turning hid on above water

[alo100]

Hello,

I have now attached an EXCEL sheet with measurements. The values are average values measured by me in a laboratory.


Greetings, Michael

Thanks Michael!!!
emm.. let me read the data.....

 

[Meng_Tze]

The "pop" of the bulbs could be due to switching/heat -- like Meng said.
Or the difference in temp --> like Michael said "current air lowered the lifetime"
heat inside the bulb builds up vs. cold media outside, on both sides of the bulb surface, maybe?

Any thoughts, comment, idea?

No not the latter.... thing is the ballast once it has ignited the bulb will control current output due to negative resistance of the arc. This negative resistance would want to draw more and more current. The ballast is set for a constant power output, and with this a constant voltage, it only can regulate the current based on the variable negative resistance.

If in this secnario, the bulb would be cooled (by flowing air, resulting in temp difference between inside and outside) if the heat exchange would also result in an increase of current demand (neg resistance), the ballast would shut off once the power demand would surpass its capacity. with a properly set ballast, it would happen before there would be damage to the bulb. If Michaels understanding of what happens here (in the example of 'cooled' bulb)is different, pls do highlight this, I am intrigued


At point of ignition, things are different. This is where we have a state of high voltage and high current to even ignite and ionize the gas. This coupled with already hot environment, still partially ionized gas will lead to pressure build up in an instance, and the glass would pop. This is why I feel that restriking shortly after shutdown is not advisable. Again, I hae no numbers of how long this is, but 30 secs seems short to me.

 

[alo100]

No not the latter.... thing is the ballast once it has ignited the bulb will control current output due to negative resistance of the arc. This negative resistance would want to draw more and more current. The ballast is set for a constant power output, and with this a constant voltage, it only can regulate the current based on the variable negative resistance.

If in this secnario, the bulb would be cooled (by flowing air, resulting in temp difference between inside and outside) if the heat exchange would also result in an increase of current demand (neg resistance), the ballast would shut off once the power demand would surpass its capacity. with a properly set ballast, it would happen before there would be damage to the bulb. If Michaels understanding of what happens here (in the example of 'cooled' bulb)is different, pls do highlight this, I am intrigued


At point of ignition, things are different. This is where we have a state of high voltage and high current to even ignite and ionize the gas. This coupled with already hot environment, still partially ionized gas will lead to pressure build up in an instance, and the glass would pop. This is why I feel that restriking shortly after shutdown is not advisable. Again, I hae no numbers of how long this is, but 30 secs seems short to me.

Let me catch up with what you are saying, then also some questions.
There are 2 types of ballast/regulators, one is current controller, the other is power controller. I am not a light maker and maybe some of you can tell me. If we look at some battery manufacturer web-site like the Dura Cell, it would tell you that, constant power control is the best use of the battery (meaning, using every bit of the battery) but then I do not know if this is the type of the ballast the industry is using for the dive light. Any comment, anybody?

Then back to the light operation, I am just trying to rephrase your words, Meng, and what I have heard before, so pls correct me if I am wrong. For the normal operation i.e. the light is warmed up already. The more the current, the higher the resistance at the arc => higher voltage => feedback to the ballast, the ballast would start to pull down the supplied current from the battery (I assume that it is a const current regulator) => lower supply current => lower heat => lower resistance => lower feedback voltage => ballast increase the supply current ..... so back and forth around the equilibrium pt.

Then for the ignite stage.... so are you telling me that the ballast itself would have this mode also just to help out the bulb to go through the whole warm up stage? Our it is done naturally because the resistance of the bulb is low at the beginning and the ballast would just feed in current??
Anyway, at the beginning of this stage the chemical inside is not ionized yet, high voltage, high current, and high pressure.
Then if the light is just off, the partially ionzed gas is still keeping the inside pressure high, at this moment if the re-ignition happens, igntion mode is used this will create even high pressure inside the bulb and "pop" -- kiss the bulb goodbye!

In a way, if I am changing the bulb to one which is of higher power and change the ballast, then if at the very beginning, the bulb is broke, that means that, the applied current is not even right, so that the high pressure is built, can I say that? So I need to adjust the ballast again.

 

[Meng_Tze]

Let me catch up with what you are saying, then also some questions.
There are 2 types of ballast/regulators, one is current controller, the other is power controller. I am not a light maker and maybe some of you can tell me. If we look at some battery manufacturer web-site like the Dura Cell, it would tell you that, constant power control is the best use of the battery (meaning, using every bit of the battery) but then I do not know if this is the type of the ballast the industry is using for the dive light. Any comment, anybody?

Then back to the light operation, I am just trying to rephrase your words, Meng, and what I have heard before, so pls correct me if I am wrong. For the normal operation i.e. the light is warmed up already. The more the current, the higher the resistance at the arc => higher voltage => feedback to the ballast, the ballast would start to pull down the supplied current from the battery (I assume that it is a const current regulator) => lower supply current => lower heat => lower resistance => lower feedback voltage => ballast increase the supply current ..... so back and forth around the equilibrium pt.

Then for the ignite stage.... so are you telling me that the ballast itself would have this mode also just to help out the bulb to go through the whole warm up stage? Our it is done naturally because the resistance of the bulb is low at the beginning and the ballast would just feed in current??
Anyway, at the beginning of this stage the chemical inside is not ionized yet, high voltage, high current, and high pressure.
Then if the light is just off, the partially ionzed gas is still keeping the inside pressure high, at this moment if the re-ignition happens, igntion mode is used this will create even high pressure inside the bulb and "pop" -- kiss the bulb goodbye!

