I have an old 6ft fluorescent lamp that works for a few minutes but then suddenly makes a loud noise whereupon I quickly switch it off.
I suspect a component in the ballast is overheating.
The ballast uses an 8.4uF capacitor in series with a 500mH choke (i.e. lead-lag power factor compensation).
The capacitor indicates it has a built in internal resistor. I measure about 2K using a DMM.

I have a few questions:-

Is the internal resistor there purely to act as a bleed resistor to discharge the capacitor after power is removed?
Does an internal resistance of 2K seem sensible? It seems low to me. Surely it will waste power and get hot?
How do you calculate the power disspation in the 2K resistor bearing in mind the tube is non-linear device?

I am thinking of replacing the capacitor with one of similar value but without the internal resistor.

 

Maybe time to re-fit with electronic ballast !
Could be an issue with the choke?

 

It's possible the problem could be with the choke. I have measured it (when cold) using an LCR bridge and it appears to be ok. It could be an intermittent shorted turn due to aging of the enamel coating of the coil that only happens when it warms up.

I would still interested to find out how much power is disspated in the internal 2K resistor.

 

Hello,

The capacitor could be bad. Short the capacitor and see how the flourescent lamp behaves.
The capacitor is there for cosine phi correction.
See this page for more info:
http://www.repairfaq.org/sam/flamp.htm

Bertus

 

I tried a ring test on the choke in an attempt to identify a shorted turn.
As I don't have a good one to compare against, the result is somewhat inconclusive.

 

Here is my attempt at approximating the power disspation in the 2K resistor...

I will use figures published in the following PDF article:-

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiRpp6s6fj5AhWOgVwKHeqZArwQFnoECAMQAQ&url=https://www.kupferinstitut.de/fileadmin/user_upload/kupferinstitut.de/de/Documents/Shop/Verlag/Downloads/Anwendung/Elektrotechnik/s180FluorescentLamps.pdf&usg=AOvVaw366LAFMLSgN8FigtYxjWIC

The article indicates that the current in a 58W lamp is 0.67A.
It doesn't explain how that figure is derived, but he seems to know what he is talking about I assume he means 0.67A rms.

I actually have an 85W lamp but I will assume 58W for the purposes of the calculation.

Referring to the schematic, we have a total current of 0.67A rms flowing though the capacitor and parallel resistor.
For simplicity I will assume the current is sinusoidal (the PDF article indicates this is a reasonable approximation).

From this, if you work out the current flowing through the resistor, you get about 170mA peak or 120mA rms (assuming a sine wave).
I verified this using LTspice using a 180 Ohm resistor to simulate the tube (that gives the required total current of 0.67A rms).

From P = I^2 * R, we get a power dissipation in the resistor of about 29 Watts.
For an 85W tube it would be more like 42 Watts.

Surely this cannot be right! It would mean the resistor wastes about the same power as a 40W tungsten bulb.
It would also get really hot!

What is wrong with my calculation?