Capacitors To Stop Fluorescent Lights From Flickering

[Crossy]

This is all well & good but if there is no voltage on the cables in the first place, this capacitive 'fix' would not be needed. So, how how did voltage get to be on the cables? Eliminate the voltage & eliminate the problem.

OK, let's try with a drawing

Consider the diagrams below:-

The upper diagram is a typical light arrangement, with a 10m drop down to the switch.

The capacitance of the switch cable (Cs) is about 1nF (1mm T+E is about 100pF per metre).

The capacitance of the cable to the light fitting (Cl) is 0.1nF (100pF).

For now ignore the contents of the ballast.

These two capacitors form a capacitive divider across the mains, Cl develops about 200V across it whilst Cs gets 20V (do the maths at 50Hz) The source impedence of this divider is VERY high, in the order of 30M ohms so even a digital meter won't read much if you put it across the lamp.

Now, inside the electronic ballast is a bridge rectifier and reservoir capacitior (Cb), the reservoir charges slowly via the 30Mohm impedence until it has enough voltage for the drive electronics to start lighting the tube (usually at about 50V), when the tube strikes the reservoir almost immediately runs out of charge and the tube goes out and Cb starts charging again, result an annoying flicker.

Ok, so now we add our extra capacitor across the lamp, lets use 1u (1000nF), our divider now has 1nF at the top (Cs) and 1000nF at the bottom (new value for Cl). This time we get 219V across Cs and 1V across Cl which is way too low for the electronics ever to start.

Hence, sticking a capacitor across the lamp fitting is the correct and only way to stop the flicker (actually you could use a 100k resistor to do the same job)

QED

EDIT. Of course the manufacturers could stop all this heartache by simply adding a resistor across the mains inside the ballast, but the 1 cent it would cost addus up to a lot of profit over a few million units.

Edited April 4, 2007 by Crossy

[elkangorito]

This is all well & good but if there is no voltage on the cables in the first place, this capacitive 'fix' would not be needed. So, how how did voltage get to be on the cables? Eliminate the voltage & eliminate the problem.

OK, let's try with a drawing

Consider the diagrams below:-

The upper diagram is a typical light arrangement, with a 10m drop down to the switch.

The capacitance of the switch cable (Cs) is about 1nF (1mm T+E is about 100pF per metre).

The capacitance of the cable to the light fitting (Cl) is 0.1nF (100pF).

For now ignore the contents of the ballast.

These two capacitors form a capacitive divider across the mains, Cl develops about 200V across it whilst Cs gets 20V (do the maths at 50Hz) The source impedence of this divider is VERY high, in the order of 30M ohms so even a digital meter won't read much if you put it across the lamp.

Now, inside the electronic ballast is a bridge rectifier and reservoir capacitior (Cb), the reservoir charges slowly via the 30Mohm impedence until it has enough voltage for the drive electronics to start lighting the tube (usually at about 50V), when the tube strikes the reservoir almost immediately runs out of charge and the tube goes out and Cb starts charging again, result an annoying flicker.

Ok, so now we add our extra capacitor across the lamp, lets use 1u (1000nF), our divider now has 1nF at the top (Cs) and 1000nF at the bottom (new value for Cl). This time we get 219V across Cs and 1V across Cl which is way too low for the electronics ever to start.

Hence, sticking a capacitor across the lamp fitting is the correct and only way to stop the flicker (actually you could use a 100k resistor to do the same job)

QED

EDIT. Of course the manufacturers could stop all this heartache by simply adding a resistor across the mains inside the ballast, but the 1 cent it would cost addus up to a lot of profit over a few million units.

Ya didn't really think I could leave this alone, did ya?

Anyway, I concede that the cable is capacitive but as for the switch (being such a small device), I don't buy it. It may have some capacitance but will be very small.

Again, instead of tampering with a system that wasn't meant to be tampered with in this way, I say eliminate all the logical possibilities first. The OP did not tell us (when asked) if the neutral had been earthed or not. Also, is the switch the type that has a neon (or similar) built into it?

To me & more than anything, I'd at least try a different electronic ballast in the fitting. If these electronic ballasts are so susceptible to this particular problem, why aren't we hearing of more cases as such?

If the electronic ballast has a high lagging power factor (>0.9), adding a capacitor could possibly make it a leading (& lower) power factor, reducing efficiency.

[Gary A]

I asked a Thai electrician about earthing my neutral line and he was horrified at the thought. I got a very quick NO answer. I thought it was a logical thing to do and have no idea why I couldn't do it. ????

[Tywais]

I asked a Thai electrician about earthing my neutral line and he was horrified at the thought. I got a very quick NO answer. I thought it was a logical thing to do and have no idea why I couldn't do it. ????

Find another electrician, this one seems to not have a clue.

[Crossy]

Ya didn't really think I could leave this alone, did ya? Of course not, but do remember we are moving into the realms of electronics rather than pure electrics

Anyway, I concede that the cable is capacitive but as for the switch (being such a small device), I don't buy it. It may have some capacitance but will be very small. The switch itself has a negligable effect, it's the drop cable that provides the upper capacitance in the divider chain

Again, instead of tampering with a system that wasn't meant to be tampered with in this way, I say eliminate all the logical possibilities first. The OP did not tell us (when asked) if the neutral had been earthed or not. Also, is the switch the type that has a neon (or similar) built into it? To me this scenario is totally logical. Agree about the neon switch making the whole problem worse (and the cap will fix that too).

