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The attached image is of a ceiling fan speed controller module I brought online. It seems the attached circuit mainly consists of some capacitors. This makes me believe that the 4 different settings on the switch alters capacitance, rather than giving out a different voltage?

enter image description here

Has anyone worked with a ceiling speed control switch like in the attached images before, that can shed some light on it's operation? I need to figure out how to connect this to a ceiling fan motor I'm fixing.

I don't have the required multimeter to measure capacitance, and taking resistance readings on the 4 different settings only delivers a reading on setting 4, and nothing on 1-3, not even continuity.

I'm not brave enough to connect this switch to 220V (with load of course to prevent a short) and measure it with my multimeter.

I guess I won't get a resistance/continuity reading on a cap since it's not charged and not letting current through, right?

enter image description here enter image description here

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    \$\begingroup\$ Not a voltage divider, but looks like it switches 3 values of capacitor to a small split phase motor. (Those look like capacitors poking out of the heatshrink, which would explain the lack of DC resistance). \$\endgroup\$ – Brian Drummond Mar 10 '16 at 18:44
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Here (from this source) is a fan speed switch that uses a capacitor and the wiring diagram:

enter image description here

Often the capacitors are all in one package, and there may be bleeder resistors across the capacitors.

This type of motor is called a permanent-split capacitor motor - permanent because the capacitor remains connected after starting.

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I got a ceiling fan with a remote control. I opened up the receiver and although the motor is single phase (2 wires), the system does indeed pass the current through one cap out of three to change speed.

I wrote the values down somewhere to use a relays module and ditch the remote for a wall control. 155k/400v and 200k/400v. The fan is 230VAC. I did not notice a fan on the motor itself.

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  • \$\begingroup\$ Welcome to EE.SE, Marc. It's important on this site to answer the actual question. Your first paragraph is a good start. The second seems to be irrelevant. \$\endgroup\$ – Transistor Jul 30 '18 at 21:15

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