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I'am doing some small IOT device and want to use power source 2 x AA elements which gives ~3V. It should be power efficient, so in schematic included TPS61025 as output requirement is fixed 3.3V. First PCB revision didn't worked, possible of bad PCB layout. Could you check if such layout is optimal for TPS, any suggestions for improvements ? FB left floating as its fixed voltage TPS.

L1 used LQH32PN6R8NN0L
C2 used GRM21BR71A225KA01L
C3 low ESR 40mΩ T520B107M006ATE040
enter image description here enter image description here

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    \$\begingroup\$ Um, if you leave FB floating, how should the boost converter work? Nothing in the TPS datasheet, as far as I was able to scan it quickly, says you can leave that floating and it'll use an internal feedback... nothing's special about 3.3V, so how is the converter supposed to know what the right output voltage is without the feedback voltage divider? \$\endgroup\$ Commented May 15, 2020 at 8:31
  • \$\begingroup\$ @MarcusMüller its TPS61025DRC model, so has fixed 3.3V output. In datasheet "Programming (continued) The output of the low battery supervisor is a simple open-drain output that goes active low if the dedicated battery voltage drops below the programmed threshold voltage on LBI. The output requires a pull up resistor with a recommended value of 1 MΩ. If not used, the LBO pin can be left floating or tied to GND." \$\endgroup\$ Commented May 15, 2020 at 8:34
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    \$\begingroup\$ LBO is not the same as FB \$\endgroup\$ Commented May 15, 2020 at 8:35
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    \$\begingroup\$ @MarcusMüller With this chip you can actually leave FB floating if you buy a fixed output version. See 10.3.1. I would then remove the resistors entirely. \$\endgroup\$ Commented May 15, 2020 at 8:45
  • \$\begingroup\$ @Vaikis2006 my apologies, Jeroen is right \$\endgroup\$ Commented May 15, 2020 at 8:52

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You've kept it pretty close to the reference layout, which is good!
There is some confusion about FB left floating, which is apparently allowed with a fixed output part in this family.
It also looks like you kept the current loops short.

The input for it is the feedback voltage on the FB pin or, at fixed output voltage versions, the voltage on the internal resistor divider.

Just remove the resistors completely, maybe only add a small testpoint. Now it's an antenna for problems.

I'd also remove C1. The characteristics of the input capacitor are important, and C1 is not recommended by TI anywhere.
You know the switching frequency is 720 kHz max, the 100 nF is for ~10 MHz, so you're off a bit. eg: Switch places C1/C5 and replace C1 for a dnp for future BOM changes.
Do the same on the output C10. This gives flexibility in changing the capacitance or impedance. (or prototyping)

I can't calculate whether your chosen caps are a right part. I'd usually just try a batch that match the recommended part.

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  • \$\begingroup\$ C3 value is 100uF (not nF) \$\endgroup\$ Commented May 15, 2020 at 9:13
  • \$\begingroup\$ @Vaikis2006 My commentary is not about C3. \$\endgroup\$ Commented May 15, 2020 at 9:23
  • \$\begingroup\$ for what C then ? or you mean that for input 2 capacitors are bad idea ? I have seen such solution in other designs as well.... \$\endgroup\$ Commented May 15, 2020 at 9:26

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