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I have a fan that I want to plug into a thermostatically controlled power outlet so that the fan comes on when the temperature reading reaches a certain set point.

The fan is currently wired through an on/off switch which also included a switch for a motor that operated the oscillation of the fan. I'm not using the oscillation motor/function.

Originally I was thinking I could just cut the three wires going to the motor and the same three wires going to the switch circuit board, match colors and solder them together. But there are a couple of capacitors on the board. Now I'm not so sure that would work. What is the correct way to bypass the fan switch for direct power on?

Here are photos of the front and back of the switch board.

The button on the left operates the fan.

Circuit board

Circuit board K1, on the right, is the fan button.

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    \$\begingroup\$ the top picture is useless ... please post a good picture \$\endgroup\$
    – jsotola
    Jun 16, 2018 at 19:42
  • \$\begingroup\$ the fan is a "big unknown" .... please describe in detail how the controls operate .... also what happens when you disconnect/reconnect power \$\endgroup\$
    – jsotola
    Jun 16, 2018 at 19:45
  • \$\begingroup\$ When I disconnect power the fan goes off. When I reconnect power the fan does not immediately come on. Pressing the power button once turns the fan on high speed. Pressing the power button a second time sets the fan to low speed. Pressing the power button a third time turns the fan off. Sorry about the bad picture. Is there a component on or section of the board you need more detail about? \$\endgroup\$
    – Brad W
    Jun 16, 2018 at 20:32

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looking at the photo there seem to be some triacs (in semi-cylindrical TO92 packages) that do the actual switching of the current.

removing the circuit board, and connecting the white wires together and connecting the red power wire to either the back high-speed or the red low speed terminal should get the fan motor running.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks Jasen! Red is actually high speed and black low but that worked. \$\endgroup\$
    – Brad W
    Jun 17, 2018 at 15:22

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