I'm currently facing a weird issue. For one of my project I had to design a buck converter circuit in order to step down the voltage of three LiIon batteries to 3.3V so I can power a microcontroller.

I'm using the TPS54240 chip. The datasheet can be found here. A typical application schematic is shown on page 31. Since said circuit outputs 3.3V I was able to more or less copy it. My schematic looks as follows:

After ordering a PCB prototype the buck converter didn't really work, unfortunately. When I applied ~7.4V (2 LiIon batteries) to VCC the output showed ~3.3V as I was expecting. However once I connect a load, in my case the microcontroller, the voltage dropped to ~3.26V. Why is this happening? I thought a buck converter utilizes a feedback loop in order to maintain a constant voltage. The TPS54240 chip itself is rated for up to 2.5A in this case and my microcontroller isn't even pulling 100mA.

Even more confusing was the scenario when I connected ~11.1V (3 LiIon batteries) to VCC. In that case I measured ~2.6V at the output, which again dropped significantly to ~2.4V just when I connected a LED that drawed maybe 20mA. My microcontroller wasn't able to power on on such a low voltage making the circuit more or less useless.

Since I don't have a ton of experience with designing a circuit (this was my first project) I don't know how to troubleshoot this problem. I would appreciate any kind of help. Thanks!

Edit: This is the PCB layout of the circuit.

Edit 2: Here is a possibly fixed PCB layout.

  • \$\begingroup\$ Something I just noticed: Is it possible that all of this is caused by the inductor not having an iron core? Apparently I chose one without one (according to the schematic symbol). Its a "Sunlord MWSA0603S-100MT" to be specific. \$\endgroup\$
    – Gereon99
    Commented Sep 25, 2020 at 15:20
  • \$\begingroup\$ Link the data sheet of the inductor. \$\endgroup\$
    – Andy aka
    Commented Sep 25, 2020 at 15:50
  • 1
    \$\begingroup\$ Oh man, that layout looks pants. You just cannot treat buck converters as if you were wiring up an LED to a transistor. Where is your 0 volts? \$\endgroup\$
    – Andy aka
    Commented Sep 25, 2020 at 16:33
  • 2
    \$\begingroup\$ yeah... I'd start from scratch there. PCBs for high speed stuff like this are crucial. Check out page 42 in the datasheet for a recommended PCB layout. Key stuff is to keep you decoupling caps (ones between VCC and GND) very close to the IC, all of your COMP stuff must be close to the IC as well. Those long thin traces are essentially antennas and pick up all of the noise in your circuit, then amplify it and wreak havoc. \$\endgroup\$
    – Stiddily
    Commented Sep 25, 2020 at 17:27
  • 1
    \$\begingroup\$ @eeintech I just didn't really think that the layout matters that much. (Afterall this was the first time I ever really designed a schematic + layout) After I got my schematic reviewed I thought that I can proceed with creating the PCB layout. I will most definitely get the next one reviewed before I place an order. At least I can learn from this mistake and hopefully not repeat it. \$\endgroup\$
    – Gereon99
    Commented Sep 25, 2020 at 18:02

2 Answers 2


Did you use an autorouter?

@Andyaka I kinda did, yes. Routing a PCB was surprisingly time consuming and actually wasn't that easy. So I used an autorouter and edited the traces afterwards to, for instance, remove 90° angles in traces.

The routing, component placement and track widths are pretty bad in very important areas. Don't use an autorouter unless it's a last resort like for a backplane etc. where there are many repeated steps in the process.


enter image description here

  1. In light blue is the tortuous path taken from the inductor around the PCB to connect to C17 and C18 (the main bulk capacitors).
  2. In purple is the far-too-long track from the switcher output to the flyback diode and inductor
  3. The track widths are far too thin in my opinion and their length makes a big loop that could emit nasty interference to any close-by sensitive circuits.
  • D5 should be right up close to the switcher and should ground directly to C17-C19 forming an island. That island should only connect to the rest of the design (not the power supply) at one unambiguous point to avoid switching currents creeping into ground areas of sensitive circuits.


  • L1 should be alongside D5
  • C17 and C18 should be butted up to the above
  • Then C19 and C20
  • The ground connection of R31 (not visible in the layout) should make a connection directly to the island of ground for D5, C17, C18, C19 and C20
  • Any other ground points associated with the switcher should be teed off from that ground island.

Regarding the inductor

I see nothing in the data sheet that rules it out but, whenever I design a switcher, I pick an inductor that has a specified self-resonant-frequency (SRF) just to ensure that it is fit for purpose; in other words, if your switching frequency is (say) 200 kHz, I would want to pick an inductor that has an SRF of at least 1 MHz. Unfortunately I didn't see this stated in the data sheet nor can I say that the supplier has a decent quality system and decent tolerances. This last bit is about being careful who you choose. For instance, if one of the major recognized electronic component vendors was offering it in their catalogue I would be assured that the original manufacturer was OK.

Something like this: -

enter image description here

  • 1
    \$\begingroup\$ The auto-router is not the first issue, the placement is... Good answer though! \$\endgroup\$
    – eeintech
    Commented Sep 25, 2020 at 17:55
  • \$\begingroup\$ Wow, what a good answer! Theres still a lot to learn I suppose... thanks for taking your time to help me. Also, you said that R31 isn't visible on the layout, however it is right below the IC. \$\endgroup\$
    – Gereon99
    Commented Sep 25, 2020 at 18:09
  • \$\begingroup\$ @Gereon99 The ground connection of R31 is not visible!! \$\endgroup\$
    – Andy aka
    Commented Sep 25, 2020 at 18:11
  • \$\begingroup\$ Oh, sorry! Next time I will not make the ground plane invisible. \$\endgroup\$
    – Gereon99
    Commented Sep 25, 2020 at 18:13
  • 1
    \$\begingroup\$ I've added a slightly modified version at the bottom of my answer. It improves on yours but yours is now a lot better. \$\endgroup\$
    – Andy aka
    Commented Sep 27, 2020 at 18:12

It could be that your coil core is saturating, you will need a bigger one physically. If you need less than 2A use 47uH with 10uF caps.


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