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Redrawn loadlines for unbypassed cathode question.

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In relationship to preamp triode stages, I am a bit confused about the load-line redrawing information regarding adapting the standard set of grid curves found on most tube characteristics data sheets which shows the negative voltage grid curves as they relate the plate voltage/current for a grounded cathode or bypassed cathode configuration, to one which shows the grid curves where they're influenced and altered by using an unbypassed biasing cathode resistor in the circuit.
As this seems to be a fairly common topography, rather than having a negative DC voltage supply to bias the grid, why do we not see the redraw charts more often?


The information that I have gathered and read concerning the method of redrawing the lines, I have taken from;

Inside The Vacuum Tube - John Rider - 1947 - page 327-32
Vacuum Tube Amplifiers - Valley, Wallman - 1948.
John.W.Gray - DESIGN PRINCIPLES - Single-ended Tnode Amplifiers.- page 424-30
Graphical Constructions for Tube Circuits - Albert Preisman - 1943
Feedback Construction - Current Feedback - page 229-31.

I understand how to adapt the standard set of grid curves, and can see and appreciate the differences, but what I am still not clear about in my head, is where the new limit of swing before crossing over into grid-current draw with positive biasing now occurs.

That the setting of the loadline Q point may have been on the -2V curve before redraw, and this point now becomes -0V is clear, but the new +2V line is no longer where the -0V used to be.

Am I to assume the 2V bias between grid and cathode still exists, its relative points renamed/designated, and the negative plate voltage swing reduced by the new position of the +2V line?

Or

Am I to assume that the bias voltages are now somehow compressed, the tube is less sensitive now that the effective plate resistance has gone up, and a larger, much more evenly spaced and symmetrical grid, and hence plate-voltage swing, are now available? Would this keep the position of the unaltered 0V line where it crosses the loadline as still defining the outer limit before fully crossing over into grid-current territory? If not, where is it? Not having a cathode bypass cap, would also make driving the grid into positive conditions easier to control and utilize...if I'm not mistaken.

I would be most pleased to hear any opinions about this subject.

Here is a copy of a redraw so you can relate to what I'm talking about.
CEwQ8J4.png
 
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This may not get you the curves you want, but it will help you to understand the effects of an unbypassed cathode.

I believe the cathode impedance of an unbiassed triode is as follows:
Cathode impedance = 1/Gm + RL/u
Where RL = plate load impedance

A 7F8 with 250V plate to cathode, Rk of 500 Ohms, and 6mA plate current (self bias of -3V) has:
Gm = 3,300micro mhos
u = 48
rp = u/GM rp = 14,545 Ohms

Suppose you use a plate load (RL) of 3 times rp, RL = 43,636 Ohms
1/GM = 303 Ohms
RL/48 = 43,636 Ohms/48 = 909 Ohms
The cathode impedance is 303 Ohms + 909 Ohms = 1,212 Ohms.

The effect is that for 1V grid signal, we have
1,212 Ohms + 500 Ohms = 1,712 Ohms
Of the 1V grid signal, 1V x (500/1,712) = 292mV across the 500 Ohm cathode resistor.
The result is that the 1V signal is reduced, 1,000mV - 292mV = 708mV grid to cathode.
We are losing gain at all frequencies (3dB less gain).

Of course the plate resistance is also increasing:
For unbypassed cathodes, plate resistance is rp + Rk(u)
the new total rp is 14,545 + (500 x 48) = 14,545 + 24,000 Ohms = 38,545 Ohms.

The plate load, RL of 43,636 Ohms is now almost equal to the new rp of 38,545 Ohms.

A little bit of thinking about the above examples may cause you to see that you really need a good reason to use an unbypassed cathode. The dynamic range will be quite reduced. Of course there are examples where you must have an unbypassed cathode, a split load inverter is one.
 
