I've been studying shunt feedback, and come across three stategies for tube output stages.
One of them is, of course, the Yves/Baby Huey approach. This one relies on the amplifier as a whole. It is not an isolated stage working on its own.
For the most basic topologies, we can:
1- use a resistor from the output tube anode to the driver's anode.
2- use the feedback resistor as the only load for the driver, so we get a I-V converter (like Browskie showed on one of his articles).
Both attached, in a simple amplifier I just put together in LTSpice to ilustrate this thread.
On this thread:
Partial Feedback EL84PP - Not quite as expected (yet)
Svein_B had trouble with a PA very similar to Baby Huey (after reading through all posts, it wasn't clear why).
At the end, Svein_B just added a separate load resistor for the driver, as in (1).
It is very easy to add to an existing amplifier.
Why do you think this is not used more often?
Many people say that it requires larger swing from driver stage, but simulation showed to me that same swing is required to get the same output power.
Take the amplifier I attached, as a testbench:
- Simulate the amplifier without feedback, then apply feedback, the driver swing AND output stage swing is reduced by the same factor (aprox 2.5).
- Increase the input signal amplitude, I can get the same power as before, with the same driver's output.
More drive IS (obviously) needed at the input (less sensitivity) or previous gain stage, but not the driver.
Attached, the simulation.
I can share the LTSpice file if someone is interested.
Anyone willing to share some thoughts about this?
One of them is, of course, the Yves/Baby Huey approach. This one relies on the amplifier as a whole. It is not an isolated stage working on its own.
For the most basic topologies, we can:
1- use a resistor from the output tube anode to the driver's anode.
2- use the feedback resistor as the only load for the driver, so we get a I-V converter (like Browskie showed on one of his articles).
Both attached, in a simple amplifier I just put together in LTSpice to ilustrate this thread.
On this thread:
Partial Feedback EL84PP - Not quite as expected (yet)
Svein_B had trouble with a PA very similar to Baby Huey (after reading through all posts, it wasn't clear why).
At the end, Svein_B just added a separate load resistor for the driver, as in (1).
It is very easy to add to an existing amplifier.
Why do you think this is not used more often?
Many people say that it requires larger swing from driver stage, but simulation showed to me that same swing is required to get the same output power.
Take the amplifier I attached, as a testbench:
- Simulate the amplifier without feedback, then apply feedback, the driver swing AND output stage swing is reduced by the same factor (aprox 2.5).
- Increase the input signal amplitude, I can get the same power as before, with the same driver's output.
More drive IS (obviously) needed at the input (less sensitivity) or previous gain stage, but not the driver.
Attached, the simulation.
I can share the LTSpice file if someone is interested.
Anyone willing to share some thoughts about this?
Attachments
The driver voltage swing will be the same but the effective impedance in the driver plate circuit will be lower and thus the current swing into the plate will be higher. If you are not careful you might end up driving the driver stage into cutoff and/or saturation.
Yes, that's it. But then it is a matter of using higher idle currents, and it has nothing to do with voltage swing, as many descriptions/explanations claim?
Attachments
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Correct, but you will also have higher distortion from the driver stage. However the feedback will offset this somewhat and it would be worth it if the output stage has the most distortion (which is usually the case)
Because (I believe) it's a relatively recent development ~ 10 years vs 70 years. There are many historical designs that don't use it.
People need time to adjust.
Parallel feedback by voltage (AKA shunt feedback in DIY circles) decreases input resistance, hence more current is needed to drive. If you have say 100k feedback around an output stage with gain=20, the effective input resistance would be 100k/20=5k only. Also, since the gain is non-linear (otherwise why would we want to linearize it, right?), the input resistance would be non-linear as well. Now, what happens when you load your triode driver on a non-linear 5k resistance?
No no, the effective resistance is just the correction action of the feedback !
If the gain gets less the feedback is less too and v-v.
It is the Ri of the driver that changes making a non fixed feedback.
Mona
If the gain gets less the feedback is less too and v-v.
It is the Ri of the driver that changes making a non fixed feedback.
Mona
If you open pretty old RCA tube data books (I am attaching one example, with nested feedbacks and regulated screen grid voltage), you can find lot of wisdom there, but the industry did not use them.
Why? Lack of education.
Designers of military and laboratory measurement equipment learned at work, while designers of consumer radio and audio followed simple rules of dumbs. There were few exceptions like Altec Lansing and Quad, and probably few more worth mentioning.
Solid state components caused fast development of designers, but they were not taught to design using vacuum tubes.
