My Buck-Boost Converter Circuit Isn't Working

Question

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.

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.

Answer 1

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.

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.

Answer 2

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.