I have a pair of old fluorescent tubes in my garage. One works fine. It has the conventional ballast arrangement with a series choke together with a power factor correction capacitor connected between L and N.

The other works for a few minutes but then makes an alarming noise. It suggests something inside is overheating. It has been this way ever since it was installed.

When I opened the lamp fitting I found that instead of the usual ballast arrangement it has a capacitor wired IN SERIES with the choke. The capacitor is 8.4uF and the markings indicate it contains an internal resistor.

I made measurements of the choke and capacitor using an LCR bridge. L = 500mH, R = 22 Ohms. C = 8.4uF, R = 2K Ohms.

I don't understand how this unusual ballast arrangement works. Has it been wired incorrectly? Does the 2K internal resistor inside the capacitor get hot? If the capacitor were connected directly across 230V as in the usual arrangement, the internal 2K resistor would disspate about 26W!

I tried an LTSpice simulation to get some idea of the power dissipation in the capacitor's internal resistor. Modelling a fluorescent tube is tricky, so to keep things simple, I'm assuming it behaves like a 100 Ohm resistor once the tube has struck.

According to the simulation, when supplied with 230V rms, the power dissipation in the 'tube' is around 63W which is probably about right. The power dissipation in the capacitor's internal resistor works out at about 40W which seems way too high. Could it be that the capacitor's internal resistor has changed over time and is much lower than it should be?

Here is an LTSpice model (the tube heaters and starter are not included) Ballast Schematic (LTSpice)

Here is the more conventional ballast arrangement including the power correction capacitor (also shows the starter and heaters):-

Conventional Ballast

I found an article below that describes the twin-tube lead-lag arrangement mentioned by Neil_UK.


In my case there is only a single tube, although information inside the fitting suggests there are twin tube variants that have a pair of sockets at each end.

Twin Tube Lead-Lag Ballast

It makes no mention of a resistor in parallel with the capacitor.

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

I use figures published in the following PDF article:-


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.

if that is correct then the resistor inside the capacitor will get really hot.

Another possible reason for the fault could be an intermittent shorted turn in the choke that only manifests itself when it gets warm. I carried out a ring test using a 'scope. The waveform decays pretty quickly after only about 3 cycles. Without a known good choke to compare against, I am unsure if it looks ok or not.

Choke Ring Test

  • \$\begingroup\$ A picture would help a little ... What length is the "tube"? \$\endgroup\$
    – Antonio51
    Commented Sep 1, 2022 at 15:30
  • \$\begingroup\$ It's a 6ft tube - probably about 80W. \$\endgroup\$
    – brian_mk
    Commented Sep 1, 2022 at 15:34
  • \$\begingroup\$ Welcome! Schematic please. Click on edit and then the schematic symbol and draw it from there. \$\endgroup\$
    – winny
    Commented Sep 1, 2022 at 15:39
  • \$\begingroup\$ It is one of the "more" difficult to "light" ... \$\endgroup\$
    – Antonio51
    Commented Sep 1, 2022 at 15:41
  • \$\begingroup\$ It was quite common for twin tube arrangements to be operated with one tube lagging (simple series L) and the other tube leading (effective series C, but an L in series as well to provide the high voltage kick for starting). Together, the power factor would be quite good. It's not common in a single tube though. Were your component measurements made at 50 Hz, or some other frequency? The L value sounds inappropriate for 60 W. The capacitor resistance sounds broken, whether a series or parallel model is assumed by the LCR bridge. \$\endgroup\$
    – Neil_UK
    Commented Sep 1, 2022 at 15:48

2 Answers 2


I'm assuming the cap is not connected "in parallel" with the starter (Starters normally have these caps internally, though). Or, the capacitor might be connected in series with both of the filaments:


simulate this circuit – Schematic created using CircuitLab

If this is the case then no problem i.e. this is the normal arrangement.

The series caps (i.e. connected in series with one of the filaments) are usually for power level control I used series capacitors for dimming purposes in my electronic ballast designs - relaying different caps for different dimming levels.

If the inductor is not a good match for the tube then the filament current can be decreased by placing a series capacitor as an impedance (remember: the capacitor dissipates zero real power).

One possible purpose is power factor correction especially when there are more than one tubes connected "in series". It's not that common/popular but will still help to correct the PF.

The parallel resistor should be there to discharge the cap as it may remain charged. But I'm not sure about the real purpose.

  • \$\begingroup\$ I would expect a parallel resistor intended to discharge the capacitor to be much larger than 2k Ohms. Something like 50k Ohms would make more sense. The power dissipated by a 2k resistor is far too high. I think the internal resistor in the capacitor must be faulty. The date stamp on the capacitor is 1966 so it is 56 years old! \$\endgroup\$
    – brian_mk
    Commented Sep 1, 2022 at 23:18
  • \$\begingroup\$ I have ordered an 8uF Power Factor Correction capacitor that has no internal resistor to replace the existing one. \$\endgroup\$
    – brian_mk
    Commented Sep 1, 2022 at 23:21
  • \$\begingroup\$ @brian_mk the voltage across the capacitor will not be that high as it's in series, the frequency is low and the current flowing through a 80W lamp is not that high. And also, since it's always good to discharge a mains-connected cap in up to a sine period, 2k is a good choice to discharge an 8u cap: The time constant for 8u-2k is about 16ms, quite close to the period of 60 Hz. \$\endgroup\$ Commented Sep 2, 2022 at 5:50
  • \$\begingroup\$ So what would the power dissipation be in the 2K resistor in this case? I am not sure how to calculate it. \$\endgroup\$
    – brian_mk
    Commented Sep 2, 2022 at 9:03
  • \$\begingroup\$ @brian_mk well it's difficult to say a number, and it's not easy to calculate because we don't know the tube voltage (it can be as high as 700 V) or the filament current. If you can measure the voltage across the capacitor it'll also tell you the dissipation across the resistor. \$\endgroup\$ Commented Sep 5, 2022 at 6:41

Don't know if this can help.
I have modeled the "tube" as a High-voltage "Zener" or "VDR".

enter image description here

I have worked, a long time ago, with fluorescent "tube", but I do not have or remember waveforms of currents & voltages ... This would help.

NB: If R4 is in parallel with C1, it can be also for discharging capacitors to avoid electrocution and physical damage to those who risk "holding the mains plug in their hand".

  • \$\begingroup\$ but I do not have or remember waveforms of currents & voltages "Negative resistance" \$\endgroup\$ Commented Sep 2, 2022 at 5:52
  • \$\begingroup\$ Right. But I would like to recover "waveforms" for these "tubes" ... when they are "working". If I remember well, current is quasi the same form as above. Voltage were a bit "different". \$\endgroup\$
    – Antonio51
    Commented Sep 2, 2022 at 7:09

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