I was trying to boost 3.7 V to 5 V @ max 1A. Used LTC3426 DC-DC converter for it. LTC3426 datasheet here.

I made a simulation before printing PCB and I get the voltage I needed. However, now I have the PCB and it is not giving the desired voltage. Simulation Circuit:

enter image description here

Graph: enter image description here

Actual Circuit: enter image description here

I dont know why, I get only 3.8 V not 5 V. My current battery voltage is 4 V also it is not different. I checked shdn pin of the converter and it is on HIGH state. I removed D3 diode on schematic and nothing has changed. I changed the inductor to 220 uH(normally it is 2.2 uH and nothing has changed.

With Vcc signal I run microcontrollers or similar 5 V operating voltage devices which means I have load also.

What is the problem? Where is my mistake?


I changed the resistors to the values which are given in the datasheet

enter image description here

But my inductor was 22uH

I got 5v finally. However after some time like 3 seconds, the regulator burned! Is it because of the 22uH inductor? Because on the reference design it is 2.2uH.

I dont think it burned because of high current consumption. Because the other part of the circuit is just Atmega2560 and SIM808 chips and some sensors. I think this devices does not need too much current to break the regulator.

The layout of the regulator is(u2) is the regulator you can check it from the schematic on above post:

enter image description here

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    \$\begingroup\$ Well, what you are seeing is what I would expect without switching : Vout = Vin - schottky diode drop. You now need an oscilloscope (though AC V measurements at SW may show something) ps. 11.9 MHz? Really? \$\endgroup\$
    – user16324
    Nov 11, 2020 at 14:17
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    \$\begingroup\$ Does your inductor have a high enough saturation current rating? \$\endgroup\$
    – Mattman944
    Nov 11, 2020 at 14:26
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    \$\begingroup\$ @ratatosk - did you mean 11.9MHz? It is supposed to be 1.2MHz. \$\endgroup\$ Nov 11, 2020 at 15:11
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    \$\begingroup\$ Can you check voltage on feedback pin and compare with datasheet feedback voltage? \$\endgroup\$
    – bobflux
    Nov 11, 2020 at 15:42
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    \$\begingroup\$ Your feedback resistor is wrong and your comment in response to that nonsensical; additionally you need to include a good photograph of the board. Did you follow the manufacturer recommendations for layout, or did you just wing it? \$\endgroup\$ Nov 11, 2020 at 15:46

1 Answer 1


Here's the layout you posted:

enter image description here

Well... Don't take it personally, but it has all the mistakes we usually find in someone's first switching converter layout. You're not the first to get burned here ;)

Layout is an essential part of any switching converter, especially a 1.2MHz one. From the datasheet:

enter image description here

So you have two solutions, depending on your goals.

If you just want it to work and need some 5V quickly, purchase a canned DC-DC converter like this one. These are readily available and cheap, and save a lot of time. You can still salvage your board if you mount it with very short wires, maybe recycling some thru hole pad (like the diode) to solder on.

If you want to learn about switching converter layout then it'll be a lot longer than what the $3.5 canned converter is worth. You'll also get some 5V, but only as a byproduct of the learning experience. If you intend to make a lot of boards then it'll also be worth the money versus a canned converter.

Problems with the layout:

First, the hot loop (in red):

enter image description here

Depending on the state of the switch inside the chip, inductor current will flow in one of the two halves of the red loop. The switching has very high di/dt, a couple amps in a couple nanoseconds. Loop area is inductance (L) and problems are proportional to Ldi/dt, so you want L to be small thus the hot loop should have the smallest possible area. In your layout it is huge due to a thru-hole diode, long traces, no ground plane, and I don't see where the output caps are. Most likely this high inductance combined with high di/dt creates voltage spikes (ie, Ldi/dt) which fry your chip. So it won't work, and even if it did, another problem that is proportional to Ldi/dt is EMI emissions.

I'm not going to write a detailed guide, because it all boils down to: minimize the area of the hot loop (red) first, then the green loop second. You can find guides and tutorials, here is a good example.

So. You picked a non synchronous converter, which means it has an extra diode that has to fit somewhere. Honestly it's easier and simpler to use a synchronous boost like TPS61023, no diode, easier layout, and it has better efficiency.

Optimizing the hot loop means a low inductance output cap, ie a SMD ceramic as close as possible to the chip:

enter image description here

There are vias for a low inductance connection to the ground plane too. This could be improved by moving the input cap in place of the inductor, so its ground pin connects directly to the output cap's ground, but this probably requires placing the inductor on the bottom side of the board.


Your layout looks like it was done with the autorouter. If this is the case, remember the autorouter has no idea what the traces actually do. It will happily route a 10 amp power rail for a motor with 0.3mm trace, and then the board will fry. It doesn't know decoupling capacitors have to be placed close to the load and routed tight. It will also tend to create long winding traces for power/ground and then it'll be happy that everything is connected together... but the extra impedance of these traces means you can get problems.

So it is important to think about placement, and put the components that need short traces where they should be. For example, a trace inductance of say 20nH is significant relative to a decoupling cap, to the point of making the decoupling cap useless but it is negligible when it's in series with a 1 kOhm resistor.

If placement is good, routing is usually easy. If routing gives you headaches, check the placement. For example I see a neat row of resistors on the screenshot, this looks nice but you get lots of hectic traces all over the place.

  • \$\begingroup\$ Yes, that was my first boost converter layout design :( Thank you for detailed answer I think I should redraw the layout this will be better for me. And I'll just follow the reference layout I didnt think it is important like this big... \$\endgroup\$
    – ratatosk
    Nov 15, 2020 at 13:32
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    \$\begingroup\$ Yeah, switching converters are very unforgiving due to high di/dt making every bit of inductance into a problem... \$\endgroup\$
    – bobflux
    Nov 15, 2020 at 13:42

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