Author Topic: Capacitive Dropper Ballast  (Read 16862 times)
Ash
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Re: Capacitive Dropper Ballast « Reply #30 on: September 04, 2017, 01:42:07 AM » Author: Ash
Yep
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Medved
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Re: Capacitive Dropper Ballast « Reply #31 on: September 04, 2017, 02:53:19 AM » Author: Medved
So once they turn on at the 280 volt (example) threshold in the sine wave, they will continue to conduct as long as current passes, ie even at the 10 volt threshold the SIDAC will be "on", but shut off at the zero crossing only to start conducting at 280 volts on the negative half of the wave.

My understanding is that if I eliminate the diode, I will loose the double voltage, but the resistor will still allow current to flow through the cap and in turn voltage across the SIDAC.     

When the sidac is on, its voltage drop is justsomething between 1..2V, it is really a hard switch.

The resistorwithout the diode willpass some current, but that will do nothing at all, compare to when the resistor is not there at all. The reactance of the capacitor is way lower than the resistor, so there would be just a small drop over the cap (before the sidac triggers, of course), so the voltage across the sidac (that matters) is practically equal to the mains. So when that is enough to trigger the lamp (e.g. a 240V sidac, 230V mains), you do not need there anything. If the sidac needs to have the Vbo higher than the mains, you need the R-D ro charge the cap in order to trigger the sidac the first time after powee on. After that, it is the sidac, what leaves the capacitor charged when it turns off,so then the peak voltage there really goes up tothe actual Vbo in both directions.
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Keyless
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Re: Capacitive Dropper Ballast « Reply #32 on: April 15, 2018, 09:14:07 PM » Author: Keyless
Came across this recently by accident. I was wondering is anyone could shed any light on this CFL or if more of them exist. It appears to only have a large metalized film capacitor and a few resistors. I see what appear to be 3 leads from each tube base so I am guessing a probe start CFL?



http://www.hko.gov.hk/education/edu06nature/images/ele-cflamp-3.jpg


http://www.hko.gov.hk/education/edu06nature/ele_fluorescent_e.htm
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Ash
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Re: Capacitive Dropper Ballast « Reply #33 on: April 16, 2018, 02:20:26 PM » Author: Ash
Yep. Typically the center wire on either end is a probe, it's just the end of the wire itself standing in the tube below the electrode. The other two hold a filament as in any lamp

Each probe is connected to the opposite side electrode through a resistor on the order of 100K

The main electrode wires on either end i'd expect that either one of them is connected and one left floating, or both shorted
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Keyless
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Re: Capacitive Dropper Ballast « Reply #34 on: April 16, 2018, 09:55:51 PM » Author: Keyless
Neat stuff  ;D

One could in theory take advantage of the 240 volt supply and just place a resistor and diode across the lamp supply for an instant start lamp. Though the gas pressure and penning mixture might have to be tweaked. 

« Last Edit: April 16, 2018, 10:01:14 PM by Keyless » Logged
Ash
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Re: Capacitive Dropper Ballast « Reply #35 on: April 17, 2018, 01:50:32 PM » Author: Ash
It is simple indeed but the lamp won't like the operation on a capacitive ballast at all (no matter what starting method would be used)
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Keyless
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Re: Capacitive Dropper Ballast « Reply #36 on: April 17, 2018, 09:00:18 PM » Author: Keyless
I remember you mentioned that the non linear current draw of the arc causes high frequency which leads to higher current draw. Bigclive mentioned something similar along those lines that  an arcing switch (for example) can kill a LED lamp because the high frequency causes current spikes through the dropper. If I could offset this by under driving the lamp? Or to let the resistor do its job by having it do 1/4 the ballasting? Or I guess I am really wondering how much life is shaved off from the lamp. 


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Ash
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Re: Capacitive Dropper Ballast « Reply #37 on: April 17, 2018, 11:55:17 PM » Author: Ash
You could, but that is only one part of the problem

The other is - the capacitor does not do anything to help, it even stands in the way, of proper lamp reignition after zero crossing of the AC. This means the lamp is extinguishing for longer periods, and then when it lights again, the peak current is higher (to still have the same RMS current). This is hard on the lamp electrodes
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Keyless
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Re: Capacitive Dropper Ballast « Reply #38 on: April 18, 2018, 12:52:24 AM » Author: Keyless
A higher current crest factor if I have the term right. 1.4-1.5 for a reactor HPS ballast, 1.6-1.8 for a CWA, 1.8 for a CWI and 2.0 for a 3 coil MAG REG. I take a cap is higher than that...
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Re: Capacitive Dropper Ballast « Reply #39 on: May 01, 2018, 06:27:59 PM » Author: Keyless
You could, but that is only one part of the problem

The other is - the capacitor does not do anything to help, it even stands in the way, of proper lamp reignition after zero crossing of the AC. This means the lamp is extinguishing for longer periods, and then when it lights again, the peak current is higher (to still have the same RMS current). This is hard on the lamp electrodes

Yahhh, no inductive kick so the strike is higher in the sine wave.

So far I have not been able to figure it out- but what is the resistor for to begin with? I see it in every capactive dropper circuit (even LEDs) but to be frank I still have no idea what its true purpose is. Obviously its an important component in that its typical to eliminate all none essential items in low cost bare-bones products.





