LTP NFB question

[Chuck H]

One potential problem (as I see it) will be ensuring there is almost zero loading or through circuit impedance for the cathode voltage source.

[Helmholtz]

Like this ?

No, just a 1M grid stopper before the grid.

[Helmholtz]

Yes. This works well also. There was a tiny shift in phase but I didn't check it for an absolute figure. The output for both redraw circuits is identical.

In you original proposal the feedback signal applied to the left side grid will depend on the impedance of the source connected to the PI input.

If you used a zero impedance spice voltage source no FB signal will reach the grid.
So only NFB.

[catalin gramada]

I tried that. Didn't work.

Both of my redrawn circuits work. Helmholtz has more technical prowess by a stretch so if he thinks the input grid leak should be grounded, do it that way. So, the second redraw. ta2

It works.

[Helmholtz]

It works.

Fine. Makes my grid leak proposal obsolete and shows that the problem was indeed asymmetrical grid termination.

But don't overlook that the signal has to pass the right side grid leak of 1M.
Means that there is a LP filter in the NFB path composed of the 1M resistor and the Miller capacitance.
As a consequence NFB will reduce at HF and amp frequency response will increase at HF.

I suggest to bypass the 1M resistor with a cap of maybe 100pF for compensation.

[catalin gramada]

Are You suggested something like this ? My circuit work pretty stable now

[Helmholtz]

Are You suggested something like this ? My circuit work pretty stable now

That's actually Chuck's altered proposal of post #13.

Forget my grid stopper test, no longer necessary.

But considering the consequences of the Miller effect at higher frequences, why not just accept that your low impedance NFB divider causes PI asymmetry?
The simplest remedy would be the use of different plate resistors.

[catalin gramada]

I tested also with the shunt resistor in only grid circuit with the source to ground. It not work, the unbalance was commin back.
So, the shunt resistor should be in cathode

[uneumann]

Are You suggested something like this ? My circuit work pretty stable now

Doesn't the electrolytic cap to -100v just shunt the NFB to gnd? I assume that supply has (near) zero AC impedance. If so, this circuit doesn't actually have any NFB.

[catalin gramada]

I wish to use different plate resistors , but I can't. The next stage is DC coupled so I should have care for minimal dc offset.

[catalin gramada]

Doesn't the electrolytic cap to -100v just shunt the NFB to gnd? I assume that supply has (near) zero AC impedance. If so, this circuit doesn't actually have any NFB.

No why ? The negative supply is floating over ground in top of shunt nfb resistor

[Helmholtz]

Doesn't the electrolytic cap to -100v just shunt the NFB to gnd? I assume that supply has (near) zero AC impedance. If so, this circuit doesn't actually have any NFB.

If wired according to schem the positive end of the ecap doesn't go to ground. The -100V supply must be floating.

[catalin gramada]

Doesn't the electrolytic cap to -100v just shunt the NFB to gnd? I assume that supply has (near) zero AC impedance. If so, this circuit doesn't actually have any NFB.

More clearly

[uneumann]

No why ? The negative supply is floating over ground in top of shunt nfb resistor

Maybe I'm missing something, but when you have a -100v supply - it must be referenced to ground, right? The -100v rail is then at AC gnd.
Otherwise, if it floats with respect gnd, it floats with respect to the whole PI circuit and doesn't actually bias the PI.
Once the supply is ref'd to ground, the + side of the cap with the NFB signal is also coupled to AC ground. Can you scope that node? Is there any NFB signal there?

[Chuck H]

It works.

Ah. I failed to include the non inverting side coupling cap to the NFB. Doh! Re did the test and yes, it works.