PLate voltage and power tube wattage.

[leadfootdriver]

So I have this Fryette Deliverance amp. It has 4-KT88's that are supposed to put out about 120 watts.

I hooked up my bias probe, and find about 420 at the power tube plates. That said, looking at the data sheet for a JJ KT88, the max wattage for this tube is 42 watts. THe amp is biased for 70% PD using the 42 watts in the math: 42/420=.10 *.7= .07 or 70 miliamps.

SO, how does Mr. Fryette figure the amp is putting out 120 watts from 4-KT88's when He biases the amp as if each tube is putting out 42 watts?

[g1]

Not sure where you got that math from. You can not determine bias based on the plate voltage. You must know the plate or cathode current which can only be determined by measurement.

[Justin Thomas]

I think the current is about right, but how you got it is fuzzy... and this isn't exactly something you should guess about.

The last dentence is the kicker - the tubes are NOT biased to dissipate 42W. If they're biased at 70% max dissipation, then that is 29.5W/tube, divided by 420V = 70mA... so start with 42W(max)*70%(supposed "ideal" safe dissipation) / Vp = current.

But measure anyway - only way to be sure, and still the actual safest way.

Also, plate dissipation ≠ output power, necessarily.

Justin

[Enzo]

yes^^^

Plate dissipation is a DC measurement at idle.

Audio output power is not related.

Think if your car, you set the idle speed of the motor at 750rpm, and that has nothing to do with how fast the car will go.

[Alan0354]

1) It's not accurate to calculate the power output using the primary condition. You really have to look at the power at the output.

2) What you calculate is idle current using the 70% rule. In another word, if you want to bias so the idle power dissipation of the tube is 70% of the maximum power capability, then you set the idle current at 70mA. That's all that means. This has nothing to do with output power.

[Dave H]

SO, how does Mr. Fryette figure the amp is putting out 120 watts from 4-KT88's when He biases the amp as if each tube is putting out 42 watts?

He probably measured it.

I think you are confusing the idle power dissipation of the tubes (heat) with the amp's audio power output. They are not directly related. It would probably still put out 120W of audio with half that bias current.

Edit, Simultaneous posting! You guys are too quick for me.

[leadfootdriver]

So then power output is proportionate to the plate voltage then? I'd get more power if my plate voltage was around 500? That data sheet gives a max voltage of 800.

[loudthud]

THe amp is biased for 70% PD using the 42 watts in the math: 42/420=.10 *.7= .07 or 70 miliamps.

Watts divided by Volts equals Amps. So 42W/420V=0.1A
Taking 70% of the current (and thus the power) gives 70mA.

SO, how does Mr. Fryette figure the amp is putting out 120 watts from 4-KT88's when He biases the amp as if each tube is putting out 42 watts?

Plate dissipation is not the same thing as output power although both are measured in Watts.

[g1]

The output power will be limited by the plate voltage. The plate voltage will be limited by the power transformer.
When you had the bias probe connected, did it say the tubes were idling at 70mA ?

[eschertron]

A quick visit to Interactive Valve Data Sheets leads me to believe I can get a solid 60W from a pair of KT88s (so 120 from a quad) while dissipating a mere 35W per tube. Note, as mentioned above, that this is 35W per tube in addition to the 120W delivered to the OT.

I made that mistake when spec'ing the PT for my 18W build and came up a few VA short when the power section needed to burn 11W per tube in order to make 18W at the speaker.

EDIT: simulpost? I must have been sleeping for the last 20 minutes!

[leadfootdriver]

Thanks for all the replies. Yes, the amp is idling at 70 ma's per tube.

[Chuck H]

One thing not addressed... At 420Vp I really don't think you could get 120 watts from a quad of kt88's. The gist of the thread is correct, idle current and max output are different things. But if the max watts are 42 and the max Vp is 800 I have to wonder, even with increased current, how anyone could get 40W per tube in this scenario with only 420 plate volts. But this is common in amplifier nomenclature. If it's commonly known that two kt88's put out 60 watts then a quad must out 120 watts. Never mind about the actual output of any given circuit. Many amps have an actual number in their moniker that corresponds to the chosen power tubes maximum output though the circuit is not designed to produce it.

[Gregg]

At 420Vp I really don't think you could get 120 watts from a quad of kt88's.

It's not even 420V because there will be a sag at full power so forget about those 120W.

[g1]

Once you get into square wave territory, the output power will increase very rapidly in a non-linear fashion .
They mention RMS in the spec., but nothing about the distortion level.

[Chuck H]

Yup. I don't know how much a square wave skews accurate reading, but I have an amp here that uses a pair of el84's at 360Vp. Cathode biased near class A but with enough drive voltage to push the tubes into push pull at full clip. I test only 14 watts clean but over 20 watts at full tilt.