I do not have much experience with boost circuits, and am having a hard time picking a suitable inductor required for my LM2733 boost circuit. I have done a lot of research on the actual boost circuit, including how to implement a "true shutoff" using the PNP, but I am unsure of what the current requirements for the inductor would be?

I guess it comes down to I don't understand of the actual current flow is through the inductor or through the LM2733? I also found some other places online that indicated I need an inductor in the 2A range, but the LM2733 is only spec'd to 1A current. Is this because of the boosted voltage, a safety margin or what?

Also, I am lost as to if I should choose a coil-type inductor as I see on a lot of cheap amazon/ebay boost circuits, or if I can use any of the "multi-layer" SMD variants I have found on mouser that look like just any other 0805 (or similar standard SMD footprint). Is cost the only factor?

For what its worth, my application will only require around 100mA of actual 12V current. My VCC is from a 500mA 3.3v source off my microcontroller board. Any other red flags or things I should know are very much welcome! :)

enter image description here

  • \$\begingroup\$ Welcome to EE.SE! What peak inductor current did you calculate or simulate? \$\endgroup\$
    – winny
    Jul 6, 2020 at 21:29
  • \$\begingroup\$ What on earth the transistor is doing there? That's a red flag for me. \$\endgroup\$
    – Justme
    Jul 6, 2020 at 21:40
  • \$\begingroup\$ @Justme EDITED: PNP is meant to prevent eg 5V supply powering output via L1 when U4 is disabled. If SMPS not running the transistor is not biased on so 12V output falls to zero. Without Q2 Vout is Vcc - D1 drop and L1 resistive drop. \$\endgroup\$
    – Russell McMahon
    Jul 6, 2020 at 22:38
  • 1
    \$\begingroup\$ @Justme Yes - it needs work. See the useful TI discussion via the link mikemacahdo provided. It's a std solution aimed more (I think) at situations where there is no shutdown control signal. Here 12V_EN could be used to drive a high side supply line switch (I'd probably use a FET) The arrangement shown has the advantage of providing output on a cycle by cycle basis and can probably eliminate D1 (with due . care re reverse transistor conduction etc. ... \$\endgroup\$
    – Russell McMahon
    Jul 7, 2020 at 9:34
  • 1
    \$\begingroup\$ ... Putting it past the feedback resistors means that they drain Vcc when off AND the Q2 drop comes off the accurate Vout.| You could probably move C2 to Q2 emitter. \$\endgroup\$
    – Russell McMahon
    Jul 7, 2020 at 9:34

1 Answer 1


The chip has a rated current of 1A, so that is the minimum switching current.

In the typical parameters, it is listed as 1.5A.

In the graphs describing switching current at different temperatures, it can go up to 1.65A.

So to choose safely, you should choose a inductor that has a saturation current rating of at least 1.65A. But it means that the inductance has already started to drop at that current, so you might as well choose an inductor with current rating of 1.65A or larger, so that the saturation current rating is even higher.

Typically, inductors with higher current rating also have lower ESR, which will make them heat less and thus it will make your design more efficient.

  • \$\begingroup\$ Ok - so it sounds like the current does pass through the inductor. I come from the thinking of projects that only ever used linear voltage regulators, where the actual current goes "through" the vreg, so thats why I wasnt sure if current was going "through" the lm2733 or the inductor. \$\endgroup\$ Jul 6, 2020 at 22:57
  • \$\begingroup\$ @mikemachado Yes, first a bunch of energy is stored up into the magnetic field of the inductor. Then this magnetic field is allowed to collapse, driving its energy through the only available path and delivering it to some capacitor. So you apply a voltage across an inductor, allow the current to build up as the magnetic field energy builds up, then you remove the voltage and force the field to collapse and deliver its energy to be stored on a capacitor. Repeat and rinse, over and over, and the capacitor voltage builds up over time and so does the stored energy present on the capacitor. \$\endgroup\$
    – jonk
    Jul 7, 2020 at 0:23

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.