1
\$\begingroup\$

I have a circuit set up similar to that shown bellow which allows readings to be taken from an infrared sensor (MLX90614) using an Adafruit Feather HUZZAH ESP8266.

enter image description here

This works fine when the battery is connected to the USB port on my computer, but not to a 5V power bank (PURIDEA 10). I get readings of 1037.55 Celsius, which is a common error. I have confirmed that the voltage of the power bank is 5V with a multimeter. The sensor needs a minimum of 4.5V.

enter image description here

This image says there is discharging protection of the power bank. could this cause the voltage to decrease to lower than 4.5V?

I'm really unsure of why the power bank cannot supply the voltage. Thanks

\$\endgroup\$
  • 11
    \$\begingroup\$ This must be because of the cutting-edge-circuitry. \$\endgroup\$ – Nick Alexeev Apr 25 '18 at 20:18
2
\$\begingroup\$

Try putting a load resistor across the power pack output lines and see if it wakes up. Try ~20 Ohm for a short while, that would waste some 1.25 W so don't burn your fingers... If it wakes up, you could easily make a periodic load that keeps the pack alive:

schematic

simulate this circuit – Schematic created using CircuitLab

How often you need to pull the load needs some experimentation. There is of course a possible flaw with this circuit: it's not self starting. But that could be a topic for a new question.

\$\endgroup\$
1
\$\begingroup\$

Since this "power pack" thing is delivering power via a USB socket, it may not be turning on the power unless it senses a proper client device to power.

If I remember right, a device has to show particular resistances between specific lines to get full power from a USB port, even when just used as a charging port. Check the USB charging spec.

\$\endgroup\$
  • 2
    \$\begingroup\$ Also, most power banks have boost converters that do not produce low ripple output. You may require significant additional filtering or a high PSRR LDO (although you may not have the voltage margin for this) to get clean enough power for your sensor and ADC. \$\endgroup\$ – Dean Franks Apr 25 '18 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.