I have boost converter powered by 2 AA batteries (3v) which increases the voltage to around 40v. I have a 12v zener diode to cap the output to 12v. Regardless of the zener diode when I connect a strip of LED lights (rated for 12v, 33 LEDs @ 20ma each) the voltage drops to 8.1v. The LEDs do light up but not to their fullest potential.

Note that the diagram shows four LEDs, but in real life there are 33 in the form of an LED strip -- they're grouped in sets of 3 (you can cut every 3). So I've got 11 sets of 3 LEDs each, in parallel, for a total of 11 × 20 mA per set = 220 mA @ 12 V.

3v to 12v boost converter

I have a few questions:

  1. How can I increase the current so the voltage doesn't drop when the lights are connected (I think this is what is happening)?

  2. Am I losing efficiency (battery draining faster) by going to 40v then capping it with zener diode at 12v?

  3. My NPN is really hot to the touch, is this normal? I'm using BC337 which has a max rating of 800ma (and BC327 for my PNP).


  • 2
    \$\begingroup\$ The answer to (2) is "yes, dramatically". You also have nothing really limiting the output current of your boost circuit (aside from the internal resistance of the batteries), so you're likely to just burn up the zener (or the diode, or the inductor) once you get a big enough battery to power your load. \$\endgroup\$
    – The Photon
    Commented Aug 16, 2019 at 23:22
  • \$\begingroup\$ Instead of going for a minimalistic design, why not go with a proper design? \$\endgroup\$ Commented Aug 16, 2019 at 23:42
  • \$\begingroup\$ "proper" such as? Form factor is important, I want something small and minimal if possible \$\endgroup\$
    – koga73
    Commented Aug 16, 2019 at 23:44

1 Answer 1


You're trying to produce 220 mA @ 12 V = 2.64 W. Even if your converter was 100% efficient (and it isn't anywhere close to that), you'd need to draw almost 1 A (average) from the batteries. Your poor BC337 needs to handle almost double that (peak), which is way beyond its rating, so of course it's running hot.

With the component values shown, and making some wild guesses about the gain of the transistors, you're putting at most maybe 1 A peak into the coil. The average current is going to be half of that, so you're only getting about half the power you need. You need a bigger transistor, and a coil that won't saturate at 2 A.

To improve efficiency, switch to a different circuit that uses feedback to control how much current it draws from the battery. The circuit you have simply runs "flat out" all the time, and any power not consumed by the LEDs gets wasted in the Zener diode. There are plenty of boost converter ICs from many manufacturers that are designed specifically for this application.

Here's one simple way to add feedback to your circuit so that you don't waste power that you don't need:


simulate this circuit – Schematic created using CircuitLab

I've added Q3, R4 and R5 to your original circuit. As the output voltage rises and Q3 starts to conduct, it steals base drive away from Q2, causing it to saturate at a lower peak current, which reduces the amount of energy stored in L1 in each switching cycle.

The TIP31 (Q2) has rather low gain, so depending on exactly what kind of transistor you end up using, you may need to reduce the values of R1 and R2 to increase the drive to an appropriate value.

  • \$\begingroup\$ ZTX1048A (mouser.com/datasheet/2/115/ZTX1048A-92228.pdf ) has saturation voltage of ~0.1V at 2A with 10mA Base current. LTspice calculated 80% efficiency at 12V*220mA output, with 143mW dissipation in Q2. \$\endgroup\$ Commented Aug 17, 2019 at 2:19
  • \$\begingroup\$ Works in simulation: everycircuit.com/circuit/5126295315546112 \$\endgroup\$
    – koga73
    Commented Aug 17, 2019 at 2:54
  • \$\begingroup\$ Ok simulation works but in real life I'm getting same issue as #1 above. Even with a more robust coil and the transistors suggested the voltage drops to 8.9v when the LED strip is connected. \$\endgroup\$
    – koga73
    Commented Aug 30, 2019 at 16:30
  • 1
    \$\begingroup\$ Is the peak current of the coil reaching 2A or more? You can infer this information by measuring the switching frequency and duty cycle at the collector of Q2 using your oscilloscope. Also look at the battery voltage to see whether it is "drooping" at high current. As I said, depending on what transistor you ended up using for Q2, you may need to adjust R1 and R2 to get sufficient base drive. Also, a large electrolytic capacitor across the battery might help with the peak current. \$\endgroup\$
    – Dave Tweed
    Commented Aug 30, 2019 at 17:24
  • \$\begingroup\$ Alright I upgraded my coil and reduced the values of R1 and R2 to 3.3k and 4.7k respectively. Now under load I'm getting 10.2v which is decent and Q2 (BC337) is pulling about 450ma. I still think there is room for improvement but idk how to get more current and the inductor is starting to hum. \$\endgroup\$
    – koga73
    Commented Sep 2, 2019 at 21:50

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