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EL34 Push Pull in Class A

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sorry, incomplete detail...

not in triode mode, but pentode.

and the real intention for this is i want to use a current regulator to bias the EL34 (ditch the power resistor and have more or less equal current between tubes). will use the LM317 or DN2540 for this.

maybe if i use 50mA constant current on the cathode, then i can force the tube into class A. With 50mA each EL34 at 400Vp (just a number I picked :D) where am I in the operating region?
 
The El34 has 25 watt dissipation. So, at 400v this would be 62.5 ma per tube. At 350v, it would be around 70ma per tube. Check the tube specs to see what the max current the EL34 cathodes can handle.
At 350v at 70ma you would probably be working in pure class "A" . Probably at the 400v plate voltage also. Again, check your tube specs / curves to make sure.


Below are typical values for a single ended class "A" amplifier. You could make this push-pull using two tubes. Plate to plate load would be 3.5-4k.

Class A Amplifier
Plate Voltage ................................. 250v
Grid No. 2 Voltage ............................ 250 V
Grid No. 1 Voltage ............................ -13.5 V
Plate Resistance (approx) ..................... 15000
Transconductance .............................. 11000 µ
Plate Current ................................. 100 mA
Grid No. 2 Current ............................ 15 mA
Load Resistance ............................... 2K Ù
Power Output (approx) ......................... 11 W


Hope this helps... Daniel
 
I recently built an EL34 PP in ultra linear mode. The driver was 6J5GT with CCS, cap couple to Electra-Print phase splitter autoformer to a pair of EL34s. Each EL34 was running at 370V plate, -26V cathode bias, 65mA. The OPT was Hammond 1650R. The power is 27W/channel. It sounds great with my Fostex 166E. Giving clear and clean sound with warm mid and full bottom end bass.
 
You can do an estimate for max class A power on the primary side by using P=I^2 x R.
Max current swing = 2 x Iq. Since 2 x Iq is peak current you have to divide it by sqrt(2) x sqrt(2) = 2.
P[Amax] = 1/2(2xIq)^2 x Ra-a/4

If Iq=70mA and Ra-a=4k, max class A power will be 9.8W.
There will be some loss in the transformer as well.
 
jane said:
You can do an estimate for max class A power on the primary side by using P=I^2 x R.
Max current swing = 2 x Iq. Since 2 x Iq is peak current you have to divide it by sqrt(2) x sqrt(2) = 2.
P[Amax] = 1/2(2xIq)^2 x Ra-a/4

If Iq=70mA and Ra-a=4k, max class A power will be 9.8W.
There will be some loss in the transformer as well.

Sure,

This may be somewhat improved using higher B+ AND higher Ra-a.
And, of course, rebias !

Anyway, the advantages of this "Classe A" are not obvious to me.
Just the fact that "A" is the first letter of our occidental alphabet does not implies that it is the "First Class" :D

Yves.
 
What exactly do you want to achieve... Is it that you want Push-Pull Class A with pentode connected EL34's???
WOuld you like to run 2 valves Push-Pull or 4 valves???
DO you want the Maximum output power possible???
I would not always trust the data sheets...they were not always done properly or acurately....
Use the Plate curves for EL34 to figure this....
I will do the numbers and get back to you..

Chris
 
Hi guys,

If you are running in Class AB right now...what are the voltages..

Right now I'm running textbook UL 420V with shared cathode of 470R. Power supply B+ is not an issue, as I can change it to suit what ever is needed.

What exactly do you want to achieve... Is it that you want Push-Pull Class A with pentode connected EL34's???
WOuld you like to run 2 valves Push-Pull or 4 valves???
DO you want the Maximum output power possible???

I want to refrain from using a cathode resistor (+cap bypass) and switch to fixed current bias (autobias in TubeCAD), and that's precisely why I have to be in Class A.

Yes I want Pentode connection, or UL but not Triode, 2 valves, I'm not after maximum power, otherwise I will be happy with where I am right now.

You can do an estimate for max class A power on the primary side by using P=I^2 x R.
Max current swing = 2 x Iq. Since 2 x Iq is peak current you have to divide it by sqrt(2) x sqrt(2) = 2.
P[Amax] = 1/2(2xIq)^2 x Ra-a/4

Jan, thanks for the formula. Sure come in handy.

Anyway, the advantages of this "Classe A" are not obvious to me.
Just the fact that "A" is the first letter of our occidental alphabet does not implies that it is the "First Class"

Yves, sonically I don't know. That what I'm trying to find out. And then again, I just want to use autobias on the cathode.
 
Arnoldc,

Since you ask, theoretically class A is better, because at low output (where most of our average listening lies) there is no cross-over distortion. You will know that this class distinction comes from the first use of tubes way back in the 20s or so. It started with class A and class B - just about totally cut-off; interstage transformers and such. Class AB came a little later.

But one must look at the whole picture. Tubes, technology and components have developed, and it is possible to design class AB cross-over distortion just about out of the picture. The advantages of higher available power output in class AB for a certain (maximum) tube dissipation, to my mind significantly outweighs class A.

To illustrate, if you get 25W from a pair of EL34s in class A but 70W is quite possible in class AB, just consider which will give the most enjoyment. At say 23W the class AB is going to be better than the class A anyway because the latter is approaching overload, even though the distortion at maximum output for class AB might be higher than at (a lower) maximum output for class A (the difference will not be much).

The design of the output transformer is slightly different because for some time during a cycle only one half of the primary will be driving. One therefore has to consider equivalent impedance and inter-winding leakage, but most good transformers do the necessary sectionalising to take care of this.

You do not have to be in class A to consider fixed bias (you are not confusing this with class C?). No grid current flows in either class A or AB mode - in a manner of speaking the control grid does not know which class you are using.

Regards,
 
The advantages of higher available power output in class AB for a certain (maximum) tube dissipation, to my mind significantly outweighs class A.
hi johan, i agree with you. though with my efficient DIY speakers (they can rock with as little as 45 power), 30W or over would be over kill.

i can not recall any class A push-pull design that ever came close to the popularity that the Dynaco ST70
tony, how's russia? i heard news that there is a mob take over on one of the tube plants there...

ST70 as popular as it is, never get my attention :D

You would run 250V plate 250V screen..Bias at -13V and a 2.4K plate load you would have about 18W to 20W output...

Thanks Chris, just enough power I need.
 
arnoldc, I understand exactly what you want to do. Have though of the same thing myself many times :) Presuming you mean tie the two cathodes together and then down to earth via a single unbypassed CCS. This setup will *only* work properly in class A. If the conduction angle of either valve is anything less than 360 deg the total cathode current will change abruptly at that point and the CCS will jump in and try and "fix" things...

Definitely a good idea to try.
 
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