LTPI balance

[nickb]

Sorry, this I don't understand (assuming that "bottom of the tail" means the junction between R6 and R8)
To my knowledge, its the junction of the 2 cathodes where half the input signal appears. The signal at the top of R8 (= bottom of the tail?) must be much smaller as the value of R8 is only a small fraction of the total tail resistance. For this same reason I can't see much local feedback or bootstrap effect from doubling the value of R8. The balance of the circuit still seems to be dominated by the value of R6.

In order for R6 to appear infinite the voltage at both ends is made the same. So if half the input voltage is at the top then the same is at the bottom.

Also there seems to be some typing error in the formula(s) for k above. E.g the result of 1/(2a) + b would be 0.616. And the result of k = 0.5*14 + 0.058 = 7.06 (not
0.094).

Yes there was a typo. I fixed it.

[Mike Sulzer]

Given how cheap Ds and ECs are these days, just add a negative supply and use a large resistor from the cathodes. Then it is easy to add in two channels, or use the second input for negative feedback.

[Helmholtz]

[QUOTE]In order for R6 to appear infinite the voltage at both ends is made the same. So if half the input voltage is at the top then the same is at the bottom./QUOTE]

Yes, but what would make R6 appear infinite in the proposed circuit (if you don't replace it with a real current source)? I don't see a measure that applies the same signal voltage at the bottom of R6 as on its top to achieve such bootstrapping effect in post #1. What am I missing?

[nickb]

What am I missing?

The input from the speaker, I would guess.

[Helmholtz]

The input from the speaker, I would guess.

Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?


Still having problems with your k formula. You say b = 0.58 but are using b = 0.058.

[nickb]

Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?


Still having problems with your k formula. You say b = 0.58 but are using b = 0.058.

It works in my simulation.

It's 0.058 (typo again). If you look as the circuit you can see that for yourself.

[uneumann]

Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?

Yes - This pic shows sim traces for the cathodes, V1 input and V2 input. The middle signal is the cathode signal and it's almost exactly the average of the two grid signals, as expected.



The next pic shows sim traces for the cathode and base of R6. They are almost identical, so the net signal voltage across R6 is virtually zero for the entire signal.

[Helmholtz]

Yes - This pic shows sim traces for the cathodes, V1 input and V2 input. The middle signal is the cathode signal and it's almost exactly the average of the two grid signals, as expected.

[ATTACH=CONFIG]52661[/ATTACH]

The next pic shows sim traces for the cathode and base of R6. They are almost identical, so the net signal voltage across R6 is virtually zero for the entire signal.

[ATTACH=CONFIG]52662[/ATTACH]

Thanks!

I just don't see how the circuit would improve balance for the forward signal especially in open loop situation - as claimed.

[nickb]

And here are my sim results. Used b = 0.1 for no particularly good reason at all.



Thanks for the idea uneumann. It's certainly the cheapest way to do this, just two $0.01 resistors.

[Helmholtz]

And here are my sim results. Used b = 0.1 for no particularly good reason at all.

[ATTACH=CONFIG]52665[/ATTACH]

Thanks for the idea uneumann. It's certainly the cheapest way to do this, just two $0.01 resistors.

Thanks.
Just to make sure: The sim results are for the open loop circuit as shown? Just wondering why you mention a feedback ratio b =0.1. Or are all "outs" connected?

[uneumann]

Thanks!

I just don't see how the circuit would improve balance for the forward signal especially in open loop situation - as claimed.

So maybe this will help. Here is a basic non-NFB output stage. The input is 0.1v and the spkr signal is ~1.5v (both peak). The nfb input to the PI is grounded by a cap (C2 in this drawing), there is no NFB applied to the system. However the tail 10K resistor has an added 29 ohm resistor added (R2) which gets some of the spkr signal through the 820 ohm resistor.



The R2 signal is set to exactly match (as close as I can get with integer resistors) the cathode signal. The image below shows those two signals at R58 and R54. These are AC signals and both referenced to ground so they line up nicely in the trace image. Note they are almost identical.



The next image shows the two plate outputs. Note how nicely balanced they are for what is a pretty small tail resistor (10K).



The last image is what the plate signals look like if R2 is replaced by a wire. This is the normally imbalanced output for this 10K tail PI.



In fact you can do ridiculous things like reduce the tail resistor to 100 ohms. You'd never do that, right?
Here are the plate outputs for the same circuit with 29 ohm R2 and a 100 ohm tail resistor. Looks pretty balanced to me.

[nickb]

Thanks.
Just to make sure: The sim results are for the open loop circuit as shown? Just wondering why you mention a feedback ratio b =0.1. Or are all "outs" connected?

All 'out' labels are the same node.

For open loop k = 1/(2a) ( pretty obvious!!!). I see already uneumann gave you a sim for this case so I won't duplicate.

[Helmholtz]

The high resistance R54 essentially prevents grounding of the global (speaker) NFB signal. So we still have a closed loop situation with global NFB. Why don't you just open the feedbck loop and maybe correct for the DC resistance?

[Helmholtz]

All 'out' labels are the same node.

For open loop k = 1/(2a) ( pretty obvious!!!). I see already uneumann gave you a sim for this case so I won't duplicate.

Sorry, but since the very beginning of this thread I have been asking about the open loop balance of the circuit.

[nickb]

Sorry, but since the very beginning of this thread I have been asking about the open loop balance of the circuit.

In the example given the loop is open. There is no NFB. Here is another example. I hope it helps clarify things