I am trying to use an AP6507 to power a Raspberry Pi from a 12V battery. In the datasheet is written that it should provide a continuous current of 3A, but when I try to power the Raspberry Pi, the output voltage drops to 1V.

The output voltage of the converter is set to 5V. I also tried to load the output with a 4.7Ω resistor, and the voltage already dropped by 0.5V, so this means that it can't sustain the voltage at 1.06A.

What can cause this and how can I fix it? I don't really have a lot of experience in debugging switch mode converters, so any help would be appreciated.

  • Used inductor: DJNR5040-3R3-S, it's saturation current is 4A.
  • Output capacitor is an electrolytic, 47uF, 6.3V one, here is the datasheet. I couldn't find its ESR in the datasheet.

I am attaching my schematic, PCB design (the buck converter part is marked with a black rectangle) and a photo of the soldered board.


PCB design

enter image description here

  • 1
    \$\begingroup\$ Welcome to the site. You said you need help but did not tell with what, and you did not ask a question (this is a Q&A site). Attaching the schematics is a very good start (not all bother to do that), but unfortunately it is not enough to see what would be wrong. It would help to know which exact inductors and capacitors you used, to see if they are suitable. It would help to see the PCB design, to see if it is suitable. It would help to see a photo how it is built, to see if there is just some soldering error. Could you edit those in and include a specific question? \$\endgroup\$
    – Justme
    Nov 21, 2021 at 10:22
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    \$\begingroup\$ +1, most of these are either bad layout or wrong choice of inductor/caps, so we need the info. All of this is important. For example the caps are polarized on the schematic, so I suspect high ESR tantalum/alu caps which won't work in this application. \$\endgroup\$
    – bobflux
    Nov 21, 2021 at 10:23
  • \$\begingroup\$ What's the saturation current rating on your inductor? \$\endgroup\$
    – John D
    Nov 21, 2021 at 18:44
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    \$\begingroup\$ Is your input remaining at a stable 12 volts, or does the input voltage drop under load? \$\endgroup\$
    – Hearth
    Nov 21, 2021 at 21:05
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    \$\begingroup\$ The electrolytic capacitors are unsuitable for this application. Scratch off some soldermask and add some ceramic caps. You’ll need 2 or 3 each for the input and output. Pay attention to the ESR and DC derating specs. If you can find a similar chip in TI’s range, use their webench tool to choose the capacitors. Otherwise you can easily spend hours working through the combinations. Do not underestimate the complexity of designing a switchmode supply! \$\endgroup\$
    – Kartman
    Nov 21, 2021 at 21:18

3 Answers 3


The most important thing in a DC-DC layout is to minimize the area of the hot loop. This minimizes both its inductance (which causes L.di/dt spikes) and emissions.

enter image description here

The hot loop is the loop with highest di/dt. In the case of a buck, input current is a square wave, output current is a triangle, so the highest di/dt is on the input side. For a boost, it's the other way around.

This means you should first place the input caps as close as possible to the chip with fat short connections. Ideally, Vin and GND should be planes or copper pours, not traces, because the wider the trace, the lower the inductance.

The closest input cap should have the lowest inductance (ie, SMD MLCC X7R/X5R). If several caps are used, the low inductance one is the physically smaller one. Note the only usefulness of 100nF caps in this case is to be tiny ; inductance depends only on package size and PCB mounting (vias, etc) not on value. If you hand solder and you don't like 0402 caps (who does?) then pick the package you're comfortable soldering, then pick the largest capacitance in this package in X7R, then put that right next to the pins. Remember MLCCs lose capacitance with voltage, so a 10µF 0603 cap could have less actual capacitance at 12V than a 4.7µF 0805 cap. When in doubt, google "murata simsurfing", click on the caps, there is a button to display C versus V and compare various caps.

Next you need input caps which can take a ripple current equal to the full inductor current at 500kHz. Unless you want to use fancy polymers, the cheapest and lowest inductance option will probably be a few 10µF MLCCs. For hobby projects, note MLCCs are much cheaper in quantity, so if you make the input and output caps the same, you save money. Basically, get a strip of one value like 10µF 25V, and if you need more µF, just put more of them.

Once the input caps are placed, then you can place the inductor and output caps to optimize the lower priority hot loop. It's better if the GND pins of the input/output caps and the chip are on the same copper pour, that makes less HF current in your ground plane.

Speaking about ground planes, there is one on your board, but there are no vias to it, so it's useless. This means the ground part of both loops will go through the closest via to the ground plane, which in this case is the GND pin of a connector:

enter image description here

It's better to just put the caps next to the chip and use wide copper pours, with plenty of vias to the ground plane. If you put vias below the chip, you can also solder the thermal pad from below with a soldering iron.

Since the SW node has high dv/dt its capacitance should be minimized, ie use a thin short trace. Wide enough for the current, but not wider.


First, as @Kartman pointed out in the comments, replace electrolytic capacitors with ceramic. Both of them, C2 and C3.

Second, I strongly recommend you to redo the PCB routing.

  1. First rule of these converters is that SW trace should be as small as possible. You have those huge copper pours sticking out beyond L3 and C3. Get rid of them, you already have copper going from chip to components, do not go past them.
  2. Same for GND zone sticking out past C4, R3, R2. Get rid of it and you will be able to place L3, C3 much closer to the chip, which is the second rule for these converters. Don't forget to move C2 too, as close as possible.
  3. Add thermal VIAs to GND trace under the chip.
  4. Remove thermal clearance on the GND, VIN and VOUT zones connected to the converter components. You are limiting thermal dissipation AND current carrying capacity of these critical traces. Make these zones separate from same nets if you want to keep thermal clearance on the rest of the circuit and connect zones outside of the converter's parts.
  5. While you at it, I recommend increasing thermal clearance and trace spacing for the rest of circuit. Top and bottom. Those whiskers between the pins are outrageous and some trace spacings are worrisome.

3.3 μH is too low for your application (You probably copied the typical application circuit from the datasheet).

The calculated value is 5.8 μH. I don't know which Pi you're using, but depending on the model, a RaspPi may draw too much current at start up. So you should use a higher inductance with higher saturation current. Also you should consider the SRF of the inductor during the selection of the inductor.


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