I'm using an LM2734 buck regulator to get +3.3V at 400mA. The datasheet is unclear on how to lay out the supply, so I tried to follow some common sense, keeping all power traces short where possible. The power supply has to be very small, which is why it uses ceramic caps and the LM2734, which operates at 3 MHz and thus requires a smaller inductor (it's still pretty massive!)

  • L1: 3.3uH 2A inductor
  • C1: 4.7uF 50V ceramic cap
  • C2: 4.7uF 50V ceramic cap
  • C3: 0.01uF 16V ceramic cap
  • C4: 22uF 10V ceramic cap
  • D1: RB160M (input protection)
  • D2: RB160M
  • D3: (mislabeled D1, the SOT-23) any small signal silicon diode
  • R1: 10k
  • R2: 31.6k
  • R3: 1k
  • LED1: any small LED
  • U1: LM2734, SOT-23-6

I'm generating the boost voltage from the output.

my PSU

  • \$\begingroup\$ AH, I figured out what I was talking about when I was saying the 3MHz frequency. That is for the LM2734Z, not the X or Y parts. Sorry for any confusion! \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 19:41

3 Answers 3


From the details below:

At 400mA and 20V in, you're running in continuous mode with a duty cycle of around 18%, with a peak inductor current of around 0.7A. At 4.8V, it's also CCM with a duty cycle near 70% and a peak inductor current of around 0.5A.

You may want to consider having a footprint on the PCB for the feed-forward capacitor Cff, just in case the internal compensation needs a speed-up. (You could also add a resistor in series with Cff, making what's referred to as type-3 compensation when combined with the ICs internal feedback). Bucks can be tricky to stabilize, especially with ceramic output capacitors (and with most of the compensation inside the chip!)

If you can afford some copper, add some around the SW and/or Vin pins. Copper here can pull some of the heat out of the internal MOSFET and improve reliability.

The 2% reference may cause a setpoint error of up to +/- 66mV, not including the tolerance of the resistors used in the feedback divider. You may want to add another footprint in parallel with the bottom resistor in the divider or add a trim pot if the setpoint is critical for your application.

Don't expect great efficiency at 20V in - the duty cycle is very small. It'll be much better than a linear regulator of course, but not great.

  • \$\begingroup\$ Input voltage range is 4.8V to 20V. \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 13:33
  • \$\begingroup\$ I might be getting mixed up with frequencies. It is only rated for 1.6 MHz you're right. \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 13:33
  • \$\begingroup\$ I am using the configuration on page 18, of the datasheet: national.com/ds/LM/LM2734.pdf \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 13:35
  • \$\begingroup\$ +1 for your edits, and thanks for doing the calcs. Accuracy is not critical - ±5% is fine. I'm not expecting great efficiency - anything greater than 50-60% is fine. \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 17:25
  • 1
    \$\begingroup\$ The copper should be as close as possible to the pin. The same layer is ideal as vias add thermal impedance. The copper should be exposed to free air (if possible) for maximum convection cooling. Any copper is better than none at all (as long as the copper isn't made hotter by something else and is heating, rather than cooling, the IC) \$\endgroup\$ Oct 13, 2010 at 16:38

Good first stab, just some thoughts:

1) The R1 / R2 feedback voltage divider has some really long traces and paths to ground. Anything you can do to keep the FB traces short will improve regulation stability and the ground used should be tightly coupled to the IC ground. I'd put R1 and R2 directly next to each other and put em where the small signal diode is, moving the diode over to where R1 is now. Ground R2 to the IC ground via (its low current). The LED can go pretty much anywhere and trace length doesn't matter so just move it to fit.

2) The input caps, C1/C2 and the catch diode D2 should be very tightly coupled, their ground references need to be very very close to each other. I'd flip D2 90 degrees and move it over next to C2, ground them all right there, you should use more than 1 via for this ground. What i normally do for this is a little copper pour on the surface that connects to all 3 pads (D2/C1/C2) and extends out enough to plop 3 decently sized vias in it.

3) the output filter cap C4 is in a really odd place It should also be quite tightly coupled to the ground of D2/C1/C2, move L1 over and you can probably put it right next to D2 and included it in the little ground pour I was talking about in 2).

4) With the above re-organization you can move C3 over just above/to the right a touch of the IC which will get rid of the problem you have with the input supply trace having to fit between C3's pads, as drawn right now that will short out, the trace is way too close to the pads.

Keep in mind that your critical path in this circuit, that is the loop where there will be high current AC and DC spike is from the input -> D1 -> C1+ -> C2+ -> ICin -> ICsw -> D2+ -> L1 -> C4+ -> C4- -> through other grounds back to C1-

Your goal is to minimize the loop area and the resistance in that loop, which moving C1,C2,D2,C4 as close as possible to each other and the IC will help with as well as tightly coupling their grounds through a small pour on the surface with multiple vias to the ground plane.

  • \$\begingroup\$ Thanks for your critiques. I will take them into account when designing the final version. \$\endgroup\$
    – Thomas O
    Oct 12, 2010 at 16:12

Looks good to me. The trace running through C3 is really tight-- maybe switch to a cap in a larger package? I might also flip C4 180 degrees so the trace to the small signal diode and R1 is shorter.


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