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Pentode Loadlines PP

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Lack of spice models is forcing me to design the old fashion way via. loadlines (this is a really good thing kind of like eating my veggies :)). It seems that I ought to be able to use the published curves without the composite curves. If I am going astray at any point please slap me upside the head. Using the following curves...

6HZ8_PLoadlines.png

I like the 5K 20mA (green line) as it gets close to the knee. In Class A this would be a 10K P-P if I understand correctly. However since 20mA is a Class AB bias point that complicates things. As long as both tubes are conducting this loadline should be correct however when one shuts off the 2x relationship becomes 4x does it not? Does this mean that the 10K xformer now behaves as a 20k? How would I represent this on the curves? Would I draw a line with the 10k slope (20k P-P) from the point on the 5k loadline where the opposing tube cuts off over to the Y axis? Do I want this Y intercept to approach the knee (meaning my 10k P-P is the wrong choice)?

Apologies if this ought to be obvious.
 
When one tube shuts off the load becomes 1/2 the Ohms for the remaining tube.

With both tubes conducting, the load of X Ohms can be seen as two 2X Ohm loads in parallel, so each 2X Ohms seen by each tube.

So class AB takes a higher Zpri OT than class A. But on balance, for class AB, when the class B load-line sections are entered, the tube has maximized gm to handle that heavier load (from higher current). At twice the tube current, a typical square law power tube has 1.4 times the gm1. So a more reasonable OT adjustment from class A would be 2/1.4 X or 1.4 X Ohms of the class A OT case.

Note:
Tube grid2 is typically near 3/2 power law. Power tube grid1 is typically near square, 2.0, power law.
(Ip = k2 Vg2^3/2 and Ip = k1 Vg1^2 ). Then gm varies as the 1st derivative, so gm2 = k3 Vg2^1/2 and gm1 = k4 Vg1^1)

Just for laughs, I recently found that the 19 Watt 6JC5 tube has a cubic, 3.0, power law for grid 1, so its gm would be the square law of Vg1. (grid 1 is placed too near to the cathode on these) (superficially looks like the 6V6 tube for specs, but very different grid 1 characteristic)
The 6JC5 tubes are on sale on Ebay currently for $0.75, a "bargain" I guess! (use -plenty- of N Fdbk)
Have no idea why these were designed for Vertical Sweep tubes, except maybe for some weird off center CRT electron gun (side scan).
 
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1. Consider a 10k Ohm plate to plate transformer. Using the whole plate to plate winding, it is 10k Ohms. As mentioned earlier in this thread, using 1/2 of the total plate windings (plate to the center tap) it is 2.5k Ohms (Z is proportional to the square of 1/2 of the turns). In class A push pull, the plates work in conjunction, so each plate effectively sees 5k Ohms.

2. Consider a single ended amp, with an air gapped transformer that is 5k Ohms. I will use voltage numbers that may not be real, but the concept is real, and with this method you will be able to approximate the output power you would get with push pull on a class A amp. Do not forget to also calculate the reduction of the output power caused by the loss of the output transformer (i.e. might be 0.5 dB or 1 dB). This method only works if no tube goes into cutoff (it would now be class AB).

Taking a hypothetical set of tube curves, and applying a 5k load line.
Suppose the quiescent plate voltage is 300V, and the grid bias is -10V.
Suppose that with 0 Volts on the grid, the the plate voltage is 100V.
Suppose that with -20V on the grid, the plate voltage is 400V.

The plate moved from 300V to 100V, a 200V change.
(200V squared)/(5k Ohm * 2) = 4 Watts rms
The plate moved from 300Vk to 400V, a 100V change.
(100V squared)/5k Ohm * 2) = 1 Watt rms

This would be extremely distorted as single ended. But for 10k plate to plate, and class A push pull we get approximately: 4 Watts + 1 Watt = 5 Watts rms, with very little 2nd harmonic distortion; and the 3rd harmonic distortion would be dominant.
 
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Or in other words, when both tubes are conducting, each tube sees 1/2 of the end to end impedance. But when one tube shuts off, the other tube sees 1/4 of the end to end primary impedance.

So there are two load lines to draw in an AB amp: the load line when both tubes are conducting, and the load line when only one tube is conducting. The slope of the load line changes based on whether one or both tubes are conducting.
 
