This was part of my basic duration when I went to RCA Institutes in 1961. We even plotted our own family of curves from scratch. Today in school vacuum tube theory is barely explained.
Thank you so much for taking the time to make this whole process clear. You are an outstanding teacher! Everywhere else I've looked it was as though there were always key parts left out.
I do love this video and the formula that you gave use when determining load lines. The only thing that I pondered on for a long time was the B+ out of the power supply. You said to use that for the maximum voltage. I finally figured out it wasn't directly out of the power supply as the Eb. Shouldn't be said to be the Ebb. The voltage supplied before the load resistor for that tube. Say your power supply B+ directly out of the power supply is 300v Eb going to the output transformers and to a second node resistor dropping it to maybe 280v Ebb, then to the load resistor. Wouldn't the maximum voltage be the 280v Ebb, not 300v Eb. I'm just learning and want to make sure that I have it right. Still, Great video! No one else explained it as good.
A 6j5 has a plate resistance of about 7k ohms. A triode wants to see, at the very least, 3 times its own impedance as a load. The load seen by the tube is the plate resistor in parallel with the grid resistor of the next stage. I would make a higher voltage power supply, increase the plate resistor value to 33k ohms and watch the harmonic distortion fall like a rock while, at the same time, the gain of the stage would go up. The more horizontal you can make the load line, the more gain you get with lower HD. To take it to the next step a Constant Current Source as a plate load can get you a gain roughly equal to the mu of the tube used, with almost no distortion at all.
Thanks Dennis, this was really explained well. I'm getting ready to do a first "ground up" design for a specialized musical instrument and I would like to be seeing how the operation of my tubes are reflected by the data sheets and the math. This video did very well to have a clear method of seeing how it all comes together. Thank you very much! - Phil Donovan
Glad I found this video. It's the most detailed yet. But I have a question. Is the maximum voltage measured with the tube out? (ref 15:33) Cause, it would be different with it in, wouldn't it?
Great teacher, but I had a question. I'm trying to find 6n3p bias that it is -2 and on 6n3p chart is 10 ma...when I'm calculate anode resistor and you need double ma that comes 20ma than on chart only shows till 16 ma.
Excellent video. Very comprehensive & informative. Quick question. Can’t we just get the grid bias voltage & plate current details from tube sheet & calculate cathode bias resistor value ?
Great,! I've been looking at schematics, and trying to figure out how to calculate that. My voltage differs a little from most used in schematics. Most are using new run tubes that can operate on higher B+. I have NOS tubes and want them to last as long as possible. Much appreciated!
I tried to construct a load line with a 12AJ6 tube running at a plate voltage of 12.6 volts and got a line that sloped down to the left instead of down to the right as shown here. What am I doing wrong?
Dennis, I don't know what your students paid to get this type of knowledge but I know we got one hellofa deal just now. I'll be watching this again to help set it in lol. Thanks as always. Great stuff. Carl
Thanks for the many educational videos i like them all. Can you make a video where you bias calculate a kt (pentode) tube instead of a triode tube or is it the same method? example a kt120 tube typically uses according to data sheets plate voltage 400v Plate current A135-160mA Gr2 225v Gr1-14 Max signal power output 3000 ohms and 30uA. Can i bias it with plate voltage 450v 3k transformer Gr1-45v and 270-300ohm khatode resistance it will give about 150ma. or is it me who has not understood it.
By biasing with cathode resistor, upping the cathode voltage at 8.5V, doesn't that effectively drops B+ by 8.5V? does that need to be compensated somehow?
If I'm wrong then someone please correct me. But from what I've learned a milliamp is 1/1000 of an ampere. If true then what was shown in the video is 0.0007 which would be 7/10,000 of an ampere and 7 milliamps is equal to 0.007 or 7/1000 . That being said then the result should have been 0.016 - 0.007 = 0.009. So 220/ 0.009 = 24,444. hence RL= 25k .
Now, what would you say if you could talk? Get to the 'effin' point! Find out what you want to say before recording and then say it while recording! Jeez!
This was part of my basic duration when I went to RCA Institutes in 1961. We even plotted our own family of curves from scratch. Today in school vacuum tube theory is barely explained.
It was mandatory for me in technical high school way back when.
Thank you so much for taking the time to make this whole process clear. You are an outstanding teacher! Everywhere else I've looked it was as though there were always key parts left out.
Thanks Dennis, this is the most comprehensive video on that topic that I have seen.
I do love this video and the formula that you gave use when determining load lines. The only thing that I pondered on for a long time was the B+ out of the power supply. You said to use that for the maximum voltage. I finally figured out it wasn't directly out of the power supply as the Eb. Shouldn't be said to be the Ebb. The voltage supplied before the load resistor for that tube. Say your power supply B+ directly out of the power supply is 300v Eb going to the output transformers and to a second node resistor dropping it to maybe 280v Ebb, then to the load resistor. Wouldn't the maximum voltage be the 280v Ebb, not 300v Eb. I'm just learning and want to make sure that I have it right. Still, Great video! No one else explained it as good.
