[Galapac]

Looks like an honest seller - he also lists rebranded Telefunken 6080 tubes clearly as RCA made.
https://www.ebay.com/itm/255313837773?hash=item3b71e3fecd:g:swsAAOSwoa1h1KMo

And here is another Telefunken branded GE tube:
https://www.ebay.com/itm/264912109312?hash=item3dadfdff00:g:KZcAAOSwunJflVJc

The back of the tube has the telltale GE dots:

Perhaps the easiest way to tell real ones apart from fakes is to look for a double top mica:

@mordy - You beat me to it by 1 minute, lol.

 

[cddc]

Yes, you have to be careful buying Telefunken 6080s.
I bought some years ago and educated myself on what was genuine.

For example here is a bad fake:
https://www.ebay.com/itm/255313837773?hash=item3b71e3fecd:g:swsAAOSwoa1h1KMo

• Bad date code
• 6AS7W marking (really?)
• no ”Made in Germany” marking
• single top mica(Telefunkens have two)
• new boxes
• and worst of all GE markings to boot

Many took Thomson France 6080WA tubes and faked them to look like Telefunkens but the easiest give away is bad date codes ormany tubes having the same date code since the forgers were lazy and didn’t bother to change the codes.

Agree. Sorry for your bad experience. There are lots of fake Telefunken 6080 tubes out there, actually the ones listed out there currently are mostly fake, only a very few of them are genuine, just like the one asked.... Mostly they are just rebranded GE 6080s, and some rebranded RCA 6080s, of course most of the fake ones come with a shiny Telefunken box which confused lots of folks, sadly...

 

[Tom-s]

Assuming the tubes measure roughly the same and will be at the same operating point in the same circuit - meaning the same plate-to-cathode voltages, grid-to-cathode voltages, and plate currents - the heaters and the plates will dissipate the same amount of power in two different models.

However, the plates have different material and physical construction, as such they are going to have different thermal resistance, °C/W, increase in material temperature for unit of power dissipated.

So the power dissipation is the same, but the temperature will different based on the thermal resistance of the internal structure.

Also, there is no "grid current", this is class A1, the grid is always negative relative to the cathode. Looking into the grid from the previous stage will present an impedance in the tens of megaohms range, it won't draw any current. The only current contributing to the increase in tube temperature will flow through the heaters and from plate to cathode.

LG, I can't believe an educated person like you with a Dr. degree would argue something that is factual...you think I'm as gullible as some uneducated person does...I'm so disappointed in you!

Okay, let's talk about thermal resistance and the law of conservation of energy!

So what is the thermal resistance that you're trying to base your argument on? It's an object's resistance to heat conductance. Yes, some materials conduct heat faster, some slower, just like the Raytheon graphite plate 6080 takes longer to warm up, and the regular plate 6080 warms up much faster. But have you considered the law of conservation of energy?!!!

As long as the same 2.5A heater current is drawn, and the same amount of plate current and grid current (yes, I intentionally list grid current here just in case someone will be nit-picky on the subject! I will elaborate later) are going through both the graphite-plate and regular-plate tubes, the same amount of total heat will be generated, which can be calculated by the following equations, Total power consumed = heater power consumption + plate power consumption + grid power consumption = I_heater x V_heater + Ip x Vp + Ig x Vg. The 6 parameters on the right hand side of the equation will be the same for both Ray graphite-plate 6080 and regular-plate 6080, hence the total power consumed will be the same for both tubes ( and total heat generated = total power consumed - power consumed by the load on tube, will also be same since load will be the same)! Yes, Ray graphite-plate 6080 will reach normal thermal operating temperature slower than regular 6080 does because of its higher thermal resistance, but it will eventually get there (some folks say graphite-plate 6080 needs 10-30mins to warm up), and what's more important, the total heater generated by both 6080 tubes will be the SAME, that's defined by the equations I listed above and the law of conservation of energy.

Now, let me elaborate why I intentionally listed grid current there previously. Yes, most tubes run on negative grid voltage on their grids, hence their grid current will be close to 0, but let's not forget that some tubes do run on positive grid voltages, so there will be measurable grid current on them! And even under the negative grid voltage mode, some electrons will go astray and hit the grid by chance, hence will generate grid current theoretically (though this current is almost 0 and is negligible). So if I did not list grid current there, some guy will be nit-picky on it. And even I did include the grid current there, someone is still trying to be nit-picky on it

IMHO there's better ways to acknowledge you don't completely get the point.
Even when there's an obvious lack of understanding of what @L0rdGwyn is actually saying.
Let yourself be educated here. The forum's here to help. Not to fight you.

It would help to educate oneself a bit on the principles of heat transfer that LG is trying to convey.
Instead of going all keyboard warrior when another member is actually trying to teach you something.

