I made power supply using TPS61025. It's a 3.3-V Output, 1.5-A Switch Boost Converter.

I have chosen relatively large capacitors as my circuit drives a power hungry WIFI chip. I have two new AA batteries (2.7V). Without load, output voltage is 3.4V. But when I use a 5.5 ohm resistor as load, the voltage drops to 330 mV and current is about 60 mA.

My inductor is rated for 1.6 A. Per IC data sheet, it can deliver over 1A. I checked soldering and it seems OK.

What's the possible cause?

TPS61025 PCBA picture. yellow lines enclosed power supple circuit. No parts on the other side.

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    \$\begingroup\$ How much inductance? What is the input voltage under load? \$\endgroup\$ – winny Jan 21 '17 at 20:42
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    \$\begingroup\$ Please supply close-up photos of the top & bottom of your circuit board (or, if this is part of a larger board, then the photos need to show the components on your schematic above). \$\endgroup\$ – SamGibson Jan 21 '17 at 23:14
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    \$\begingroup\$ 1)Inductance is 6.8 uH (Murata LQH5BPN6R8NT0L). 2)EN is connected to BATT. 3)Section of datasheet recommends that ESR of output capacitor (C3) be larger 33 mOhm. I have used a ceramic capacitor ( SAMSUNG CL32A227MQVNNNE). As ceramic caps have low ESR, I added a resistance in series to be on the safe side. \$\endgroup\$ – Amir Samakar Jan 22 '17 at 1:28
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    \$\begingroup\$ I measured BATT voltage. When 5.5 ohm resistor is connected as load, BATT voltage is 2.12 V. When there's no load, BATT voltage is 2.41 V. \$\endgroup\$ – Amir Samakar Jan 22 '17 at 2:15
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    \$\begingroup\$ Would be really nice to see scope captures of Vin, SW, and inductor current... \$\endgroup\$ – Matt Young Jan 22 '17 at 5:04

Circuit looks OK. I am going to guess that you have not soldered the ground slug (datasheet calls it "PowerPad") at the bottom of the IC to GND at all. Besides the thermal connection, the IC may be relying on a low impedance electrical ground on this ground slug.

Best would be to redo the IC soldering and use a hot air reflow station to get the ground slug soldered down to the GND plane.

An alternate hack would be to remove the IC, scrape off the soldermask so you get exposed GND plane copper next to the IC all the way to the slug area, cleaning the board and putting flux on the board and IC bottom, and then use a really huge tip on a soldering iron to reflow the solder so it creeps underneath to the IC's ground slug. You will need a huge tip and the iron temp cranked up to heat up the ground plane. You need good heat, a clean board with flux to help wick the solder under there.

  • \$\begingroup\$ There's a large pad under the IC (PowerPad). I have a corresponding copper pad in my layout but it's not soldered. (Because of hand soldering limitations.) Per section 6 of datasheet it "Must be soldered to achieve appropriate power dissipation. Should be connected to PGND." In my application the normal MCU current is 80 mA with power spikes up to 600 mA. So I thought I can live without "PowerPad" soldering. \$\endgroup\$ – Amir Samakar Jan 22 '17 at 18:37
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    \$\begingroup\$ Do you have an oscilloscope? A good way to check is to look at the SW node relative to GND. The switching should have a nice even duty cycle. If the duty cycle is erratic, then a common cause is a poor ground issue which injects noise into the error amp loop or current limiting loop. I've had this issue and fixing the layout corrected the problem. Read the article in this link. Figure 2 is what an erratic control loop looks like: eetimes.com/author.asp?section_id=183&doc_id=1273264 The pulses should not be fat/skinny/fat/skinny like this. \$\endgroup\$ – Vince Patron Jan 23 '17 at 1:58

A standard AA battery is assumed. Short answer is: revise your expectations.

With an output voltage of 3.4 volts and a load of 5.5 ohms you should get a current of 619 mA and this will translate to a current from the 2.7 volt input of about 900 mA if power transfer efficiences are taken into account (approximately 15%).

When this much current is drawn from an AA battery the terminal voltage drops possibly as much as 0.2 volts per cell and the result of this is that the input voltage instead of being 2.7 volts is now more like 2.3 volts.

With 2.3 volts at the input to the boost regulator, what current can you expect to be taken if the output is sustained at 3.4 volts. It will be more like 1.06 amps and this in turn modifies the the terminal battery a bit more and there is an iterative process that ends up with possibly only 2.1 or 2.2 volts at the input and rapidly the batteries are running out of steam to sustain things.

Given also that the output current of the TPS61025 is about 200 mA (front page circuit of the data sheet), I think you are expecting too much: -

enter image description here

  • \$\begingroup\$ Although I agree the current capability of the battery could well be limiting the output voltage in this case (+1), imho it isn't as clear-cut as "the output current of the TPS61025 is about 200 mA". That limit only applies at the 0.9V end of the battery voltage range (1 cell) shown on the image you linked, but I see no indication that the OP's battery voltage is so low (they are using 2 cells). Looking at Figure 1 (datasheet page 6) shows that for a target Vout of 3.3V and Vin ~2.1V, the IC's max current is ~700mA. Therefore imho something else may be limiting Vout with Vin ~2.1V. [cont'd] \$\endgroup\$ – SamGibson Jan 22 '17 at 13:35
  • \$\begingroup\$ Of course when the battery voltage drops, then as seen in Figure 1, the IC's max current also reduces and this may be a problem for the OP's planned usage - but I think not in this test. Or have I misunderstood? \$\endgroup\$ – SamGibson Jan 22 '17 at 13:35

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