Wintersweet highvividity driver circuit

[HighVividity]

Linear General Driver circuit for multiple power Triodes, derived from the Wintersweet Highvividity cathode follower power amplifier.



Please criticize. I will modify it to better fit your needs.

[Sch3mat1c]

Why not take op-amp feedback from the cathode output? Or for that matter, from the output tube plate, or better still, the output itself?

I think you forgot the cathode follower's load resistor. :P

For that matter, negative supply to the first triode's cathode would give useful output range; I suppose the output tube is actually standing on like 100V or so (this is not indicated).

The 10k "grid leak" on the NE5532 is a bit small relative to the "10-100k" input pot -- 100k or even 1M would be better. Alternately, you can nix the coupling capacitors altogether and apply a small negative bias to the feedback node, same thing.

Tim

[HighVividity]

Question answered:

"Why not take op-amp feedback from the cathode output? Or for that matter, from the output tube plate, or better still, the output itself?"

This is because I do not want to include the coupling capacitor in the feedback loop. The feedback loop operates down to DC, like any operational amplifier. Maybe it is fine to do what you suggested, but I have not done any experiment with that.

"I think you forgot the cathode follower's load resistor. :P
For that matter, negative supply to the first triode's cathode would give useful output range; I suppose the output tube is actually standing on like 100V or so (this is not indicated)."

I am aware of that :P. It is deliberate. I heard that this arrangement "sounds better" and from theoretical point of view, maybe it is fine too. Think of the "infinite equivalent leak resistor". I need to think it over in depth, but it is the amplifier builder's problem (It is part of the output stage and not part of the driver) and not mine.

"The 10k "grid leak" on the NE5532 is a bit small relative to the "10-100k" input pot -- 100k or even 1M would be better. "

the "leak resistor" was used to make the linear input pot more "exponential like". If you use an exponential input pot, then yes, a higher "leak resistor" will have less influence on the pot.

"Alternately, you can nix the coupling capacitors altogether and apply a small negative bias to the feedback node, same thing."

I did not do this because I did not want a capacitor in the feedback loop. Yes, a resistor-capacitor network on the feedback node is part of the feedback path. In this sense, an input operation amplifier should use the positive input node to take input, and the negative input node to take feedback. In my case, negative for input and positive for feedback.


OK, maybe I am too defensive. Thank you for your insights, my answers are but some clarifications of my design trade-offs and they do not mean to "fight back".

[Sch3mat1c]

Originally Posted by HighVividity

This is because I do not want to include the coupling capacitor in the feedback loop. The feedback loop operates down to DC, like any operational amplifier. Maybe it is fine to do what you suggested, but I have not done any experiment with that.

You could remove that coupling capacitor then -- the first triode (and op-amp) will have to run from -V, but they'll also set the output tube bias, which will be very stable. You even have the opportunity to split feedback (oh noez, caps in the path!), so it's AC feedback to the grid or plate, DC feedback to the cathode current, so it's current servoed. It's still stable down to DC, the caps just move it to a slightly different method which works better.

the "leak resistor" was used to make the linear input pot more "exponential like". If you use an exponential input pot, then yes, a higher "leak resistor" will have less influence on the pot.

Eww, that doesn't work. You get a big jump at the first couple degrees, then a whole lot of nothing, then the rest of the range. It doesn't feel right at all. The only thing that works right is a log pot with a high resistance load (at least 10 times the end-to-end resistance).

I did not do this because I did not want a capacitor in the feedback loop.

No, you don't need a cap, the bias divider can be supplied from regulated voltage alone. Don't need the 4.7uF bypass (but it's a good idea for noise anyway).

Tim

[HighVividity]

Originally Posted by Sch3mat1c

You could remove that coupling capacitor then -- the first triode (and op-amp) will have to run from -V, but they'll also set the output tube bias, which will be very stable. You even have the opportunity to split feedback (oh noez, caps in the path!), so it's AC feedback to the grid or plate, DC feedback to the cathode current, so it's current servoed. It's still stable down to DC, the caps just move it to a slightly different method which works better.

You could be right. My design is to demonstrate that linear driver can be the best driver. The power stage itself can solely be responsible for producing the "tube sound". Your suggestions serve that purpose equally well.

Eww, that doesn't work. You get a big jump at the first couple degrees, then a whole lot of nothing, then the rest of the range. It doesn't feel right at all. The only thing that works right is a log pot with a high resistance load (at least 10 times the end-to-end resistance).

I followed the design on this page: Better Volume Controls, ESP - A Better Volume Control

No, you don't need a cap, the bias divider can be supplied from regulated voltage alone. Don't need the 4.7uF bypass (but it's a good idea for noise anyway).

You could be right.