Author Topic: MV lamp question?  (Read 3398 times)
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MV lamp question? « on: February 04, 2020, 09:46:32 PM » Author: HPS_250
Since MV lamps have a tungsten filament for arc tube ballasting, what would happen if you were to run a 160W SBMV lamp on say a 175W ballast? Or any other ballast?
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Re: MV lamp question? « Reply #1 on: February 04, 2020, 09:50:50 PM » Author: sol
On a 175W MV (I assume H39 only) ballast, the ~200V OCV would significantly overdrive the filament upon startup, and thus shorten the SBMV lamp's life.

On a M57 ballast (which operates H39 lamps as well), the ~300V OCV would exacerbate my first point.
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Re: MV lamp question? « Reply #2 on: February 04, 2020, 09:52:31 PM » Author: wide-lite 1000
I'm not sure about a "good" SBMV , but I tried doing that years ago with one with a burned out filament and NOTHING !
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Re: MV lamp question? « Reply #3 on: February 04, 2020, 09:54:57 PM » Author: sol
Well, the filament is in series with the arc tube, so if it is broken, the current will not flow to the arc tube rendering the lamp useless.
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Re: MV lamp question? « Reply #4 on: February 04, 2020, 09:55:39 PM » Author: HPS_250
I'm not sure about a "good" SBMV , but I tried doing that years ago with one with a burned out filament and NOTHING !

Odd! I though that since the filament was broken and couldn’t be used as a ballast anymore, then the arc tube would at least operate on an external ballast. Maybe the filament was a path for current and the lamp went open circuit when it broke.
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Re: MV lamp question? « Reply #5 on: February 04, 2020, 09:58:16 PM » Author: HPS_250
Well, the filament is in series with the arc tube, so if it is broken, the current will not flow to the arc tube rendering the lamp useless.

That’s sad. So when the filament goes EOL the whole lamp is rendered useless.
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Re: MV lamp question? « Reply #6 on: February 04, 2020, 09:58:35 PM » Author: sol
Yes, unfortunately.
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Re: MV lamp question? « Reply #7 on: February 05, 2020, 06:28:37 AM » Author: Medved
Odd! I though that since the filament was broken and couldn’t be used as a ballast anymore, then the arc tube would at least operate on an external ballast. Maybe the filament was a path for current and the lamp went open circuit when it broke.

In order to work as the ballast, there is no other option than to be in series with the arctube. So then once it breaks, its all over.

But normally the SBMVs use to be designed so the filament breaks only after so many hours the arctube has blackened too, so it is EOL as well, so no benefir to allow it to work elsewhere... Of course, "no benefit" means for normal use, not for collectors purpose...
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Re: MV lamp question? « Reply #8 on: February 05, 2020, 10:43:41 AM » Author: HomeBrewLamps
You can always smash the lamp open and salvage the (likely still clear) arc tube. Then try running it on a 100 watt MV ballast.
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Re: MV lamp question? « Reply #9 on: February 06, 2020, 03:32:35 AM » Author: HomeBrewLamps
Would you have to put the arc tube inside a new glass bulb before running it on the ballast?

If you can find someone to do it or do it yourself yeah.

But what I usually do is attach it to wires then place it inside a lidded jar and light it that way to block the UVC.
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Re: MV lamp question? « Reply #10 on: February 06, 2020, 06:18:43 PM » Author: BT25
On a 175W MV (I assume H39 only) ballast, the ~200V OCV would significantly overdrive the filament upon startup, and thus shorten the SBMV lamp's life.

On a M57 ballast (which operates H39 lamps as well), the ~300V OCV would exacerbate my first point.
With an SBMV rated for 120V, hooked up to the secondary of an H39 ballast (that has an OCV of 240V), the SBMV would last a few seconds if that before self destructing, and releasing the magic smoke.
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Re: MV lamp question? « Reply #11 on: February 06, 2020, 08:07:47 PM » Author: HPS_250
If you can find someone to do it or do it yourself yeah.

But what I usually do is attach it to wires then place it inside a lidded jar and light it that way to block the UVC.

That’s cool! Does it light like a normal MV lamp?
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Re: MV lamp question? « Reply #12 on: February 07, 2020, 12:31:41 AM » Author: Medved
With an SBMV rated for 120V, hooked up to the secondary of an H39 ballast (that has an OCV of 240V), the SBMV would last a few seconds if that before self destructing, and releasing the magic smoke.


