I am building a mu follower with a 31 tube as v1(lower tube) and 12GN7A pentode as the follower. this stage will drive a E80L triode strapped. I am trying to understand the 12GN7A pentode operating point. I've studied Alan Kimmel's pdfs and looked at Triode Dick's designs. My question is, doesn't the follower grid 1 get too high of a signal to operate properly? What is going on here? Please help me understand this! I can share my schematic if that helps.
Thanks!
Thanks!
It's not the absolute grid voltage itself that drives it, it is the voltage between grid and cathode Vgk that drives it.
Jan
Jan
Thanks for the reply. I understand that. What I'm curious about is the fact that in the mu follower, you are taking the ac signal (some of it) and feeding it to the grid of the pentode. But, my DAC puts out 2.8 volts pp and that gets amplifier at the triode plate. That means that I'm getting at least a 6 volt swing on the pentode grid. If you look at the pentode load line chart, there isn't that much grid voltage to work with. The grid lines go from zero to 3 or four volts negative with respect to the cathode. I'm holding -2 volts as an operating point. So at least on the positive swing, the grid will go past zero. That's the problem. I probably didn't ask the right question. I guess what I need to know is, in a mu follower, does the tube operate as a amplifier or is the css doing something different? I can see that it isn't loaded like a tube normally would be. That means that there isn't really a load line. I can see an operating point at 21 mA and 172 volts. I can see that both the cathode and grid experience the amplfied signal. I need to know how that transltes into a css and what the pentode is doing exactly. How does that bootstrap the plate resistor of the triode?? Kimmel and Dick don't ever go into enough detail for my little brain. Other discussions just assume you know. I don't know. It's hard to design something when you don't REALLY understand what it is doing.
Nice choice of valves - I have found the 12GN7A and 12HL7 very useful as the upper valve in a mu stage. Have you had a chance to listen yet?
Christopher Paul published a very nice article on mu followers and mu stages in Linear Audio (here: https://linearaudio.net/so-whats-mu-you) that is well worth the purchase price. He explains how the circuits work and the role of both valves. And he reports on his own experiments.
Christopher Paul published a very nice article on mu followers and mu stages in Linear Audio (here: https://linearaudio.net/so-whats-mu-you) that is well worth the purchase price. He explains how the circuits work and the role of both valves. And he reports on his own experiments.
And Merlin Blencow has a nice explanation and worked example here: http://www.valvewizard.co.uk/mufollower.html
Jan's point is that the upper valve works as a cathode follower, so its cathode has almost the same excursions with the signal as its grid, so the voltage between grid and cathode varies much less than the voltage between grid and ground.
Thanks for the clarification Marcel. I tend to be a little bit terse sometimes.
My hope is to get people to think about it ;-)
Too many people forget that a multimeter or scope has two probes.
You're not measuring 'a voltage', you are measuring 'the voltage between two points'.
We often assume one of the two points is ground.
A tube reacts to it's Vak voltage, the voltage between grid and kathode.
So the voltage of the grid to ground is pretty irrelevant.
Its the voltage between grid and cathode that determines the tube operating point (with the anode voltage of course).
I hope I have sufficiently confused everyone now 😎
Jan
My hope is to get people to think about it ;-)
Too many people forget that a multimeter or scope has two probes.
You're not measuring 'a voltage', you are measuring 'the voltage between two points'.
We often assume one of the two points is ground.
A tube reacts to it's Vak voltage, the voltage between grid and kathode.
So the voltage of the grid to ground is pretty irrelevant.
Its the voltage between grid and cathode that determines the tube operating point (with the anode voltage of course).
I hope I have sufficiently confused everyone now 😎
Jan
It would probably help to know that each valve actually does operate on a loadline, independent of its flying up in the air above another valve or any other circuit peculiarity. The valve must operate on its characteristic curves - for any particular voltages between cathode, various grids and anode, a characteristic curve defined current will flow, and vice-versa. The slope of the loadline may not be obvious (especially when local feedback is operating) but there still is one that applies to each valve.
Looks like Marcel and Jan have covered the important points. The mechanism of "bootstrapping" can be derived from their comments, but a clue is to think in signal currents rather than voltage.
All good fortune,
Chris
Looks like Marcel and Jan have covered the important points. The mechanism of "bootstrapping" can be derived from their comments, but a clue is to think in signal currents rather than voltage.
All good fortune,
Chris
Great! Thank you for the link. I didn't see that in my search. I am just about to mount the iron and wire it up. I can't wait to hear it. I figure that I'll have a fair bit of measurements and testing different components before its "done". Thanks for your help.
Marcel, that is exactly what I needed to know. Now I'm not so worried. I've build a Triode Dick amp before and I've convinced myself that his source didn't have nearly the voltage that modern DACs have. I built monoblocks from his Top Cap design and had to use 6SN7s instead of his 6sl7s because the output from the gain stage was way too high for the 807s. Heck, it is still too high. so anyway, thank you, I was thinking of the grid voltage as absolute rather than relative to the cathode.
Chris/Jan, I thought about what Jan said in the first reply and the pentode load line this morning and that Kimmel referred to it as a current amplifier. That would be the case if the load line was more or less vertical due to the lack of load on the plate. At least that's my thinking. So the small difference between the grid and the cathode will make for a pretty good current swing to hold the triode steady. And, I don't need 10 volts or more of grid available. Thank you both for clarifying.
The filament transformer for the 01A DC supply may require a Faraday Shield.
Otherwise, AC leakage currents from the AC line may be enough to cause modulation
of the cathode currents of the 01As. You will hear that in the output of the amp,
especially during quiet passages.🙂
Thanks for looking at the amp! I put an aluminum plate between the power supply section and the amp circuit section. I don't know if you can see that in the pictures. The whole power supply section is in a Faraday box. I learned that the hard way on a preamp project. I had to unwire a lot of it and put a wall in and then rewire. Never do that again! I saw on Cascade tubes that he always does that. My 201a amp is finished now and is dead quiet. I really like the Rod Coleman filliment supplies. Thanks for the advice though.