Acceleration voltage effect on VFD phosphor lifetime

[f36]

I'm thinking about using a small VFD (a funky Soviet single-pixel tube; kinda like a magic eye, but without variable shadow) in a future project. It's rated for 50V anode voltage, but I already have a ~200V rail in the project for other purposes. Assuming the tube doesn't arc over inside, the 200V should be able to light it up just fine, and I should be able to reduce the heater power and grid voltage to get the emission current back in spec.

My concern is regarding the lifetime of the phosphor (maybe ZnS or CdS?) and how the higher-energy electrons might affect it. It seems like the main (?) mechanism for phosphor degradation is due to the electrons converting the metal cations back into their neutral, metallic form, which reduces the efficacy of the phosphor. However, I haven't really found much information regarding the factors influencing the rate of this process. Is it related to electron energy (anode voltage), rate of electrons/accumulation of charge (current), or a thermal thing related to power?

My intuition says it would have more to do with emission current than acceleration voltage. I would think the difference between 50eV and 200eV electrons isn't going to be the difference between breaking ionic bonds and not, but more free electrons available could accelerate damage. If so, limiting the emission current should be sufficient to retain the lifetime of the tube, and the max voltage would be limited by arcover or your geiger counter starting to tick...

[Ian.M]

Underrunning your heater is asking for trouble with cathode poisoning, or if the grid voltage ever spikes, cathode stripping.  Why not simply include a dropper resistor in series with the anode, calculated to bring the anode voltage down to spec at the nominal anode current?  You'll still need to verify it doesn't flashover or fail to blank due to the increased E field when the grid is grounded.

[f36]

Yeah, that would be the obvious solution, but I was initially concerned that that resistor would cause the power consumption of the whole thing to increase by 4x (this thing is going to be battery powered). But, if the same current is required for the same brightness regardless of voltage, then it would make no difference, apart from possibly working at a lower heater temperature, but then that brings the problems you mentioned.

I am also planning on turning the heater on and off to save power, so I would think a lower temperature could improve lifetime there. I suppose I could program the heater to increase in temperature with the anode voltage disabled to clean it periodically if that would help. Or I could just not care and put a 10% bigger battery in

[David Hess]

My concern is regarding the lifetime of the phosphor (maybe ZnS or CdS?) and how the higher-energy electrons might affect it. It seems like the main (?) mechanism for phosphor degradation is due to the electrons converting the metal cations back into their neutral, metallic form, which reduces the efficacy of the phosphor.

I thought phosphor degradation was primarily from heating, which explains why aluminized phosphors are so much tougher.

[f36]

I thought phosphor degradation was primarily from heating, which explains why aluminized phosphors are so much tougher.

I was mainly going off of this section in Wikipedia which mentions a number of different factors, most of which presumably could be accelerated by increased temperature.
https://en.wikipedia.org/wiki/Phosphor#Phosphor_degradation

[David Hess]

The Tektronix Circuit Concepts book about CRTs has some discussion about burning of CRT phosphors.  It is the first book here:

https://w140.com/tekwiki/wiki/Concepts_Series

I know that high voltage CRT phosphors last longer than low voltage ones.  This was a problem with CRT flat panels (emission displays?) which had to use low voltage phosphors.

[f36]

That Tek book is an interesting read, and here are a few relevant excerpts:

If beam current is increased or the same amount of beam current is concentrated in less area by reducing the spot size, there will be an increase in luminance. An increase in accelerating potential will yield an increase in luminance. (page 67)

When a phosphor is excited by an electron beam having an excessively high current density, a permanent loss of phosphor efficiency may occur. ... Darkening or burning occurs when the heat developed by electron bombardment cannot be dissipated rapidly enough by the phosphor. (page 77)

They assert that the brightness is a function of both the current density and the electron energy, and support your statement about heat being the factor to control regarding burn in.

In that case, maybe the tube would be more of a constant-power device, and the current should be reduced with a higher anode voltage. Adjust the grid to give the same power/brightness as the standard config? I might try that later to see how it behaves...

[David Hess]

On the other hand there are materials which degrade in proportion to the number of electrons which strike them.  The materials used for secondary emission work this way, so no matter what the current density or acceleration voltage, only a finite number of secondary electrons are available, and the material has a limited life based on the number of electrons that hit it.

So I guess this is going to depend on the chemical composition of the phosphor.  The commonly used P31 phosphor has no such limitation and when aluminized is incredibly resistant to burning.  I have seen burned P31 on display CRTs, but I do not think those are aluminized.

[f36]

If aluminization makes a big difference regarding lifetime, then there might be a distinction to be made between CRTs and VFDs. In VFDs the electrons strike the exposed surface of the phosphor, while in aluminized CRTs, the electrons hit the phosphor that is in direct contact with the aluminum first. If the heat from the electron beam is generated in just the very surface of the phosphor, this might make a difference, but otherwise VFDs might behave more like aluminized CRTs since the phosphor is usually deposited on a metal electrode instead of glass.