Why do you think?
H39 is a 1.5A ballast, 160W 120V lamp is about 1.4A nominal lamp, so barely 7% overvoltage. Detrimental for life for sure, but not any immediate death. See the incandescent characteristics (the ballast filament is the weak point there)...
And that meant when the filament is still there (although I see no reason why to use the ballast in that case)
Connecting just the arctube (from the 160W SBMV) to the H39 means just about the right current, it is just a question if it will start, because without the filament ballast heat I doubt the preheat system will work, so may need disconnecting, so what remains is an arctube without starting probe.
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Re: MV lamp question? « Reply #13 on: February 07, 2020, 12:54:18 AM » Author: BT25

Why do you think?
H39 is a 1.5A ballast, 160W 120V lamp is about 1.4A nominal lamp, so barely 7% overvoltage. Detrimental for life for sure, but not any immediate death. See the incandescent characteristics (the ballast filament is the weak point there)...
And that meant when the filament is still there (although I see no reason why to use the ballast in that case)
Connecting just the arctube (from the 160W SBMV) to the H39 means just about the right current, it is just a question if it will start, because without the filament ballast heat I doubt the preheat system will work, so may need disconnecting, so what remains is an arctube without starting probe.

You're missing the point...it's about the OCV (open circuit voltage) which is 240V nominal. This will blow the 120V lamp when power is applied. A 7% over-current is not the same.
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Re: MV lamp question? « Reply #14 on: February 07, 2020, 07:09:35 AM » Author: Medved
You're missing the point...it's about the OCV (open circuit voltage) which is 240V nominal. This will blow the 120V lamp when power is applied. A 7% over-current is not the same.

You may be surprised, but the OCV is irrelevant, once the current is limited and it is within the range of the load, if the load is stable with a constant current supply.
In other words: The voltage would never go that high because of the load presented by the lamp.

It is the same as thinking about the maximum current the mains may deliver vs what your incandescent is designed for and panicking "a 30A outlet will blow a 0.5A lamp instantly". You pretty agree it wont, if that lamp is designed to the voltage in the mains. And the same is with the loads of a constant current source: 240V OCV can not blow a 120V lamp, if the lamp can take the source output current (the 1.5A of the ballast, vs 1.4A of the 160W SBMV)

So because of the high OCV, what we have is a constant current source supply.
That is a component, which feeds all the same current, regardless what voltage drop the load presents. So an ideal constant current source may go to an infinite OCV, yet still no problem at all.
The voltage is then all the voltage drops across both components of the lamp, summed together: The arc tube drop, plus the filament drop.
The arc tube drop is rather insensitive on the current, so it will be the same as when the whole assembly is operated at 120V.
The filament is an incandescent lamp, so in fact a resistor A resistor means, the voltage drop is Resistance times Current. So we should know, what the resistance would be when fed by the current source, so what the voltage drop would be, so how much the filament would be overloaded. If you look into any incandescent characteristics, once the major power dissipation is the radiation, the resistance changes only a little (because only small temperature difference has big impact on the radiated power), what means the thing is only mildly overloaded with that small (7%) overcurrent (the voltage drop would be about 15% higher than designed because of the extra current).

For many people is hard to grab the concept of constant current supply and its duality nature compare to the well understood constant voltage supply:
The constant voltage maintains the voltage constant, so not changing, whatever the load does with the current.
With constant current the current remains constant, regardless what the load does with the voltage.
With a constant voltage, the voltage is dictated by the voltage source and the current by the load conductivity (so 1/resistance).
With a constant current the current is dictated by the source, the voltage then by the load resistance (so 1/conductivity).
With constant voltage source system, multiple loads are connected in parallel, if they are supposed to not influence each other. The maximum is given by the maximum current the source is capable to deliver. That capability could be then described e.g. as a Short Circuit Current.
With constant current source system, multiple loads are to be connected in series, if they are supposed to not influence each other. The maximum number of loads is given by the maximum voltage the source is capable to deliver. That capability could be then describer as e.g. Open Circuit Voltage.
In both cases the Power = Voltage * Current * PowerFactor.
With constant voltage, higher load conductivity, so lower resistance, means higher current, so higher power delivered to the load and vice versa.
With constant current, higher load resistance, so lower conductivity, means higher voltage, so higher power delivered to the load and vice versa.
With a constant voltage, the load voltage rating has to match the source voltage output.
With a constant current, the load current rating has to match the source current output.

The SBMV is an incandescent with a series discharge. So in order to judge if it is overloaded and how much on a constant current source supply, the determining parameter is the source output current.
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