   
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Ash
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Re: Capacitive Dropper Ballast « Reply #40 on: May 02, 2018, 03:46:57 PM » Author: Ash
I assume you mean the resistor that's in series with everything else, not the probe-start resistors

Let's see what happens when the capacitor is connected to the AC power (think of some capacitor-ballasted LED nightlight as example. Since the voltage drop on the LED is small, let's ignore it altogether and assume that the capacitor is just getting the full AC voltage as is)

The AC voltage is a sine wave, there is a finite speed at which the voltage changes with time (dV/dt). The change is the fastest in the times where voltage goes through 0V, and slowest (down to a halt) in the peaks of the wave

The capacitor (according to our assumption) is connected straight to the source, the voltage that's charged in the capacitor follows the sine wave. So the capacitor is charging and discharging accordingly

Since
Q (charge) = C (capacity) * V (voltage)

Then the speed at which the capacitor is charging/discharging matches the speed of the voltage change :
dQ/dt (speed of charging or discharging) = C (capacity) * dV/dt (speed of voltage change)

The speed of flow of electrical charge is none other than the current
dQ/dt = I

So, the current that goes through the capacitor is determined by the voltage change, which ultimately means, by the shape of the supplied voltage (sine wave)

I = C * dV/dt



You switch the capacitive ballasted LED on. This happens at a random time. It have all the odds to happen in a moment, when the voltage is non zero. So, you give the caoacitor a sudden jump in supplied voltage, from 0 to whatever the position of the sine wave was at the moment. dV/dt = infinity, a jump in voltage level in no time...

This would mean, infinite current through the capacitor...

In reality the current won't be infinite, it will be limited by the capacitor ESR, wire resistance and such. But it can reach many 10's A even for a small capacitor. In some LED ballast, it will after just a few turn on events (if not right from the first one) fuse the LEDs. To provide some more reasonable limit of this peak current at the moment of switch on, there is the resistor

The worst case current peak with normal supply voltage will be, if you unplug the lamp at the moment when the capacitor voltage is max in one polrity, and plug it back when the line voltage is max in the other polarity. As then, it must change its voltage by double the peak of the line voltage in no time. But there are also voltage transients, momentary voltage drops, etc on the line, which mean that similar current peaks might come anytime when the thing is just powered on and not touched



Also, the resistor is of such value that will burn out very quickly if the capacitor shorted out. If the correct type of resistor is used (a metal film resistor), it will successfully act as a fuse : "pop" open cleanly without catching fire, and limiting the current at that moment of the short low enough, to limit the size of resulting arc (so reduce the chance of blowing apart the enclosure of the device by the expanding gases from the arcing)

In cheap devices the fuse function doesn't always work well because the wrong type of resistor could be used, or the clearance around it (or between it's soldering pads on the PCB) is inadequate, and so on. In addition, in many of the cheap devices the capacitor is some cheap "basic" film capacitor, and not the proper X2 type capacitor that's designed to work safely in across-the-AC-supply applications
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Keyless
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Re: Capacitive Dropper Ballast « Reply #41 on: May 03, 2018, 12:15:34 AM » Author: Keyless
Yes, a series resistor.


In other words (simplified terms on my part), or rather one might say: Xc = 1 / (2PiFC) where F is 50 or 60Hz, however arcing or transients can be much higher that 50 or 60HZ, causing current spikes. More frequency = more current. And yup, shape of the wave also effects the current. Now that you have described it, it really is easy to picture. As well as the flame proof resistor acting as a fuse should the cap fail- though a metalized film class X and Y capacitors won't fail short circuit... so I guess that spares me from using a flame proof resistor  ;)



But going back to the fusible resistor, in many ways that is cheaper and better than a fuse. Technically a fuse would need to be sand filled (guessing) to cover installations with high short circuit current at the socket. A resistor on the other hand naturally provides current limiting properties will fusing open and does two jobs in one package.     


Are there any rules on sizing this resistor? I plan on sizing it such that it will dissipate 3 watts per 10 watts of tube, and a separate glass fuse sized to blow with the cap shorted. However, I don't know if that rule will protect my CFLs (or LEDs) from failing. Also, if I use a capacitive voltage doubling rectifier, I still need the resistor to limit inrush? See the first part of this schematic:

https://www.lighting-gallery.net/gallery/displayimage.php?album=2496&pos=15&pid=95991




   

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Re: Capacitive Dropper Ballast « Reply #42 on: May 03, 2018, 12:28:49 AM » Author: Keyless
Perhaps for another thread, but found this quite interesting. Is DC more efficient then 50/60Hz AC? I know that when you drive a tube in the 25000-75000hz range there is a dramatic spike in light output for the same wattage.


https://www.youtube.com/watch?v=rloKY8K9KuM


My thinking: The capacitive voltage rectifier provides about ¿648 volts peak DC? (230 x 1.41 x 2) on a 230 volt supply to strike, and then the two caps limit the voltage down to a low level.   
« Last Edit: May 03, 2018, 12:32:44 AM by Keyless » Logged
Keyless
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Re: Capacitive Dropper Ballast « Reply #43 on: May 03, 2018, 01:23:57 AM » Author: Keyless


Then the speed at which the capacitor is charging/discharging matches the speed of the voltage change :
dQ/dt (speed of charging or discharging) = C (capacity) * dV/dt (speed of voltage change)

The speed of flow of electrical charge is none other than the current
dQ/dt = I



Not to nit pick, but isn't it technically the amount of electrons in a given defined period of time (one second)? I've always assumed that electrons themselves flow at the same speed. Maybe I am viewing it the wrong way- I don't know much about electricity as compared to others here. 
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Medved
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Re: Capacitive Dropper Ballast « Reply #44 on: May 03, 2018, 04:09:27 AM » Author: Medved

Not to nit pick, but isn't it technically the amount of electrons in a given defined period of time (one second)? I've always assumed that electrons themselves flow at the same speed. Maybe I am viewing it the wrong way- I don't know much about electricity as compared to others here. 

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.
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