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It seems that many "Joule Thief" circuits depend on a clunky (bulky and heavy) toroid or "donut" that has to be carefully wound with copper wire. But now there are several very small 4 legged ICs available that do the job using only a simple inductor, single cell battery and a LED. In effect, the 4 legged IC replaces the clunky toroid.

I came across these ICs when I disassembled some solar powered yard lights. I looked for a toroid but only found a four legged IC and a part that looked like a resistor but actually was a very physically small inductor (coil). Both of these parts along with wire attachment points were soldered to a small circuit board. I was able to remove parts, attach wires to them and assemble them on a Radio Shack type of "Breadboard" to test and better understand this circuit.

But then I created a very crude and minimal circuit to better understand some of the key parts of a "Joule Thief."
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Step 1:

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I used a Yellow LED that requires 2 volts (or a little more). My 1.5 volt "Rocket Battery" has been worn down to 1.4 volts. As a result, the LED is off and is not even close to conducting any current. Points A and B on the inductor coil L are at pretty much the same voltage, 1.4 volts compared to "ground" or the minus of the battery.

When the switch is pushed and and held ON, briefly, current flows through the coil and creates a magnetic field around the coil. Points A and B are still positive with point A being slightly more positive than point B.

But when the switch is released and turned OFF, the magnetic field suddenly collapses and creates a 1.4 volt voltage with a reverse polarity. This means that point B is now 1.4 volts higher (more positive) than point A. It is as if the coil has become like a temporary battery connected in series with the actual battery, presenting 2.8 volts to the LED. The LED reacts to this by flashing on for a very short moment. Pushing the switch again repeats this cycle. If I could push the switch rapidly enough, the LED would appear to be solidly ON.

The pictures that follow will reveal how simple it would be to recreate this. The coil or inductor is 12 feet of 24 gage wire wrapped (200 turns) around a 1/4 inch diameter soft iron nail.
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leosedf2 months ago

There is no need for huge coils or toroids, search ebay for "joule thief" and you will know what i mean.

Dave Kruschke (author)  leosedf2 months ago

You are really right about ebay. They sure have some interesting parts there. Thanks for pointing me there...

My Instructable titled "Joule Thief" circuit in a Flashlight Bulb? Yes!!! revealed a "Joule Thief" inside a ordinary looking flashlight bulb.

Any ideas what the efficiency of this asic is? There are dc-dc converters out there that outperform chips like that or joule thiefs with a much cleaner output. Still a joule thief is kinda fun.

harmhero2 years ago
I build a tiny Joule Thief, it works grea on one AA battery!

But what I don't understand is why the output voltage is the same as the input, 1.5 volts?
Dave Kruschke (author)  harmhero2 months ago

Try measuring the voltage with your multimeter set to AC instead of DC...

The voltage is doubled by oscillation and cutoff. The extra voltage cannot be read on a DC meter, but a scope will show it as AC. 3 volt LED's can be run. Be careful with yellow and red colors.

Dave Kruschke (author)  harmhero1 year ago
Ha! I just happened to think: What did you use to measure the voltage when you measured 1.5 volts? The output of the Joule Thief is not a steady DC voltage but a voltage that fluctuates rapidly. Some meters only measure the average value of a fluctuating voltage so while the average reading might be 1.5 volts but the peak voltage (that is high enough to light the LED) is maybe 2.0 volts or higher. However, this momentary peak voltage might not be seen on the meter display. I suspect that using an oscilloscope would reveal these peak voltages I'm referring to here.
Dave Kruschke
Dave Kruschke (author)  harmhero2 years ago
Could you describe your circuit or better yet, provide a schematic?

On the other hand, without further info, I wonder what your output voltage would be when your battery voltage depletes to 1.0 volts. Sooner or later, the battery voltage will drop but if you are still getting 1.5 volts, something is working right to at least some degree...
kc8hps4 months ago

HI Dave and the group.

I just found the 5252F chips at this link. it appears they are 100pcs for $11 + $2 shipping. 12 cents a piece didn't seem to expensive to me.

