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I want to build a circuit that powers 2 standard red LEDs (forward voltage of 2V each) from 1 AA battery. I would like to learn how to design and build a boost converter that can take 1.5V in and output 4V for the LEDs. I am not sure what circuit would be best for this job.

I have tried building a Joule thief (using axial inductors,) however it did not light up both LEDs as brightly as it did with just one LED.

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It uses a 2N2222A transistor. The inductors are 220uH with DC resistances of 2.9 ohms each. The battery is an alkaline AA LR6 1.5V. How can I optimise this circuit to produce 4V over the LEDs?

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Ideally, I would build the circuit without using any specific boost-converter ICs.

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    \$\begingroup\$ ”Ideally, I would build the circuit without using any specific boost-converter ICs.” Why? \$\endgroup\$
    – winny
    Commented Jul 2, 2021 at 17:38
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    \$\begingroup\$ The 555 will not work, it needs 4.5V minimum per the data sheet. You state your joule thief (using axial inductors) did not light up both LEDs to full brightness. Was it designed to do that. Axial inductors would be fine. \$\endgroup\$
    – Gil
    Commented Jul 2, 2021 at 18:08
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    \$\begingroup\$ Isn't a requirement for the joule thief that the inductors are magnetically coupled? In the design as I know it, both inductors are wound on the same core, and in a particular direction. \$\endgroup\$
    – marcelm
    Commented Jul 2, 2021 at 20:56
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    \$\begingroup\$ "however it did not light up both LEDs to full brightness" - how much current do they draw at 'full' brightness? \$\endgroup\$ Commented Jul 2, 2021 at 22:19
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    \$\begingroup\$ @marcelm the joule thief definitely can work with axial inductors, as long as they are close enough so that their magnetic fields can interact. Although perhaps a double wound toroidal core would be more efficient \$\endgroup\$
    – Adam
    Commented Jul 3, 2021 at 20:38

3 Answers 3

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First let's try to deal with what you've got. Let's get better coupling for our coupled-inductors, if we can. Try to move the inductors closer together in different positions. The LED's should get brighter -- make them as bright as you can. Once you've found the best coupling position you can get, use one or two small zip-ties to hold together the inductors in the good-coupling-position with a little bit of pressure, and then add a dab of hot-melt glue to join the two inductors together semi-permanently. Once you're sure you're happy, add more hot-melt to join them together for good.

Then replace the 1K resistor with a 100 ohm resistor in series with a 1K potentiometer or trimmer pot, so you can dial in the resistance (brightness) that you want. When you find the right resistance with the pot, turn the circuit off and measure what you've got with the pot and 100 ohms combined, and then you can replace both the 100 ohm resistor and the pot with a single resistor having the value you have determined with the help of the pot.

If you don't have a pot, then you can keep swapping in resistors for the 1K according to the E24 series until you find what you need, best found with a binary search.

Quick E24 Ref for following discussion:

1.0 1.1 1.2 1.3 1.5 1.6 1.8 2.0 2.2 2.4 2.7 3.0

3.3 3.6 3.9 4.3 4.7 5.1 5.6 6.2 6.8 7.5 8.2 9.1

Example of a binary search heuristic: Since you want it brighter, divide the search-space below in half (average the low and high values of the target range). So 1K+100=1100/2=550, and the closest standard E24 value is 560, so first try 560 ohms (replace the 1K resistor with a 560 ohm resistor). The Joule Thief should get brighter. If you want it even brighter, 560+100=660/2=330 so replace the 560 ohms with 330 ohms next. if that's too bright now, go up, not down, 330+560=890/2=445 try 470 ohms next. if that's still too bright, go up again, 470+560=1030/2=515 try 510 ohms next. That's the general idea.

Keep in mind, though, that your circuit will have poor efficiency with the 2.9 ohm inductor as the power inductor, so your battery may not last as long.

If you want to make a better Joule Thief, start with a decent power inductor, like this Wurth 7447020 power toroid for $1.64 from Digikey (220uH 1A Fixed inductor having 160 mOhms DC resistance),

A good quality 1 Amp power toroid that you can add a secondary to, thereby making a very good Joule Thief coupled-inductor

then add the sense winding to it. The sense winding can be much thinner wire, but use magnet wire, as you may need all the space you can get at the toroid center. The power inductor is what has to be thick wire, and that's where the real "magic" happens. Notice that this toroid has lots of windings (turns) on it. You'll have to add at least the same number of turns to achieve the same inductance on the secondary.

As you wrap the secondary, you can use an inexpensive LC (inductance and capacitance) meter to guide you. My general rule is a little more than the primary, so maybe 270uH. Even though they're not as durable as a good multimeter, I have had good experiences with the LC200A meter ordered from Amazon, and they're generally less than $70 USD.

LC200A clone for measuring inductance as you wind secondary

Hope this helps. Please come back and let us know how it went, since we helped you, it would be nice if you help others in return. Please upvote this post if you liked it, and use the green check-mark to mark this answer as your accepted-answer, if you think it is best.

Finally, keep in mind that a Joule Thief does not offer all of the protections that a professional chip does, which protects the LED's, the power source, and the circuit, both while you are working with it, and while in service. That professionalism is kind of like insurance for you and your company, and also for your customer, especially when you use the better, more expensive LED's. A bike light, for instance, may have enough jarring that an intermittent connection to the LED's develops (let's say that this is early in the product life-cycle, before you've worked out all of the bugs in the product). A chip will protect the $4 worth of red LED's in the product when it keeps going open. In contrast, the Joule Thief will probably allow the voltage to rise to dangerous levels and damage the transistor, and then when contact is again established with the LED's, they too can be damaged by the high voltage. Usually, the chips just turn things off when they sense an open on the LED connection, protecting everything. And in this intermittent case, a chip will do the best it can do (sometimes unsuccessfully, but it's better than nothing, which is what the Joule Thief gives you in the way of protections).

I hope this glimpse of what a chip does for you helps you to see their added value for your more "important" or "valuable" products.

Here is a simulation that I did in LTSpice for you to compare:

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There are several makers who make ICs to do this sort of boosting. I suggest you go look at one or two of those and read the datasheets carefully. All such boosters basically work on the fact that the average voltage across a perfect inductor is zero. The joule thief switches on end of an inductor with a transistor.

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Hi I've been using the BC547 - 47ohm resistor - 30turns winding on toroidal from an old computer mother board and it works well as a garden light with solar charge on a AA 1.5v battery. I have added from 1 led to 6 LEDs. Still working very well

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    \$\begingroup\$ Please describe your circuit more fully - preferably with a circuit diagram. Are you using the circuits shown above. \$\endgroup\$
    – Russell McMahon
    Commented Nov 26, 2021 at 5:39

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