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
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
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.
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.
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 !!
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 !!
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
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
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
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
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.
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.
bozole said:
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"
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.
Hi
I think I must learn and understand some theory before I'll understand all of your posts guys !!
Or build an output stage, and put an oscilloscope with probes on each anode and observe the waveform.
I didn't wanted at the base to build a simple push pull class B amplifier, I just wanted to understand why we can take so much power with a pair of EL34. And if we assume no more than 50W dissipation on each tube on a half period (resulting 25W dissipation on a period), 100W would be the result of a 100% efficiency ? I must be wrong one more time ...
And what max power in class B with a 6L6GC push pull could we hope ? The max anode voltage in this case is just 500V in theory ...
I think I must learn and understand some theory before I'll understand all of your posts guys !!
Or build an output stage, and put an oscilloscope with probes on each anode and observe the waveform.
I didn't wanted at the base to build a simple push pull class B amplifier, I just wanted to understand why we can take so much power with a pair of EL34. And if we assume no more than 50W dissipation on each tube on a half period (resulting 25W dissipation on a period), 100W would be the result of a 100% efficiency ? I must be wrong one more time ...
And what max power in class B with a 6L6GC push pull could we hope ? The max anode voltage in this case is just 500V in theory ...
Bozole, if this confuses you, you are not alone! Efficiency is determined as the proportion of power consumption that appears as output. You will never get 100% efficiency because that would mean that no power is dissipated in the tubes, it would all go to the load (the speaker).
Very approximately, Class A offers about 25% efficiency and typical Class AB about 40-50%. The highest efficiency you're likely to get from Class B is about 68%. That means that for 100w output the total power input will be 100w x (100 / 68) = 147w. This means that the tubes themselves will dissipate 47w between them, or 23.5w per tube.
Mullard's figures for Class B operation are rather extreme. They use the EL34 almost to the limit of its plate and screen voltages and take no account of the crossover distortion likely to be incurred. The capabilities in Class B were published to show what the tubes could do if power output was the primary concern and quality of sound was of lesser importance, such as in a public address (PA) amplifier used at full output. It would be unsuitable for serious music listening, especially at low volumes when the crossover distortion, which is not signal level dependent, would dominate.
Mullard's own EL34s (and those of its main rivals) might have been able to perfom saafely under these conditions, but modern "clones" would probably not survive the ordeal!
As explained above, the anodes being the two ends of two identical windings on the transformer, you cannot observe something else that two identical signals.bozole said:
Just out of phase in reason of the direction of the windings.
This is what the transformer is expected to do !
And it does its job easily since, as Douglas pointed, the "inactive" half primary is unloaded.
Exemples in datasheets claim for 60W at "only" 350V anode but ... this was for the "original" (non GC) 6L6 wich was limited to 20W anode dissipation.
Moreover, the grids must be driven positive : AB2 class and this calls for a specifically designed driver.
Yves.
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