I'm trying to repair a floor fan that seized up.
I've got all of the galling removed, and the motor spins freely.
The fan still doesn't work, and I suspect the capacitor, it is a CBB61.
How do I test it?

 

What is it's capacitance rating? eg, 10uF, 4.7mF, etc.

Do you have a multimeter?

Do you happen to have a selection of resistors around? A 1 MEG resistor would be handy about now.

 

Have a look at this chapter in our E-book, if you're wondering where I'm going with this:
http://www.allaboutcircuits.com/vol_1/chpt_16/2.html

 

It's a 2 uF +- 5%, 300 volt AC, 50/60 hz.
Yes, I have a multimeter, and I do have some resistors, I'll have to look for a 1 meg.

I'll also check out the link and your test method (tomorrow).

Thanks,
Gary

 

I found a 1 Meg resistor, it's actual reading is .999.
What's next?

 

Do you have a DC power supply?

See the attached.

Short across the capacitor leads to discharge it completely.

When the short is removed, it should take approximately 3 seconds to get within 4/5 of the supply voltage. If it's a lot less than 3 seconds, the cap is open. If it never comes up near the supply voltage, it may have excessive leakage current.

 

SgtWookie,

I'm not sure what I'm looking at in the drawing.
My fan capacitor has two wires coming out of it.

I can see that V1 is 10 volts positive, coming in to the resistor, and then the resistor hooks up to one wire of the capacitor. I'm guessing the 10 volts negative of the power supply then hooks into the other wire of the capacitor?

I don't have a switch, should I add one into my circuit?

Thanks,
Gary

 

SgtWookie,

I'm not sure what I'm looking at in the drawing.
My fan capacitor has two wires coming out of it.

I can see that V1 is 10 volts positive, coming in to the resistor, and then the resistor hooks up to one wire of the capacitor. I'm guessing the 10 volts negative of the power supply then hooks into the other wire of the capacitor?

That is correct.

I don't have a switch, should I add one into my circuit?

It's OK if you don't use a switch. It might be easier if you did.

You want to make certain that the capacitor is completely discharged, and then see how long it takes to get to 4/5 of the supply voltage.

 

Using a 12 volt PS with 17 v actual reading, and using a 5uF cap,

The readings were
3 seconds = 5v
10 sec = 14v
11-12 sec = 15v
15 volts was the max.

Would you say the cap is good?

I'm now going to try the 2uF at 9v, to see how close it is to your table.

Thanks,
Gary

 

SgtWookie,
Here are the new stats, using the 2uF, with 9v PS
I set my meter to read 10ths instead of just whole numbers.
These are interesting enough that I will go back and retest the first cap later.
1 sec = 4.8v
2 sec = 6.8v
3 sec = 7.8v
4 sec = 8.4v
5 sec = 9.3v
6 sec = 9.6v
7 sec = 9.7v
8 sec = 9.8v
9 sec = 9.8v
10 sec = 9.9v

with 9.9 volts being the max.

A great lesson, Thank you so much.
Gary

 

They sound pretty close to what they should be. It's hard to be accurate for the shorter time periods, particularly if you're flipping switches manually.

One thing you really can't test them for without more equipment is leakage at the rated voltage.

 

When checking capacitors, a general rule is that 5 time constants (RxC) will take you to about 98% of the final charge or discharge figure. This also applies to inductors as well. With a votmeter or a scope, you can easily test this rule.