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kward,

Yes, good point! You teach me a better way to look at that. I like the idea of the second load line. But . . . There is not a single exact point where one tube is cut off. It is gradual. Is it cutoff at 10mA, 1mA, 100uA, etc.? For push pull, you might say the one tube is cut off when it is perhaps 1/4, or 1/10 of its quiescent current. That is because at the same time, the other tube's plate current may be 1.4 times or 2 times its quiescent current.

For Pentode operation, it is when the dictated plate current is no longer effectively controlled by the grid voltage (you see that on the family of curves).

For Triode operation, you might consider when the slope of the curves (plate resistance) is much lower (higher plate resistance), than the quiescent plate resistance. At the same time the other tube's plate resistance is now lower than it's quiescent plate resistance.

Where does class A quit, and where does class AB begin? This is an analog world, not a digital world. "Some things nearly so, others nearly not" (King of Siam?).
 
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Naturally my talent for getting things exactly backwards showed itself again. Using Merlin's method (assuming I didn't get that turned about) I get the dark blue line for 10 P-P 300V 20mA bias. If so it looks reasonable.

6HZ8PentodeLoadlines.png

Then the driver triode with 51K load and 560 Rk lands us at about Vg of 1.5V and Ia of 2.8 mA.

6HZ8TriodeLoadline.png

For each I look at the output voltage swing for a 1V p-p input. the gain of 50 on the triode seems a bit optimistic for a tube with mu of 70. of course there will be degeneration from the un-bypassed cathode resistor (planning for P-K feedback). The output stage swings about 40V for a 1V input. So the voltage gain from driver grid to output plate should be a bit less than 2000. If so then we want the feedback to drop the gain to somewhere around 200.

If it were an op-amp the feedback resistor would be 560*200=115k. So perhaps a 120K feedback resistor might be a reasonable starting point.
 
Assuming the triode is driving an infinite load (which it is not), the load line you drew will have a gain of about 50x if cathode is fully bypassed, and about 30x if the cathode is not bypassed. Actual gains will be less than that depending on the load the stage is driving.

The gain reduction you wanted (from your quoted 2000x to 200x) is a factor of 10, or 20 dB. But if you bring the feedback back to the unbypassed cathode (where in that condition stage gain will be some amount less than 30x), you will need maybe 14 dB of feedback to achieve your targeted forward voltage gain, and that would require a feedback resistor of approximately 135KΩ.

But we're not done yet. If feedback is DC coupled, that condition will inject about 2 mA of additional current into the 560Ω cathode resistor, which will skew bias of the first stage by 1.2V or so. You would have to account for that in the first stage biasing. Alternatively you could look into AC coupling the feedback but that brings with it an R/C time constant that will need to be dealt with.
 
So something on this order is my first shot at it.

6HZ8_PP.png

Starting from the output we are using cross coupled bias because the OPTs that I intend to use are some of those 70V line transformers and want to limit current imbalance. According to the curves 300V and 20mA comes at about -4.5V leading to a bias resistor of 225 so 220 is probably close enough.

For the screen at 170V we need to drop 130V in the screen resistor. According to the data sheet with the screen at 170 and grid at -4.5V we get a screen current of 4.9mA giving a dropping resistor value of just under 27k (26.53).

6HZ8_ScreenCurrent.png

As discussed previously the driver triode is biased at 2.9mA with -1.5V on the grid and 150V on the plate. The 120K Rf and 560 Rk form the feedback network. after building R23 and R24 can be tweaked for desired sensitivity.

I anticipate the preamp having sufficient output such that additional gain will be unnecessary so the input is just a 12AU7 based Cathodyne PI using a simple voltage divider bias to set the cathode at about 1/4 B+.
 
mashaffer,

During turn-on . . . R8, R18, and C2 may pump charge current into your signal source.

You may want to calculate the cathode impedance of the 6HZ8P. (1/Gm + RL/u). (that is more complicated for a Pentode than a Triode). The 100uF capacitor has to properly bypass the 220 + 220 (440) Ohm cathode resistors, but it also has to bypass the cathode impedance (looking into the cathode). When Xc = 0.1 of (440 Ohms in parallel with the cathode Z), the response is -1 dB at that frequency.
 
What did you mean by: better to self bias the PI? Did you mean to not use the Gartner method, and switch to simple self bias?

Calculation of the bypass capacitor requirement: It is a little less simple than that, because the cathode impedance is driving the capacitive reactance that is in parallel with sum of the two cathode resistors that are in series. Being the conservative person I am, I take the worst case, and also simplify, I just mentally "parallel the cathode impedance and series cathode resistors" and calculate a capacitor that has capacitive reactance 1/10 of the cathode/self bias resistor combination. For frequency, I set a -3dB frequency one octave below where I want to be only 1 dB down. -3dB at 10 Hz will be -1dB at 10Hz.