 

[Galapac]

Agree. Sorry for your bad experience. There are lots of fake Telefunken 6080 tubes out there, actually the ones listed out there currently are mostly fake, only a very few of them are genuine, just like the one asked.... Mostly they are just rebranded GE 6080s, and some rebranded RCA 6080s, of course most of the fake ones come with a shiny Telefunken box which confused lots of folks, sadly...

Oh no bad experience here, sorry if I implied that. I have genuine ones but it was this site and jacmusic.com that educated me on what was the real deal before I bought mine.

 

[Renexx]

Here is my pair of Telefunken 6080s.

For the real deal take a look for double O getter at the top and Telefunken code printed with 7 numbers or a German label like that.

 

[Galapac]

Here is my pair of Telefunken 6080s.

Take a look for double O getter at the top and Telefunken code printed with 7 numbers or a German label like that.

Nice tube examples you have there and one of my favorite work horse tubes in my Euphoria. They are a true gem.

 

[Renexx]

Nice tube examples you have there and one of my favorite work horse tubes in my Euphoria. They are a true gem.

Thanks, at first I was a little dissapointed but I'm starting to find good combinations for it on my Euphoria. Yesterday I did listen to them with Sylvania 53' Bad boys and they come through beautifully with the Telefunkens.

 

[L0rdGwyn]

LG, I can't believe an educated person like you with a Dr. degree would argue something that is factual...you think I'm as gullible as some uneducated person does...I'm so disappointed in you!

Okay, let's talk about thermal resistance and the law of conservation of energy!

So what is the thermal resistance that you're trying to base your argument on? It's an object's resistance to heat conductance. Yes, some materials conduct heat faster, some slower, just like the Raytheon graphite plate 6080 takes longer to warm up, and the regular plate 6080 warms up much faster. But have you considered the law of conservation of energy?!!!

As long as the same 2.5A heater current is drawn, and the same amount of plate current and grid current (yes, I intentionally list grid current here just in case someone will be nit-picky on the subject! I will elaborate later) are going through both the graphite-plate and regular-plate tubes, the same amount of total heat will be generated, which can be calculated by the following equations, Total power consumed = heater power consumption + plate power consumption + grid power consumption = I_heater x V_heater + Ip x Vp + Ig x Vg. The 6 parameters on the right hand side of the equation will be the same for both Ray graphite-plate 6080 and regular-plate 6080, hence the total power consumed will be the same for both tubes ( and total heat generated = total power consumed - power consumed by the load on tube, will also be same since load will be the same)! Yes, Ray graphite-plate 6080 will reach normal thermal operating temperature slower than regular 6080 does because of its higher thermal resistance, but it will eventually get there (some folks say graphite-plate 6080 needs 10-30mins to warm up), and what's more important, the total heater generated by both 6080 tubes will be the SAME, that's defined by the equations I listed above and the law of conservation of energy.

Now, let me elaborate why I intentionally listed grid current there previously. Yes, most tubes run on negative grid voltage on their grids, hence their grid current will be close to 0, but let's not forget that some tubes do run on positive grid voltages, so there will be measurable grid current on them! And even under the negative grid voltage mode, some electrons will go astray and hit the grid by chance, hence will generate grid current theoretically (though this current is almost 0 and is negligible). So if I did not list grid current there, some guy will be nit-picky on it. And even I did include the grid current there, someone is still trying to be nit-picky on it

As I said, the power dissipated in both tubes will be the same. The thermal resistance of the tubes will be different, so their measured temperatures will be different. This is the reason you can attach a resistor dissipating 10W to a 1 oz. heat sink and the heat sink will get hotter than if the same resistor were attached to a 10 lbs. heat sink. Both heat sinks dissipate 10W (minus the dissipation by the resistor body and leads), but the 1 oz. heat sink will rise to a higher temperature as it will have higher thermal resistance. Power dissipation is the same, temperature is not.

I was simply clarifying why you and mordy were having a disagreement. Next time I won't bother.

 

[cddc]

As I said, the power dissipated in both tubes will be the same. The thermal resistance of the tubes will be different, so their measured temperatures will be different. This is the reason you can attach a resistor dissipating 10W to a 1 oz. heat sink and the heat sink will get hotter than if the same resistor were attached to a 10 lbs. heat sink. Both heat sinks dissipate 10W (minus the dissipation by the resistor body and leads), but the 1 oz. heat sink will rise to a higher temperature as it will have higher thermal resistance. Power dissipation is the same, temperature is not.

I was simply clarifying why you and mordy were having a disagreement. Next time I won't bother.

The heat sink or thermal resistance has nothing to do with the total heat generated, which will be the same for both Ray graphite-plate 6080 and regular plate 6080, the room temperature will be raised to the same level.

