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I have this circuit from a flashlight that runs an LED with just 2xAA batteries. Initially I thought it used a boost converter or a joule thief but when I opened it up I found that it did not. It depended merely on the 3V those two Alkaline or Zinc Carbon cells output in series to light up the LED at it's minimum voltage at 3V. Problems arise when the voltage of these primary cells drop after a while making the light very weak.

So I decided I want to modify it and make it run on two NiMh cells at 2.4V. I have attached the circuit below. Can I add components at just these 2 green points to make the joule thief work? Also there is not much space inside the flashlight to place a toroid, just enough space to squeeze a transistor and a couple of tiny components. I hear there exists Joule thief that uses discrete axial inductors and or ceramic caps?

enter image description here

Update:

Here is the latest Schematic. There is an additional point where I can add a component, that is the resistor point. I can de-solder the resistor (orange point) and add a component there if needed. The green points remain unused so anything can be added there.

enter image description here

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    \$\begingroup\$ There are a variety of circuits now that appear to go under that name. It should be trivial to find them on the web, where you can also readily see their size as well as the stuff you need to do. You cannot add things to your green dots, though, to make this work. It's a different "black box" setup. \$\endgroup\$ – jonk Oct 2 '17 at 13:02
  • \$\begingroup\$ L=v*dt/dI, so if you have boost converter , you need a diode, L, C and logic level fET such that di>=20mA and determine f >1/dt such that L/dt>=2.4V/20mA and DCR of L is small, making part big. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 2 '17 at 13:16
  • \$\begingroup\$ I will update my schematic soon since there is also another point where I can add a component. \$\endgroup\$ – Kokachi Oct 2 '17 at 13:25
  • \$\begingroup\$ "joule thief" ha.. I learned a new term today .. thanks \$\endgroup\$ – Trevor_G Oct 2 '17 at 14:25
  • \$\begingroup\$ @Trevor Glad to hear that. :) \$\endgroup\$ – Kokachi Oct 2 '17 at 14:40
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enter image description here

TI.com suggests this solution. They have complete BOM and layout. Good luck.

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  • \$\begingroup\$ That looks like a solar garden light circuit \$\endgroup\$ – Kokachi Oct 2 '17 at 13:41
  • \$\begingroup\$ could be, but can fit on 36mm² PCB with SMD , Ven is just the battery switch. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 2 '17 at 13:48
  • \$\begingroup\$ I also have a charge pump in mind. Do you think this is a better choice than a step up converter? \$\endgroup\$ – Kokachi Oct 2 '17 at 14:41
  • \$\begingroup\$ Only if you can beat 90% efficiency, which I doubt. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 2 '17 at 18:14
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Using inductive kickback, it should be possible to simply add a circuit between those two points. But I do NOT recommend it. It would be better to use a better solution.

The battery won't like this, but there is no way to protect it from between the two points.

schematic

simulate this circuit – Schematic created using CircuitLab

Not drawn is the clock source, but it can get its power from the same two points, but it needs to be able to run on low voltage.

The zener diode is there to limit the voltage for the clock source.


If you want to test this circuit, you should add a protection diode to the battery. And then you can go ahead.

L1 charges through D1, R2 and M1. And then discharges through D2, D3 and the load. While capacitors store voltage, inductors store current. If you let 15 mA (or something) flow through the inductor when it's fully charged, that same 15 mA will flow through D3, R1, LED and D2 as soon as M1 turns off.

The value for the inductor depends on the clock frequency and the load and the current. If you want good results you need to do some numbers, but pretty much anything will make the LED flash in a dark room.

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  • \$\begingroup\$ looks good, I'm up-voting. I'm planning of using a charge pump since it's small and uses fewer components. \$\endgroup\$ – Kokachi Oct 2 '17 at 19:44

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