Using an unbuffered concertina phase splitter at the input will cause a charging current to go out the input connector. At one time I thought about using 5V zener diodes and 1N914 or similar diodes at the input side of the input capacitor (one regular diode in series with the zener diode prevents the zener from conducting in its forward direction).

At the input connector, connect a zener diode anode to signal 'hot', and the zener diode cathode to the cathode of a regular signal diode, and the anode of the regular diode to signal 'common'. And . . . At the input connector, connect a zener diode cathode to signal 'hot', and the zener diode anode to the anode of a regular signal diode, and the cathode of the regular diode to signal 'common'. If you use low leakage zener diodes (do not even begin to turn on at 3V), you can apply the output of most CD players, and not load the signal, and not distort the signal. I tested 5V zener diodes, and found a type that was very low leakage at 3V. A typical CD player has 2.1Vrms output (2.97V). You have low leakage at 3V on the zener, and maybe 0.2V on the 1N914 before you have a problem of loading the signal source.

But the charge from the concertina through the input coupling cap has now been limited to about 5.6V (and is only a transient at warm-up).
 
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Oh, btw I meant use cathode bias for the cathodyne instead of voltage divider so that direct B+ to input is avoided but I gather that is not the issue. A series input resistor would slow the charge down but not eliminate it.

I do have a pair of.Cinemag input splitting trannies that I could use instead (would save some current on the PT in the process). I was saving them for a snazzier amp but it would solve the issue neatly. I also have a pair of splitting transformers from solid state power amps but I suspect they may not like the highish impedence.
 
1. Suppose the concertina bottom resistor has 80 Volts across it, the tube has 80 volts across it, and the top resistor has 80 Volts across it. That means the the cathode will be at 80 Volts; the grid will be a little lower than that, perhaps 80 - 4 = 76 Volts. That requires a resistive divider to put the grid near 76 Volts. That voltage can go back out the input cap.

That is a hard one to solve, even if you have a negative power supply to power the cathode resistor (it is not a current source), and use a lower B+ voltage to make the total voltage be 3 x 80 Volts..

2. Instead there are other circuits that will not put significant voltage back out the input cap. The key is to use a negative voltage power supply to power the cathode(s).

A. If you have a cathode follower with a cathode current source, you can connect the grid resistor to ground. The grid will be at 0 Volts.

B. If you have a LTP cathode coupled phase splitter, you can connect the grid resistor to ground. The grid will be at 0 Volts.

3. You can also use a common cathode stage with self bias, and use a grid resistor to ground (grid at 0 Volts). This can drive the concertina spitter, with DC coupling from the input tube plate to the concertina grid.

4. When analyzing these circuits for potential problems, remember that tubes tend to warm up slowly, and they tend to cool down slowly. Look at the rest of the circuit to see what happens at turn-on and turn-off sequences.
 
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The 6JC5 tubes are on sale on Ebay currently for $0.75......Have no idea why these were designed for Vertical Sweep tubes

The 6HE5 is a rebottled 6EZ5. What's being sold today is a single tube with 3 type numbers, 6JB5 / 6HE5 / 6JC5. I bought about 20 6EZ5's when they were on the dollar menu and ran them in my SSE boards. They worked well with minimal cathode feedback and seemed to fall somewhere between a 6V6 and a 6W6 in their performance. They also worked OK in a P-P amp but there are better choices for amps in the 15 to 30 WPC range where these work well.

I have collected about 15 or so of the compactron flavors, 6HE5's, 6JB5's and 6JB5 / 6HE5 / 6JC5's. There are a few different constructions of each depending on who made them and when. I modified one of Pete's bit red boards to run high voltage on G2 and found these tubes to be all over the place audio performance and bias wise. I suspect that they may be at least two different types of guts being stuffed into this glass. I decided to pass on these for audio amps.
 
mashaffer,

I think you might have too much gain using a 12AX7 as an anode follower. Is your power supply too limited to power one more tube? Can your power supply support a 12AU7, one triode per channel of a stereo amp. If this is a mono-block, you already have 1/2 of a 12AU7 that is not used. That would be a 12AU7 anode follower DC coupled to the grid of a 12AU7 concertina.
 
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