It is not meant as a real "speed", but rather a flow. But the "speed" is sometimes quite a convenient word for that, even when it is not technically correct.

X and Y caps should not fail shorted, but capacitors really do fail shorted, it's a matter of how they react (and how fast) to the fault to stay within the requirements

Have a look at part from a datasheet :


Interesting, right ?

The capacitor is X2 SH, ok to connect across the line, but not ok to connect across the line with some LED in series. They don't say why. My guess is, that this capacitor requires some minimum short circuit current in order to blow off damaged plate areas cleanly, which is not guaranteed in a "ballast" application

If this guess is correct, then what could happen when this capacitor is in series with some 22 Ohm resistor, and it fails ? The short circtuit current would be (for 230V) up to 10A. This is not sufficient to trip any breakers, it might be too low for the capacitor SH function too. Your last line of defense was the resistor.. except that you used one of a non fusible type

Do consider such cases when you choose the components (for something that's going for use outside of the testbench environment)

One of the specifics of fusible resistors is, that (the proper ones) don't arc across when they blow. If the resistor arcs across, then it's resistance is not useful to limit the current...

The main rules for sizing the resistor are :

 - Sufficient resistance, to stay within the peak current withstand capability of the device to be ballasted, at least under worst case condition (unplugged and plugged back in after half cycle). Consider that there may be a few successive peaks in a fast sequence, such as when a bad connection in series with the device is arcing

 - Not any higher resistance, to not dissipate power excessively

 - Fusible resistor power (external size) according to the expected load current, with some margin to not be on the edge, including at the highest ambient temp you expect

 - Not too high current rating, so it still will blow in case of catastrophic failure

I think HF AC is the most efficient way to drive Fluorescents :

 - In DC there is Mercury migration. That means, that only in a small area of the tube there is the optimal concentration of Mercury, in the rest of the tube there is too much or too little

 - In HF AC i'd expect that some of the current is conducted capacitively from the electrode to the discharge. This is good because this current is not subject to the resistive cathode drop

 - Inductive, capacitive, etc. "losses" - As induction and capacitance are reactive impedances, the current going through them does not carry power, so there is no actual power loss involved

Yep, that isn't about speed of moving of the individual electrons. That speed is on the order of millimeter/sec IIRC. The speed of propagation of electrical field in the conductor (what we would normally consider as the speed of electrical current reaching along the conductor) is on the order of the speed of light

However, it's close enough to describe the current as "how fast is the charge moving" (in terms of electrons/sec, or in metric units, Coulomb/sec)

So what about capacitors in switch mode power supplies that connect the primary and secondary together?

My plan is to mitigate this via a line fuse. The resistor will limit the current and heat up only to the point which the fuse blows- time current curve below the resistor "short circuit" current. Because the resistor limits current to a dozen amps, I do not have to worry about exceeding the low interrupting rating of the glass fuse nor do I have to worry about the resistor over heating. At least this is my current plan.

I'd imagine a fluorescent tube can take one heck of a wallop- but unsure about LEDs.

For worse case I calculate the peak current let through duration for 1/2 to one cycle, as though the cap is shorted and only the resistor is considered in the equation? I plan on ignoring arcing fauls (other than normal switching), in that my understanding is that unless the resistor makes the bulk of the ballasting Z (impedance) the LEDs will fail regardless.

Veering of but still somewhat related, does a higher Z (R) resistor relative to capacitor Z (Jx) in the total impedance (Z=R+JX) lower the current crest factor or it remains the same?

If you had to choose: you would pick 50/60Hz over DC when driving a CFL or T5? I am only considering this circuit due to the simplicity, however, if the life or lumens take to much of a dip, then I will scratch it off my list. I can live with multi color tube though     

There is one very important difference between "X" vs "Y" rated capacitors:
The "X" should be fire safe in case they break down (due to any cause), that means e.g. featuring the selfhealing property is good enough.
Of course, the rating is valid only for certain use, so is limited to only certain conditions (e.g. no extra series impedance to mains voltage, presence of an external fuse,...)

