I want to mount the "Pololu Carrier with Sharp GP2Y0A60SZLF" IR distance sensor with hot glue to my robot. I can't mount it with the hole.

Is there any problem (like heat) to add hot glue on the sensor to mount it?

enter image description here Pololu Carrier - Schematic diagram

  • \$\begingroup\$ Plus one for providing schematics and PCB picture. No, I don't see any issues. Components are build to withstand a hot environment for longer and hotter then your glue will be. \$\endgroup\$ – Oldfart Apr 8 '18 at 10:20
  • \$\begingroup\$ Thanks for your comment. Do you see any problem that the hot glue prevent heat from disappearing (because resistors are isolated)? \$\endgroup\$ – Sascha Apr 8 '18 at 10:26
  • \$\begingroup\$ No, the thing uses 50mA max at 5V that is 1/4 watt distributed over the whole PCB. (And I suspect 90% of the energy goes into the IR transmitter) \$\endgroup\$ – Oldfart Apr 8 '18 at 10:37
  • \$\begingroup\$ I read in the datasheet "Please use an electric source with an output current of 400mA or more because LED pulse current is more than 300mA.". It seems the burst current draw is 300mA and the max average are 50mA. Make this any difference to your answer? Or is there still no problem because the IR transmitter uses most energy? Maybe you can add an answer, so I can mark it as the answer of the question. \$\endgroup\$ – Sascha Apr 8 '18 at 10:56

Hot glue melts at ~120C. By the time you have put it on the surface and pushed the PCB against it will cooled rapidly. Electric components are build to withstand 125C for short periods. The data sheet says: Soldering temperature Tsol 260 °C 5s or less/time up 2 times

I looked at the Electro-optical Characteristics in the datasheet which mentions 50mA. I found the note you mentioned: Please use an electric source with an output current of 400mA or more because LED pulse current is more than 300mA. That energy will not be at the bottom of the PCB as it is the led, sticking out at the top, which even has a plastic housing.

Sticking with 5V and the full 50mA of the chip (on the PCB) the energy is 0.25 watts. The energy will go mostly in the PCB which works as a heat sink. Your glue will not make a noticeable difference.

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  • \$\begingroup\$ I testet it for a few days and it works without any problem, thanks again. \$\endgroup\$ – Sascha May 3 '18 at 17:38

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