For a project that I've been working on, I want to drive a couple of ICs and several LEDs from an ~12V supply (lead-acid battery). I think the whole thing should draw about 100mA, so I didn't really want to use a linear regulator: I shouldn't need to dissipate 1.2 watts just to drive a couple of LEDs!

So I designed things with a buck converter based on an MC33063 chip and I've been breadboarding to see how everything works. Unfortunately, what I should have realised before is that I can't avoid discontinuous mode with such low current draw and such a large step down in voltage. Trying 3.3V output, the regulator switches on and off at about 500Hz giving an unpleasant sounding buzz from the inductor and a spectacularly poor output waveform (~1V ripple). I'm using a 50 Ohm load, which should draw about 50mA.

Am I just "Doing It Wrong"? If so, what's the sensible solution I've missed? I guess I could follow the buck regulator with an LDO to filter the power rail, but that's getting a bit ridiculous and won't help with the unpleasant sound from the inductor being switched on and off. Ideas?

There's not much to see, but the buck portion of the circuit is just this:

Buck regulator part of schematic.

When breadboarding, I've replaced R2 with a dead short because I don't have any low value through-hole resistors, but I shouldn't imagine that makes much difference (since it's a high current cut-out). D1 is an SR104 Schottky.

  • \$\begingroup\$ Smaller inductor -> store less energy -> more frequent switching? How have you calculated sizes of L1 and C3? \$\endgroup\$
    – pjc50
    Jul 5, 2013 at 11:03
  • \$\begingroup\$ 1. What Rupert says. 2. R2 will limit cycle by cycle switch current and move you towards continuous operation. 3. With a little more work you could dynamically alter R2 to guarantee you remain in continous mode. \$\endgroup\$
    – Russell McMahon
    Jul 5, 2013 at 11:38
  • \$\begingroup\$ Russell: Presumably pjc50, not me :-) I hadn't thought of altering R2. But won't that effectively be the same as sticking a linear supply on the front? \$\endgroup\$ Jul 5, 2013 at 11:52
  • \$\begingroup\$ I realise I wasn't completely clear in my question: The regulator chip seems to be going into an intermittent mode with two or three fast pulses and then a cool-off period to let the charge in the capacitor drop off? Anyway, I actually replaced this with a linear regulator in the end, after doing some thermal resistance calculations and realising that it didn't matter. \$\endgroup\$ Jul 5, 2013 at 12:39

1 Answer 1


Basically, your problem is that one impulse of energy through the inductor/capacitor is far more than you need to keep the output voltage "in regulation". You have chosen to short out the current limit resistor but, this is fundamental to controlling the energy per pulse.

Energy per pulse is I^2*L/2 and you have no control on what I limits at. The power delivered to the load is this energy multiplied by the number of times you pulse per second.

Because it is only "doing" 500Hz it certainly is in discontinuous mode and this, given what I've just said, shouldn't surprise you any more. You need to choose a value of current limit resistor that stops too much energy being delivered to the load.

You might also try reducing the value of L to about 100uH as well - this reduces the energy per pulse by a factor of 6.8.

  • \$\begingroup\$ Ahah. Thank you very much, that makes sense to me. I didn't properly understand what the current limit resistor did - I assumed it was only used by the regulator to sense over-current. Brill! \$\endgroup\$ Jul 5, 2013 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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