The "Y" rated should be made so, no single internal point failure could lead to leakage greater than allowed as a PE (ground) leakage. That means a self-healing feature does not help at all, but an internal construction using multiple, space separated internal capacitors in series does work, provided each of them has dielectric strength corresponding to the overall strength requirements. Plus the construction must be so, the short circuit or elevated leakage failure modes are made very unlikely (thick enough dielectric layer, the material quality control, sufficient spacing from the metalized part towards the edge of the dielectric plate, surface protection and treatment, presence of a nondestructive current bleed when overvoltage above the rated dielectric strength happens,...)
Again, the rating may be conditioned, e.g. by the need for static electricity bleeding resistor, spark gap, need to have multiple in series ()...
The consequence is, you can not make more than few nF of "Y" rating...

So these types can not be interchanged, where an "X" is specified, you should use an "X" type with particular rating, where an "Y" is specified, you can not use "X" and vice versa.

And regarding the fusible resistors: These are rated as fusible only when certain mounting conditions are met, namely no conductive metal in the proximity, need for heat resistant sleeve, robustness of the surrounding against flying yellow hot fragments (all I've seen had this limitation mentioned in their datasheet), etc.

All these you find in the corresponding datasheet and application notes. And when there is no limitations mentioned with some unknown brand, usually it means the component heavily lacks on real quality, as it is practically impossible for the component maker to avoid all of the limitations and still be compliant with the standards...

Should not fail shorted. This does not change the fact, that i found one with a hole blown through it in a PC power supply that was tripping the RCD. In this case, no dangerous situation happened as the secondary of the power supply is connected to PE. In a SELV power supply this would make the output live at line voltage


 
There is some range of current, which is too close or exceed the fuse rating, but still not high enough to blow the fuse for sure (and in short enough time, before the resistor overheats enough to pose a danger by itself). So there may happen (not necessarily will) a situation, where you choose a fuse rating, lower you can't go because of the normal operation current draw (and deratings for ambient temp, and some headroom to not load the fuse on the edge all the time), yet it allready is too high to protect the resistor. And if you choose higher power resistor (bigger size with bigger thermal mass), it might be too big to fit where you want it

And how would the fuse vs. the resistor curves work together if the current is limited by something else ? For example, a non dead short in the capacitor (cause of failed SH event), damaged LEDs that failed into some resistive mode, etc ?


You might want to avoid this, as it will happen on cold cathodes and sputter them quite badly

LEDs have some rating for current spike. If you exceed that, they might start degrading or outright failing


This is too bad even compared to the real world worst case. You calculate the time as if it is half cycle, but for resistor/capacitor values like what's found in such devices normally, the capacitor charges up (and the current spike ends) way faster than that

Arcing is not necessarily a result of fault, some "silent" switches, plugging a plug into a receptacle with loose contacts, etc can produce a series of current spikes... In an experiment i done once this could really be seen : Connect a ~25W GLS lamp in series with ~0.5..1uF capacitor and see how the lamp brightens up if you hold the plug in arcing position

It sure does. Question is, maybe by the point where it really improves significantly, the resistor is so much dominant that you could omit the capacitor altogether...

The Mercury migration means that part of the tube that is Mercury starved, lights less efficiently

With DC there is also, that one cathode have the losses on it all the time, so to prevent it from overheating (in a lamp designed for AC) you might want to underdrive the lamp. And that might have its own effect on lamp efficacy too (for example by lowering even more the available Mercury vapor quantity), i dont know how significant this effect will be

I'd use 50/60 Hz AC and try to provide as good starting as possible (so electronic starter or RS etc). But it's just my guess

Well, can't add much other than I agree with your thinking. If wondering the compartment I have to work with is the size of a typical CFL (think Sylvania straight tube or Philips Marathon) at worst so room is generous

I will when designing a high power factor dropper for LED... In theory one could make such a beast for commercial applications- it would meet PF requirements and still use less power than two F40T12s with magnetic ballast. Troffer would dissipate the heat. Then again maybe to much heat for that type of lumen output.

