Flourescent Lighting Fixture Starters

[LarryFine]

From an old Advance troubleshooting guide, for a preheat ballast using starters the minimum open circuit voltage is 176V for a F30T12 or a F40T12 lamp.

Where would you get that voltage with a 120v light with a starter?

 

[gar]

180703-1700 EDT

Larry:

I would expect that ballast is actually a transformer. Assuming the voltage is a sine wave read with either an RMS meter or an average reading meter calibrated with an RMS scale.

The peak voltage of a 120 V sine wave is about 170 V.

To drive a fluorescent bulb one has to use a source that limits current. A high impedance between a voltage source and a load can achieve this. A transformer can be designed to have a high internal impedance or a standard transformer can be used with an external high impedance.

.

 

[kwired]

Where would you get that voltage with a 120v light with a starter?

I am with you when it comes to single lamp preheat circuit like the wiring diagram early in this thread. Open the circuit at any point and you should read 120 volts across the "open part" if supply voltage is 120 volts.

 

[gar]

180703-2415 EDT

The initiation voltage to start conduction in a fluorescent bulb with a heated filament will be dependent upon bulb length, filament temperature, bulb gas temperature, gas pressure, gas mixture, and stray radiation.

In a 4' T12 tube with a heated filament I saw initiation at about 150 to 175 V. Too high for straight line voltage of 120 with a simple inductor for current limiting. A shorter tube or differently designed tube might work without a transformer.

.

 

[MTW]

Where would you get that voltage with a 120v light with a starter?

I would think that you get it, 176V and higher the same way you get a 10KV spark ignition from 12VDC, interruption of the transformer coil current.

Almost every other ballast type I looked at for a 48" lamp had a much higher open circuit voltage listed in the specs 200-300V.
Rapid start and Instant start lack the pre-heat emissions, so they must strike at a higher voltage. All provided by the 120V source.

It's not a number I picked, It's Advance's number for the preheat ballast in question. There is a link for substantiation.

 

[Besoeker]

Just recently I had an E&I mechanic experience an electrical shock while reinstalling a fluorescent lighting fixture starter. The starter is a FS-4 30-40W. The mechanic was replacing the bulbs for the fixture, and noticed the starter seemed not to be installed correctly since the fixture did not illuminate once the bulbs were replaced. So the mechanic removed it, inspected the starter, and reinstalled it after noticing no damage. When he twisted the starter into place he felt an electrical shock, and described the sensation as "it lit me up". He was standing on a fiberglass ladder, and stated "that only his right hand touched the starter". When a voltage test was performed with a low voltage proximity tester on the starter and fixture, the tester did light up only in proximity of the starter not the fixture. When the starter was tested with a Tegam 110 voltage tester, the voltage measured was 5V to ground. There was several different grounds used to include the light fixture. The starter was removed and damage was noticed. During a continuity test between the starter's male terminals to its casing, one male terminal was shorted to the starter's casing. The starter was replaced, and the light fixture began operating. So my question is this since I'm not very familiar with fluorescent lighting starters, what is the voltage conducted through the starter? Is it the normal line voltage of 120V, or does it operate at a lower voltage? I am trying to understand why when tested, the voltage on the starter's casing was measured only at 5V, yet the mechanic stated he felt the shock enter his right side and exited his left while standing on a fiberglass ladder, i.e. note he also stated his left hand was touching nothing, not grounded. Is it due to the current of through the starter, or something else? Any help would be greatly appreciated. Also, the mechanic is fine and was released by medical personnel with no restrictions. Thanks

Should he have been working on it live?

 

[gar]

180704-1053 EDT

I doubt that any ordinary fluorescent fixtures start bulb conduction by an inductive kick method.

Conduction is started in one of two ways --- heating the cathode to a sufficient temperature to produce a large space charge of electrons (comes from a heated object) (note the cathode switches tube ends every half cycle), or from a large enough voltage gradient at the cathode to generate enough electron flow.