Yes, 280V. You are correct.
Thank you, this is the best explanation on TH-cam and worked perfectly for my amp.
A 6j5 has a plate resistance of about 7k ohms. A triode wants to see, at the very least, 3 times its own impedance as a load. The load seen by the tube is the plate resistor in parallel with the grid resistor of the next stage. I would make a higher voltage power supply, increase the plate resistor value to 33k ohms and watch the harmonic distortion fall like a rock while, at the same time, the gain of the stage would go up. The more horizontal you can make the load line, the more gain you get with lower HD. To take it to the next step a Constant Current Source as a plate load can get you a gain roughly equal to the mu of the tube used, with almost no distortion at all.
Thanks Dennis, this was really explained well. I'm getting ready to do a first "ground up" design for a specialized musical instrument and I would like to be seeing how the operation of my tubes are reflected by the data sheets and the math. This video did very well to have a clear method of seeing how it all comes together. Thank you very much! - Phil Donovan
Very informative. Why use 2*Io? Curious.
you are the ULTIMATE JEDI MASTER OF ELECTRONIC VACUUM TUBES, i love your videos
Very clear demonstration, thanks Dennis.
great job,Dennis,looking forward to your upcoming videos,Thanks.
Looking forward for more. Thanks Dennis.
Glad I found this video. It's the most detailed yet.
But I have a question.
Is the maximum voltage measured with the tube out? (ref 15:33) Cause, it would be different with it in, wouldn't it?
Very educational...thank s a lot for this video tutorial
very informative tutorial, I watched more than 5 times thanks!
you are realy good teacher..bravo
Excellent presentation and explanation. Where does the .68 come from?
Great teacher, but I had a question. I'm trying to find 6n3p bias that it is -2 and on 6n3p chart is 10 ma...when I'm calculate anode resistor and you need double ma that comes 20ma than on chart only shows till 16 ma.
DENNIS CARTER, how do you bias the power tubes in amplifiers compression point? biasing the compression of the power tubes?
hi sir ..thankyou for make this video...sir i have a question.. how can i calculate input and output impedance..please lead me
How would you determine the AC gain of that stage?
Excellent video. Very comprehensive & informative. Quick question. Can’t we just get the grid bias voltage & plate current details from tube sheet & calculate cathode bias resistor value ?
Great,! I've been looking at schematics, and trying to figure out how to calculate that. My voltage differs a little from most used in schematics. Most are using new run tubes that can operate on higher B+. I have NOS tubes and want them to last as long as possible. Much appreciated!
I tried to construct a load line with a 12AJ6 tube running at a plate voltage of 12.6 volts and got a line that sloped down to the left instead of down to the right as shown here. What am I doing wrong?
Great video
Dennis, I don't know what your students paid to get this type of knowledge but I know we got one hellofa deal just now.
I'll be watching this again to help set it in lol.
Thanks as always. Great stuff.
Carl
+skycarl Thanks Carl for stopping by. I'm glad you enjoyed it.
Very useful info Dennis... Do you actually remember these formulas, or do you have them written down?
Thanks for the many educational videos i like them all. Can you make a video where you bias calculate a kt (pentode) tube instead of a triode tube or is it the same method? example a kt120 tube typically uses according to data sheets plate voltage 400v Plate current A135-160mA Gr2 225v Gr1-14 Max signal power output 3000 ohms and 30uA. Can i bias it with plate voltage 450v 3k transformer Gr1-45v and 270-300ohm khatode resistance it will give about 150ma. or is it me who has not understood it.
Good morning, can you or someone in field help me with 6n3p? I'm little confused... I want doing this graph for 6n3p. Thanks
By biasing with cathode resistor, upping the cathode voltage at 8.5V, doesn't that effectively drops B+ by 8.5V? does that need to be compensated somehow?
The same current that flows through the cathode is the same as the plate!
Thanks for sharing information.
If I'm wrong then someone please correct me. But from what I've learned a milliamp is 1/1000 of an ampere. If true then what was shown in the video is 0.0007 which would be 7/10,000 of an ampere and 7 milliamps is equal to 0.007 or 7/1000 . That being said then the result should have been 0.016 - 0.007 = 0.009. So 220/ 0.009 = 24,444. hence RL= 25k .
Never mind. I mistook what he said. 0.7 milliamps would be 0.0007.
Thanks
Thank you.21
Sniff
Now, what would you say if you could talk? Get to the 'effin' point! Find out what you want to say before recording and then say it while recording! Jeez!