I found this data which has me addled on one (actually two) thing:

https://products.currentbyge.com/sites/products.currentbyge.com/files/documents/document_file/20504_GE_CFL_Biax_T_E_SellSheet.pdf

On the last page under starting requirements, you have "Minimum OCV at +10*C" and then "Maximum OCV (none-ignition)" What do they mean by this...? IS this a cold cathode value? Or once the electrodes have heated up? And what is Rsub (ohm)?

Inductor acts as impedance for AC. So yep, it could. It's mainly not done because it's cheaper to use capacitor for the low currents used mostly with low power LEDs (so pretty much anything that uses 5mm LEDs)

When current through inductor is interrupted, it generates voltage such, to try to maintain the same current. With loads that dont conduct below a minimal voltage : Discharge lamps, strings of LEDs, anything that rectifies the line voltage with a with full bridge rectifier and capacitor without PFC, etc. When the line voltage drops below the minimum voltage, the current stops. If there is inductor added in series, the current will continue for longer and the stops will become shorter, so the crest factor gets improved

One application is in PC power supplies. There are power supplies with boost-converter PFC (active PFC) - a boost converter between the bridge rectifier and the big electrolitic capacitor (that can be seen as a constant voltage component). And there are power supplies with no PFC, that draw a spike of current only near the peak of the input sine wave (so, high crest factor and low power factor on the order of 1/2). In the latter, adding an inductor in series with the AC input, without changing anything else in the PS design, improves both very considerably - The power factor can improve to as high as 3/4, and allready pass EU regulation. The addition of inductor have been done in many PS's and got called Passive PFC for it's PFC feature

A proper resonant circuit won't ballast the lamp... The designs of ballasts like SRS or like frequency-sweep start (as in CFLs) are much about getting a not really resonant circuit to do what we want and not what a resonant circuit from the schoolbook does. This is engineering in the whole meaning of it...

But these spikes will not kill the LED string, correct? At least not under normal conditions. I could see an intermittent failing LED creating some huge voltage spikes much like a glow starter dropping out and striking a tube.

I will Google these terms and do my research if I ever get bored and want to do a high power LED ballast. Even if I don't, its worth knowing. Its kind of funny though, even though we have moved into a semiconductor world and LEDs, I am thinking about how to drive them with magnetic ballasts. That would kind of be neat, a cobra head with a magnetic driver and the open board igniter being a bridge rectifier.

A true resonant circuit- but there is a key to the madness- or at least not having it. I have seen 120 volt US shop with a series capacitor and inductor driving the T12 40 watt tubes. Had a Sidac starter... A few members will instantly know what I am referring to.

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Voltage-wise, the bad connection in series with capacitor and the glow starter do very different things :

The capacitor can, at most, make double the peak line voltage appear on any open circuit area in series with it. (peak line voltage coming from the line + peak line voltage charged in the capacitor from previous time)

The inductor ballast, generates voltage which is equal to L di/dt. In theory, if the circuit is abruptly opened then dt=0 and so the voltage would be infinite. In reality, capacitances between conductors make dt non-zero, arcing across opening contacts limits the voltage, and so on. So you get something on the order of 1..1.5kV out of a Preheat ballast. This have little to do with the line voltage - you could get similar voltage spikes from a Fluorescent ballast even when powering it with a 9V battery

Well, I guess you could say I have learned what I already knew but really did not if that makes sense... An inductor generates voltage based on the collapsing field- the rate of current change- and not dependent on the input voltage. 

So if lets say a LED goes instantly open, the inductor could in theory slam a kilovolt or more of voltage across the array?

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Why go that far. Most emergency lighting (atleast here) is powered in standby with plain simple transformer working at 50 or 60Hz. The load on this transformer is float charging of the (fairly small) battery + lighting up an indicator LED

In other words ballasting a LED through a very high impedance transformer? I've seen NEON signs where the ballast was literally nothing more than a 2 winding transformer. Bascially a transformer that could operate with the secondary short circuited for some time. Then you have CWI ballasts... However I have no idea how to design one.

