"Joule Thief" Circuits, Crude to Modern...

Intro: "Joule Thief" Circuits, Crude to Modern...

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."

Step 1:

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.

Step 2:

A crude "Joule Thief" circuit was created and hooked up as shown but no current is flowing.

Step 3:

Now the switch is held down and current is flowing through the coil. The switch only has to be turned on momentarily to create a magnetic field. Also, the switch is shorting out the LED.

Step 4:

But when the switch is released and turned off, the LED lights very briefly.

Step 5:

Again, if I could only turn on and off the switch in a very rapid manner, I would have a working "Joule Thief" like circuit.

And it turns out that the IC I mentioned at the beginning of all of this does just that. This IC turns the LED on and off somewhere between 50,000 and over 100,000 times a second, making the LED appear solidly ON.

With the QX5252F IC connected as shown, Pin 4 on the IC very rapidly connects and disconnects to Pin 3, the negative on the battery, causing the LED to repeatedly flash similar to the crude "Joule Thief" already mentioned. But here the LED flashes so rapidly that the LED appears completely ON, more so than some florescent lights that have a flicker to them. 

Step 6:

The picture here reveals the QX5252F IC at work, keeping the LED flashing so rapidly that it appears to be solidly ON.

The bulky, handwound coil-on-a-nail can be replaced with the tiny inductor shown between the IC and an LED.

Step 7:

Although the IC "Joule Thief" seems to work with my crude, handwound coil on a nail, I actually use a very small 330 (orange-orange-brown) inductor that has the size and appearance of a small 1/2 watt resistor.

Step 8:

So you can see that all of the parts of the circuit using the QX5252F IC would fit on a small circuit board with the whole apparatus taking up much less room and having much less weight than a circuit using a Toroid with two copper windings on it.

Step 9:

Although the toroid or "iron donut" has served us well and deserves much respect, the tiny inductor along with the QX5252F IC may very well serve as an advancement to some of the readers here.

Note: If the battery is replaced by a rechargeable battery, a solar cell can be connected to "Unused" Pin 1 and Pin 3 on the IC. During the day, the IC will turn off the LED while also trying to charge the battery, depending on sunlight. During the night, the IC will turn on the LED. But this is another story since this operation takes us away from our "Joule Thief" discussion... 

5 People Made This Project!

Recommendations

  • Audio Contest 2018

    Audio Contest 2018
  • Electronics Tips & Tricks Challenge

    Electronics Tips & Tricks Challenge
  • Optics Contest

    Optics Contest

85 Discussions

0
None
AlejandroM224

6 months ago on Step 9

I post in spanish, my language .

Pensé que era un joven con mucho ánimo.
Pero veo que eres eso y más, has dado respuesta a todas mis dudas con este excelente artículo.
Estoy muy agradecido. :-)

Searching by internet all kinds of articles, i found this!
Easy and explicit.
Thank you so much.

Imagine dismantle an old PC power source searching for parts to this perfomamce.
This proyect was full of expreince, and now i realize how it works and i complete to fix all my garden sun lights, and now all is ok :-)

0
None
uc1pruthvi-in-instructs

Reply 11 months ago

According to theory and also the datasheet of the QX5252 you can reduce the inductance of the inductor (L), as the inductance reduces the maximum current through the inductor increases and the maximum current through the diodes can also be increased. Look at the datasheet please. Also another factor in increasing the power output is to increase the battery voltage, causes similar effect to decreasing the inductance.

MAYBE, you could take several separate joule thief circuits, each with an added output diode and connect the diodes to a common point for power output. The voltage wouldn't increase here but the current might, hence an increase in power out. I have never tried to increase power out so all I'm offering here is just talk....

0
None
JohnO163

2 years ago

Dave,

My wife got a little cheapy, solar light (2 led) cement animal from Menards. It has never worked. It is a seasonal thing so Menards have no more. I thought I would see if I could fix it. I replaced the 1.2 volt watch type battery. Nope, no help there. The solar panel puts out a uniform sign wave at 50 mV p-p. Is that too low? Is the solar panel bad? This was checked with an oscilloscope. With the new battery in place, I measure 1.2 volts dc at the leds with the switch turned on. 0 volts with the switch turned off. I tried a digital Fluke meter and a scope. I see no sign of any kind of AC waveform present at the leds. Is the QZ5252F chip bad? I know that I am knuckle head to spend time on a $3.00 cement porcupine, but I am the curious kind. Thanks for any light you can shed on this.

John

Lowell, Indiana

20160630_065843.jpg
2 replies
0
None
Dave KruschkeJohnO163

Reply 2 years ago

First thing, I took a 1 inch+ by 1 inch+ solar cell from one of those $1 Walmart garden lights and measured the unloaded voltage while holding the cell one foot from a 60 watt incandescent light bulb. My multimeter measured 1.5 volts, well above 50 mv. I believe that the 5252 IC requires at least 0.9 volts from the solar cell to charge the battery. If the voltage is less than 0.9 volts, the IC acts like it is dark outside and turns on the light connected to the charged battery. If the battery is not charged then there is no light. If you took a 1.5 volt alkaline battery, size AAA or larger and connected it directly to your small 1.2 volt nicad, it might charge up in less than an hour, enough to power the IC that will turn on the LED(s) for a while. Some of the solar cells have poor connections on the backs. The wires look "soldered" to the cell but I suspect that what looks like solder might be some kind of conductive glue that may not be reliable. Anyhow, go to Walmart and get one of their $1 solar garden lights to help nail down your problem. I bet you will make progress without even turning on an oscilloscope...