In a way, if I am changing the bulb to one which is of higher power and change the ballast, then if at the very beginning, the bulb is broke, that means that, the applied current is not even right, so that the high pressure is built, can I say that? So I need to adjust the ballast again.

Almost, but in reverse

V=I*R and P=V*I (view the bulb as a resistivel oad for the ballast)

Here is a cut from a post I did on another site, maybe this clarifies things better:

What does a ballast do:
First, the ballast system provides the necessary voltage required to start and stabilize the lamp. However, once the arc tube is ionized, and the arc strikes, it is necessary to control and limit the current flow through the lamp because of its negative resistance characteristics that want to draw more and more current. So the second function of a ballast is to limit the amount of current the lamp draws while at the same time controlling the lamp wattage at its design center while the incoming line voltage (of the bulb) varies.

In other words:

A ballast looks for minimum constant input voltage to deliver the higher output voltage and controled amperage (fixed voltage x controled amperage = controled Wattage -Power-) to maintain the arc.

Ignition: High voltage (say 80V) to 'charge the gas' Ioninze it, this then sparks and the arc is pulled between the electrodes. Immediately ater that..... current starts rising (negative resistance - 'conductance' if you like in Siemens (unit of conductance)). I = 21/80 = 2.6mA. It is the voltage that causes the gas to ignite (similar to lightning)

Operation: In order to keep the bulb within operating parameters (lets say 21W), the current needs to be controlled..... With an operating voltage of say 18V (down from 80V) and a power req. of 21W, the operating current would be 21/18= 1.16amps...... (these are just examples to illustrate) It is the current that maintains the arc. This is at the lamp end.

At the battery end, the ballast will pull the required amount of current to deliver the output power. The voltage is contstant (12 Battery) this this means that the current is the only thing it can vary according to the 21W output requirement. (minus a small drop out voltage/power wastage in the ballast itself).

Hope this calrifies it...

 

[micbu]

@Meng-Tze

Hey, what are you talking about????

The ignition voltage is several thousand volts!!!
The ignition voltage is between 6000V and 20000V depending on the system you are using!
The burning voltage is between 65V and 85V, also depending on the system you are using.
It is not only the current that maintains the arc. It is an interaction of voltage and current depending on the Q of the bulb.
A ballast converts the battery voltage to the desired burning voltage, but before that an ignition circuit has to provide the needed ignition voltage.
The arc has nearly no resistance. Therefore a ballast is called ballast, because it determines the resistance and so the wattage. Some systems have a power control function (e.g. Brightstar and Xenlight) other systems not (e.g. Welch Allyn).

Greetings, Michael

 

[Meng_Tze]

@Meng-Tze

Hey, what are you talking about????

The ignition voltage is several thousand volts!!!
The ignition voltage is between 6000V and 20000V depending on the system you are using!
The burning voltage is between 65V and 85V, also depending on the system you are using.
It is not only the current that maintains the arc. It is an interaction of voltage and current depending on the Q of the bulb.
A ballast converts the battery voltage to the desired burning voltage, but before that an ignition circuit has to provide the needed ignition voltage.
The arc has nearly no resistance. Therefore a ballast is called ballast, because it determines the resistance and so the wattage. Some systems have a power control function (e.g. Brightstar and Xenlight) other systems not (e.g. Welch Allyn).

Greetings, Michael

My example was just to illustrate the function. What you say above is exactly what what I describe below.....(read it carefully) so we dont disagree on that.

 

[micbu]

Hi,

o.k. I agree, but then you should write that it is just example with fictional voltages, otherwise you make the systems more harmless as they are. Also you write about a negative resistance characteristics of the bulb?????? what do you mean by that??? When there is an arc then the arc has nearly no resistance.
I can not follow what you mean by that.


Greetings, Michael

 

[alo100]

Hi,

o.k. I agree, but then you should write that it is just example with fictional voltages, otherwise you make the systems more harmless as they are. Also you write about a negative resistance characteristics of the bulb?????? what do you mean by that??? When there is an arc then the arc has nearly no resistance.
I can not follow what you mean by that.


Greetings, Michael

It's ok that we are divers but not engineers, the society needs different types
of professions, each serve for it's purpose:
Conductance instead of "negative resistance", which is 1/Resistance (in Ohms)
Power (in Watts) instead of wattage.

 

[alo100]

@Meng-Tze

Hey, what are you talking about????

The ignition voltage is several thousand volts!!!
The ignition voltage is between 6000V and 20000V depending on the system you are using!
The burning voltage is between 65V and 85V, also depending on the system you are using.
It is not only the current that maintains the arc. It is an interaction of voltage and current depending on the Q of the bulb.
A ballast converts the battery voltage to the desired burning voltage, but before that an ignition circuit has to provide the needed ignition voltage.
The arc has nearly no resistance. Therefore a ballast is called ballast, because it determines the resistance and so the wattage. Some systems have a power control function (e.g. Brightstar and Xenlight) other systems not (e.g. Welch Allyn).

Greetings, Michael

Michael, got you, so that Brightstar and Xenlight is using constant power control, and Welch Allyn is using constant current control.
Then besides the above control circuitries, part of the ballast/controller also includes ignition circuit....
I have heard about the thermal feedback control from other lights but I guess it may not be neccessary for your product..... somehow I have more confidence about the design.. the reason is, I have seen some DIY projects... but I guess your engineering team has taken care of the issues.

 

[micbu]

Hi alo100

I think there is a misunderstanding. These devices are not our devices!!
I have just tested them deeply and I sell them in Germany.
The WA products are not controlled at all. The resistance is fixed and so the current decreases when the input battery voltage is decreasing. This will result in a light that is getting more and more blue.



Greetings, Michael