This occurs every half cycle as can be seen when looking at the fluorescent tube voltage drop on a scope.

Where a manual pushbutton or starter is used to initiate conduction it is used to provide current flow thru the heaters (cathodes) in the fluorescent tube to make them hot (temperature) to get a large amount of electron emission. When the pushbutton or starter opens, then conduction is initiated by adequate voltage and the electron space charge. The cathodes remain hot (temperature) from thermal storage and current flow thru the fluorescent tube.

The ballast design, internal fixture wiring, and wiring to the fixture will determine what voltages may be present at the starter. Anything from near zero to full line voltage. When open there is considerable voltage across the two starter pins. Basically the voltage drop across the fluorescent bulb. During the start time the voltage across the starter is near zero. But this does not mean relative to earth.

Because a 5 V voltage source can provide at least 1 A does not mean that if you touch your hands across that 5 V source that 1 A will flow thru you. On a low humidity day I would see about 5/500,000 A (10 micro amps) thru me.

.

 

[gar]

180704-1609 EDT

In post #18 MTW provided a very good reference, "European paper". Study this.

But I do not agree with MTW that inductive kick is what is normally used to initially start conduction. Separately note that every half cycle conduction has to restart but in the opposite direction. The electrode that is the cathode has to switch from one end of the tube to the other.

Too many of the responses to this thread are based on someones imagination.

Get a scope and other instruments, some components, and run controlled experiments. Then you may be able to accurately talk about what actually happens.

.

 

[kwired]

I would think that you get it, 176V and higher the same way you get a 10KV spark ignition from 12VDC, interruption of the transformer coil current.

But that 10kV is only momentary as the magnetic field collapses, open circuit voltage after that collapse is done is only 12 VDC.

 

[hbiss]

180704-1609 EDT

In post #18 MTW provided a very good reference, "European paper". Study this.

But I do not agree with MTW that inductive kick is what is normally used to initially start conduction.

Gar, even your reference states otherwise:

When applying the mains voltage a glow discharge is initiated inside the glow starter (Fig. 3.4) which heats up the bi-metallic contacts and causes them to close (Fig. 3.5). Now current flows from the mains via the ballast, the cathode filament, the starter and the second filament. This way the cathodes are pre-heated. But since the glow discharge has solely been shorted by the bimetallic contact, the bimetallic contact cools down and opens again few seconds after closing. By interrupting the current through the (relatively great) inductance of the ballast a substantial voltage surge is generated across the ends of the fluorescent lamp, starting a current flow through the tube.

-Hal

 

[gar]

180704-2401 EDT

hbiss:

The article may suggest that inductive kick starts the lamp arc when the starter contact opens, and there may be such a kick. But I would not agree that this is of any importance.

What is important is that the filaments are hot and thus electron emitters, and that there is sufficient supply voltage to cause arc conduction in the fluorescent tube. This has to and does reoccur every 1/2 cycle. You can see a high initial voltage drop across the tube that drops substantially as current increases thru the half cycle. When voltage drops below holding voltage then the arc stops. On the next half cycle the cathode shifts to the other end of the tube, and a new arc has to be started. After this initial heating, and on a continuing basis this all occurs with no opening and closing of the starter.

The purpose of the starter is to initially heat the filaments so that they are good emitters of electrons. Current thru the heaters to support the arc keeps the heaters hot from cycle to cycle after the filaments have been initially heated, and the heater thermal time constant is long enough that too much cooling does not occur from cycle to cycle.

.

 

[gar]

180705-0720 EDT

In my previous post heater and filament mean exacly the same thing.

In the early radio days a tube had at least a filament and a plate (a diode) (Edison effect). With DC used to power the filament there was no hum introduced. When AC was used to power the heating element, then it became necessary to separate the heating function from the emitting function, and the indirectly heated cathode was developed.