Mordy treated a tube's max heat rating as actual heat generated, and thinks a graphite plate Ray 6080 (rated at 300c) will be 100c hotter than a regular plate 6080 (rated at 200c), you think it's a correct point to argue for...

 

[cddc]

IMHO there's better ways to acknowledge you don't completely get the point.
Even when there's an obvious lack of understanding of what @L0rdGwyn is actually saying.
Let yourself be educated here. The forum's here to help. Not to fight you.

It would help to educate oneself a bit on the principles of heat transfer that LG is trying to convey.
Instead of going all keyboard warrior when another member is actually trying to teach you something.

LOL...Tom, do you really need to jump in here just because I said you are trying to sell Crack modding services for a fee in the Crack thread?

 

[cddc]

Oh no bad experience here, sorry if I implied that. I have genuine ones but it was this site and jacmusic.com that educated me on what was the real deal before I bought mine.

Glad to hear you had no bad experience.

 

[L0rdGwyn]

The heat sink or thermal resistance has nothing to do with the total heat generated, which will be the same for both Ray graphite-plate 6080 and regular plate 6080, the room temperature will be raised to the same level.

Mordy treated a tube's max heat rating as actual heat generated, and thinks a graphite plate Ray 6080 (rated at 300c) will be 100c hotter than a regular plate 6080 (rated at 200c), you think it's a correct point to argue for...

Mordy only stated that his tubes run at different temperatures based on internal structure and plate material, which is true, not that the temperature rise in the room or power dissipation will be unequal when using two different tubes. You called him uneducated for making an accurate observation, which can be explained via thermal resistance.

The tube with the lower thermal resistance will transfer heat to air in the room more efficiently than the one with a higher thermal resistance, the net temperature rise of the room will be the same.

As has already been stated, the max temperature specification of the tubes has nothing to do with their operating temperature, power, or heat transfer.

So he is saying tube envelopes have different operating temperatures, you are saying power dissipation and net room temperature rise will be the same. All of these things are true, which is why I clarified.

 

[cddc]

Mordy only stated that his tubes run at different temperatures based on internal structure and plate material, which is true, not that the temperature rise in the room or power dissipation will be unequal when using two different tubes. You called him uneducated for making an accurate observation, which can be explained via thermal resistance.

The tube with the lower thermal resistance will transfer heat to air in the room more efficiently than the one with a higher thermal resistance, the net temperature rise of the room will be the same.

As has already been stated, the max temperature specification of the tubes has nothing to do with their operating temperature, power, or heat transfer.

So he is saying tube envelopes have different operating temperatures, you are saying power dissipation and net room temperature rise will be the same. All of these things are true, which is why I clarified.

Mordy were trying to prove Ray graphite 6080 will be 100c hotter than regular plate 6080 by using the max heat ratings from tube data sheets.

You can verify that in Apr 24, 2022 at 1:22 AM Post #8,623

 

[L0rdGwyn]

Mordy were trying to prove Ray graphite 6080 will be 100c hotter than regular plate 6080 by using the max heat ratings from tube data sheets.

You can verify that in Apr 24, 2022 at 1:22 AM Post #8,623

Of course that isn't true, but he is correct that the tubes operate at different temperatures and was looking for an explanation as to why. The specific words were "gut feeling", that doesn't convey a declaration of fact. Data was then provided, which showed a temperature difference, which he states he would "leave for others to figure out."

In this post right here.

I don't think any of that warrants calling someone uneducated. Let me remind you that you tried to convince me that a 7802 could bias the same as a 6080 not too long ago, which is false, and could not interpret the characteristic curves which clearly showed the 7802 would operate in cutoff. Everyone can learn, myself included, you do not need to denigrate someone's else's knowledge while trying to impart your own.

 

[cddc]

I

Of course that isn't true, but he is correct that the tubes operate at different temperatures and was looking for an explanation as to why. The specific words were "gut feeling", that doesn't convey a declaration of fact. Data was then provided, which showed a temperature difference, which he states he will "leave for others to figure out."

In this post right here.

I don't think any of that warrants calling someone uneducated. Let me remind you that you tried to convince me that a 7802 could bias the same as a 6080 not too long ago, which is false, and could not interpret the characteristic curves which clearly showed the 7802 would operate in cutoff. Everyone can learn, myself included, you do not need to denigrate someone's knowledge while trying to impart your own.

Of course, I've read his post, he was just still trying to prove graphite plate 6080 will be 100c hotter than regular plate 6080, that's obviously ridiculous, there is nothing to prove an error, he mistakenly treated a tube's max heat rating as its actual heat generated and now he wanted to use a thermal gun to prove he's correct...I was like what?!

Then you stepped in and tried to say we were looking at the different things, and came up with the nit-picky thing on grid current, what do you expect me to respond...