0
None
JohnO163Dave Kruschke

Reply 2 years ago

Thanks Dave!

Great minds think alike.. We had a old $1.00 solar pole light with a failed battery. I brought into work to examine it and than I read your email this morning. Not knowing any better, I used the o-scope on the direct output of the little panel and only looked at AC. I used the Fluke meter on every thing else. The 60 mV p-p is normal when comparing the two little solar panels. Had I checked the DC right at the panel, I would have found 1.5 to 2 volts (depending how close my light source was to the panel). Both panels checked the same. I am not familiar with solar panels (as you can tell) so this was a fun learning experience. The four pin IC chips used on the solar products had different numbers. I could not find the pin layout for the HW012. Both solar units had different style 1.2 volt batteries. I threw dice and installed the chip with the different numbers. One of the posts on your site said to be careful on chip replacing because the writing on these Chinese chips could be on either side; no marking for Pin #1. I followed your sites advise and just did a quick ohm job between the two different style boards. I checked to see what pins were connected to what part of the circuit. Sure enough the writing was on the opposite side of the HW012 as compared to the QX5252F. Its works like a champ now! Thank you so much for sharing your technical expertise without hesitation. Have a Safe and Happy Fourth!

John with the lit up porcupine....

0
None
Dave Kruschke

2 years ago

Nah, I haven't really gone further than you. And the "design" you refer to isn't even mine but something I saw somewhere on the internet and simply tried out. But your post made me think more about why a diode is needed to make the capacitor/flashing LED combination work. Remove the "diode" and it doesn't work. Looking at the schematic, one sees that pin 2 of the IC grounds out the capacitor every cycle, IF THERE ISN'T A DIODE. This would prevent the capacitor from building up a charge large enough to operate the flashing LED. So some kind of diode is needed in this circuit. Perhaps using a Schottky diode would make the circuit more efficient but I don't really know...

0
None
DenisR1

2 years ago

Hi Dave many thanks it seems that you have gone further with your design I havn't tried your method with a red led instead of a Schottky diode but i'm sure others have found ways of making a very bright led from a aaa battery, I didn't think that I found something that hadn't been done before after all I'm quite new to electronics.

Kind Regards

Denis

something

0
None
MarcelG6

2 years ago

pls note that the Paulmann solar cube contains this chip, nimh battery,solar panel and 4x led. i have a few of these in the garden, and on 2 locations, there is too much shadow, so decided to modify 2 lamp units for wired power supply. after opening them, i found out it uses the 5252F chip.

1 reply
0
None
Dave KruschkeMarcelG6

Reply 2 years ago

Yes. So far, every solar "garden light" I've taken apart uses the 5252F chip OR IT'S EQUIVALENT. The bolded text is to emphasize that the pinout of this four legged IC can vary from that of the 5252F chip. Before removing this IC, it is helpful to trace the runs on the small circuit board and make a schematic that reveals the true function of each of the four pins of the IC...

0
None
GeorgiR1

3 years ago on Introduction

Because LED is supplied with a pulsing voltage around 200KHz DVM (Digital Voltage Meter) will only show voltage close to a battery level voltage. Around 1.3-1.4v. Take electrolytic capacitor 10uF 20vdc ( or close to that ) and connect across LED. Positive (+) side of a capacitor to a positive side of LED. And now you can measure 5v to 7v (or whatever) on your LED.

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.

5252pcb.JPG5252FAC-CDS.GIF

Hi, I'm curious about this power failure circuit -- how will it protect the battery from over-charging? Based on the QX5252F datasheet, while it has over-discharge protection, it doesn't say how it would handle over-charging. I assume it doesn't (though I hope I'm wrong). And since this circuit will always be connected to the AC mains, how would it affect the battery in the long run?

I don't really know for sure about overcharging the battery but I can relate my experience with a cheap solar light over more than the last three years. This light ran over a year when it was first deployed. But then it sort of quit. I took it apart and replaced the small Nicad battery with an unused battery from an identical solar light and now It has been running for a couple of years. So what is happening? This light gets indirect sunlight all day long and it gets direct sunlight for at least 1/2 day. After sundown, the light stays on for at least several hours. In other words, the solar light continues to operate. Maybe the ic protects the battery from overcharging or maybe the solar cell output isn't strong enough to overcharge the battery. Either way, I don't see any "overcharging" issue at this time...

Thanks, Dave. Yes, I was also thinking that the solar cell output wasn't strong enough to overcharge the batter. The reason I asked is I found BurgersBytes' power failure circuit quite interesting. It uses AC as an alternative to charging the battery via QX5252. It also opens up the possibility of charging it via USB (5vdc).

Still waiting for my QX5252F's and solar cells from Aliexpress :-(

qx5252f-ac.gif