Little theory please

[Helmholtz]

stacked plate resistors with a tap in between.

"Tapping" the plate resistor creates a voltage divider with HT being AC ground. So available output voltage will be reduced. The RL in the gain formula means total plate load. As long as this doesn't change, gain at the plate output stays the same.
What is the idea behind this question?

Bypassing the cathode resistor in the input stage reduces heater hum.

[catalin gramada]

Hi. Input impedance should be huge at circuit input, so guitar signal should not be affected by pickups impedance,no matter how much volume pot is rolling off...I suppose. How big the input capacitor should be , please ?
I already changed grid resistor with 2M for test and is still quiet.

[Helmholtz]

Hi. Input impedance should be huge at circuit input, so guitar signal should not be affected by pickups impedance,no matter how much volume pot is rolling off...I suppose. How big the input capacitor should be , please ?

1M input impedance is typically considered high enough. A coupling cap of 10nF with an input impedance of 1M results in a cut-off frequency of 16Hz. A coupling cap is only necessary if the input grid is not at ground potential.

[catalin gramada]

Thanks. The grid resistor is already bootstrapped not ground referenced so I think it can be reduced in value if noise is concerned.

[catalin gramada]

One drawback of the circuit is very hard to over drive, too much local feedback, need 10v at input to clip and when it do , it clip very hard..But as first gain stage it is not a concern in my project.

[Dave H]

Hi. Input impedance should be huge at circuit input, so guitar signal should not be affected by pickups impedance,no matter how much volume pot is rolling off...I suppose

I don't think input impedance higher than 1M is going to help much in reducing guitar vol pot (HF) roll off because there is still the capacitance of the cable.

[eschertron]

"Tapping" the plate resistor creates a voltage divider with HT being AC ground. So available output voltage will be reduced. The RL in the gain formula means total plate load. As long as this doesn't change, gain at the plate output stays the same.
What is the idea behind this question?

if we call stage gain = apparent output voltage / input voltage, then both techniques reduce gain. However, gain of the tube for the two methods is different and the stage's clipping behavior will be different. I see that now, thanks for that.

[Helmholtz]

then both techniques reduce gain. However, gain of the tube for the two methods is different and the stage's clipping behavior will be different.

Not sure what you mean with "both" techniques/methods?
What is the idea behind tapping the plate resistor?

[Helmholtz]

Input impedance should be huge at circuit input, so guitar signal should not be affected by pickups impedance,no matter how much volume pot is rolling off

As Dave H wrote, high input impedance has no benefit regarding HF roll-off caused by turning back the volume. This effect is due to the low pass filter effect caused by the effective series resistance of the vol pot and the cable capacitance.

Higher input impedance reduces the damping of the PUs resonance. But as the PUs are already loaded by the controls, an input impedance > 1M won't be an audible improvement.
I see no reason for bootstrapping the 1M input resistor, which would increase impedance to several Ms.

[catalin gramada]

As Dave H wrote, high input impedance has no benefit regarding HF roll-off caused by turning back the volume. This effect is due to the low pass filter effect caused by the effective series resistance of the vol pot and the cable capacitance.

Higher input impedance reduces the damping of the PUs resonance. But as the PUs are already loaded by the controls, an input impedance > 1M won't be an audible improvement.
I see no reason for bootstrapping the 1M input resistor, which would increase impedance to several Ms.

Alles klar!