I also would expect that the phosphor in CRTs is stressed much more than that in VFDs, depending on the thermal capacitance/conductivity of the phosphor layer. You have a very small beam lighting up only a small area of the screen at a time and in order to get a good average brightness, that spot has to be quite high intensity. The Tek book says it can cause immediate burn in if the deflection circuitry is not working.

But what if the horizontal amplifier inoperative is inoperative? The horizontal plates will not sweep receive a signal under that condition and the beam will not be deflected but it will be turned on by the unblanking pulse. Result?- possibly a burn mark on the screen! (page 77)

According to this: http://www.labguysworld.com/crt_phosphor_research.pdf
P31 phosphor is ZnS:Cu. The phosphor in VFDs is commonly ZnO:Zn according to Wikipedia, which is apparently P15. The Tek book lists both as "hard to burn."

Also in the Tek book they show running aluminized P15 phosphor at an acceleration voltage of 10kV. They also say

Some of the effective energy of the beam is used to penetrate the aluminum coating, reducing the effective acceleration potential. The effective deceleration of the beam depends on the thickness of the aluminum and may range from 1 kV to 3 kV or more. (page 80)

Which means the P15 phosphor is getting hit with, say, 8keV electrons. That doesn't entirely answer my original question, but it certainly indicates that the phosphor in VFDs can be hit with far more than 200eV electrons without being excessively damaged as long as the current density is kept low, which is really all I needed to know.

[andy2000]

CRTs are aluminized to prevent ion burns.  Early CRTs lacked this, and needed an ion trap in the electron gun.  Without this, the central area of the screen would quickly degrade.  A side benefit of aluminized screens was that it acted as a mirror to direct more light toward the front.   

[SeanB]

Put the resistor in, you either dissipate power in the resistor, or do it in the tube itself. Better to run the tube within ratings, that run it at around 4 times the voltage, where it is extremely likely to suffer from arcing over between the anode and the other electrodes, as they have been built to withstand 50V only, likely fine to 100V, but above that arcing, or just simple slow build up of ion deposits, is very likely.  Resistor and a zener diode to clamp to 47V will work. Heater run at the specified, if you want longer life put a series resistor to drop the voltage to the lower end of the allowed filament voltage range, and thus also have a bonus of inrush limiting when cold. Running at lower heater current, and thus making the heater current limited, will strip it badly within a short time, and kill the tube. You want full emission with a VFD, it will otherwise band, as the limited current is going to flow to the closest portions o the anode.

[f36]

Put the resistor in, you either dissipate power in the resistor, or do it in the tube itself. Better to run the tube within ratings, that run it at around 4 times the voltage, where it is extremely likely to suffer from arcing over between the anode and the other electrodes, as they have been built to withstand 50V only, likely fine to 100V, but above that arcing, or just simple slow build up of ion deposits, is very likely.  Resistor and a zener diode to clamp to 47V will work. Heater run at the specified, if you want longer life put a series resistor to drop the voltage to the lower end of the allowed filament voltage range, and thus also have a bonus of inrush limiting when cold. Running at lower heater current, and thus making the heater current limited, will strip it badly within a short time, and kill the tube. You want full emission with a VFD, it will otherwise band, as the limited current is going to flow to the closest portions o the anode.

You say that when running the tube at a higher than designed voltage, other than arcing, there may be movement of ions between the electrodes that will cause damage. Although I did find reference to CRTs with the same phosphor being run at a much higher voltage in the Tektronix book, I understand that there are many other considerations that the engineers who designed the tube made when specifying the anode voltage. From what I'm getting, these ions will come from the cathode or the material coating it, and so this is the material that will be (directly) damaged by the higher voltage. From my experience, the exact composition of the cathode seems less readily apparent and is less commonly discussed.

The main reason I started this thread was to understand the effects high voltage has on VFDs, and I assumed it was the phosphor itself that we should care about since burn-in is clearly an effect on the phosphor. So, perhaps my question should be reformulated to include the rest of the components of the tube.

When you say that there will be ion movement across the tube that will damage it, what exactly moderates this process and causes it to become a problem beyond a certain voltage? Is there an E-field strength that, beyond which, said ions will have a much lower chance of returning to the cathode and instead will be launched towards the anode?

People sometimes report successfully "repairing" VFDs by overdriving the heater temporarily. Is the aging mechanism that leads to this related?

Obviously the ideal solution would be to run the tube at its rated voltage, and I can easily design a circuit to do so. I am wondering, even if it ends up being for purely educational purposes due to the impracticality of it, what exactly is the damage mechanism associated with overvolting a VFD. Perhaps if it is a quantifiable "risk," some engineering tradeoff decisions can be made regarding it, instead of blindly declaring the datasheet values absolute maximums (though, admittedly, that's what we EEs should do 99.9% of the time...)