Then it's a crapshoot. You might get a good idea about low frequency performance from weight, comparing to weight of similarly spec'd respected brand comparable model. Lundall has excellent info on theirs easily available, for example. But high frequency performance isn't brute-force, so all you'll know that way is LF.
All good fortune,
Chris
All good fortune,
Chris
For most projects, I find that a quiescent Idc point equivalent for 50% power output works universally for most customers. I figured this by my own, so it could vary by manufacturer. Sometimes I'd go for 40%.
For example, to get 10W from an 3500R primary transformer, you need 187Vrms and Idc of 76mA (54mA) Rms. 15W by Idc results into 65,4mA Rms and an idle point of 92.3mA. In a practical way, some folks like to run their 300Bs at 90mA.
Rules of thumb to serve as a starting point always help. I'll note that. Thanks.
I would have to settle on a design for a "forever" amplifier to spend that kind of money. Maybe one day. For now I'm just experimenting and learning. The little 6H8C-EL34 pentode SE amp I am building is an experiment and a learning tool. I'll be asking questions whenever I actually build it. The power supply is in another thread.
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So, for single ended, you would recommend idling at the point where peak current (idling plus signal swing) just touches the 0 VDC grid voltage line? Otherwise, there's still more output to be had. This is normally the lowest distortion area for triodes, but the most demanding for OPTs.
If I've misunderstood you, please correct me. The 50% part is unclear.
All good fortune,
Chris
If I've misunderstood you, please correct me. The 50% part is unclear.
All good fortune,
Chris
I'm not recommending anything, but some SE amplifier builders like to idle their point closer to the 0V line, so that less swing occurs into the compressing knee regions (bottom-right).
The 50% statement is just a way to explain overcurrent (Idc) headroom for such users. Because my OPT calculation spreadsheet automatically converts Idc bias point to equivalent power output, as if you were to swing the equivalent current on the OPT primary.
All the best,
Alexander.
I’m not sure if it was asked and/or annswered already, already but is this a matter of LF response vs power? I.e. at relatively low power, say 1-2w out, can you still get good low frequency response from a small output transformer with relatively low primary inductance? Or is the primary inductance a limiting factor regardless of the power an amp is producing?
No: the primary inductance dictates the FR at all levels.
In addition, the core capacity limits the VLF amplitude.
anchorman,
Let's use a Triode in a single ended stage, that makes the below example realistic, and simple to understand.
You can get good LF if the plate impedance is low, versus the inductive reactance, rp, of the tube.
If not, there are problems . . .
Consider a 4k transformer (at 1kHz), and a tube with a plate impedance, rp, of 1k Ohm.
The load line at 1kHz is 4k Ohms.
Draw a 4k Ohm load line on that tube's plate curves.
The 2nd Harmonic distortion is Moderate.
The output stage gain is the tube u x (4k/(4k + 1k)) = 0.8 x the tube's u (mu).
Now, suppose the inductance of that 4k transformer is only 8 Henrys.
XL of 8 Henrys @ 20Hz = 1k Ohm.
Now, draw a 1k Ohm load line on that tube's plate curves . . .
There will be Lots of 2nd Harmonic distortion.
The output stage gain is the tube u x 1k/(1k + 1k) = 0.5 x the tube's u (mu).
The gain at 20Hz is 20 (Log (0.5 / 0.8)) = -4.1 dB less than the gain at 1kHz.
More than 3db rolled off, and high distortion.
Did that answer your question regarding a small output transformer?
And, I did not even talk about the possibility of the saturation of the laminations.
Just the 2nd Harmonic distortion at low frequencies alone is enough of a problem by itself.
Let's use a Triode in a single ended stage, that makes the below example realistic, and simple to understand.
You can get good LF if the plate impedance is low, versus the inductive reactance, rp, of the tube.
If not, there are problems . . .
Consider a 4k transformer (at 1kHz), and a tube with a plate impedance, rp, of 1k Ohm.
The load line at 1kHz is 4k Ohms.
Draw a 4k Ohm load line on that tube's plate curves.
The 2nd Harmonic distortion is Moderate.
The output stage gain is the tube u x (4k/(4k + 1k)) = 0.8 x the tube's u (mu).
Now, suppose the inductance of that 4k transformer is only 8 Henrys.
XL of 8 Henrys @ 20Hz = 1k Ohm.
Now, draw a 1k Ohm load line on that tube's plate curves . . .
There will be Lots of 2nd Harmonic distortion.
The output stage gain is the tube u x 1k/(1k + 1k) = 0.5 x the tube's u (mu).
The gain at 20Hz is 20 (Log (0.5 / 0.8)) = -4.1 dB less than the gain at 1kHz.