Thanks, that does go some way to helping understand the situation. I am hoping to get the full graphical representation work out...just for my own interest. And yes, the Cathodyne lower side is what I am working towards as well. I have the rest of the loadline drawing for them worked out and on paper, but the true representation of the grid curves is still something alluding me. Not that I need them, or that it really matters...it's just nice to know how it could..or should, be done.
 
Some tubes have a 90V cathode to filament, and some have 180V cathode to filament
rating (and other values too).
Keep in mind that for a cathodyne inverter, there is the DC cathode to filament voltage rating, and there is the DC + Peak Signal cathode to filament voltage rating.

The cathodyne invertor gain from grid to the in-phase and out-of-phase signals are less than unity and - unity (<1, and <-1).

A cathode coupled invertor, with a current source in the cathodes, has a gain from the grid to the in-phase and out-of-phase signals is less than 1/2 mu; less than 1/2 -mu
(<1/2 * u; <1/2 -u).

The 2nd harmonic distortion of a cathode coupled invertor tends to cancel by the nature of the circuit topology.

You can use a negative supply to power the current source.

Or, you can use batteries in series from the signal going to the grid stopper resistors.
Signal, battery -, battery +, grid stopper resistor, grid 1.
Ground, battery -, battery +, grid stopper resistor, grid 2.
The tube will self bias, so that the cathode voltage is larger than the grid voltage (no grid current). I have used 3V and 6V lithium cells.

I am going back to cathode coupled invertors for input stage/drivers.

In another amplifier:
I have also used a single ended driver to drive a cathode coupled push pull output stage.
I used a TIP42C (C = 100V breakdown), much larger than the 60V breakdown of a LM317T.
Suppose there is 35V across the TIP42C (cathode bias), and then the driver has 80V peak out (because the gain is turned up too far). Then the cathodes and current source will be more than 70V (too much for a LM317T).
 
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Joined 2011
As this seems to be a fairly common topography, rather than having a negative DC voltage supply to bias the grid, why do we not see the redraw charts more often?
Because it is too much work to do it manually...;) Of course, it is quite easy to do it if you use a circuit simulation program. In any case, since the u of the tube remains constant, the cathode feedback lowers the stage's gain. So it's easier to use the chart from the datasheet then recalculate the gain - much simpler than making up a new chart for each design.

p.s. some of the above linked images are not showing up.
 
Those extra links were my bumbling efforts to get a picture in place, then the resulting misslinks can't be removed by editing for some reason.

It's not working out the gain that i'm concerned with, I can do that.
I was just interested in the graphical approach, without the math ( other than what is needed to culculate the graphic. ) in showing how what was set up as a 2V bias, might now need a 4V swing upwards to reach the original 0V point, as is exemplified in the graph I posted, and if this can really be the case that the current drop over an unbypassed resistor can somehow be less (half) than the normally calculated ohms law voltage drop
icon9.gif
..or if the upwards swing now requires so much grid current positive swing to reach the same place.
 
When you use self bias that is unbypassed, you not only reduce the gain, you increase the plate resistance of the tube.
Suppose Rk = 100 Ohms, and u = 4, and rp was 1 k Ohm.
Then the new rp is now (rp original + (100 * 4)) =
1k Ohms + 400 Ohms = 1,400 Ohms.

The cathode impedance and unbypassed self bias create a voltage divider (of the grid signal voltage). That, and the increased rp are the reasons that the gain is reduced.
If this is an output stage, the increased output impedance reduces the damping factor
(a lower DF number, which is less damping).
 
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Joined 2011
Those extra links were my bumbling efforts to get a picture in place, then the resulting misslinks can't be removed by editing for some reason.
:cop: I removed them for you.

or if the upwards swing now requires so much grid current positive swing to reach the same place.
There is no grid current. The graph's labeling is a bit mis-leading: for bypassed cathode, egk is just grid-to-ground (from the datasheet), while for the unbypassed cathode, egk is actually grid-to-cathode, where ek is above ground.
 
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