I am currently working on a heated glove project using a Buck-Boost converter to adjust the temperature of the glove. I'm using the LTC3112EDHD#TRPBF QFN Buck-Boost IC to control the Buck-Boost functionality. I just finished soldering my test PCB, and my Buck-Boost Converter IC isn't outputting a voltage. My multi-meter is reading a voltage output of 0V. I've included my schematic design below. The PCB layout is exactly the same as the schematic. Some of the ports on the IC are controlled by a micro-controller.

enter image description here

I honestly do not know what could be the issue with this circuit. My two guesses are that either some of the passive components got damaged during soldering (which I doubt. I confirm the resistors work properly, but can not confirm the capacitors functionality due to my multi-meters limitations), the Buck-Boost IC isn't solder correctly, or my design layout is incorrect. Any help would be greatly appreciated.

Edit: I've added the PCB layout of the schematic. It may be hard to see due to all of the connections, but I've included it as a reference.

PCB Design

  • \$\begingroup\$ You can check the individual capacitor using resistance mode. Check if it's high or low \$\endgroup\$
    – Aparna B
    Commented Jul 7, 2019 at 3:49
  • 1
    \$\begingroup\$ "I just finished soldering my test PCB" - hand-soldered QFN? If so then you may have a dry joint or solder bridge. \$\endgroup\$ Commented Jul 7, 2019 at 7:58
  • 1
    \$\begingroup\$ Jay - Welcome :-) You said: "The PCB layout is exactly the same as the schematic." While it may not be related to this question, it's not clear what you mean by the quoted text. The PCB layout requires certain characteristics (e.g. component locations, trace widths etc.) which are not shown in the schematic, and so the layout cannot be assumed to be valid, even if matches the schematic. With any SMPS question, it's a good idea to include the PCB layout (e.g. from your EDA software) and some photos of the completed PCB, in the question. \$\endgroup\$
    – SamGibson
    Commented Jul 7, 2019 at 16:49
  • \$\begingroup\$ Where is C10 and C13 located? \$\endgroup\$
    – winny
    Commented Jul 7, 2019 at 21:03
  • \$\begingroup\$ C13 and C10 are in the screenshot. C13 is right below the Buck-Boost, and C10 is to the left of the big via. \$\endgroup\$
    – Jay
    Commented Jul 7, 2019 at 23:31

2 Answers 2


Your DC-DC won't work at all without at least one of the feedback transistors being turned on.
When they are turned on the VCE(sat) is directly in series with your feedback voltage. While VCE(sat) will likely only be a few mV in this case it still has an impact on the accuracy of you V(out)

You should be extra careful that the transistors used in the feedback loop are extremely low leakage and not too high an Hfe. You don't seem to specificity the part number in your schematic so I can't check.

The CE leakage current when these transistors are off (no Base current) is typically higher than that quoted for the CBO leakage specified in the datasheet. The feedback resistors indicate that the feedback current is very small at only about 5uA. With only one of the three feedback switches on, it means you are dealing with 2x the leakage current.

If you are NOT driving the transistor bases, then the CE leakage will be approximately Hfe*I(CBO). With high Hfe devices this can be troublesome. You should ensure that the drive point for your base resistors is always driven by a logic zero or one, this ensures that the base is terminated to close to ground when turned off.

You might find this answer helpful.

Update: Now you have added the feedback switching transistor data it is clear that they are unsuitable for this task. They are power devices with a max I(CBO) of 1uA so even when off will impact the feedback loop.

If you can find something in the same package layout, then you are looking for characteristics similar to these ONSemi small signal devices, NST3904DP6T5G.

enter image description here

With I(CEX) of only 50nA these are a much better choice, and you can find transistors with I(CEX) down to 10nA. I(CEX) is the CE current with the base open circuit so is a much better indicator of leakage performance than I(CBO).

Ideally IMO the best choice choice for a switched feedback loop would be a small signal Depletion mode N-Channel FET switching from the output.

  • \$\begingroup\$ rohmfs.rohm.com/en/products/databook/datasheet/discrete/… Here is the datasheet to the BJT's. In my program, the desired transistor to be turned ON at Vbe is set to logic HIGH while the rest are set to logic LOW. So are you saying because of CE leakage, that can create some issues with the feedback input? \$\endgroup\$
    – Jay
    Commented Jul 7, 2019 at 12:36
  • \$\begingroup\$ Understood. Please excuse my lack of knowledge in FETs, but what is the difference between a N-Channel MOSFET and the small signal FET you described? \$\endgroup\$
    – Jay
    Commented Jul 8, 2019 at 2:22

I would add a comment instead of an answer but my "reputation" is not high enough.

Check the following:

(1) Vin, Run and PWM pins. (Start with PWM of 0% duty)

(2) Check the FB pin is connected to the voltage divider [This needs feed back to work]

(3) Check if SW1 and SW2, are doing anything.

Never mind. Check the PWM pin; it should be <0.5V.

enter image description here

  • \$\begingroup\$ Shouldn't the PWM/SYNC voltage >1.5V still work, just in fixed frequency mode? \$\endgroup\$ Commented Jul 7, 2019 at 23:54

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