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On my TC-15 the plate voltage on the el84's are measuring 366 and 362. The idle current is 32.8 and 31.5. This is on some old Mullards which measure 6250Gm. Is this too hot?

You need ti calculate what each tube is dissipating, allowing for the screen current.
Subtract the screen current from the cathode current to get the true anode current.
Then multiply the anode current by the difference between the anode voltage and cathode voltage to get the true dissipation.
This should be no higher than 12W, or tube life time will be impacted. Bear in mind that Matchless amps have a reputation for wearing out tubes, so don't be too surprised if the TC-15 also has its power tubes dissipating fairly high.

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You can increase the cathode resistor is you want. They do run pretty hot as Zaphod mentioned. I dont think I'd use those Mullards in there!

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Stephen
Web: www.trinityamps.com. Facebook: facebook.com/trinityamps. Twitter: @trinityamps

Phil,
Im sorry, but you totally lost me. Can you direct me to an explanation on how to do this?

BTW, I got the readings I posted by using a BBQ Bias-Rite.

I have no idea how the Bias-Rite works, or how accurate it is. I always prefer to calculate dissipation the real way.

Measure the voltage across your cathode resistor. Divide that by the resistance value, and then divide it again by two for the cathode current of a single power tube. Measure the voltage across either one of the screen grid resistors (100 ohms in a TC-15) and divide the voltage by the resistance value to get your screen current. Subtract the screen current from the cathode current, to get the true anode current. Multiply that number by the *difference* between anode and cathode voltage. That is each power tube's dissipation calculated accurately.

HTH

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Outstanding explanation! This should be in the Resources forum!

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Ok, got it. Here goes,

Across cathode resistor, 7.75/120=0.065/2=0.03

Cathode current=0.03

VDC across screen grid resistors, .29/100=.0029

Screen current=.0029

Anode voltage 365-Cathode voltage 7.75=357.25

357.25x0.027(anode current)=9.64

Is this right? Only 9.64 watts?

If I can go to 12 watts with an El84 should I reduce the value of the cathode resistor?

Nice, and with cathode bias you bias to just about 100% plate dissipation, yeah?

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It depends on which approach you prefer to take with cathode biasing. One approach is to bias hot (100%) and then use a small cathode cap, typically around 47uF or 25uF. This low cap value lets the cathode-bias voltage swing freely, so that the amp biases itself progressively colder as it works harder. So at full throttle peaks, it's probably running at close to 50% dissipation.

The other approach is to bias cooler, say to around 80% or 85% of max dissipation, and then use a huge cathode cap, in the 1000uF to 2200uF region. This helps extend the life of tubes, which may be useful if you're running NOS. It also makes for a bit more of a stiffer feel, taking the amp a bit more like a fixed-biased amp.



So your actual cathode current is 30mA, which is slightly less than the Bias-Rite showed. Interesting....
And your anode (aka "plate") current is 27mA. Multiplied by the difference between anode and cathode voltage, that gives 9.68W by my calculations. Then you can chose whether you want to bias warmer or not, based on the two approaches I explained above.

However, I am really puzzled at how you only got 7.75V across the cathode resistor, with around 360V on the EL84s. I would normally have expected to see more around 12V. How much does it change with a different set of tubes?

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It doesnt change much at all. With NOS Mullards it measures 7.75 with 358 on the plates. With the New Sensor Mullards its 7.42 with 359 on the plates.

I have 120.5 AC at the wall outlet.

What do you think?

And what about Trinity who use a 220mF bypass cap. That would be around the 90% region no?

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I find it a little surprising, but if that's the way it is, I guess that's the way it is. Just check that the EL84 grids are actually sitting at zero volts. If so, then there's nothing to really worry about.


As a quick approximation that sounds about right.

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As for the grids. Im not sure which one you want me to measure. Between pin 2 and ground there is .005 vdc. Pin 9 to ground there is 209 vdc. Between pins 2 and 9 there is 182 vdc. Is this ok?

Also, is there any reason I shouldnt use a trimpot and temporarily put in place of the cathode resistor (beginning at the value of the resistor), and then tweak it a little bit as I play through the amp. When I find the sound I like check in a dark room to make sure the plates arent red. If plates arent red, remove pot, measure value, and replace with resistor of that value?

Pin 9 is the screen grid. So that is ok, it should have a HV. How does 209 - 0.005 = 182 though?

A trimpot will burn up very quick in the cathode resistor position. Most trim pots are 1/4W or 1/2W at best. Note that power rating is for the max resistance, so from outer lug to outer lug. As you decrease the resistance the power rating goes down. (not a linear relation despite the taper being linear).
Anyways you said you had something like 7.5V across a 120R resistor. That's close to 1/2W right there, so yeah a trimpot will burn up.
Maybe try an old Allen Bradley pot or a PEC, both are 2W.

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Sorry Phil, I wasnt clear. When I said "between pins 2 and 9" what I mean is with one lead on pin 2 and the other on pin 9 I get 182vdc on my meter.

Is that ok?

When referring to "grid" or "grids", the generally understood default is the control grid - ie pin 2 on an EL84. That should be very close to zero volts. Otherwise, you would say "screen grids" or "screens' for short.

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