Cathodyne phase inverter - altering tail value

[Quatro]

What happens when one alters the value of the resistor closest to ground in the cathodyne? Commonly this value is 56k in the 5e3. What happens (or is supposed to happen) if this value changes up or down?

Thanks

[tonewood]

I think making both top and bottom bigger may allow more voltage swing. Making them different may cause a slight unbalance witch may be good or bad.
I think I read the above here (it's been a while):
http://www.freewebs.com/valvewizard/cathodyne.html

[sluckey]

What happens when one alters the value of the resistor closest to ground in the cathodyne? Commonly this value is 56k in the 5e3. What happens (or is supposed to happen) if this value changes up or down?

Since the cathodyne operates like a cathode follower, changing the value of that resistor will not change the signal voltage that appears across that resistor. The cathode signal will always be about 80% to 90% of the grid signal. But, making it smaller will allow the tube to conduct more plate current, and making it larger will decrease the tube plate current. This causes the signal on the plate to become larger or smaller.

Notice that the plate load resistor is always the same value as that cathode resistor? Using resistors that are not the same value will result in different value signals sent to the output tubes. It's usual design practice to keep these two signals equal amplitude but 180 degrees out of phase.

[Quatro]

Ok. i just went ahead and tried it out. I clipped out the 56k and put a decade box in it's place. I am going to say the results were interesting but simple. Bigger resistor value = less volume. Smaller value = more volume. My testing, with the human ear as the only device to evaluate results, seemed to indicate that volume change was the only change. The signal did not seem to get more or less dirty (not to any noteworthy degree anyway) nor did the signal quality change in any other way that I could perceive.

[Willabe]

But, making it smaller will allow the tube to conduct more plate current, and making it larger will decrease the tube plate current. This causes the signal on the plate to become larger or smaller.

Like sluckey said, smaller R in the cathode, = more plate current = more out put current = more volume.

I'd say your ears are working fine, you don't need test gear to know/see/hear the difference.    


                Brad    

[tubeswell]

Also, the more imbalance there is between the load resistors for a cathodyne, the more the output impedance of the stage increases.

[Quatro]

is there any reason this couldn't/shouldn't be used as an attenuator type device as the imbalance between the top and bottom doesn't seem to affect the performance to any large degree?

[HotBluePlates]

I clipped out the 56k and put a decade box in it's place. I am going to say the results were interesting but simple. Bigger resistor value = less volume. Smaller value = more volume.

Question 1: How did you listen to the output? At the speaker of the push-pull amp? At plate only?

Question 2: Did you change the input signal level feeding the phase inverter? Crank up the volume knob and see at what volume distortion sets in?

Sluckey said:

Since the cathodyne operates like a cathode follower, changing the value of that resistor will not change the signal voltage that appears across that resistor. The cathode signal will always be about 80% to 90% of the grid signal. But, making it smaller will ... causes the signal on the plate to become larger ...

The split-load normally has the plate and cathode load resistors large enough that the stage has a gain to each output of a little less than 1. As you make the cathode load smaller, it will allow the plate output to become a little bit larger. Eventually, the cathode load becomes small enough to approximate a normal common-cathode gain stage. At that point, the plate output might be up to a gain of 50-60 with a 12AX7. That could account for the volume increase you heard, under certain circumstances...

However, at the point where the cathode resistor starts allowing a gain above 1 at the plate output, the cathode output also starts changing. If a 25vac input resulted in a bit less than 25vac on the cathode output, there will come a point (as the cathode load gets smaller) where Rk * Ik doesn't equal 25vac. Meaning, the cathode resistance gets so small that the current required to get the same 25vac output at the cathode is impossible for the tube to manage. So, when you get to the point where plate output is noticeably up, the cathode output will be noticeably down.

What happens to the push-pull output stage? Well, if you listened with a routinely small signal input to the phase inverter, you probably didn't notice your overall clean power output kept going down.

Thought Experiment: What happens if you go so far to make the cathode load Zero ohms? This would be the same as the limit for your "volume boost" situation, right? If cathode load is zero, plate output has gone up, but cathode output is zero also. If you're using a 5E3 Deluxe, this is the same as yanking the 6V6 fed by the phase inverter output from its socket. Try it. The amp will still play.

You won't notice an issue at low signals. As you crank up, you'll find the amp distorts much earlier and has less power output, less volume. Instead of 2 output tubes, you only have 1 driving the speaker. The OT also goes from looking like R_a-a/2 to R_a-a/4 for the remaining tube. That will be well under the designed load, which probably allowed for close to maximum clean power. So 1 output tube + wrong load = much less power than the original circuit.

So something don't add up: you get more volume by reducing the cathode load, but you are assured of less volume by taking that to its logical conclusion. I'm guessing you tested with a guitar at a fairly quiet speaking-volume.

