3
\$\begingroup\$

I built the schematic from this link: http://www.learningelectronics.net/extremecircuits/led-torch-uses-blocking-oscillator_13.html

It is working, but the LED is not illuminating at full brightness. I used a 5 mm white led. I know that the current is controlled from the inductor value. I powered the circuit using a 1.5V LM317 power supply and using a AA battery.

  1. What value should I use for the inductor in order to make the LED to light at full brightness and not to destroy the LED ?
  2. Can I use series inductors ?

The voltage across the LED is 2.79V. The current through the LED si 3.12mA. The power supply voltage is 1.49V.

Circuit diagram:

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Please add the schematic into your question so that it makes sense if the link dies. You can credit the author. What voltage do you get across the LED? What current do you read through the LED? What is the actual input voltage under these conditions? Pop all the info into your question. \$\endgroup\$ – Transistor Jul 1 '18 at 13:22
  • \$\begingroup\$ do you really want to optimize this circuit instead of using a simpler, yet better, alternative? There's ready-to-buy step-up controllers that are way way more efficient than this could even theoretically be \$\endgroup\$ – Marcus Müller Jul 1 '18 at 13:22
  • 1
    \$\begingroup\$ You can just click the "add image" button when editing your question. I've gone ahead and fixed it for you. \$\endgroup\$ – JRE Jul 1 '18 at 13:38
  • 2
    \$\begingroup\$ what is the specified forward current of that LED and what is the specified forward voltage of that LED? Without knowing what your LED needs, you can't progress here. \$\endgroup\$ – Marcus Müller Jul 1 '18 at 13:58
  • 2
    \$\begingroup\$ To begin, you should replace the 1N4148 with some Schottky-based diode. You might gain few mA from that. \$\endgroup\$ – Ale..chenski Jul 1 '18 at 14:46
4
\$\begingroup\$

Now that you've posted the schematic, we can see that you are trying to build a boost converter:

Unfortunately, this is a particularly crappy one. Run away. Find something better. Don't waste time trying to make this turkey hop along.

The basic problem with this circuit is that it relies on the non-ideal and unspecified characteristics of the transistors to work.

It's tough to follow the exact thought (if we can even call it that) process of whoever designed this mess, but here is how this circuit seems to work:

Start with everything off and quiescent. The switch is then closed. Q1 is turned on due to base current provided by the resistor. That turns on Q2, which pulls down on the bottom end of the inductor. That inductor voltage goes low. That turns on Q1 even more solidly thru C1 (Argh, add component designators to all components).

For some transistors, Q1 would stay on indefinitely, which keeps Q2 on indefinitely, which keeps voltage across the inductor. That can end several different ways, including burning out the inductor, blowing up Q2, or even blowing up Q1 since there is nothing limiting the current thru E-C of Q1 and then B-E of Q2.

What this circuit apparently relies on is the right combination of gain of the transistors, maximum current handling capability of Q2, the saturation current of the inductor, and the DC resistance of the inductor. If all these happen to work out, then the inductor saturates, Q2 can't keep up with the current required to keep voltage across the inductor, and the voltage on the C of Q2 goes up. This turns off Q1 thru the capacitor, hopefully before Q2 gets too hot and fries.

That switches off Q2, and the rest of the circuit is just a regular boost converter. The current thru the inductor can't stop immediately, so for a short time it squirts thru D1. That charges the output cap, which eventually builds up to enough voltage to run the LED, at least somewhat.

When the inductor current dies down, the C voltage of Q2 drops, which turns on Q1 thru the C1, and the process repeats.

Hopefully you can see how there are many things that can go wrong, and how this circuit relies on parameters not guaranteed in the datasheets of the transistors.

Trying to tweak this circuit for more LED current requires trial and error, and will likely leave a trail of burnt parts. Go find a real boost converter circuit that wasn't designed in the fit of a hallucinagenic stupor.

