Voltage regulator output increases significantly when step-up converter runs

Question

I'm using the LTC3108 step-up converter to charge a supercapacitor from a thermoelectric generator. The output voltage drives a 1.8V regulator (S-1318) which is enabled once the voltage reaches 4V (MIC2779 is used to enable at 4V and disable at 3.3V). When enabled, the regulator drives a microcontroller and sensors.

If I directly power the board using a 4.1V external input applied directly across the output capacitor, the regulator functions as expected. However, when the input is coming from the step-up converter input, the regulated output increases to about 3.8V. It is not clear to me at all why this is happening, I'm not really sure where to look. I don't see any obvious coupling between the two. The output voltage is pretty steady (see plot) so I wouldn't think it is due to high frequency interference from the converter operation.

Any ideas on where to look?

Edit 1:

Here is a schematic, top copper layer, and board CAD rendering (PDF in link, images below). The board is a four layer design stacked from top to bottom as Signal, GND, Vout, Signal.

Edit 2:

A bit about the test setup: in order to test board functionality, the input is not actually a thermoelectric device, but rather a power supply. A voltage divider is used to supply a ~60mV input.

Edit 3:

I think this has something to do with whether or not there is a substantial load. When the MCU is asleep, the voltage increases as per the previous graph. When it is awake, the output is a stable 1.8V. I actually noticed that the output regulation seems to work even while asleep when I had the programmer adaptor connected (see picture). I did this to facilitate measurement (instead of holding a lead to the test point) and noticed it was gone. I imagine the additional resistance in the wiring was enough of a load to stabilize the voltage, there are no other components (see photo below).

So it seems when the capacitor is fully charged (the step-up converter should want to hold it at around 4.1V) maybe the converter is still extracting energy and that charge is building up in, e.g., C12 on the regulator output? I'm not really sure how that would be possible, but it seems like having a large resistor on the output of the regulator would solve that issue. This sounds like a hack though.

Edit 4:

When unloaded (to induce the 1.8V signal failure mode) the Vstore remains near 0V, maybe getting up to 0.5V max. Maybe somehow charge is instead accumulating elsewhere (like the 1.8V line). Also of note from the manual:

An internal shunt regulator limits the maximum voltage on VAUX to 5.25V typical. It shunts to GND any excess current into VAUX when there is no load on the converter or the input source is generating more power than is required by the load.

Which is interesting: where is this charge really going if the device is floating? If it's generating electricity but doesn't have anything to do with it, is charge just accumulating on the board then?

Edit 5:

With no input to the boost regulator, but a charged Vout capacitor, connecting and disconnecting the programmer cable (my "load") causes the system to toggle between the proper 1.8V regulation (connected) and the floating to 3.8V failure mode (disconnected). The regulator datasheet doesn't indicate this should happen, although load regulation is only rated down to 1µA current (the MCU sleep should draw 300nA). Maybe this is just what the regulator is going to do with such as small load. o_O'

Edit 6:

(At this point I'm just rambling and thinking out loud:) In the edit 5 tests, the connector wasn't being used to measure the voltage, the test point was, so it's not clear there's really any additional "loading". Unless the programming board is acting as a larger ground plane, disrupting capacitive behavior of the board's ground plane and signals, or discharging static buildup, something mysterious is still going on...

Answer 1

I see a possible leakage current path here:

1. MIC2779L has a push pull output that delivers the voltage of Vout to EN_1.8V if the range of Vout is valid.
2. TP3 will be Vout - one Schottky diode drop (D1)
3. A reverse leakage current through D2 flows into MCU_EN_1.8V
4. The internal protection diode or an enabled output of MCU pin 29 feeds this leakage current to the 1.8V MCU supply.
5. This small leakage current needs 10 s to charge C10 up to Vout - one Schottky diode drop (D1) = 3.8 V

If you connect a load of e.g. 1 Mohm to the 1.8 V supply, this leakage can flow to GND. The used programmer probably checks the MCU supply and connects a high impedance load there.

I have no explanation why this is prohibited using an external DC supply instead of the harvester circuit.

Which diodes are used for D1 and D2?

Answer 2

the regulated output increases to about 3.8V

The assumption here seems to be that the 1.8V regulator is somehow malfunctioning. I doubt that. I bet if you attached a series Schottky diode to the regulator output, the regulator output pin voltage woul never be above 1.8V.

Any ideas on where to look?

Most likely something feeds back from the battery to the 1.8V circuitry, but it's hard to tell. Look at anything that is connected to both the 1.8V logic supply and the 4V battery voltage. Especially look at is connected between the step-up regulator, and perhaps its input, and the 1.8V side.

Other than that, this answer can only go so far without OP providing the information needed to avoid this kind of tea leaf divination.