negative feedback input question?

[xmark]

Taking the example of a Fender Long Tail Pair for the Bassman 5F6-A.Does the negative feedback signal input into the grid of the non inverting half? If so, I'm unclear about the grid capacitor on the non-inverting grid.Isn't it supposed to keep the grid at AC ground?

[cdw2000]

I'll take a shot at this. The long tail pair phase splitter is really a differential amplifier.

The grid on the 2nd tube is the non-inverting input (relative to the output of the first triode stage. The capacitor is there to provide a low impedance DC path to ground so that the non-inverting input can't pickup noise.

Here is the schematic of the 5F6-A:

http://www.schematicheaven.com/fende...5f6a_schem.pdf

Looking at the schematic you will see that the negative feedback signal connects at the node between this capacitor and the presence control.

Comparing the Fender schematic to the schematic from the tube wizard site, you will see that Fender inserted the presence resistor at the ground point of a typical long tail pair. This forms a voltage divider with the negative feedback resistor and allows control of the amount of negative feedback. The level of negative feedback controls the damping factor of the power amp section, giving it stiffer (less "presence") control or looser (more "presence") control of the signal to the speaker. Looser control allows the speaker impedance resonances to interact more with the amplifier which translates to more audible frequency resonances to be generated from the speaker. This is perceived as adding richness or "presence" to the sound.

[xmark]

Thanks.I am not as far along in electronics as many and I appreciate your knowledge.I'm not sure that I have this correct.It seems only logical that both the 1st half tube and 2nd half tube of the LTP need to receive the negative feedback signal therefore the n.f.b is inserted into the common cathode circuit,thereby affecting both sides.In other words the feedback signal does not appear as a signal voltage on the grid itself of tube half 2?Causing the grid 2 voltage to vary.If that were the case it would seem that only tube 2 half would have Neg.fb.I understand that tube half 2 is a grounded grid amplifier,tube 1 half is a grounded cathode amplifier.

[cdw2000]

Think of the long tail pair as an amplifier with two inputs and two outputs. The circuit is configured intentionally so the two triodes interact. A signal applied to the input grid one of the two triodes imparts an amplified voltage onto BOTH plates. In an ideal circuit these voltages will be equal and opposite in phase.

So here's how applying the negative feedback signal to the second input works:

First, for clarity, I prefer to think of the 12AX7 as two separate triode amplifiers, since other than being in the same glass envelope, they are completeley separate in operation. So looking at the diagram from valve wizard, I will refer to them as triode 1 on the left and triode 2 on the right:



A signal provided through capacitor Cg1 to triode 1 will generate an amplified out-of-phase signal at the node between resistor Ra1 and the plate of triode 1 and an amplified in-phase signal at the node between resistor Ra2 and the plate of triode 2.

So given:
Vin1 = the input signal to triode 1
Vout1 = the output signal to triode 1
Vin2 = the input signal to triode 2
Vout2 = the output signal to triode 2

Then:
Vout1 = -G x Vin1;
Vout2 = +G x Vin1

Where G is a gain factor and the sign indicates signal phase relative to the input.

Notice Vin2 has not been used in the above equations. The grid of triode 2 is grounded (DC-wise) through Cg2. This is equivalent to having a 0V signal source on the second input to the long tail pair, so only Vin1 has an affect on the outputs.

If you were to ground Cg1 and lift Cg2 and apply a signal Vin2 to the second triode, the circuit would work equally as well except the output signal phases would be reversed. The formulas become:

Vout1 = +G x Vin2
Vout2 = -G x Vin2

Notice how the signs have changed. Vout1 is in-phase with Vin2 while Vout2 is out-of-phase with Vin2, the exact opposite of how the outputs react to Vin1. Therefore one input can be thought of as the non-inverting input (relative to one of the outputs) and the other input can be though of as the inverting input. If both inputs are used (I.e. both Cg1 and Cg2 are lifted from ground and attached to signal sources Vin1 and Vin2) the formula becomes:

Vout1 = - G x (Vin1 - Vin2)
Vout 2 = + G x (Vin1 - Vin2)

So looking at the Fender schematic again - The main pre-amplified signal from the tone stack is applied as Vin1 (through the 0.02uF capacitor) to the first triode grid and the negative feedback signal is effectively applied as Vin2 (through the 0.1uF capacitor) to the second triode grid (If you look closely at the schematic the negative feedback signal IS NOT connected to first input). Since the output waveforms are proportional to the difference between the two input signals, you have affectively provided negative feedback.

This may be more technical than you were looking for, but I like to explain how things work

[xmark]

Thanks, that is not too technical. A great explanation on your part.That really clears up for me, what Aikenamps had to say about the negative feedback input.