    Commented Dec 22, 2012 at 20:41
  • \$\begingroup\$ Please somebody answer this question. This is a real question. I'm too waiting for the answer. \$\endgroup\$
    – Standard Sandun
    Commented Dec 24, 2012 at 19:41

2 Answers 2

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The typical conventional Joule thief can power a LED at 50 to 70 milliwatts, or about 20 milliamps at 3.3 volts. The output can be shunt regulated by a 5.1V zener, and if your uC uses 50 milliwatts, for example, then the JT will put out enough power. If you need more, then it can be made more powerful, and a zener and feedback can regulate it so that it wastes less power. One example is here. See my blog for more.

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  • \$\begingroup\$ I don't like down-voting , but this too does not address the question. It's not about regulating the voltage. There are many ways to regulate. But OP do not mark this as an accepted answer. \$\endgroup\$
    – Standard Sandun
    Commented Dec 25, 2012 at 2:17
  • \$\begingroup\$ The Joule Thief converts to a higher voltage which could be dozens of volts if not regulated. As long as the device being powered does not use more power, the JT will convert to somewhere around the voltage required. But the amount of power converted is directly related to the battery voltage, which changes as the battery discharges, so there has to be something to compensate for the change. Simplest is to stop the voltage from rising above 5V for example, by adding a zener diode. The more complicated way is to use one of the circuits in the links I gave. Or use an IC made for this. \$\endgroup\$
    – Watson
    Commented Dec 27, 2012 at 12:49
  • \$\begingroup\$ I too given a answer electronics.stackexchange.com/questions/52007/… on how to stabilize it. But that's not the point. The question is how to calculate that voltage , no matter it's HV or LV. \$\endgroup\$
    – Standard Sandun
    Commented Dec 27, 2012 at 22:22
  • \$\begingroup\$ Nice to see you, Watson A. Name (Rusty Bolt). Appreciated your blog greatly. \$\endgroup\$
    – MicroservicesOnDDD
    Commented Oct 27, 2023 at 6:21
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The JT is essentially a very simple flyback converter, and thus the output voltage would raise indefinitely should there be no losses. Of course there are - so voltage will stop rising at around 50-60V. Of course this at open circuit, because if you connect any load the voltage will drop. So your question about output voltage is incorrectly put - you should talk about power, not voltage. And that depends on the power your battery is still capable of providing. If your battery is capable of sourcing 10mW (i.e. 10mA@1v for example) then you will be able to draw around 3-5mW from the output, so around 1-1.5mA@3V. If you need more, voltage will drop, if you need less voltage will rise. If your battery has more, you will obtain more.

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