You can assume this vacuum tube forum as a rare laboratory similar to what RCA had in 1930'Th, where enthusiasts familiar with both solid state and tube design participate in discussions, some of us are even educated. 😉
Enjoy and contribute!
Back to school Mona; Ohm laws and math, Ohm laws and math! 😀
Without math you can use anode load lines. A first, draw a load line for an output tube, and calculate voltage gain in several points of a grid voltage. Divide the value of the feedback resistor on the gain and get a reflected load of the triode in each points.
Then by this points draw a load line of the triode stage. It would not be a straight line. It would be a curve, that makes the result even worse than just a triode loaded on a low resistance. It would be the result of non-linearities of both tubes.
That's why a local feedback around output tube only, especially when driven my a triode, is insufficient, to make an amp more linear. However, it decreases it's output resistance pretty well.
Well that's a long time ago, I am 77 by now 😱
Yes, but the non-linearities of the final tube is within the feedback loop and the one from the driver isn't.
It makes the output stage better at the expence of the driver, not because of the change of gain of the final but the heavy load.
Mona
Attachments
Are you sure about that formula?
Can we assume the input Z of the output stage to be:
voltage swing at driver's anode / (current going into C1 + R9) ?
(for the circuit attached in OP)
If so, the input impedance gets reduced by 17 when adding the feedback resistor R9.
As R9 is 1 Meg, and output stage's voltage gain (measured in the simulation) is 34:
1000k / 34 = 29.4
Which doesn't match 17 (almost doubles it).
So could one not route the FB directly to the grid and then drive it with a follower of their choice?
I have built an amplifier that is loosely based on that RCA design. The main difference was that I used the transformer's CFB winding to return to driver cathodes, instead of using plate-to-plate feedback. The nested feedback concept definitely works well. There is an interesting paper by Bruno Putzeys that a nested feedback loop is basically a second order system (kind of like TMC compensation the soiled-state guys rave about): https://linearaudio.net/sites/linearaudio.net/files/volume1bp.pdf
It is challenging to compensate and stabilize the loops but worth the effort as you will have higher loop gain in the audio band while avoiding peaking at the frequency extremes.
Absolutely! 🙂
Current change through a resistor equals voltage change on it, divided by it's resistance. 😀
For a resistor between input and output of a stage, voltage change would be equal to input plus output voltage changes. Output voltage change would be equal to input voltage change multiplied by the stage gain.
Similarly to Miller capacitance that is the same parallel feedback that increases apparent input capacitance, in this case it equally decreases input resistance.
I mean Input resistance, guys, not an internal resistance of the stage!
It is non-linear since the gain is non-linear!
And this non-linear resistance loads the triode driver stage, so the resulting non-linearities of both stages combine!
For triodes contrary we want as high as possible load resistance. That's why I use pentodes in amps with nested feedbacks, when drive output stages with parallel feedback by voltage.
Since the output is in oposite fase the voltage over the feedback resistors is input voltage+output voltage=(gain+1)x input voltage.The resistor looks (gain+1) smaller.
But the non-linearitie of the output stage is (partly) corrected by the feedback.
Or a cascode, or a cathode follower with a series resistance (best).
Mona
I did that here: https://www.diyaudio.com/forums/tub...dback-output-via-channel-fet.html#post5024674
Using a p-channel fet follower can make the parts count really low, but you could also do it with an n-channel device or a tube. I think you'd need a coupling cap somewhere to make it all work, though.
😱 Of course! I confused myself with the simulation. Let me please correct it.
The input Z (using feedback) for the stage, measures 28 kohm (which is 470/28 = 16.8, indeed, but this value is not relevant).
1000k/(33.5+1)=29k
Which is close to simulated value.
I used the factor gain+1 as Ketje suggested.
Very nice. I find, though, the need for such a high voltage a serious drawback. From 450 to -390 is 840 V difference.
Sure, if you were to grab the 450V with one hand and the -390V with the other, you would have a real bad day. However, the voltage with respect to ground is at most 500V in the amp. It doesn't require any special insulators or precautions beyond what is normal in tube amps. Keep large negative voltages away from large positive ones and hands away from both, etc.
The real danger in an amp is the output tube anodes anyway since they can swing 2X B+ with respect to ground. That's where you need to be really careful, and that's a problem in pretty much all tube output stages of this power class.
The mosfet source needs to swing 150V in each direction to drive the output stage to clipping, so you would need -(170+150+20)V = -340V minimum to keep at least 20V across the fet during drive conditions up to full power. As you can see, I targeted some more minimum voltage across the fet to keep Crss in the flat zone.
If less feedback is used (I used 30%), a lower negative supply can be used.
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