I've never thought too hard about CFs but I'm playing with the idea of DC coupling power tube grids for bias stability on peaks. Since there's now a negative rail the Ak voltage can get big pretty quick so it looks valid to set up a sort of bipolar supply for follower stage. So, I'm trying to work out how big of a positive/negative rail is needed and not excessive. From articles like this is seems that the load line is drawn normally but I'm still a bit perplexed.
Can a follower typically swing all the way to the positive rail as suggested by the load line? Like, If I need to pass a 75Vp signal is a 75V anode voltage (in theory) sufficient? Putting aside for second that the grid signal will be shifted down by the bias voltage. If one was unconcerned with supplying grid current to the power stage could the positive rail be, effectively, 0v since the power tube grid would clamp here as well?
It seems easy enough to determine how close to the negative rail the follower can go but is there a way to determine how low one needs to go into cutoff in a class B output stage? Does it make sense to have a follower capable of reproducing the the entire negative swing when the power tube will be in cutoff at some point? Let's say a -75v waveform peak plus a -50v DC bias, and then another 50 volts of clearance for where the follower's load line crosses 0vgk? -175v?
So, how close can you cut it?
Can a follower typically swing all the way to the positive rail as suggested by the load line? Like, If I need to pass a 75Vp signal is a 75V anode voltage (in theory) sufficient? Putting aside for second that the grid signal will be shifted down by the bias voltage. If one was unconcerned with supplying grid current to the power stage could the positive rail be, effectively, 0v since the power tube grid would clamp here as well?
It seems easy enough to determine how close to the negative rail the follower can go but is there a way to determine how low one needs to go into cutoff in a class B output stage? Does it make sense to have a follower capable of reproducing the the entire negative swing when the power tube will be in cutoff at some point? Let's say a -75v waveform peak plus a -50v DC bias, and then another 50 volts of clearance for where the follower's load line crosses 0vgk? -175v?
So, how close can you cut it?
No. The maximum allowable input swing on a cathode follower is a function of the load to plate resistance ratio (commonly called 'k'), the tube amplification factor 'µ' at the operating point, and the operational plate voltage 'Ebo' of the stage. This is a clip from my design notebook.
For a more detailed explanation consult "Theory and Operation of Vacuum Tubes", Reich, H., 2nd ed, 1944, Sections 6-11 through 6-11E.
Thanks for that, Suncalc. I ran the numbers using a practical case and got what I thought had to be an incorrect answer, then found the reference to Fig 6-20c and cheated a little. With a lowish mu tube like 12AU7 the typical case with a modest load resistance was about 60% of Ebo. Ebo looks like the voltage from anode to cathode at the quiescent point.
So, a 12AU7, mu of 17, Rp of 7k7, a load resistor of 47k, say +/-125V rails. Assume the cathode is at -50v for biasing the power tube
k = 47/7.7 = 6.1
Ebo = 125 + 50v
I get (110.8/207.4)175v = 93.4v crest?
Changing the load R to 100k and things improve to ~92% of 175v.
Am I following? Not sure I'm treating Q point Plate voltage (not B+) correctly. And to confirm, the crest peak is referencing the cathode quiescent voltage?
So, a 12AU7, mu of 17, Rp of 7k7, a load resistor of 47k, say +/-125V rails. Assume the cathode is at -50v for biasing the power tube
k = 47/7.7 = 6.1
Ebo = 125 + 50v
I get (110.8/207.4)175v = 93.4v crest?
Changing the load R to 100k and things improve to ~92% of 175v.
Am I following? Not sure I'm treating Q point Plate voltage (not B+) correctly. And to confirm, the crest peak is referencing the cathode quiescent voltage?
Last edited:
You are on the right track.
In your example however i'm not following your derivation of the quiescent plate voltage on the tube. A load line or schematic would help.
In your example however i'm not following your derivation of the quiescent plate voltage on the tube. A load line or schematic would help.
In the Valve Wizard article it is clearly visible the AC loadline ( 10K load ) from about 110V to 170V ... so signal swing is smaller than DC loadline if you have a load . Even without a load , of course it is not rail to rail like an op amp 🙂 max swing ( undistorted ) is about from 90V to 190V
Negative voltage is not important , you translate it to a total positive total Va on the loadline .
Negative voltage is not important , you translate it to a total positive total Va on the loadline .
Last edited:
I was kind of hoping you would tell me since the terminology being used seems so specific.
If you had a follower with a 250vDC b+, no plate load and the cathode biased to 1/2b+ then the idle voltage from cathode to anode is 125 volts. I thought Ebo would be this voltage since that is what the stage has available for theoretical positive swings. Since the notes say “NOT B+” there’s only so many other choices.
In my example I used +/-125v instead of 250 with the cathode at 0v. This would be the same Ebo I presume. But the cathode needs to be a little further negative to provide bias to the output stage so I pulled it 50v negative. Now the plate is at 125 and the cathode at -50 for a 175 Ebo.
That is, assuming I understand what Ebo actually is.
Okay. With that explanation you are correct. It was your terminology "+/- 125V rails" which was confusing. But it works much better with an actual example.
I've attached a load line example of a 12AU7 cathode follower design with a 50kΩ load resistor (RL), a 1kΩ biasing resistor (Rk), and a B+ of 250v. There is a 250kΩ resistor on the output (Ro) and a grid resistor of 1mΩ (Rg). It looks like this.
And here is the load line.
In this example the tube is biased to a plate voltage of 93v. This is Ebo from my notes. That's the drop across the plate-cathode space. So the cathode voltage is 250v-93v or 157v. Note that the load resistance (determining the load of the DC load line) is 51kΩ but the total AC plate load is 51kΩ in parallel with the 250kΩ output impedance. This assumes that the 250kΩ load is capacitively coupled. The calculated parameters are listed in the figure.
It is important to understand that the stage is not limited by the 49v interval between the 93V quiescent operating voltage and the 44V intercept on the Vg=0 plate counter. The peak swing this stage can support is actually ≈66V. So an AC waveform of ≈46.7V-rms. This is calculated with the formula I supplied above which is also contained in the Reich reference.
The cathode follower is simply a normal common cathode amplifier stage with 100% local feedback accomplished by placing the plate load within the grid circuit. It is not appropriate to look at its operation like the AC waveform is superimposed on the DC load line. This is because at the same time that the plate voltage would be varying, its inverse (i.e. the voltage drop across the load resistor in the grid circuit) is being applied to the grid. The operation of the stage is better understood in the terms of basic feedback amplifier theory rather than gain stage theory.
Does this make sense?
Indeed.
I think this at the core of why I asked. The zero crossing point is such a clear cut part of determining headroom in a common cathode stage - but when the cathode follows the grid how exactly would that really happen? You have answered that question for me. There is a function of mu and the ratio of the plate resistance and load that limits positive swings. Thanks.