You might need an SPL meter to really know what volume you get when distortion kicks in. 

[Quatro]

Question 1: How did you listen to the output? At the speaker of the push-pull amp? At plate only?

At the speaker of the push pull amp (5E3 type power section - pretty close to 5e3 - no cap across pwr tubes cathode resistor - a few other minor changes).

Question 2: Did you change the input signal level feeding the phase inverter? Crank up the volume knob and see at what volume distortion sets in?

Played soft and loud on the guitar and soft and loud on the amp volume knob. Also both ways with gain cut switch engaged.

I'm guessing you tested with a guitar at a fairly quiet speaking-volume.

Tested with guitar at speaking level  and at wide open blistering  loud and all in between. Also after reading this post tried all scenarios with moderate music from an Ipod.

I used the full range of the knob on my "decade" box. The knob in question goes from 10k to 10meg. Pretty consistent through the whole range. Kind of like having a graduated volume knob. Quite a volume increase as one ratchets down.

Maybe someone with better ears/intellect than me wants to try it and see how it works for them?

[HotBluePlates]

I used the full range of the knob on my "decade" box. The knob in question goes from 10k to 10meg. Pretty consistent through the whole range. Kind of like having a graduated volume knob. Quite a volume increase as one ratchets down.

Good info!!

Nothing wrong with your ears/intellect. The missing piece of the puzzle was the bottom limit of 10k from your decade box.

10k is gong to be too big to exhibit the effect I was describing, because it's too large to behave like a normal cathode resistor (one used for biasing). Instead, it still tends to look like part of the load for a cathode follower/split-load inverter.

As a result, the extreme I was describing never happened, and what Sluckey described happened.

To clarify, if you design a split-load inverter using 56k resistors for the plate and cathode, you draw a loadline for 112kΩ. The total load is the sum of the two individual load resistors. Your lower limit of 10k is like drawing a 66kΩ loadline.

Tube current will slightly increase due to the different load. If this was the real goal, you'd do it much easier by varying the bias resistor. But what does happen is whatever current flows through the tube goes through both load resistors. At the low end of your range, the result is your plate output has 5 times the signal output of your cathode output (56k vs 10k).

As a result of the disparity of output signals, you're effectively creating even order distortion at the phase inverter and relaying that to the output stage. This would normally be due to reaching some limit of the tube on one side of the signal swing, but your scenario is creating it externally.

Low-order even harmonic distortion (say, 2nd and 4th harmonics) tend to be heard as "more volume" rather than as distortion, because they're 1 and 2 octaves up from the signal, respectively.

So you probably had enough added distortion to make up whatever power loss occurred due to mismatch drive to the output tubes.

You could try this exercise again, but use a 50k pot in place of the cathode load. I'm guessing the bottom-most range of that pot (0Ω - 10kΩ) will gives results counter to what you experienced. For a wired control, place a 5.6k-10kΩ resistor in series from the pot to ground to avoid turning off the cathode output.

To prove the theoretical scenario above, it would help to inject a sine wave, monitor the output tube grids (from grid-to-grid, not grid to ground) or speaker output with an o'scope, and fiddle your new distortion control. I'm thinking we'd see an increase of apparent distortion at about the same actual volume level.

[J Rindt]

Can I "piggy-back" on this thread.....
Is it a LTP PI (like in a BF Fender) that often had a 100k on one plate, and an 82k on the other.?
Was Fender trying to create an imbalance, or correct one.?
Thank You

[jazbo8]

Can I "piggy-back" on this thread.....
Is it a LTP PI (like in a BF Fender) that often had a 100k on one plate, and an 82k on the other.?
Was Fender trying to create an imbalance, or correct one.?
Thank You

They used to correct an imbalance, because the gains are slightly different, you may refer to Merlin's page for the details.

http://www.freewebs.com/valvewizard/acltp.html

Jaz

[HotBluePlates]

Can I "piggy-back" on this thread.....
Is it a LTP PI (like in a BF Fender) that often had a 100k on one plate, and an 82k on the other.?
Was Fender trying to create an imbalance, or correct one.?
Thank You

What Jazbo said. Correct an imbalance because there is a grounded grid instead of two driven grids in the long-tail.

But the Vox AC30 uses a pair of 100k resisotrs, because each grid is driven, so the imbalance doesn't happen.

To keep the analogy to what I was describing regarding the split-load, imagine a long-tail with a 100k plate load and a 20k plate load...

[Quatro]

I had a chance to try this experiment on two identical amps  10k on the tail of the experimental amp and a/b them through the same speaker. Indeed unpleasant artifacts were created by making the value smaller that i had not noticed with the decade box. The volume seemed to go up ( or ambient noise for sure), but wide open volume didn't seem to be much louder when on the a/b switch. No magic bullet. thanks for all the help.