An additional comment on starting the arc. Change the inductor in the ballast to a resistor. You still need the AC peak voltage to at least exceed the breakdown voltage of the bulb to start the arc discharge. And this has to occur every half cycle. The very first cycle and every one thereafter.

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[Nardie]

A starter is closed at first, so each filament sees 60v. After the lamp fires,the starter is open, so it has 120v across it. The ballast is a simple 1-coil inductor. See the diagram in post #2.

The fixture has two starters, so does that make a difference in the voltage the starter sees. Since the supply voltage is 120V, I would think that during the replacement of the starter the mechanic experienced 120V shock. If otherwise, please let me know. Thanks.

 

[Nardie]

You say yourself you measured 5 volts so you have to believe that. It's limited by the ballast of which there are many different designs. So you can't say that, under the same conditions they all will cause the same voltage to be present on the starter to ground.

But don't go by what I say. Take the fixture, remove the starter and measure the voltage to ground from each of the two starter contacts in the socket. Remember too that the starter can be inserted either of two ways (there is no polarity) and you say that the starter had one side shorted to the case. So you have to think about whether the starter was removed and re-inserted before you took your voltage readings.

So you can answer your own question by assuming that the shorted side of the starter was connected to the highest of whatever you measure at the socket.

Now whether it caused the shock that was reported is another matter.

-Hal

Please understand, the 5V was measured from the starter's casing to ground with a Tegam 110. The mechanic used different grounds, such as the fixture, conduit, and receptacle ground. All readings taken stabilized at 5 volts. When the starter was removed and checked with a Fluke from the prong to the casing, it measured a short. So I'm still finding it hard to believe a 5V measurement.

 

[LarryFine]

The fixture has two starters, so does that make a difference in the voltage the starter sees. Since the supply voltage is 120V, I would think that during the replacement of the starter the mechanic experienced 120V shock. If otherwise, please let me know. Thanks.

If the fixture has two starters, it also has two tubes. Each starter-and-tube acts as an independent fixture, just sharing the line and neutral electrical supply, so everything we've said here still applies.

In my opinion, with an internal short between one starter pin and the starter case, the voltage to ground is most likely to be 60v during start, or 120v but through one tube-end filament, so somewhat lower.

However, the steady 5v described in the above post suggests that the pin that was shorted to the starter case was the pin connected to the filament that was in turn connected to the neutral, and the voltage came through the gas in the tube.

 

[hbiss]

There are inconsistencies in your story or at least the story that was related to you that cause some doubts. One is that both lamps failed to operate until the one "offending" starter was replaced. As Larry states, with those fixtures with two lamps, each lamp is independent of the other with its own ballast and starter. So the that starter couldn't affect the other lamps operation.

Second, you never went back and tested the starter socket to see what voltage was present on both of the pins. Assuming that the starter was shorted one pin to case, that would have told you the two voltages possible.

Third, did you ever disassemble the starter to see where the short was? In those days when they made starters with aluminum cans, there was a cardboard sleeve inside to insulate the can from the live parts.

Fourth, the mechanic reported an electrical shock that entered his right side and exited his left side while his left hand was free and while standing on a fiberglass ladder. At most, with 120v to ground on the starter can he could have felt something in his right hand. That's it. His story is suspect because of the absence of the possibility of conduction from arm to arm and it's an old story often told by those cautioning about electric shock. "Electric shock from hand to hand travels across the chest can cause your heart to stop, so be careful and keep one hand in your pocket when working on live wiring."

Not that anybody is allowed to work live these days.

