2
\$\begingroup\$

I am designing a circuit to control a multiplexed led matrix and for that I am testing two boost circuits. The TPS61046 and the TPS61080.

My idea is that the input voltage is via USB or a Li+ coin cell (probably [email protected]).

The attached schematic shows the power supply stage, the ground is common between both boosts but only one is energized at a time and the other is kept disconnected. (possible first error)

enter image description here

Since my priority is to run the TPS61040 due to its size and better performance in the current range I require (10 - 30 mA) I will focus this question to the error I get from it.

When I test its operation supplying it with a laboratory power supply set to 3.7V max 0.1A I get the consumption of [email protected], that is, the TPS61046 is consuming -IMO- an excessive amount of current. I must say that the loads so far have been three leds in parallel, each with a 400 ohms resistor and the other was an Arduino mega. I havent conected the whole circuit due to potential smoke risk.

In the attached picture you can see the PCB layout I made, R10 and R20 are not connected and R21 is a 0Ohm 1/4W resistance, this way FB with Vin ar directly connected, thus 12V can be obtained only with 3 components C_in(C6&C7)= 2x1uF, L(L2)= 10uH, C_out(C5)= 10uF (Datasheet 9.3.1).

enter image description here

Something that can cause another failure (possible second error) is the distance of C_out (C5), which maybe should be closer to Vout, but even so, I think the effect should be reflected in the ripple and not in the difference of the voltage I get to the desired one.

I have checked the continuity between all the pins and the TPS is well soldered. GND-Vin diode shows 0.5V (possible third error). I will also do a re-work on the inductor to be sure there is no short underneath.

Wrap up: Using the TPS61046 I want to get 12V output but I am getting a lower value with a very high current consumption.


SOLUTION: IC wasn't able to start up. By increasing the current limit a little it worked flawlessly.

Thanks, @Andy aka

\$\endgroup\$
4
  • \$\begingroup\$ on the schematic on the left, you have INPUT voltage fed into feedback. They're NOT supposed to be connected. EVER. Feedback is meant to feed output back into the chip so the chip can actually see and control what it's producing. Remove the 0 Ohm resistor ENTIRELY. Also, 1MOhm resistor is too high, may catch some noise. Go down a few orders of magnitude, at least one. Better not to have them higher than 100k \$\endgroup\$
    – Ilya
    Commented Mar 13, 2021 at 10:54
  • 1
    \$\begingroup\$ @Ilya Usually that is true. However this particular chip has an internal feedback arrangement that will output 12V when FB is connected to VIN. \$\endgroup\$
    – Justme
    Commented Mar 13, 2021 at 10:56
  • \$\begingroup\$ Why not just use normal configuration? There already is a voltage divider on the output, so remove 0Ohm and it will work correctly (as long as there are correct resistors to produce correct FB voltage). There is input connected to output through the resistor now. Some weird overcomplication \$\endgroup\$
    – Ilya
    Commented Mar 13, 2021 at 10:58
  • 1
    \$\begingroup\$ @Ilya, as Justme said, what you state would be true, however, this IC comes with an internal voltage resistor that automatically outputs 12V (which is the voltage I need), In consequence, the system implementation is simpler. also, as you can see from the PCB image I provide, The physical voltage divider is not connected but is there. \$\endgroup\$ Commented Mar 13, 2021 at 11:54

1 Answer 1

2
\$\begingroup\$

When I test its operation supplying it with a laboratory power supply set to 3.7V max 0.1A I get the consumption of [email protected], that is, the TPS61046 is consuming -IMO- an excessive amount of current.

If your output voltage is meant to be 12 volts and feeding 3 parallel LEDs each with a current limiting resistor of 400 Ω then, each LED will take a current of approximately 10 volts ÷ 400 Ω = 25 mA. With three parallel circuits, that's a loading current of 75 mA.

So, that's an output power of 12 volts x 75 mA = 300 mW.

That's going to require an input power of about 333 mW (taking into account an efficiency of 90%).

So, if your input voltage is 2.4 volts, the current needed from that supply is going to have to be at least 139 mA. Hence, you have set your current limit too low in value for this circuit to start up properly.

Maybe you need to re-examine you current limit setting and be a bit braver.

Also, your L2 inductor; it doesn't look like you fitted the correct one as per your data sheet link. The DS implies a footprint size of 2 mm x 2.5 mm yet, in your picture it looks a little small.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Thanks @Andy aka. I completely forgot the notion of power. I changed the load for a 1.2k resistor and a led (to get 10mA) and it works perfect, even though the consumption to my taste is a bit high (40mA) I still have to work on that. thanks for the push! \$\endgroup\$ Commented Mar 13, 2021 at 12:11
  • \$\begingroup\$ With respect to the inductor, since it might be the most important component in the boost circuit, I tried to make "one footprint to fit several". Since my main restriction is the overall area, I gave priority to small chip inductors at the cost of DC_resistance. In case the outcome wasn't good, I can fit a bigger one without the need to make a new PCB. The implemented one is this and the big one is this one between others. \$\endgroup\$ Commented Mar 13, 2021 at 12:31
  • 2
    \$\begingroup\$ @Smart_Celery unfortunately, the saturation current rating for the small inductor only appears to be 100 mA and so you mill likely have problems with that part. In future, try and link directly to English data sheets. The bigger inductor is certainly suitable so I would definitely try for an inductor that has a saturation current rating of twice the DC current expectation into the input pins of the chip. \$\endgroup\$
    – Andy aka
    Commented Mar 13, 2021 at 12:46
  • \$\begingroup\$ Sorry for the datasheet in Japanese, I trusted that mouser would have the correct one. Regarding the inductance value, what an eye you have. In total I ordered 5 different inductances with an Isat >800mA, and just the one I soldered mouser published a wrong value. I trust too much in the vendors :/ Thank you very much for your help. \$\endgroup\$ Commented Mar 13, 2021 at 18:26
  • \$\begingroup\$ Always read the data sheet is 100 % good advice. \$\endgroup\$
    – Andy aka
    Commented Mar 13, 2021 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.