Paraphase inverter question

[catalin gramada]

Hello,
What's happen if I couple R4 and R5 resistors to plates with capacitors and remove C2 and tie second grid directly to the junction instead, in the circuit below, please ? The coupling caps to power tubes remains into position. Thanks






Just like this;

[Dave H]

What's happen if I couple R4 and R5 resistors to plates with capacitors and remove C2 and tie second grid directly to the junction

I guess it'll work the same. But why spoil the elegant simplicity of the original circuit by adding an extra capacitor?

[eschertron]

I don't expect that R4 and R5 have the same value. Do they?

[Old Tele man]

Adding a C to an existing R will create an RC-filter, you wanting to roll-off / reshape the bandwidth or something?

[Helmholtz]

I don't expect that R4 and R5 have the same value. Do they?

Why not? I would expect the circuit to have excellent self-balancing properties.

[catalin gramada]

Why not? I would expect the circuit to have excellent self-balancing properties.

It have, already tested.

[Dave H]

I don't expect that R4 and R5 have the same value. Do they?

The same value is close enough. For perfect balance using a 12AX7 R5 should be about 2% bigger than R4

[eschertron]

Why not? I would expect the circuit to have excellent self-balancing properties.

All right, I was thinking that like a conventional paraphase, there's a voltage division. But I see this is a little different.
I see that a voltage will develop at the node of R4 and R5 only when the voltages at the plates are not exactly equal in magnitude and opposite in phase. An error signal. But with "perfect symmetry" the voltage is zero, and then the voltage reaching the plate is zero... I still find myself wanting the resistors to be unbalanced in order to generate a drive signal to the grid that nets an ideal opposite plate voltage.

edit: I see Dave has put a number on my vague itch. Thanks!

[Helmholtz]

But with "perfect symmetry" the voltage is zero, and then the voltage reaching the plate is zero.

The second triode will still be driven from its cathode input, even if grid signal is zero.

[catalin gramada]

Interesting materials about circuit

The See-Saw circuit - A self balancing phase splitter (5).pdf

and different ways of thinking about nfb aplication

150_UL_6V6s (5).pdf

[eschertron]

The second triode will still be driven from its cathode input, even if grid signal is zero.

True. But even that depends on a signal that is not balanced; if the cathode currents are opposite and equal then there's no cathode voltage deviation to drive a signal. If the grid error signal is doing its job then the cathode deviation should be minimal.

But I confess, there are more servo loops than what I initially thought.

[eschertron]

Interesting materials about circuit

I like
And in the example, the cathode is bypassed.

[Dave H]

Interesting materials about circuit

[ATTACH]54954[/ATTACH]

That paper lost me
I simply calculated the open loop gain (A) of a 12AX7 driving a 1M resistor and put that in the NFB equation.

A = uRL/(RL+ra) = 100*90/(90+62) = 59

G = A/(1+AB) where B is the feedback factor (1 in this case)
G = 59/(1+59) = 0.98

Closed loop gain G should be 1 for perfect balance therefore the feedback resistor should be 2% bigger.

[Helmholtz]

What does simulation say? Any difference with bypassed cathode resistor?

[Dave H]

What does simulation say? Any difference with bypassed cathode resistor?

I suspect not but I'll check it out after I get another beer