\$\endgroup\$
  • \$\begingroup\$ you're a little harsh. The original author of that circuit (a Peter Goodwin, according to the site OP links to) probably designed this with very much the transistors he could buy at the time and probably wrote an extensive application note explaining all the parametrics and how to get a coil with a low enough sat current and so on. Then the classical "I copy only 'the important' part" set in and the linked page copied the schematic incompletely (including leaving out component designators). \$\endgroup\$ – Marcus Müller Jul 1 '18 at 14:15
  • \$\begingroup\$ (Having googled the copied author, and the magazine quoted: I'm not sure Mr. Goodwin came up with this himself – the magazine was founded in 1987, and I'm pretty sure people had better methods of stepping up voltages than that back then. So this was probably also a lossy copy already) \$\endgroup\$ – Marcus Müller Jul 1 '18 at 14:18
  • 2
    \$\begingroup\$ @Marcus: Even if this came with extensive explanation, it is still a crappy circuit. It relies on the maximum gain of the transistors, which can vary widely. It also relies on how much you can abuse Q2 to force it out of saturation with collector current that exceeds the base current times its gain. This strikes me as a circuit where one copy was made to work on a bench after much trial and error by someone that didn't really understand what was going on. \$\endgroup\$ – Olin Lathrop Jul 1 '18 at 14:27
  • \$\begingroup\$ Yeah, I see your point there; don't know if I agree that this someone had no idea what they were doing – I'd presume they had different goals: minimum component count, working with exactly the particular box of transistors they had on that bench, building the most-surprisingly-working circuit, demonstrating temperature instability of gain to students … certainly everything but building a good step-up converter, let alone a LED driver. \$\endgroup\$ – Marcus Müller Jul 1 '18 at 14:31
1
\$\begingroup\$

Better than 90% efficiency through the entire lifespan of an AA battery.
For 10mA - 30 mA LEDs
An inexpensive (50¢ single qty) high efficiency battery powered solution Zetex ZXSC310 can be used to drive an LED with an AA battery from full charge down to its cutoff of 0.8V

Input range is 0.8V - 8.0V so works with most batteries.

The ZXSC310 is a single or multi cell LED driver designed for LCD backlighting applications. The input voltage range of the device is between 0.8V and 8V. This means the ZXSC310 is compatible with single NiMH, NiCd or Alkaline cells, as well as multi-cell or LiIon batteries.


enter image description here
100µH Inductor


enter image description here



Microchip MCP1643
Very simple inexpensive ($1 single qty) can drive a single white or two red LEDs.
0.65V minimum input, 5.0V max output.
Works well with mid and high power LEDs.

MCP1643 is a compact, high-efficiency, fixed frequency, synchronous step-up converter optimized to drive one LED with constant current, that operates from one and two-cell alkaline and NiMH/NiCd batteries. The device can also drive two red/orange/yellow series connection LEDs.

enter image description here
enter image description here

\$\endgroup\$
0
\$\begingroup\$

What I am thinking is that you may not need a boost converter. The voltage you are providing to your LM317 is probably high enough to drive the LED directly.

Here is a sketch of the basic idea (this is not a finished schematic.. for example no capacitors). Select R3 and R2 so that the base of Q1 is around 1V. Making R1 smaller will increase the LED current. Making R1 larger will decrease the LED current.

LED current will not vary much over a wide range of Vin.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 2
    \$\begingroup\$ I read the question as, "I want it to work with an AA cell but tested it with a bench PSU (LM317) at 1.5 V to make sure that the problem wasn't some limitation of the battery." \$\endgroup\$ – Transistor Jul 1 '18 at 18:32
  • \$\begingroup\$ @Transistor, that is definitely a valid reading. Initially I didn't think of it, because the LM317 is an IC, not a bench supply. But someone who doesn't have a bench supply might use an LM317 in place of one. \$\endgroup\$ – mkeith Jul 1 '18 at 20:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.