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I have integrated a BME680 temperature, pressure and humidity sensor on a PCB with NL27WZ17 Schmitt triggers (U3 & U4), a RT9193-33GB LDO regulator (U1) and a CM1224-04SO ESD diode (D2). I also have 33 Ohms series resistors and 100 Ohms termination resistors.

The system starts reading temperatures at around 20°C which is what two other different temperature sensors just next to it are giving me as well, but the temperature reading from the BME680 increases and stabilizes around 21 or 22°C linearly and systematically over a couple of minutes.

The BME680 sensor is geographically isolated from the ground and +5V planes. Gas measurement is turned off and I have tried triggering the BME680 every 10 seconds and every 60 seconds, with the same result. I have also removed the sensor from its enclosure which has holes anyways, and the result is the same.

Here is the current consumed by each component according to their data sheet:

  • BME680 : 3.7 μA at 1 Hz humidity, pressure and temperature
  • CM1224-04SO ESD diode: 8uA
  • RT9193-33GB: 90uA
  • NL27WZ17 Schmitt trigger: 100mA *2
  • resistors: Unknown

Unfortunately, I do not own an infrared camera. Do you have an idea what might be causing the +1.5°C rise in temperature? Is it the Schmitt triggers? The termination resistors? Or the diode/LDO regulator? Would I be better off powering the board directly with 3.3V (over 2 meters with 22AWG wire) instead of 5V?

Alternatively, do you see a firmware solution to avoid this overheating and that doesn't involve compensating?

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    \$\begingroup\$ Those powers are not the actual power dissipation - I suspect some of them are the absolute maximum dissipation before damage may occur. Measure the supply current. \$\endgroup\$
    – Kevin White
    Commented Jun 14 at 19:54
  • \$\begingroup\$ Those are not the power consumptions of the components, but the maximum they can possibly withstand before damaging and frying up. You have not stated the actual power or current consumption for any chip. Also the RT9193 is not a buck converter by any metric. \$\endgroup\$
    – Justme
    Commented Jun 14 at 19:57
  • \$\begingroup\$ Thanks for clarifying this confusion, I will edit my post to give the current consumption of the components instead \$\endgroup\$
    – Tommy95
    Commented Jun 14 at 20:04
  • \$\begingroup\$ Now use caution and common sense with what I am going to say. What components are hot / warm when you touch them? I'd start with my fingers close to components. \$\endgroup\$
    – StainlessSteelRat
    Commented Jun 15 at 13:44
  • \$\begingroup\$ It doesn't heat enough for that method \$\endgroup\$
    – Tommy95
    Commented Jun 15 at 14:08

2 Answers 2

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R1, R2 and R7 appear to have some logic voltage across them and they are very low value so they would be my first suspicion. I don't see any reason offhand why they could not be 100kΩ or even 1MΩ. For each one that has 3.3V across it you'll have about 110mW.

If you really need them for "termination" then you can try an AC termination scheme (put a ceramic capacitor in series with each) but I question the need for them at all.

The ST chips don't draw much current when they have a voltage near GND or Vdd applied to the inputs, maybe 100nA or 1uA.

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    \$\begingroup\$ You were right. I removed the 100 Ohm termination resistors, the system still works and the temperature is now stable and reliable. Thanks! \$\endgroup\$
    – Tommy95
    Commented Jun 15 at 15:02
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your Schmitt trigger's power dissipation depend on the load they're driving, and that is not part of your schematic; so you can't know what they actually consume. Note that the current they sink into that load has to come from your voltage regulator, so if the load is strong, that's probably the part that's getting warmer the fastest.

All in all, 2 K is not much.

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  • \$\begingroup\$ Thanks for your answer. So when no signal is going through the Schmitt trigger, it will not dissipate any heat? \$\endgroup\$
    – Tommy95
    Commented Jun 14 at 21:36
  • \$\begingroup\$ Since that thing doesn't have a "high-Z" output mode, the output is always driven; no such thing as "no signal passing through". It fully depends on your load / bus characteristic which output state has the lower quiescent current. \$\endgroup\$
    – Marcus Müller
    Commented Jun 14 at 22:37
  • \$\begingroup\$ The heat generated by the Schmitt triggers and possibly other components may be conducted through the PCB or radiated to the BME680 sensor, causing it to read higher temperatures. \$\endgroup\$
    – liaifat85
    Commented Jun 15 at 10:19

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