Why is the gain control a pot feeding the signal to earth

[Numpticious]

Why is the gain and / or vol control a potentiometer sending the signal to the earth and not a potentiometer in series with the next stage acting as a varyable resistor?

[Robert M. Martinelli]

Why is the gain and / or vol control a potentiometer sending the signal to the earth and not a potentiometer in series with the next stage acting as a varyable resistor?

Because it is not designed to work as a variable resistor, but to create and act like a voltage divider - imagine you have two equal resistors in series, whatever signal you apply you' ll find it' s amplitude to be half at the center because the resistance between the center and the external lugs is the same ( this is much like having a linear potentiometer with the wiper centered ) - now imagine you have two resistors in series, one being ten times the other, if the lower resistor is on the ground side you' ll have 1/10 of the input signal at the center lug ( much like a linear potentiometer at 10% of its travel ), if the lower resistor is on the signal side you' ll have 9/10 of the input voltage at the center lug ( much like a linear potentiometer at 90% of its travel ). That is how voltage dividers work, and a potentiometer creates a voltage divider made of two resistors connected to the wiper.
Hope I' ve managed to be clear enough.
Best regards
Bob

[BackwardsBoB]

When you hook up a pot as a voltage divider to provide volume control, you can turn it to "0" and ground the input to the next stage, turning the volume off totally.

You can just wire the pot as a series resistor, creating a divider with the input impedance of the following driven stage, but the value of the pot needs to be much larger than the input impedance of the following stage to get much attenuation, and you can never get the input level at the next stage to zero. For example, if you had a 500K grid leak/bias resistor after the cap on the next stage, and a 5M pot, at the maximum pot resistance, about 10% of the signal would still make it through to the grid of the driven tube, and, due to the Miller capacitance of the tube, the signal would be bassy and attenuated.

The other side of the coin is that wiring the pot as a resistive divider to ground loads the previous driving stage, so you can't get 100% of the signal through, but people expect their volume controls to go to "0", and you can't get there if you wire the pot as a varible series resistor.

[Enzo]

"Voltage divider" is a findamental electronic circuit concept. And the classic volume control is a voltage divider. Look up voltage divider and learn how it works. It is not a complex theory or anything.

A simple series resistance won't do much. To reduce volume, we need to reduce the signal voltage. VOltage dropped across a resistor is determined by the current through the resistor. That is "Ohm's Law," another fundamental we use every dau. if you learn nothing else in electronics, learn Ohm's Law.

SInce there is almost zero current into the grid of a tube, there would be almost zero current flowing through your series resistance. Zero current results in zero voltage drop. So a lone series resistance would not drop much voltage, if any.

In fact, amp makers rely on this effect as part of their channel switching circuits in some amp models. If you disconnect the ground end of a standard volume control, it essentially disappears.

[Guitarist]

Perhaps getting a switching pot to attach the wiper to the top would give you both functions.

1) Volume control: side effect being interaction with tone, treble mostly.

2) Impedance control: but pulling the grid closer to zero must reduce the signal too. Bigger resistor = more bass. You hear of these at the inputs of some amps but have diminished effect thanks to your guitar's pots.

I am by far a guitarist over knowing what is really going on, please excuse any and all errors ;

[CarlZ]

As an aside to what's already been mentioned, another option that's rarely used in tube amp designs is to use the pot as the feedback resistor in an inverting feedback amplifier stage. Setting the pot to 0 ohms gives you 100 percent feedback and therefore no gain or output. The down side here is that the circuit can get a little complicated and as such a plain jane wiper to ground is generally the easiest and most cost effective solution. Therefore that's pretty much what everybody uses.

-Carl Z