Dave Kruschke (author)  kc8hps4 months ago

I have many more 5252F ICs than I'll ever need. If any readers here would like two of these ICs at no cost beyond the cost of a Self Addresses Stamped Envelope, send me a "post" or "patch"...

Dave Kruschke

I would very much enjoy one of those IC's . what would be the process to obtain a very small # of them from you?

Dave Kruschke (author)  Fearce13 months ago

Send me your mailing address to me using Instructables mail or my email:

When I receive your address, I'll send you an IC...

Dave Kruschke (author)  Dave Kruschke3 months ago

PS; Using my email address works best!

I would like to extend my gratitude to you for sending me those Ic's.

I received them over the weekend. Thank you greatly!

The world could use more generous ppl like you!

Dave Kruschke (author)  Fearce12 months ago

You're Welcome. And thanks for letting me know you received them...

Cheaper than I could sell them. cost me $30 for 100 from China months ago and the next order would have cost me double. Great price!

BurgersBytes7 months ago
That 5252F chip does more than just drive the LED. It also limits battery drain to .8 volts to lessen the chance of damage. I am currently waiting on an order of them from China if anybody is interested.

A photocell resistor can be run from + battery to pin 1 to turn the LED on only at night instead of using a solar cell to do it. Any device can be hooked to pin 1 to bring it high for off and low for on. The pin does not require a resistor to take the voltage low either. No connection brings it low so your Joule thief worked.

This battery drain limit is actually a real problem, because you can't drain the battery down below 0.8v or so. A "traditional" Joule Thief will run for ages, taking the battery to almost nothing. One based on the QX5252 will just shut off when it gets to 0.8.

I actually discovered this when I made a few circuits with the 5252 using old dead AAs (non rechargeable). I came back in a few days to find all my lights off - because the batteries had drained to 0.8v or so.

Interestingly, I can make them turn on for a few minutes simply by touching one of the LED leads. It then stays on for several minutes before going back off again.

I'm definitely not an EE-type, but this makes me think there'd be a way to modify the QX5252 based circuit to fool it into thinking the battery isn't dead.

The lower battery limit is to keep the circuit from destroying the rechargeable battery. A lower limit would damage a battery so that it may not recharge at all. I was able to use the chips in power failure circuits with an LDR to have the light only come on when the area is dark too. Pin 1 monitors DC voltages so that voltages below .9 volts will trip the chip on. The LDR connected to the + battery and pin 1 will also keep the chip off in daylight. Pin 1 voltage up to 2 volts will also be used to recharge the battery through the chip.

If you want to run a Joule theif battery down to about .6 volts, use a transistor circuit with a wire wound coil or 2 pre-made coils from circuits here:

Yup I get the reason for the cutoff, but I'm using cast off AA's for power so am not using rechargeable batteries - also skipping the solar cells.

I've made plenty of "standard" Joule Thief circuits, but in this particular case I'd like to make 100 of them for a festival. Using the QX5252 is a really cheap and efficient way to make a lot of them, except for the power cutoff. If it could be disabled, these would be ideal.

Basically I can make a QX5252 based circuit, complete with battery holder and all for about $0.65 (quantity 100) and they go together really fast. Toroid based ones, much longer assembly.

And if you put a QX5252 one side by side with a toroid style, the toroid one will keep putting out useful (enough for me) light for days longer.

Dave Kruschke (author)  holtt2 months ago

As you can see below, I hooked up the IC without using Pin 1 at all and now the LED is on all the time. This might work for you and therefore it might not be necessary to tie Pin 1 to Negative , either directly or with a resistor. The wiring below is hardly neat but reveals what I'm talking about here...

IC schematic.jpgIC hookup a.jpgIC hookup b.jpg

Yes that's exactly how I wired mine up. My only comment is that the 0.8-0.9v shutoff happens with this when the battery goes low. It clearly could keep lighting the LED for longer if you didn't mind it going lower.

Pin 1 is pulled low internally when nothing is connected. It can also be turned on or off using digital circuits. The battery voltage is monitored on pin 2 and that will also stop it when the voltage goes below .9 volts. If the battery is removed or completely dead, the charging voltage on pin 1 may turn the LED on also.

