I have an old ceiling fan motor that runs with a 1.5µF run capacitor, at what I believe is, its full intended speed.

With the help of some folks here at StackExchange I've wired everything up as in the image below, also refer to the schematic further down this question. The motor seems to be running well.

enter image description here

The next step is to add a 4 way rotary switch that selects between off and 3 different speed settings. I need to figure out the capacitor values I can use in C3 and C4 in the schematic below to achieve the slightly slower speeds for speed settings 1 and 2, with 3 running at full speed.


enter image description here

I do believe that the capacitors need to be in parallel for the capacitance to add up, and slow the motor down, please do correct me if I'm wrong. As far as I can understand the capacitance shifts the phases apart further in the two coils.

The simplest approach might be to go and buy some 1µF to 1.5µF caps and play around with different configurations. I already know I need a 1.5µF to get and keep the motor running.

Is it safe to say I can just add another +/-1µF/1.5µF in parallel for my 1st speed setting, and another 1µF for the 2nd setting to slow the motor down? Thus, I have the following configurations:

Speed = 1 : 1.5µF

Speed = 2 : 1.5µF + (C2)1µF = 2.5µF

Speed = 3 : 1.5µF + (C3)2µF = 3.5µF

If this bit of information helps, our supply is 220V at 50Hz, and the motor has, what appears to be 14 coils, according to my counting through the holes in the motor casing.

  • \$\begingroup\$ I think your best bet is experimentation, it would be difficult to do an analysis on a circuit such as this. The reason for this is you don't know the inductance and the inductance is also dependent on the back EMF of the motor. Don't burn anything up :) \$\endgroup\$
    – Voltage Spike
    Feb 17 '16 at 21:03
  • \$\begingroup\$ Your profile lists you as living in South Africa (lekker ;), so your mains supply is actually 50Hz, not 60Hz. I would have guessed that reducing the capacitance (not increasing it) would slow the motor down. And I think there's an error in your diagram where you've linked the switch common point to mains neutral (bypassing the switch, extra caps & run coil). \$\endgroup\$
    – brhans
    Feb 17 '16 at 21:09
  • \$\begingroup\$ Thanks @brhans. I've edited my post to at least fix the supply frequency, before I look like an even bigger poepol. Not too sure about the circuit myself, that's why I'm asking. ;) \$\endgroup\$ Feb 17 '16 at 21:19

The circuit I believe you're looking for is something like this:
(please excuse the odd symbols for the switch & motor ...)


simulate this circuit – Schematic created using CircuitLab

With the switch in the Low seting, only C1 is connected in the circuit.
When the switch is set to Medium or High, either C2 or C3 will be connected in parallel with C1, giving you a larger capacitance which (I believe ...) will produce a faster fan speed.
It would not be uncommon for the capacitance value for the Low speed setting to be low enough that the motor won't start spinning by itself.
Ceiling fan speed control switches are usually wired such that the switching sequence runs Off - High - medium - Low - Off, so that the fan starts up with the full-speed capacitance to get it going.
You'll probably need to do a bit of experimenting with capacitor values, but a rough guess would be to choose a full-speed total capacitance of about 10x the value which just barely allows it to start.

  • 1
    \$\begingroup\$ +1 for the circle! There's a real circle at the bottom of the Arial font near the ▲▼►◄ symbols - both hollow and solid - ○●. You can pick them up using CharMap from the Windows Run prompt. \$\endgroup\$
    – Transistor
    Feb 17 '16 at 22:11

Full marks for persistence on this one, Josef.


simulate this circuit – Schematic created using CircuitLab

Figure 1 and 2 - depending on switch availability.

I think you'll find that the more capacitance you add the faster the motor will run. Due to limitations in the schematic editor I've drawn the contacts of the one four-position switch as three separate contacts. For Figure 1 the switch sequence would be:

SW1   SW2   SW3   Speed
Off   Off   Off   Off
On    Off   Off   Low
Off   On    Off   Medium
Off   Off   On    High

The original device probably had a fancy contact arrangement like this:

Posn  SWA   SWB   Speed
0     Off   Off   Off
1     On    On    High    <-- best starting torque on switch-on.
2     Off   On    Medium
3     On    Off   Low

As far as I can understand the capacitance shifts the phases apart further in the two coils.

Not quite right. The phase shift is primarily required to determine direction of rotation (as covered in your original question). This is an induction motor and it rotates by currents induced in the rotor interacting with the rotating field. The rotor doesn't quite keep up with the rotating field in full speed mode and the smaller the capacitor the more slip there is. Reduce the capacitor size enough and the motor won't start reliably.

Borrow some capacitors from your local washing machine repair man.


More capacitance increases both the current and the phase shift of the auxiliary winding. Both the increased phase shift and the increased current allow the motor to produce more torque (make the motor stronger). With no fan blades attached, there is no load connected to the motor, so it will start and operate at nearly full speed with very little capacitance connected. It will be difficult to see any speed change without any blades attached. Reducing the capacitance makes the motor weaker, allowing the load to slow it down.

The usual switch arrangement is to use a 4 position switch that has an off position and three speed connections as shown below. The switch connects first both capacitors for high speed, then just the larger one for medium speed then the smaller one for low speed.

If you use capacitor values that are too high, the auxiliary winding may draw too much current and overheat.

enter image description here


"when you increase the capacitance, the voltage across the capacitor decreases but that across the fan motor increases. Accordingly, the speed of the fan increases.

In other words, you need to increase the capacitor value to increase the fan speed. However, since there is no power loss in the capacitors, there is no heat generated, and consequently, no extra expense.


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