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    Building, troubleshooting and testing of these amplifiers should only be
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Need to know OT loads for S.E. EL34 and KT77

KT77 is a pentode. Treat it as an EL34.
The load depends upon the HT.

Looking at the Genalex KT77 data sheet and for PP UL, it shows Va=500V with Rload =5.5k, then Va=600V with Rload=9k.


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2016-06-07 6:50 am

You are correct, a KT77 is a Beam Power tube with Beam Formers;
an EL34 is a true Pentode tube with a real Suppressor Grid.

SE Single Ended:

SE can have Triodes, Pentodes, and Beam Power output tubes.
The plate is connected to the primary and the other end of the primary connected to B+. The transformer needs to be Air gapped.

The pentode and beam power tube can be in pentode mode and beam power mode.
Or they can be wired in triode mode (screen connected to the plate, often with a 100 Ohm resistor instead of a wire).
Which mode will you use for your KT77 or EL34?

Pentode mode and Beam Power mode generally requires negative feedback.
Triode connected mode often may be used without negative feedback, gives lower power out.

Can you post a schematic of what you will build?

SE can also operate in UL, primary winding connected with the plate at one end, B+ at the other end, and a UL tap somewhere in-between connected to the screen. This needs to be an air gapped transformer.


2019-02-08 12:16 am

If you google "Universal loadline calculator for vacuum tubes" you can play simulator. But with given data its look like you are always better thd with growth of OT load. I was always wonder do that really make sense to use 8-9k with kt77 or 7k with el34? Based on this simulator thd it has. But its only thd in real life could be other than that. I understand higher load sacrifices some power output, but what if its enough, should I go so far? (with OT load)​

As a rule of thumb the optimum load for maximum power output of true power pentodes is given by the voltage between cathode and plate divided by the plate current (under no signal condition).

This diagram for the EL41 shows load impedance versus output power for five levels of second harmonic distortion at Va = Vg2 = 250 V (curve "e" represents the power output at the start of control grid current). From it one can deduce that the second harmonic distortion at a certain output power level increases when the load impedance deviates (in both directions) from the optimum load (7K).


The following diagrams for the EL5 at Va = 250 V and Vg2 = 275 V show that at the cost of some output power, a somewhat lower load impedance (2K5) than the optimum load impedance (3K5) makes that the total harmonic distortion is made up of a bit less third harmonic distortion and some more second harmonic distortion. If you compare the two for 6.5 Watt output power you get:
  • 3K5: d2 = 4 % and d3 = 5 %
  • 2K5: d2 = 9 % and d3 = 3 %

EL5 (Ra = 3K5).jpg

EL5 (Ra = 2K5).jpg


Paid Member
2003-06-12 7:04 pm
Maine USA
I divided ra by five and I get:
Ra is almost meaningless in non-triodes.
The load depends upon the HT.
Right, +and+ the current. Simple Ohms Law is your best guide. With soft-shoulder power tubes, 10%-20% lower will reduce 3rd harmonic at maximum output, which looks good on a datasheet. Higher may reduce THD, but we don't care a lot about the even harmonics. For a specified power output, a BIGGER HOTTER tube always works better. But humans always turn it up until it comes back bent, so it may be moot.