I have a Pololu S18V20ALV SEPIC voltage regulator board that is wired to a high-amperage 5V USB power source on the input and a DC brushless fan on the output. The output is set to about 5.5V. Contrary to the typical behavior of a power supply where higher temperatures reduce maximum output, I've noticed that it delivers more power when hot:

  • If I heat up the regulator by shorting the output pins for about ten seconds, then remove the short, the fan runs faster than it otherwise would.
  • If I hook up a 10A ammeter to the output, the meter readout progressively increases from less than 1A to as high as 3A as the voltage regulator heats up. The limit is due to the thermal shutdown function on the TPS55340 regulator chip.

Why is my voltage regulator doing this? Is this behavior typical for a SEPIC converter? What implications does this have? (Does this mean that an overloaded SEPIC converter without thermal protection can experience thermal runaway and consequently catastrophic failure?)

  • \$\begingroup\$ It would help if you put a multimeter on the output and actually measured the VOLTAGE of your voltage regulator. It could be as simple as the feedback resistor network heating up and drifting. \$\endgroup\$ – Daniel Jul 13 '16 at 2:35
  • \$\begingroup\$ Oddly enough, output voltage under load does not measurably change with temperature. However, with the load disconnected, if I short the output for several seconds and remove the short, the no-load voltage spikes by several tenths of a volt before settling to the set voltage. The voltage after settling isn't any different due to temperature. \$\endgroup\$ – bwDraco Jul 13 '16 at 2:43
  • \$\begingroup\$ Also, the measured voltage at the output pins is extremely stable whether under load or otherwise, unless the supply is overloaded. However, as I don't have an oscilloscope, I won't be able to get a closer look at noise or other output characteristics. \$\endgroup\$ – bwDraco Jul 13 '16 at 2:52
  • \$\begingroup\$ We don't necessarily know that it's heat ... with the output shorted, are there feedback signals with time constants (capacitors) in them? The error signal may be winding up to compensate for the low output voltage, and taking time to return to nominal condition, during which the output is overvoltage. \$\endgroup\$ – user_1818839 Jul 13 '16 at 13:27

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