How to slow down a 220V fan?
Ok, when you slow down a motor in generally draws more current. At stall most motor types draw maximum current... which they generally are not designed to do for any significant length of time.

side note: If the motor is a 2 phase AC induction motor a then one phase must be 180 degree's out of phase. All 1 phase motors actually are 2 phase and the 2nd phase acts as a small "Starter". The second phase is hooked up to a capacitor to provide a voltage 180 degrees out of phase with the original and it provides a sort of "kick start" to the motor. If you don't hook up this phase then the motor just vibrates back and forth and you have to turn it to get it to start(once it gets going it's own momentum keeps it going).

The post you refer to says use a capacitor because at ac a capacitor has impedance. This can reduce the voltage to the motor. This also causes a phase shift. This is a not a great way to do it but is quick and efficient. A resistor will do the exact same thing except it will waste power and reduce the overall torque since it will reduce current(of course for a simple fan it doesn't matter that much).

Your motor has an effective resistance of 220/0.6 ~= 350ohms.

If you add a resistor of 50 ohms then you will have an effective load of 400 ohms which will reduce the current by 220/400 = 0.55 and the voltage across the fan will be 190V. The resistor will dissipate 0.55*50 = 25W. I'm not sure how slow the fan will go though.

But notice that you are wasting 15W of power. 

A capacitor has an impedance to ac of 1/2pifC. f = 60(or 50). 

If 

C = 1/2/pi/60/50 = 50uF

then if you use a 50uF cap you'll have the same effective resistance as a resistor but dissipate 0W(wel, almost) in it. It is much more efficient to do this. A 50uF in series with the fan(220V---Cap--- + fan () fan - ----GND) will have an impedance of 50Ohms. That is, it will reduce the voltage across the fan to 190V and the current through it to 0.55A.

The cap has to be rated for the appropriate voltage of a few hundred V's. If you increase the capacitance you reduce the impedance. A 150uF cap will have 1/3 the impedance and a 5uF cap will have 10 times. 

i.e., a 5uF cap will have 500Ohm impedance and will drop the voltage and current to over 1/2(since the fan was about 350ohms).

Note that a motor is a non-linear load so it is not possible to calculate exactly what it will be without knowing the exact behavior of the load.

It looks like you want to play around capacitors of about 50uF, 30uF, 15uF, and 5uF. Just make sure they are probably rated for the voltage or they may go poof. You should be able to find starter caps(used for the things in the side note) that will work fine and come in these ranges and voltage ratings.

Just in case you don't know:

Impedance is like a frequency dependent resistance. It is measured in ohms. Since our frequency is fixed, we can treat all impedances like resistances mentally. Capacitors and Inductors have non-constant impedance curves. A cap's "resistance" is 1/(2*pi*freq*C). So, you can think of a cap like a resistor but it doesn't reduce the voltage of all frequencies evenly. Since we are only dealing with a single frequency here that doesn't matter and 1/C is like R(but we have a proportionality factor of 1/2pif = 0.0026 in front of it).

So 1uF cap is like the same as a 0.0026*1/10^(-6) = 2.6kOhm resistor(@ 60hz AC). 

The other big difference is that capacitors and inductors do not dissipate power(because a cap has infinite "resistance" at DC and an inductor has 0 "resistance" at DC),

So if you stuck a 1uF cap in series with something that has 60Hz AC flowing through it then it would be identical to using a 2.6kOhm resistor EXCEPT that the cap will dissipate no power(ideally of course).

Cap Networks

Suppose you do not have the right capacitors to do the job.

There are two configurations to use


Parallel: combing caps in parallel adds there voltage but does not change the voltage rating. Two 10uf caps rated at 100V in parallel is equivalent to one 20uF cap @ 100V.
Series: combing caps in series divides the voltage but adds the voltage rating. Two 10uF caps in series rated at 100V is equivalent to one 5uF cap @ 200V.


BUT! There is a caveat to series caps. You have to put a resistor in parallel with each cap(across it's terminals) so that the cap can discharge or drain. If you don't do that strange things can happen do to capacitor leakage and ESR and your capacitors could prematurely fail(basically each cap won't have 100V across it but maybe 120V and 80V or something else). 

The resistors need to be small enough to prevent this unbalancing and it depends on the cap size and is hard to define accurately. A few k each is probably ok. Another way to think about it is that you have a resistive voltage divider and capacitor voltage divider but you use the resistive voltage divider to make sure the capacitive voltage divider is actually dividing the voltage in half.

 R C 
 | |
 +-+ 
 | | 
 R C


The +'s should be Vcc/2 but without the resistors across the caps it may slowly drift off to something not even close. The resistors, if they are low enough resistance, prevents this.