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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?

Failure plot

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.

Board CAD

Top copper layer

Schematic

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.

enter image description here

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...

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    \$\begingroup\$ Schematic AND detailed photos of your build \$\endgroup\$
    – Kyle B
    Commented Aug 25, 2022 at 19:55
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    \$\begingroup\$ Since we have no idea how every thing is connected, we have no idea. \$\endgroup\$ Commented Aug 25, 2022 at 21:20
  • \$\begingroup\$ I added a schematic and top copper layer layout. I can add a photo of the board itself when I go to the lab, but for now I just included the CAD output. \$\endgroup\$
    – Anthony
    Commented Aug 26, 2022 at 14:04
  • \$\begingroup\$ What's the crossover between EN1.8V, R2, D5? \$\endgroup\$
    – winny
    Commented Aug 26, 2022 at 14:24
  • \$\begingroup\$ @winny The 1.8V regulator is normally disabled (pulled down with R2) but can be pulled high either by the voltage monitor (EN_1.8V through D1) or the MCU (MCU_EN_1.8V through D2) to enable the regulator. This is a holdover from the original design that used the PGOOD output from the boost converter which had a very small hysteresis, not enough to run the entire sensing and transmitting process, so it required allowing the MCU to keep the supply enabled until it finished. (Now that I have such a large hysteresis (4V->3.3V) it's probably not necessary.) \$\endgroup\$
    – Anthony
    Commented Aug 26, 2022 at 15:30

2 Answers 2

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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?

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  • \$\begingroup\$ This may be the problem... The pin is indeed configured to be an output. I removed D2 and the problem disappeared. The external power supply exception was probably invalid: I only connected it when I programmed the board, so I always had it connected to the debugger. (This was before I knew the plug was affecting things.) Now, the programmer exception is not quite right, because what I connected in these tests was just an adaptor board I made which maps the 10 pin connector to a 20 pin J-Link, so no load is actually connected. But this was obviously the wrong design for two enable inputs... \$\endgroup\$
    – Anthony
    Commented Aug 26, 2022 at 21:05
  • \$\begingroup\$ The diode in question is here: digikey.com/en/products/detail/comchip-technology/… \$\endgroup\$
    – Anthony
    Commented Aug 26, 2022 at 21:16
  • \$\begingroup\$ Yes, this diode is trimmed for low forward voltage and a bit leaky, the datasheet diagram shows 1-2 uA for your operating conditions. So you need a more complex circuit or special trick to enable the regulator via MCU. How about some HCT gate? \$\endgroup\$
    – Jens
    Commented Aug 26, 2022 at 21:27
  • \$\begingroup\$ You could simply add a third of these diodes in reverse direction between MCU pin and GND as temporary solution to lower the voltage there. A kind of "leak voltage divider". \$\endgroup\$
    – Jens
    Commented Aug 26, 2022 at 21:36
  • \$\begingroup\$ A gate was my first thought, but then I figured I could just use diodes which I already had to make it easy. This insane ground loop trajectory was not on my radar, but clearly I need to spend a little time looking at some representative GPIO pin circuits to fully understand why this is happening. Thanks for the great catch! \$\endgroup\$
    – Anthony
    Commented Aug 27, 2022 at 2:25
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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.

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  • \$\begingroup\$ My original assumption was the regulator was bad, so I swapped it and got the same results. Then I noticed connected to an external power supply the regulator functioned properly. When there's an input to the step-up converter the problem is when there's a problem, but aside from ground and the voltage input, there's no obvious (to me) coupling to the regulator output. I added a schematic and layout details. \$\endgroup\$
    – Anthony
    Commented Aug 26, 2022 at 14:01

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