Master volume and treble loss

[pdf64]

...Assuming all values are (averaged) DC...

Yes, but many apologies, due to a somewhat messy tagstrip layout, my previous readings were between the cathode and the non-inverting grid
Please see amended readings above.

...What are the peak-to peak signal voltages at the PI cathode?

Looks to be about 5Vp-p. The red trace is at the cathodes, the blue is at other end of their 470ohm bias resistor.
The HT VDC at the LTP supply node (via a 10k dropper) drops about 2.5V between idle and full overdrive into the short (type 3 at min).

[Helmholtz]

It only takes 2 wires to move from simulations to practice.

Actually only one.

So, why don't you show your own measurements?
Or don't you have a scope?

[pdf64]

I don't understand what you meant with measuring and simulations to say (prove).
To remind you, the question was like in # 14

https://music-electronics-forum.com/forum/music-electronics/920070-master-volume-and-treble-loss?p=920246#post920246

What do you get if g1 power tubes mutually shorted.
It only takes 2 wires to move from simulations to practice.

Look at practical solutions
http://bee.mif.pg.gda.pl/ciasteczkowypotwor/%23SM_scena/VOX/Vox_AC30CC2_AC30CC2X%20(2005)%20SM.pdf
page 6

http://dealers.korgusa.com/svcfiles/TB18C1.pdf
page 3

The pic I posted is a screenshot of my scope traces taken from the amp mentioned, my scope being a USB picoscope. It's real, not a simulation
With EL34 now in circuit and the type 3 master vol turned to min, their (dc average) cathode current rises about 50%, eg from 32mA to 45, when an input signal is applied, as per the scope trace pic previously posted.
So the common mode effect does seem to be, in this case, real.
I'm pretty sure the finding is reproducible.
Here are traces of the signals at the EL43 1 ohm cathode resistors (I can't find my grounding clip for the red probe ).

[Helmholtz]

Oops, sorry about the pic resolution, should be lots better now.
Re cathode current, the VDC across the 470ohm resistor at idle is 1.1VDC, fully overdriven it rises to 2.95V master vol at max, 3.5V with the master vol at min.
The corresponding voltages between cathode and 0V common are37VDC, 36.8V and 42.2V (!).

Thanks!

I'm not sure how to interprete your measurements.

Assuming all values are (averaged) DC, the increase from 1.1V to 3.5V across the 470R would correspond to more than tripling the plate currents. From 1.17mA to 3.7mA per tube.
But as the same current must run through the total cathode resistance, the cathode voltage should increase by the same factor?

What are the peak-to peak signal voltages at the PI cathode?

[vintagekiki]

I don't understand what you meant with measuring and simulations to say (prove).
To remind you, the question was like in # 14

https://music-electronics-forum.com/forum/music-electronics/920070-master-volume-and-treble-loss?p=920246#post920246

What do you get if g1 power tubes mutually shorted.
It only takes 2 wires to move from simulations to practice.

Look at practical solutions
http://bee.mif.pg.gda.pl/ciasteczkowypotwor/%23SM_scena/VOX/Vox_AC30CC2_AC30CC2X%20(2005)%20SM.pdf
page 6

http://dealers.korgusa.com/svcfiles/TB18C1.pdf
page 3

[pdf64]

That's about 2mA peak and less than I expected. But I should have considered that a triode can't deliver more than its saturation current, which depends on Vak.
I wonder what was the cathode voltage during the measurement. It might have increased, especially during current peaks, because of in phase/common mode currents, thus lowering effective Vak.
Common mode tube currents will not show in your measurement but will make the cathode voltage swing.

Unfortunately I can't read the scales in your low resolution scope picture.

Oops, sorry about the pic resolution, should be lots better now.
EDIT Re cathode current, the VDC across the 470ohm resistor at idle is 1.16VDC, fully overdriven it rises to 1.185V master vol at max, 1.38V with the master vol at min.
The corresponding voltages between cathode and 0V common are 36.6VDC, 36.9V and 42.5V (!).

[Helmholtz]

If the master volume is turned back up and a current meter connected across it (ie replicating the master volume set to min), the current is 1.426mA, no change to the waveform etc.

That's about 2mA peak and less than I expected. But I should have considered that a triode can't deliver more than its saturation current, which depends on Vak.
I wonder what was the cathode voltage during the measurement. It might have increased, especially during current peaks, because of in phase/common mode currents, thus lowering effective Vak.
Common mode tube currents will not show in your measurement but will make the cathode voltage swing.

