1. ## Paraphase inverter question

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;

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2. Originally Posted by catalin gramada
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?

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

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

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5. Originally Posted by eschertron
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.

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6. Originally Posted by Helmholtz
Why not? I would expect the circuit to have excellent self-balancing properties.

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7. Originally Posted by eschertron
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

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8. Originally Posted by Helmholtz
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!

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9. 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.

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

and different ways of thinking about nfb aplication

150_UL_6V6s (5).pdf

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11. Originally Posted by Helmholtz
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.

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12. Originally Posted by catalin gramada
I like
And in the example, the cathode is bypassed.

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13. Originally Posted by catalin gramada
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.

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

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15. Originally Posted by Helmholtz
What does simulation say? Any difference with bypassed cathode resistor?
I suspect not but I'll check it out after I get another beer

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16. Originally Posted by Helmholtz
What does simulation say? Any difference with bypassed cathode resistor?
not bypassed 100mV input
First triode 5.7V. Second triode 5.45V. Gain 57. Imbalance 4%

Bypassed 100mV input
First triode 5.84V. Second triode 5.56V. Gain 58. Imbalance 4.7%

I calculated 2% imbalance. I think that's because I didn't allow for the output impedance of the first stage being much greater than the output impedance of the second stage.

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17. Interesting, thanks.

because I didn't allow for the output impedance of the first stage being much less than the output impedance of the second stage.
Wouldn't you expect the second stage to have the lower output impedance because of its total voltage NFB?

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18. Yes, that thought struck me too.

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19. Originally Posted by Helmholtz
Wouldn't you expect the second stage to have the lower output impedance because of its total voltage NFB?
Thanks, you're right. I had it backwards and still didn't notice after reading it again
I've fixed it now.

I think the second stage output impedance would be reduced by the loop gain to something like 30k/60=500R which is negligible compared to the 1M feedback resistor. If it was 30k the same as the first stage it cancels out. There would be 30k added to the feed on resistor and 30k added to the feedback resistor so the feedback factor is still unity.

EDIT: I ran the sim. again with the feedback resistor 5% larger (1050k) and the outputs are balanced.

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20. Originally Posted by Helmholtz
What does simulation say? Any difference with bypassed cathode resistor?
"If the two triodes are really operating in push pull,the current in R3 should not contain any alternating component,and C has no decoupling function."- N.Crowhurst

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21. More material:

Crowhurst_Cooper_1956_High_Fidelity_Circuit_Design_text.pdf

from p.126 and the follow

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22. Originally Posted by catalin gramada
More material:

Crowhurst_Cooper_1956_High_Fidelity_Circuit_Design_text.pdf

from p.126 and the follow
We think we are cleaver with our computers and simulations but it's true, we are standing on the shoulders of giants. I've only just worked out that the feedback resistor needs to be 5% bigger for perfect balance but Crowhurst had done it back in 1956 with just a slide rule.

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23. When I've tinkered with both simple and floating paraphase circuits with a type 1 master vol after it, I found that if the cathode resistor was shared and unbypassed, horrid blips of parasitic oscillation were generated when it was overdriven. Of course it could have been something else but bypassing or splitting the cathodes got rid of the issue.
Note that the Vox AC50 variants that used a paraphase had its shared cathode bypassed, also the Ampeg B15, so perhaps they noticed it too?
https://el34world.com/charts/Schemat...Vox_ac50_1.pdf
https://el34world.com/charts/Schemat...eg_b_15_nc.gif

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24. Originally Posted by pdf64
When I've tinkered with both simple and floating paraphase circuits with a type 1 master vol after it, I found that if the cathode resistor was shared and unbypassed, horrid blips of parasitic oscillation were generated when it was overdriven. Of course it could have been something else but bypassing or splitting the cathodes got rid of the issue.
Note that the Vox AC50 variants that used a paraphase had its shared cathode bypassed, also the Ampeg B15, so perhaps they noticed it too?
https://el34world.com/charts/Schemat...Vox_ac50_1.pdf
https://el34world.com/charts/Schemat...eg_b_15_nc.gif
I don't think so but who knows.... I have not You mentioned issue in my experiments. And think the decoupling cap is useful just to can apply the global nfb to the first section. It is a bit tricky due to the double function as bias resistor and component of voltage divider of nfb but can be done. I much prefer the Grommes solution for the moment.

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25. With equal values resistors, not precisely balanced but didn't looks so bad.

Of course the values should be tailored a bit...

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26. I much prefer the Grommes solution for the moment.
Does someone have a good explanation of the strange asymmetrical cathodes' wiring of the PI in the Grommes circuit?

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27. I cannot provide a good explanation, there are not, but who cares ? (meant why should be so important as time as are decoupled?). The circuit should work with separate cathode resistors as same, but think economy reasons...

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28. And just separate the cathode and tested: it works as same, with same degree of imbalance. Think the more important think in circuit are anode resistors, the second triode should be compensate in any circumstance for precise balanced results. Nfb can be applied in the bottom of first cathode resistor in more traditional way.
But Grommes shows very elegant simply solution

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29. What I found is the circuit ( in my case) support a limited amount of nfb. Over 12 db sign of instability appears. Think is more related by shifting phase and for this reason I asked at beginning of topic about. Of course think is possible to shift the pole changing the cap value but just thinking at blocking distortion opportunities...

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30. O.K., meanwhile I understand that the asymmetrical cathodes' wiring (additional unbypassed 100 Ohm resistor at the first tubes' cathode) is necessary to allow feeding NFB only to first tube's cathode.

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32. Originally Posted by pdf64
I found that if the cathode resistor was shared and unbypassed, horrid blips of parasitic oscillation were generated when it was overdriven.
It was originally designed for hi-fi. You're not supposed to overdrive it

My guess is that in a perfect world with perfectly matched outputs which are unloaded the bypass cap isn't needed but in the real word with equal feed on/feedback resistors and 150k loads (Vox) it's better to have the capacitor.

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33. Originally Posted by catalin gramada
Paraphase inverter question
The best way to satisfy your curiosity is to do everything you described, perform DC measurements, and attach the circuit to the signal source and oscilloscope. Its implies that you share your experience with us.

EDIT 190902

It work do the same, only you have two unnecessary capacitors more, whose value can affect to (low) frequency.
From ratio R4 vs R5 will depend on the symmetrical excitation of the outputs.

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34. Also the balancing resistors contribute to the effective PI load.

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35. The Paraphase is great in simple /minimalist circuits - easy to add a 1 knob tone control
See the GA40 for example of this

Cheers,
Ian

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