EL34 simple push pull class B : why 100W ?

[bozole]

Hi anybody

First of all, I'm a new user on this forum, and my english spoken is not very goods, I apologize ...

I have a technical question I can't understand with my actual knowledge, so if you could explain me, I'd really apreciate !!

When you look at the EL34 datasheet, you read that you can take out 100W of power with a simple push pull stage in class B, using a 11K primary plate to plate transformer. B+ is 775V in the example

Here is something I really don't understand :
Since we have a 11K transformer at the primary (impedance reflected), we can calculate the max voltage at the secondary, please correct me if I make some mistakes :

Suppose we use a 11K to 4 ohms transformers : we can calculate the voltage ratio which in this case is 52:1

With a B+ of 775V, we couldn't have more than 775V at each anode at the max ? Taking that, and the OT characteristics, we couldn't have more than 14,9Vmax at the secondary with a 4 ohms load ? which gives us a power output of 55W ?
We can make the calcul with other secondaries, we'll find the same result

I really don't understand how we can have more power with this output transformer voltage ratio, and when I trace loadlines, effectively I can't go other than a 11K loadline (2,75K, which is 11K/4 in reality since we're in class B) otherwise the max dissipation is not respected (50W in class B because the tube conducts only on 180° ?)

So I don't understand "where the other 45W come from ?"

If you have some explanation, please let me know. I don't know how to use spice programs, maybe if I'd Knew, I'd see where is my mistake ?

Best regards

[Yvesm]

Hi you !

Say that each anode swings (roughly) from 50 to 1500 volts around the 750V idle point.

This means that there is some 3000 volts peak to peak accross the whole primary, thus power peak to peak is:
3000 squared divided by 11000 = 818 W, and rms power is 818 / 8 = 102 Watts... roughly !

In class B each tube (ideally) conducts half the time, so it can dissipate twice the rated power during one alternance and nothing during the other one.

Is my English fluent enough

I suggest not to try that with current production tubes and to use hi quality sockets

Yves.

[bozole]

Hi

Many thanks for your answer
But I still don't understand : if each tube conducts for half the time, it can't swing between 50 and 1500V ? just between 50 and 750V ?
From what I understand, when the anode reaches 750V (and it's the bias point, stage biased at cutt off), the tube comes at cut off, so it can't go higher ? When the tube conducts for half the time, the anode voltage first decreases to 50V and secondly increases back to 750V ?
I'm sorry, but that's definitivly not clear in my head !!

[hemgjord]

Think of the opt as an autotransformer. When one tube cuts off the other goes from 750v to 50v it will induce a 700v increase in potential on the other half of the winding, this is true even if no current is flowing through that half.

/Olof

[Yvesm]

Originally posted by hemgjord
Think of the opt as an autotransformer. When one tube cuts off the other goes from 750v to 50v it will induce a 700v increase in potential on the other half of the winding, this is true even if no current is flowing through that half.

/Olof

Yeap !

Like that:
(apologies to not French speaking guys, I feel it's understandable anyway)

Yves.

[Bandersnatch]

hey-Hey!!!,
With the 750-50V swing each tube can execute while the other is cut off it is working into 1/4 of the 11k a-a load presented in your example. the other side, with no current flowing in it does nothing...save for loading the tube with its parasitics( capacitance for the most part ).
cheers,
Douglas

[bozole]

Hi

Thanks to both of you I begin to understand better, even if the fact that the voltage rises on the other tube without current when 1 tube conducts is still confuse ...

And if we take a class AB stage, how can we estimate the max voltage swing ? and so the output power ? A lot of questions, sorry !!

Best regards

PS : Thanks Yvesm for the picture

[ray_moth]

Class AB is rather vague, because its behaviour depends on how deep into CLass AB it is biased. At some point with increasingly high signal it will start to curt off. Indeed, even the so-called Class B in the Mullard datasheets is actually very deep Class AB. Quiescent current is slightly more than zero with no signal.

There is arguably no such thing as Class B in practice, because you'll never get precisely zero quiescent current. You're just as likely to end up in Class C if you try to do it.

[Bandersnatch]

hey-Hey!!!,
If I had to do a 100W EL34 amp, I'd go for two pairs in parallel PP, and choose a Citation II output TX to deliver a 3200 OHm load a-a.
cheers,
Douglas

[Yvesm]

Originally posted by bozole
Hi

Thanks to both of you I begin to understand better, even if the fact that the voltage rises on the other tube without current when 1 tube conducts is still confuse ...

It's the normal behaviour of any transformer !
The voltage accros any winding is alway proportional to the number of turns of this winding, so each half primary (where turns number are equal) must develop the same voltage accross them ... what could be the source of that voltage, including the other half or itself ! !

If something try to oppose to that (eg a very heavy load), the current in the "active" winding increases to force the current and the voltage in the "slave" winding to remain equal.
The transformer may suicide itself to acheive that !

Ask Google for "Lenz law"

You're just as likely to end up in Class C if you try to do it.

And have considerable distortion when both tubes are simultaneously "cut off" that no feed back can reduce because the gain is then zero !
Nothing to feed ... back

Yves.