-Hal

 

[Nardie]

So, the other day I disassembled the starter and this iswhat I found. The filament was completely black, which tells me a shortoccurred. One of the leads to the capacitor (I think this a capacitor, not surenot familiar with starters) and pin terminal was missing. The casing had asmall indention on the internal side. The starter is a Damar 40W starter. Whenthe starter was removed, we noticed obvious damage since the external pins andphenolic base was awkward. When the continuity check was performed between theexternal pins to the casing, one pin was shorted to the casing. I believe thispin was the one with the missing lead and caused the indention. This is basedon the external pin configuration during removal. During the inspection, itappears the casing of the Damar starter is weaker compared to some otherstarters I pressed against. So, I’m thinking that possibility duringre-installing the starter, the casing collapsed due to the pressure applied bythe mechanic. Please remember he noticed the starter didn’t seem alignedproperly and that is why the mechanic removed it. Once removed, he didn’tnotice any damage. Yet once removed after the event, we noticed damageimmediately. So, I am confident this damage happened during re-installation. I’mwondering has anyone else experienced this type of event with a Damar starter?Also, I believe based on the internal damage that is why we measured 5 volts fromthe casing to ground. Any suggestions?

 

[gar]

180714-2427 EDT

Nardie:

Look at the circuit diagram for a starter type fluorescent light.

The fluorescent tube has a filament at each end of the tube. When excited from a current thru the filament to produce a low red glow the voltage across the filament is about 4 V on one bulb I tested.

Before an arc (plasma) is established in the tube the current path in the starter type fixture is neutral to one end of a first filament. Next current goes thru the filament to its other terminal. This terminal goes to one terminal of the starter. Initially the starter has a normally closed contact between its two terminals, thus it is a short. The second terminal of the starter goes to one end of the second filament. The other end of the second filament goes to one end of the ballast inductor. The other end of the ballast inductor goes to 120 hot or some higher voltage.

The ballast inductor is designed to provide a desired current thru the fluorescent tube after the plasma is established.

In turn the filaments are designed to heat to a desired temperature when the current thru the two filaments is that which results from the series combination of two filaments and the current limiting inductor. It appears this happens when the voltage across one filament is about 4 to 5 volts.

The starter has a voltage sensor that initially closes a shunting contact when voltage is applied and there is no plasma conduction..

Note that the starter is a short across where the plasma will develop, and the plasma cannot develop until that short is removed. Internal to the starter is a time delay mechanism (thermal) to delay opening the shunting contact to allow time for the filaments to heat.

When the contact opens, and the filaments are adequately heated, then when the voltage across the the tube reaches voltage breakdown the plasma is established. The starter is designed so that tube voltage is not great enough to reclose the shunting contact.

Once the tube is in plasma mode, then the current thru the plasma is sufficient to keep the filaments heated. What voltage is observed across a filament under self heating I don't know.

The voltages within the started are dependent upon where in the sequence you are. While preheating within a few volts of neutral (common or ground). Before preheat, no starter, one terminal is near neutral, the other near hot or something higher. After plasma is present no greater than fluorescent tube voltage drop.

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[LarryFine]

In turn the filaments are designed to heat to a desired temperature when the current thru the two filaments is that which results from the series combination of two filaments and the current limiting inductor. It appears this happens when the voltage across one filament is about 4 to 5 volts.

Wouldn't the filaments each see close to 60v? I don't believe the inductor drops that much voltage, which you're saying is 110v.

 

[gar]

180715-1015 EDT

Larry Fine:

You need to do some experiments.

A fluorescent tube is a gaseous discharge device. As such it is an approximate constant voltage load with a negative resistance characteristic. This type of load must be driven by an approximate constant current source. A voltage source with a high internal impedance compared to the load impedance can be viewed as an approximate current source.

Experiments you should do. In a starter type fixture with the fluorescent tube installed replace the starter with a short. Measure the voltage drop across one filament, and measure true RMS current thru the filaments. Do this just long enough to get your readings. This current should be higher than when the tube is driven without the short. If held too long might burnout the ballast (like an hour or faction of).

Next remove the short and install the starter. Apply power to the fixture. After the tube illuminates measure the voltage across one filament. Measure the true RMS current thru the tube.

If you put even 8 to 10 V on a filament it would burnout quite quickly.

.