To me & more than anything, I'd at least try a different electronic ballast in the fitting. If these electronic ballasts are so susceptible to this particular problem, why aren't we hearing of more cases as such? It certainly is not an uncommon problem, but, as with all these things you need a lot of coincidences to actually make it happen, such as a particularly long drop or two-way cable (as was noted by our OP)

If the electronic ballast has a high lagging power factor (>0.9), adding a capacitor could possibly make it a leading (& lower) power factor, reducing efficiency. Electronic ballasts present a pretty pure resistive load much like a switch-mode PSU does (they work on the same principle). If you don't like using a capacitor then a 100k or so resistor across the ballast input will achieve the same result.

Edited April 11, 2007 by Crossy

[chownah]

Crossy,

Are you sure that a 100kohm resistor will work? If it is current/voltage mini spike that is slowly charging the electronic ballast wouldn't a 100kohm reisitor still allow the spike to develop since it would show a high resistence to the spike while a capacitor might act as a pass filter for the spike and show low impedence to the spike but high impedence to the 50 Hz during operation of the light....this would be sort of a like a high pass filter...the spike acting like a high frequency (relative to 50 Hz) signal.

Chownah

[Crossy]

Crossy,

Are you sure that a 100kohm resistor will work? If it is current/voltage mini spike that is slowly charging the electronic ballast wouldn't a 100kohm reisitor still allow the spike to develop since it would show a high resistence to the spike while a capacitor might act as a pass filter for the spike and show low impedence to the spike but high impedence to the 50 Hz during operation of the light....this would be sort of a like a high pass filter...the spike acting like a high frequency (relative to 50 Hz) signal.

Chownah

I agree, if it were spikes or noise charging the reservoir a resistor may not fix it, the sums assume fairly clean sinusoidal mains of course .

Without some fairly extensive testing with a scope etc we really are never going to determine the exact mechanism, each case is going to be different and even what look like massive spikes on the scope contain very little actual energy.

Point is, if adding a resistor or capacitor fixes the problem, no problem, provided you've verified the installation is electrically correct and you're not masking the effect of poor installation

For the couple of Baht for a resistor give it a try. Use 2x47k in series so there cannot be a single point of failure, full mains across a single 1/4W resistor is quite close to the allowable limit (500V IIRC)

Edited April 11, 2007 by Crossy

[chownah]

Crossy,

Are you sure that a 100kohm resistor will work? If it is current/voltage mini spike that is slowly charging the electronic ballast wouldn't a 100kohm reisitor still allow the spike to develop since it would show a high resistence to the spike while a capacitor might act as a pass filter for the spike and show low impedence to the spike but high impedence to the 50 Hz during operation of the light....this would be sort of a like a high pass filter...the spike acting like a high frequency (relative to 50 Hz) signal.

Chownah

I agree, if it were spikes or noise charging the reservoir a resistor may not fix it, the sums assume fairly clean sinusoidal mains of course .

Without some fairly extensive testing with a scope etc we really are never going to determine the exact mechanism, each case is going to be different and even what look like massive spikes on the scope contain very little actual energy.

Point is, if adding a resistor or capacitor fixes the problem, no problem, provided you've verified the installation is electrically correct and you're not masking the effect of poor installation

For the couple of Baht for a resistor give it a try. Use 2x47k in series so there cannot be a single point of failure, full mains across a single 1/4W resistor is quite close to the allowable limit (500V IIRC)

I doubt that the clean sinusoidal 50 Hz signal from the mains is the culprit because the flickering was sometimes more and sometimes less....if it were primarily from a clean signal then I would expect it to be fairly regular which it was not.

It is true that voltage spikes can be impressive when viewed on a scope and still have little energy but isn't the idea here that the electronics in the ballast act as an accumulator and many little bits of energy get stored there until a threshold is reached and it then empties through the tube?

I don't understand alot of your last paragraph....I understand 2x47k in series provides the required (approx) 100k ohms but I don't understand how it eliminates "single point failure" nor do I understand what is meant by this or why it should be avoided. Also what does "full mains across a single 1/4W resistor" mean....doe it mean the same thing as "a single 1/4W resistor across the mains"?...and what allowable limit is this close to?...and what does 500V IIRC mean?

Chownah

[Crossy]

I doubt that the clean sinusoidal 50 Hz signal from the mains is the culprit because the flickering was sometimes more and sometimes less....if it were primarily from a clean signal then I would expect it to be fairly regular which it was not. Could well be noise spikes then.

It is true that voltage spikes can be impressive when viewed on a scope and still have little energy but isn't the idea here that the electronics in the ballast act as an accumulator and many little bits of energy get stored there until a threshold is reached and it then empties through the tube? That's what is happening yes.

I don't understand alot of your last paragraph....I understand 2x47k in series provides the required (approx) 100k ohms but I don't understand how it eliminates "single point failure" nor do I understand what is meant by this or why it should be avoided. Also what does "full mains across a single 1/4W resistor" mean....doe it mean the same thing as "a single 1/4W resistor across the mains"?...and what allowable limit is this close to?...and what does 500V IIRC mean? OK, small (1/4W = quarter Watt) resistors have a maximum rated voltage of 500V so in order for them to be happy across the mains you should use two in series. 'Single point of failure' is not really relevant in this case, ignore