More than 3db rolled off, and high distortion.
Did that answer your question regarding a small output transformer?
And, I did not even talk about the possibility of the saturation of the laminations.
Just the 2nd Harmonic distortion at low frequencies alone is enough of a problem by itself.
Yes, thanks for the detailed solving of the maths too. Makes it easier to follow and easier to learn to do it.
Oh, for my Post # 69, I forgot an important factor,
You can not draw a 'Load Line' of 8 Henries on a tube's plate curves.
That is because 8 Henries is Reactive, and it is described by an Ellipse, not a straight line.
The Ellipse complicates harmonic distortion even more than a straight load line.
You can not draw a 'Load Line' of 8 Henries on a tube's plate curves.
That is because 8 Henries is Reactive, and it is described by an Ellipse, not a straight line.
The Ellipse complicates harmonic distortion even more than a straight load line.
Consider a Pentode that has a plate impedance of 10k Ohms.
Consider a Triode that has a plate impedance of 1k Ohms.
Triode
At mid frequency, a 4k Ohm primary can be driven by 1k Ohm plate impedance.
Pentode
At mid frequency, a 4k Ohm primary, and a 10k Ohm plate impedance, not so good, so negative feedback will be required . . .
Schade, or Global, as two examples of negative feedback that take the negative feedback signal from the output transformer primary, or from the output transformer secondary, respectively.
And, if the output transformer laminations are saturated at 40Hz, 30Hz, or 20Hz, for example . . .
Negative feedback will only make the saturation worse; negative feedback can not fix the saturation.
Consider a Triode that has a plate impedance of 1k Ohms.
Triode
At mid frequency, a 4k Ohm primary can be driven by 1k Ohm plate impedance.
Pentode
At mid frequency, a 4k Ohm primary, and a 10k Ohm plate impedance, not so good, so negative feedback will be required . . .
Schade, or Global, as two examples of negative feedback that take the negative feedback signal from the output transformer primary, or from the output transformer secondary, respectively.
And, if the output transformer laminations are saturated at 40Hz, 30Hz, or 20Hz, for example . . .
Negative feedback will only make the saturation worse; negative feedback can not fix the saturation.
Glad to see this thread is still serving its purpose after many months. My life took two unexpected turns in February, halting both amplifiers that were/are in progress, one single-ended and one push-pull. I still have not ordered the OPTs for the single-ended. Maybe this discussion will inspire me to do so. I did receive two nice, big push-pull transformers from Primary Windings in the UK back in February. I hope construction on my projects will resume this fall. The two amps are sitting on my "workbench" (living room table!) partially finished, and it is driving me nuts. I have three others sitting here to use, and like so many people here, the last thing I "need" is two more, LOL. 😀
One cannot compare se output to pushpull output.they work in completely different way. Se has magnetic gap to tickle the bh curve.pushpull has each phase winding in opposite direction to cancell core magnetisation at standing Ip with a resultant stronger magnetic circuit which equals higher prim xl as no gap.its all very complex.this thread will go on for years.
This thread was focused on the EL34 single-ended pentode amplifier (yes I said pentode, with global NFB) that I am working on. It needs OPTs that I still have not purchased yet. Some push-pull discussion sneaked in just for comparison.
For purchasing SE xformers from China, it seems to be a matter of finding one of the more established sellers and then buying "by the pound" for the best bottom end.
For purchasing SE xformers from China, it seems to be a matter of finding one of the more established sellers and then buying "by the pound" for the best bottom end.
Michael C,
There is No question about it.
There are some definite characteristics of single ended outputs versus push pull outputs that you can successfully compare.
In another thread that is currently running in Tubes/Valves, I compared the output of single ended triode amplifiers, to Class A push pull triode amplifiers, with No negative feedback for either output type, SE nor PP.
The characteristic that I compared, gave the following conclusions:
Single ended has a Non-Symmetrical Damping Factor.
Push Pull has a Symmetrical Damping Factor.
By the way, for those who do not know, single ended is also Class A (unless it is an SE guitar amp driven beyond clipping to make Grunge).
There is No question about it.
There are some definite characteristics of single ended outputs versus push pull outputs that you can successfully compare.
In another thread that is currently running in Tubes/Valves, I compared the output of single ended triode amplifiers, to Class A push pull triode amplifiers, with No negative feedback for either output type, SE nor PP.
The characteristic that I compared, gave the following conclusions:
Single ended has a Non-Symmetrical Damping Factor.
Push Pull has a Symmetrical Damping Factor.
By the way, for those who do not know, single ended is also Class A (unless it is an SE guitar amp driven beyond clipping to make Grunge).
Interesting. What does the single-ended asymmetry look like on a graph? What is the shape? How does that compare to the push-pull?