Here is a 2 inductor circuit which uses 470uH pre-made coils, but with any transistor there will be a voltage drainage point where it will just refuse to conduct anymore.

brhodewalt1 year ago
Fascinating. How did you decide that that was an inductor and not a resistor?
Dave Kruschke (author)  brhodewalt1 year ago
I reversed drew the schematic from a solar light "joule thief" and was stuck when I view what look just like a resistor and nothing else. I had never seen an inductor that looked like a resistor. I then made a crude "equivalent" circuit using only a battery, LED, switch and the mystery part. My crude "equivalent" circuit was able to momentarily flash the LED that requires about three volts. I asked myself, "if L were really a resistor, how could the LED ever get 3 volts?" Against what I thought I knew, I decided that maybe L was an inductor. Then I looked up inductors in parts catalogs and sure enuf, they had inductors for sale that looked like resistors...
Joule Thief IC.jpgSmall 330 uH Inductor.jpgJoule Thief Crude.jpg

The 5252F coil size determines the output current to the LED's

brhodewalt1 year ago
Also, in your schematic in Step 5, what is the purpose of the connection to pin 2 on the IC?
Dave Kruschke (author)  brhodewalt1 year ago
Pin 2 is for the connection to the positive terminal of the 1.5 volt battery.
Pin 3 is for the connection to the negative terminal of the 1.5 volt battery.
Pin 2 and Pin 3 are basically for the battery power supply.
Pin 4 gets repeatedly grounded to negative, connecting the battery directly to coil L momentarily. When pin 4 ungrounds coil L, the magnetic field created collapses and and results in voltage in series with the battery to flash the LED...
Joule Thief IC.jpg

Pin 1 monitors voltage from a solar panel or other battery charging source. No connection or a low DC voltage will enable the LED driver chip. An LDR (light dependent resistor, transistor or diode) from + battery to pin 1 will turn the chip off during the day.

Here are some ways to use the chip. The first schematics do not use pin 1 at all. The second one is for a power failure circuit.

crazypj3 months ago

Late to the party as usual :) Great re-use for the multiple garden lights I have where the solar panel isn't working

Dave Kruschke (author)  crazypj3 months ago

I agree. Not only do you get an IC but you get the inductor and circuit board as well.

If for some reason you decide to remove the IC from its board, verify the functions of each of the IC leads as they might not be the same as the IC used in this Instructable...

Dave Kruschke (author) 3 months ago

About 18 hours ago I received and email announcement that I had a request for some of these ICs. But when I tried to find the "email" or "patch" in my Instructable "inbox," there was nothing there that was recent. So now there is a reader that isn't getting an expected reply. I had this trouble once before.

That said, I recommend that readers wanting one of these ICs, for free, simply send me your address to my email: and I'll mail you the IC directly. I should have simply stated this earlier...


Dave Kruschke (author) 4 months ago

You are right about the inductor replacing the toroid but ONLY if the toroid inductor has a single winding. Other joule thief circuits with only one transistor seem to require a toroid with TWO windings.

The IC is an integrated circuit that does much more than a single transistor could do. If you search online for the 5252F IC, you might be able to download or view a pdf that explains this much better than I could. But let me know if you are looking for a 5252F IC...

dudes4 months ago

actually the inductor is what replaces the toroid "donut". the four pronged ic replaces a transistor, which is how it switches on and off fast.

correct me if im wrong

Joe Blogs10 months ago
I found this IC in a cheap garden light so it was not necessary to buy it in from another country.
Can the inductor replace the toroid even if you don't have the 4 pin IC, and instead have the 3 pin transistor?
Dave Kruschke (author)  manicmonday1 year ago
Please check out my Instructable that features a "Joule Thief" that uses two transistors, two resistors, one capacitor, one inductor and one LED.

Dave Kruschke (author)  manicmonday1 year ago
Yes! Please check out my Instructable that features a "Joule Thief" that uses two transistors, two resistors, one capacitor, one inductor and one LED.

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