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Tube plate dissipation vs power output

Hello all, I am very new to all of this. My question is one that was an old thread. I did not want to reopen the old thread. So here goes; it seems almost all of the time, the plate dissipation is greater than power output with the discrepancy as heat loss, and that not always by the same percentage. Very well, the question remaining concerns tubes that have outputs greater than plate dissipation, how can that be? I was going to mention the 6BQ5 but that is not a good example. It has a plate dissipation of 12w but in P-P, the power output is 17w. That figure is greater that half of the total 24 watts but not all tubes can accomplish such a percentage. The 6550 has a plate dissipation of 42w and when in P-P, it can achieve 100w. For me, all of this is tube datasheet material, none from bench work. I'm terribly confused about all of this. Here is a good question: Why don't all pentodes show power output on the datasheets? Thank you in advance.

BillyBob
 
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Some datasheets of old included specific numbers for specific use cases (SE, PP, different B+, biasing, OPT impedance etc). But there are many pentodes/beam tetrodes that were not designed for audio, therefore there was no need for that.
Almost all datasheets include anode curves, with that in hand you can easily calculate an estimate of the expected output power.
 
It depends on the type of operation. In a class A amp, where the output tube(s) never stop conduction during a signal cycle, maximum output power is half of the plate dissipation (plate voltage times plate current) in theory, practically it's less than that. Actual dissipation calculates to plate voltage times plate current minus output power, hence is maximum with no signal In class AB and B amps, when tubes don't always conduct during a cycle, mean dissipation counts due to the plate's thermal inertia, while the instantaneous dissipation may be well beyond the ratings.

I see your question about the missing output power statements in datasheets has already been answered. I asked just the same question as a 12 yr old kid in 1971, when I had my first tube data book in my hands (still own it) and was brave and unexperienced enough to send a postcard to the publisher ;).

Best regards!
 
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Amplifier efficiency is universal for transistors or tubes ... of course in PP the output power is greater than one tube dissipation , when is not , the tubes are not working with the optimal load , for example in an OTL amplifier
 
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Amplifier efficiency is universal for transistors or tubes ... of course in PP the output power is greater than one tube dissipation , when is not , the tubes are not working with the optimal load , for example in an OTL amplifier
this!

Also a transformer coupled Class-A amplifier, has a higher efficiency compared to a non-transformer coupled amplifier.

So for Class-A tube amplifiers (which are basically always transformer coupled), the max theoretic efficiency is 50%.
So half of the plate power dissipation (minus the losses obviously).

For a tube that has a plate dissipation of say 12W, the absolute theoretical max is 6W of output power.

For a Class-B amplifier (push-pull) the max theoretical efficiency is 78.5%
So for two tubes with 6W plate dissipation, the absolute theoretical max will be 2 x 6 x 0.785 = 9.42W

As a bonus, two tubes parallel used in a Class-A amplifier:
2 x 6 x 0.5 = 6W

You can't get more out of them without breaking the laws of physics. ;)

https://www.tutorialspoint.com/amplifiers/transformer_coupled_class_a_power_amplifier.htm
https://www.tutorialspoint.com/amplifiers/class_b_power_amplifier.htm
 
Efficiency in percentage is from the power supply consumed power , so from 100W consumed , class B could output 78,5W with only 21,5W dissipation in both output transistors . I doubt this could be achieved with tubes , or very rarely .
A real number is about 50% in class AB when output is 50W with 25W + 25W in power tubes , anyway output is double than one tube dissipation

For single ended the real efficiency is much lower than 50% , but we use 50% max power from tube dissipation because it is easier to understand
 
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nigelwright7557,

You said:
"If its cathode biased then some power will be lost in cathode resistor."
That is true, but the power loss is Outside of the tube.
It is a voltage loss that effectively 'robs voltage from the B+ voltage to ground, which reduces the plate to cathode voltage, True.

Plate dissipation is power lost Inside the tube.
Plate dissipation is (Plate voltage - Cathode voltage) x plate current.

A Plate that is at 310V, cathode at 10V, has 300V plate to cathode. If the Plate current is 30mA . . .
Then the plate dissipation is 300V x 0.03A = 9.0 Watts.
Never mind the 10V x 0.03A = 0.3 Watts dissipated in the self bias resistor . . . that is separate from the plate dissipation.

Similarly, if the output transformer has an insertion loss of 0.5dB, then . . .
11.2 Watts from the output tube to the output transformer primary, will only give 10.0 Watts from the output transformer secondary.

My opinions and experience.
 
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For a Class-B amplifier (push-pull) the max theoretical efficiency is 78.5%
So for two tubes with 6W plate dissipation, the absolute theoretical max will be 2 x 6 x 0.785 = 9.42W
Not correct. In your example 12 W total dissipation relates to 100 - 78.5 = 21.5 %. Hence, the output power calculates to 12 W x 78.5/21.5 = 43,8 W.

Best regards!
 
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Efficiency is not output power divided by total plate dissipation.
Rather it is output power divided by total power consumed.

E.g.: Two EL34s can produce an output of 100W (see datasheet).
Assuming that total plate dissipation at full output is 2x25W, the total power consumed would be 150W.
So efficiency results as 100/150 = 0.67.
 
Not correct. In your example 12 W total dissipation relates to 100 - 78.5 = 21.5 %. Hence, the output power calculates to 12 W x 78.5/21.5 = 43,8 W.

Best regards!
ehm sorry, but you just simply take 78.5% from the maximum possible power that can be theoretically delivered without those losses.

So if a system can deliver a total power of 100W (no losses) and the total system efficiency is 78.5%, that will result in a maximum of 78.5W. The rest 21.5W is being put into waste (heat).
 
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Efficiency is not output power divided by total plate dissipation.
Well you wanna get an estimate how much power can be pulled from a certain system (tubes in this case) with a fixed max power dissipation.

In a system with a 100% efficiency factor, this means that Pout = Pin.
Since there are losses, we get Pout = Pin * Loss factor (or system efficiency)
Because: system efficiency (loss factor) = Pout / Pin

See links provided above (but I guess nobody even takes to time to read those anyway .............. :( )
 
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