Unfortunately I can't read the scales in your low resolution scope picture.

[pdf64]

That's right colleague. Output from PI at point ? (X) will be zero¹⁾ and graphic form at point ? (X) will be horizontal line, because equal half-periods are canceled.

When the output from PI is zero¹⁾, there is no drive at the g1 output tubes, through power tubes only idle (bias) current will flow.

²⁾ I don't understand how you found the negative half-cycle at point X, when PI gives a symmetrical output voltage ???

...PI triodes will work at their limit to supply short circuit currents...

I'm having a tinker with my poor old Sound City 50 Plus (modding it to be similar to a JTM50), so as it's on the bench I thought I'd investigate this. When a Type 3 master vol is set very low (below about 5k), the common mode thing does seem to be real; with the LTP fully overdriven, a 24.6Vp-p common mode signal passes to the power valve control grids. Red trace is channel A, the 1st/inverting/82k LTP output, blue trace is the (100k) inverting output.

If the master volume is turned back up and a current meter connected across it (ie replicating the master volume set to min), the current is 1.426mA, no change to the waveform etc.
So far I've not got the EL34 fitted. The amp's HT voltage during the testing at the reservoir cap node is 420V, at the LTP HT node it's 350V.

[vintagekiki]

Because PI is not perfect.
Does not need to be complete half cycle, that was just an example. Can you guarantee no signal can possibly get to point 'X' ?

Ultimately, and if PI is not perfect.
You have a perfectly paired 12AX7. All components in PI are 1%.
Both 12AX7 anodes and both g1 6L6s are short-circuited to each other (see schematic).
How much signal will appear at point 'X'? 10, 20, 30% or <1%

You don't need a simulation. Practical check requires only 10cm of wire and 5 minutes.
Let's forget the philosophizing these 32 posts. How to explain that some certain famous brands that use CLMV in their guitar amps (#12).

[g1]

I don't understand how you found the negative half-cycle at point X, when PI gives a symmetrical output voltage ???

Because PI is not perfect.
Does not need to be complete half cycle, that was just an example. Can you guarantee no signal can possibly get to point 'X' ?

[vintagekiki]

I think we can all agree the output will be zero¹⁾, yet the power tubes are doing work.
So the question becomes "can there be signal appearing at point X"²⁾ ?

That's right colleague. Output from PI at point ? (X) will be zero¹⁾ and graphic form at point ? (X) will be horizontal line, because equal half-periods are canceled.

When the output from PI is zero¹⁾, there is no drive at the g1 output tubes, through power tubes only idle (bias) current will flow.

²⁾ I don't understand how you found the negative half-cycle at point X, when PI gives a symmetrical output voltage ???

[Helmholtz]

So the question becomes "can there be signal appearing at point X" ?

The 2 grid signals with 2k between the grids in Chuck's simulation here:
https://music-electronics-forum.com/...679#post806679
are almost identical.
So that would essentially correspond to the common mode grid signal at point X with the CLMV adjusted to 2k.
As the CLMV can only reduce the differential voltage, I expect the common mode signal with the CLMV at zero to be about the same.

I have no reason to doubt the sims. Also I understand that nickb could confirm Chuck's results.

[g1]

CLMV Question
If signal at point A as at the picture, what will be the signal at points ?

I have changed the picture and the question a bit. If there is signal at point 'X', what will be at the output?




I think we can all agree the output will be zero, yet the power tubes are doing work.
So the question becomes "can there be signal appearing at point X" ?
The PI is not 100% perfect. How can we say it is not possible for some common mode signal to get through to X ?

I know that the road to hell has paved with good intentions,

I think the road to hell may be paved with google translator.

[vintagekiki]

Dear colleague
Obviously you either don’t understand, or you don’t want to understand about I’m writing. I don’t know, maybe the cause is in Mercury retrograde, or a difference in interlingual syntax.
To avoid unnecessary misunderstanding in communication as happened in post #19, I will refrain from comment until some better times.

[Helmholtz]

I'm free to notice that you didn't understand
https://music-electronics-forum.com/forum/music-electronics/920070-master-volume-and-treble-loss?p=920246#post920246
CLMV Question
If the signal at point A as at the picture and level is U = 1V, what will be the signal at points ? (graphic form and how much is his the level).

I very well understood your question and was hoping that my above reply would answer it.

With a perfectly balanced PI the speaker output will be zero. But the signal wrt ground at the grids will depend on the common mode component of the PI output.

If you need real data use your scope and measure. Don't expect me to do it for you.

I don't like this CLMV so I don't use it.