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I do know of an attempt to commercialize magnetic LED drivers at one of our local lighting manufacturers a few years ago. They were quite powerful, 60W in about the same size as a 58W Fluorescent lamp ballast. There were a few batches made but it never took up and ended up being dismissed

Do you have any more info or knowledge about this or their design?

Do you have more info on this?

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Any L+C circuit is a resonant circuit for some frequency, but L+C circuits used for lamp ballasting are not resonating at 50/60Hz (if they would, the ballast would not perform any ballasting)

The capacitor and inductor are 2 storage components (the inductor can store current, in much the same way as a capacitor stores voltage). If connected together and some initial energy provided, this energy will travel between them forward and back forever (subject to losses). If external energy is supplied at the same timings, as the energy allready present in the storage moves, then it will add up. (Swinging yourself on a swing set)

The impedance of L is :

XL = 2pi f L

The impedance of C is :

XC = 1 / ( 2pi f C )

At some frequency, XL = XC exactly (and the phase is opposite). That means, that if they are connected in a closed loop, the voltage and current of one of them match the voltage and current of the other all the time, so each one can exactly discharge into the other, enabling the infinite energy travel of resonance forward and back

In the lamp ballast you mentioned, at 60Hz XC is bigger than XL (by 2 times or more). The overall impedance that ballasts the lamp is XC-XL. The inductor is there to provide the starting and to improve the crest factor to be atleast to some extent acceptable

By any chance does this (the shop light) circuit increase the voltage? What always mystified me is that American 40 watt T12s require a 240 volt source (reactor) or a 120 volt ballast with a voltage step up scheme (HX). Yet the shoplight appears to drive these without that at first glance.

Well, I guess you could say I have learned what I already knew but really did not if that makes sense... An inductor generates voltage based on the collapsing field- the rate of current change- and not dependent on the input voltage. 

So if lets say a LED goes instantly open, the inductor could in theory slam a kilovolt or more of voltage across the array?

In other words ballasting a LED through a very high impedance transformer? I've seen NEON signs where the ballast was literally nothing more than a 2 winding transformer. Bascially a transformer that could operate with the secondary short circuited for some time. Then you have CWI ballasts... However I have no idea how to design one.

Do you have any more info or knowledge about this or their design?

Do you have more info on this?

By any chance does this (the shop light) circuit increase the voltage? What always mystified me is that American 40 watt T12s require a 240 volt source (reactor) or a 120 volt ballast with a voltage step up scheme (HX). Yet the shoplight appears to drive these without that at first glance.

LC circuits can step up voltage. Example of how this. Think of series LC connected to AC source :

Vin = 1 V
Xl = 1000 Ohm
Xc = 900 Ohm

X = Xl - Xc = 1000 - 900 = 100 Ohm

I = Vin / X = 1V / 100 Ohm = 10 mA

Vl = I Xl = 10 mA * 1000 Ohm = 10 V

Vc = I Xc = 10 mA * 900 Ohm = 9 V

Yep, between the ends is the input 1V, but across each component alone the voltage is much higher, according to the choice of L and C values

In the shoplight, if i imagine it's internals correctly, the lamp is not connected to the mid point between the L C so this does not take place. The story is just how to drop voltage on the ballast without taking up too much voltage headroom from the lamp. Since the voltage drop on the lamp and on the ballast (mostly capacitive) are near 90deg out of phase, there is room to drop several 10's V on the capacitor (well, the inductor + capacitor together), even when the voltage of the lamp is close to that of the line

But managing 103V lamp reliably (enough so it works most of the time !) on a voltage source that could go as low as 110V still surprises me. I tried dimming a 36W T8 (with plain choke ballast, which is few 100's Ohm inductive + 45 Ohm or so resistive) on a variac - At 160V variac output or so it will go out right away