Need help identifying a component (pic attached)

[Casper The Ghoul]

First time posting on the forum, so I have no idea if this is the correct section for a post like this, so my bad if it isn't.
My family's old Samsung LCD flatscreen TV from 2007 finally gave out on us (busted relay), so we decided to get a more modern TV, and I was allowed to scrap the old one for parts.
The parts inside were your standard affair, Mosfets for driving plasma screens, high wattage resistors, what have you.
But there's one that's been bugging me for a while because I can't figure out what it is.
As you can see in the attached image, it's enclosed in glass and is about the same diameter as a 1/4" audio jack.
The labels on the glass are 452, 0, and 9H.
Inside are two parallel leads in an H shape with a small plastic tab holding the leads apart.
I think it might be a halogen light bulb, but I don't work frequently enough with them to recognize every type, and I've never seen one exactly like this.
Thanks in advance.
(Edit: Corrected image orientation)

[bob91343]

In what section of the unit was this?  It could be a glow lamp or a thermal cutout.  Have measured its resistance?

[free_electron]

Thats a gas discharge surge arrester.

overvoltage protection. it flashes over at a certain voltage,

[bob91343]

Judging from its size, it won't eliminate surges of very high energy.

[Casper The Ghoul]

In what section of the unit was this?  It could be a glow lamp or a thermal cutout.  Have measured its resistance?

It was in the power supply section. I didn't see much of a point in measuring because I did a continuity test with my multimeter and didn't get a reading. I can't insert images directly into posts from a chromebook, so I'll just attach a pic of the circuit board. The part where it was soldered is labeled RY801S.

Thats a gas discharge surge arrester.

overvoltage protection. it flashes over at a certain voltage,

I had a feeling it might have been for some form of voltage protection, thanks for confirming. Is there any way I can determine the voltage rating from the labeling, or would I just have to put it in a circuit and increase the voltage until it discharges?

[sokoloff]

https://en.wikipedia.org/wiki/Neon_lamp#Voltage_surge_suppression

If you have a variac, you can just dial the voltage up until it conducts. Start around 50V and go up from there.

[Casper The Ghoul]

https://en.wikipedia.org/wiki/Neon_lamp#Voltage_surge_suppression

If you have a variac, you can just dial the voltage up until it conducts. Start around 50V and go up from there.

Unfortunately, I don't have one. I'm more of a casual hobbyist, and the few times I've ever messed around with a transformer, I've triggered our breaker, so I'm a little hesitant to mess around with a big boy toy like a variac haha

[sokoloff]

Get another neon lamp and test this component in series with that neon lamp across 120VAC.

I think 120VAC might be too much for it, or at least I wouldn't start there.

Another option is to test it across an old-school telephone line (about 48VDC when on-hook and 75-100VAC when ringing [but highly variable]).

[bob91343]

Do NOT use a variac unless you put current limiting in series.  Otherwise, when it conducts, it will destroy itself.

[ArthurDent]

As mentioned, this is a neon lamp that can be used as a spike eliminator. A way one would test it is to wire a 56K-150K resistor in series and connect this to a variac, and as you increase the voltage, at some level, maybe 70-90 volts, the neon will ionize and the lamp lights. If you were to then measure the voltage drop across the lamp when lit you will find that it has dropped to some lower voltage, perhaps 50 volts. if a resistor isn't used to limit the current and it is placed across 120 vac, you could have a small 'pop' as the lamp explodes.

[bob91343]

Another point is that if you measure the voltage across a lit neon lamp energized from AC the reading will be misleading.  It will not be sinusoidal and your meter is calibrated for sinusoidal waves.

If you do the same test with a dc power supply you will measure dc across the lamp.  That's a more useful number.